Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Mathematics PYQs
  3. /Differential Equations

Differential Equations — JEE Previous Year Questions

Every Differential Equations question asked in JEE Main and JEE Advanced across the last 186 papers — 240 questions, each with its correct answer. Free to read, no account needed.

Questions

240

Papers it appeared in

167/186

Appearance rate

90%

All 240 Differential Equations questions

Most recent papers first.

Q1·MathematicsSingle correctJEE Advanced 2026
Let y:(−∞,∞)→(0,∞)y : (-\infty, \infty) \to (0, \infty)y:(−∞,∞)→(0,∞) be the solution of the differential equation dydx=e5xy3+y3ex+exy4\dfrac{dy}{dx} = \dfrac{e^{5x} y^{3} + y^{3}}{e^{x} + e^{x} y^{4}}dxdy​=ex+exy4e5xy3+y3​, satisfying y(0)=12y(0) = \dfrac{1}{\sqrt{2}}y(0)=2​1​. Then the value of y(log⁡e2)y(\log_{e} 2)y(loge​2) is
  1. (A)5+352\sqrt{\dfrac{5 + \sqrt{35}}{2}}25+35​​​
  2. (B)7+532\sqrt{\dfrac{7 + \sqrt{53}}{2}}27+53​​​
  3. (C)7+532\dfrac{7 + \sqrt{53}}{2}27+53​​
  4. (D)5+352\dfrac{5 + \sqrt{35}}{2}25+35​​

Correct answer: (B)

Step-by-step solution →
Q2·MathematicsMultiple correctJEE Advanced 2026
Let y=f(x)y = f(x)y=f(x) be the real valued function defined on the interval (0,∞)(0, \infty)(0,∞), satisfying y(1)=0y(1) = 0y(1)=0 and the differential equation xdydx=y−x3x\dfrac{dy}{dx} = y - x^{3}xdxdy​=y−x3. Then which of the following statements is (are) TRUE ?
  1. (A)The function fff has a local minimum at x=13x = \dfrac{1}{\sqrt{3}}x=3​1​
  2. (B)The function fff has a local maximum at x=13x = \dfrac{1}{\sqrt{3}}x=3​1​
  3. (C)The function fff is increasing in the interval (1,2)(1, 2)(1,2)
  4. (D)If g(x)=4x3−5x2+32xg(x) = 4x^{3} - 5x^{2} + \dfrac{3}{2}xg(x)=4x3−5x2+23​x for x>0x > 0x>0, then the number of elements in the set {x∈(0,∞):f(x)=g(x)}\{x \in (0, \infty) : f(x) = g(x)\}{x∈(0,∞):f(x)=g(x)} is 2

Correct answer: (B), (D)

Step-by-step solution →
Q3·MathematicsSingle correctJEE Main 2026
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation x1−x2 dy+(y1−x2−xcos⁡−1x)dx=0x\sqrt{1-x^2}\,dy + \left(y\sqrt{1-x^2} - x\cos^{-1}x\right)dx = 0x1−x2​dy+(y1−x2​−xcos−1x)dx=0, x∈(0,1)x \in (0,1)x∈(0,1), lim⁡x→1−y(x)=1\displaystyle\lim_{x \to 1^-} y(x) = 1x→1−lim​y(x)=1. Then y(12)y\left(\dfrac{1}{2}\right)y(21​) equals:
  1. (A)3−π33 - \dfrac{\pi}{\sqrt{3}}3−3​π​
  2. (B)4−3π4 - \sqrt{3}\pi4−3​π
  3. (C)4−2π34 - \dfrac{2\pi}{\sqrt{3}}4−3​2π​
  4. (D)3−π233 - \dfrac{\pi}{2\sqrt{3}}3−23​π​

Correct answer: (A)

Step-by-step solution →
Q4·MathematicsNumericalJEE Main 2026
Let y = y(x) be the solution of the differential equation (x2−xx2−1)dy+(y(x−x2−1)−x)dx=0(x^2 - x\sqrt{x^2-1})dy + (y(x - \sqrt{x^2-1}) - x)dx = 0(x2−xx2−1​)dy+(y(x−x2−1​)−x)dx=0, x ≥ 1. If y(1) = 1, then the greatest integer less than y(5)y(\sqrt{5})y(5​) is ______.

Correct answer: 3

Step-by-step solution →
Q5·MathematicsSingle correctJEE Main 2026
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be such that f(xy)=f(x)f(y)f(xy) = f(x)f(y)f(xy)=f(x)f(y), for all x,y∈Rx, y \in \mathbb{R}x,y∈R and f(0)≠0f(0) \neq 0f(0)=0. Let g:[1,∞)→Rg : [1, \infty) \to \mathbb{R}g:[1,∞)→R be a differentiable function such that x2g(x)=∫1x(t2f(t)−tg(t)) dt.x^{2}g(x) = \int_{1}^{x} (t^{2}f(t) - tg(t))\, dt.x2g(x)=∫1x​(t2f(t)−tg(t))dt. Then g(2)g(2)g(2) is equal to :
  1. (A)138\frac{13}{8}813​
  2. (B)1116\frac{11}{16}1611​
  3. (C)1532\frac{15}{32}3215​
  4. (D)1764\frac{17}{64}6417​

Correct answer: (C)

Step-by-step solution →
Q6·MathematicsNumericalJEE Main 2026
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation xsin⁡(yx)dy=(ysin⁡(yx)−x)dxx\sin\left(\frac{y}{x}\right) dy = \left(y\sin\left(\frac{y}{x}\right) - x\right) dxxsin(xy​)dy=(ysin(xy​)−x)dx, y(1)=π2y(1) = \frac{\pi}{2}y(1)=2π​ and let α=cos⁡(y(e12)e12)\alpha = \cos\left(\frac{y(e^{12})}{e^{12}}\right)α=cos(e12y(e12)​). Then the number of integral values of ppp, for which the equation x2+y2−2px+2py+α+2=0x^2 + y^2 - 2px + 2py + \alpha + 2 = 0x2+y2−2px+2py+α+2=0 represents a circle of radius r≤6r \leq 6r≤6, is __________.

Correct answer: 6

Step-by-step solution →
Q7·MathematicsNumericalJEE Main 2026
Let y = y(x) be the solution of the differential equation (tan⁡x)1/2 dy=(sec⁡3x−(tan⁡x)3/2y) dx(\tan x)^{1/2}\, dy = (\sec^{3} x - (\tan x)^{3/2} y)\, dx(tanx)1/2dy=(sec3x−(tanx)3/2y)dx, 0<x<π20 < x < \frac{\pi}{2}0<x<2π​, y(π4)=625y\left(\frac{\pi}{4}\right) = \frac{6\sqrt{2}}{5}y(4π​)=562​​. If y(π3)=45αy\left(\frac{\pi}{3}\right) = \frac{4}{5}\alphay(3π​)=54​α, then α4\alpha^{4}α4 equals _______.

Correct answer: 48

Step-by-step solution →
Q8·MathematicsSingle correctJEE Main 2026
Let f : [1, ∞) → ℝ be a differentiable function defined as f(x)=∫1xf(t) dt+(1−x)(log⁡ex−1)+ef(x) = \int_{1}^{x} f(t)\, dt + (1 - x)(\log_e x - 1) + ef(x)=∫1x​f(t)dt+(1−x)(loge​x−1)+e. Then the value of f(f(1)) is :
  1. (A)(1+ee)(1 + e^{e})(1+ee)
  2. (B)(1 + e)
  3. (C)(1+e+ee)(1 + e + e^{e})(1+e+ee)
  4. (D)1 + 2e

Correct answer: (A)

Step-by-step solution →
Q9·MathematicsSingle correctJEE Main 2026
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation: dydx+(6x2+(3x2+2x3+4)e−2x(x3+2)(2+e−2x))y=2+e−2x\frac{dy}{dx} + \left(\frac{6x^{2} + (3x^{2} + 2x^{3} + 4)e^{-2x}}{(x^{3} + 2)(2 + e^{-2x})}\right)y = 2 + e^{-2x}dxdy​+((x3+2)(2+e−2x)6x2+(3x2+2x3+4)e−2x​)y=2+e−2x, x∈(−1,2)x \in (-1, 2)x∈(−1,2), satisfying y(0)=32y(0) = \frac{3}{2}y(0)=23​. If y(1)=α(2+e−2)y(1) = \alpha(2 + e^{-2})y(1)=α(2+e−2), then α is equal to:
  1. (A)138\frac{13}{8}813​
  2. (B)613\frac{6}{13}136​
  3. (C)1213\frac{12}{13}1312​
  4. (D)1312\frac{13}{12}1213​

Correct answer: (D)

Step-by-step solution →
Q10·MathematicsSingle correctJEE Main 2026
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation dydx=(1+x+x2)(1−y+y2)\frac{dy}{dx} = (1 + x + x^2)(1 - y + y^2)dxdy​=(1+x+x2)(1−y+y2), y(0)=12y(0) = \frac{1}{2}y(0)=21​. Then (2y(1)−1)(2y(1) - 1)(2y(1)−1) is equal to:
  1. (A)3tan⁡(1136)\sqrt{3} \tan\left(\frac{11\sqrt{3}}{6}\right)3​tan(6113​​)
  2. (B)32tan⁡(11312)\frac{\sqrt{3}}{2} \tan\left(\frac{11\sqrt{3}}{12}\right)23​​tan(12113​​)
  3. (C)3tan⁡(11312)\sqrt{3} \tan\left(\frac{11\sqrt{3}}{12}\right)3​tan(12113​​)
  4. (D)32tan⁡(1136)\frac{\sqrt{3}}{2} \tan\left(\frac{11\sqrt{3}}{6}\right)23​​tan(6113​​)

Correct answer: (C)

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2026
Let x=x(y)x = x(y)x=x(y) be the solution of the differential equation 2y2dxdy−2xy+x2=02y^2 \frac{dx}{dy} - 2xy + x^2 = 02y2dydx​−2xy+x2=0, y>1y > 1y>1, x(e)=ex(e) = ex(e)=e. Then x(e2)x(e^2)x(e2) is equal to:
  1. (A)32e2\frac{3}{2}e^223​e2
  2. (B)23e2\frac{2}{3}e^232​e2
  3. (C)e2e^2e2
  4. (D)2e22e^22e2

Correct answer: (B)

Step-by-step solution →
Q12·MathematicsSingle correctJEE Main 2026
Let y=y(x)y = y(x)y=y(x) be the solution curve of the differential equation (1+sin⁡x)dydx+(y+1)cos⁡x=0(1 + \sin x)\frac{dy}{dx} + (y + 1)\cos x = 0(1+sinx)dxdy​+(y+1)cosx=0, y(0)=0y(0) = 0y(0)=0. If the curve y=y(x)y = y(x)y=y(x) passes through the point (α,−12)\left(\alpha, \frac{-1}{2}\right)(α,2−1​), then a value of α\alphaα is :
  1. (A)π6\frac{\pi}{6}6π​
  2. (B)π4\frac{\pi}{4}4π​
  3. (C)π3\frac{\pi}{3}3π​
  4. (D)π2\frac{\pi}{2}2π​

Correct answer: (D)

Step-by-step solution →
Q13·MathematicsSingle correctJEE Main 2026
Let y = y(x) be the solution of the differential equation xdydx−y=x2cot⁡x,x∈(0,π)x\frac{dy}{dx} - y = x^2 \cot x, x \in (0, \pi)xdxdy​−y=x2cotx,x∈(0,π). If y(π2)=π2y\left(\frac{\pi}{2}\right) = \frac{\pi}{2}y(2π​)=2π​, then 6y(π6)−8y(π4)6y\left(\frac{\pi}{6}\right) - 8y\left(\frac{\pi}{4}\right)6y(6π​)−8y(4π​) is equal to :
  1. (A)3π3\pi3π
  2. (B)- 3π3\pi3π
  3. (C)- π\piπ
  4. (D)π\piπ

Correct answer: (C)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2026
Let y=y(x)y = y(x)y=y(x) be a differentiable function in the interval (0,∞)(0, \infty)(0,∞) such that y(1)=2y(1) = 2y(1)=2. and lim⁡t→x(t2y(x)−x2y(t)x−t)=3\lim\limits_{t \to x} \left( \dfrac{t^{2} y(x) - x^{2} y(t)}{x - t} \right) = 3t→xlim​(x−tt2y(x)−x2y(t)​)=3 for each x>0x > 0x>0. Then 2y(2)2y(2)2y(2) is equal to
  1. (A)18
  2. (B)23
  3. (C)27
  4. (D)12

Correct answer: (B)

Step-by-step solution →
Q15·MathematicsNumericalJEE Main 2026
If the solution curve y = f(x) of the differential equation (x2−4)y′−2xy+2x(4−x2)2=0(x^2 - 4)y' - 2xy + 2x(4 - x^2)^2 = 0(x2−4)y′−2xy+2x(4−x2)2=0 x > 2, passes through the point (3, 15), then the local maximum value of f is ………

Correct answer: 16

Step-by-step solution →
Q16·MathematicsNumericalJEE Main 2026
Let f be a twice differentiable non-negative function such that (f(x))2=25+∫0x((f(t))2+(f′(t))2)dt(f(x))^{2} = 25 + \int\limits_{0}^{x}\left((f(t))^{2} + (f'(t))^{2}\right) dt(f(x))2=25+0∫x​((f(t))2+(f′(t))2)dt. Then the mean of f(log⁡e(1))f(\log_{e}(1))f(loge​(1)), f(log⁡e(2))f(\log_{e}(2))f(loge​(2)), ....., f(log⁡e(625))f(\log_{e}(625))f(loge​(625)) is equal to _________ .

Correct answer: 1565

Step-by-step solution →
Q17·MathematicsSingle correctJEE Main 2026
Let y = y(x) be the solution of the differential equation x4dy+(4x3y+2sin⁡x)dx=0x^4 dy + (4x^3 y + 2\sin x)dx = 0x4dy+(4x3y+2sinx)dx=0, x>0x > 0x>0, y(π2)=0y\left(\frac{\pi}{2}\right) = 0y(2π​)=0. Then π4y(π3)\pi^4 y\left(\frac{\pi}{3}\right)π4y(3π​) is equal to :
  1. (A)81
  2. (B)92
  3. (C)64
  4. (D)72

Correct answer: (A)

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2026
Let the solution curve of the differential equation xdy−ydx=x2+y2 dxxdy - ydx = \sqrt{x^{2} + y^{2}}\,dxxdy−ydx=x2+y2​dx, x>0x > 0x>0, y(1)=0y(1) = 0y(1)=0, be y=y(x)y = y(x)y=y(x). Then y(3)y(3)y(3) is equal to
  1. (A)4
  2. (B)6
  3. (C)1
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q19·MathematicsSingle correctJEE Main 2026
Let f:[1,∞)→Rf : [1, \infty) \rightarrow \mathbb{R}f:[1,∞)→R be a differentiable function, If 6∫1xf(t) dt=3xf(x)+x3−46\int_1^x f(t)\,dt = 3xf(x) + x^3 - 46∫1x​f(t)dt=3xf(x)+x3−4 for all x≥1x \ge 1x≥1, then the value of f(2) − f(3) is
  1. (A)−4
  2. (B)−3
  3. (C)4
  4. (D)3

Correct answer: (D)

Step-by-step solution →
Q20·MathematicsSingle correctJEE Main 2026
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation sec⁡xdydx−2y=2+3sin⁡x\sec x \frac{dy}{dx} - 2y = 2 + 3\sin xsecxdxdy​−2y=2+3sinx, x∈(−π2,π2)x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)x∈(−2π​,2π​), y(0)=−74y(0) = -\frac{7}{4}y(0)=−47​. Then y(π6)y\left(\frac{\pi}{6}\right)y(6π​) is equal to:
  1. (A)−52-\frac{5}{2}−25​
  2. (B)−54-\frac{5}{4}−45​
  3. (C)−33−7-3\sqrt{3} - 7−33​−7
  4. (D)−32−7-3\sqrt{2} - 7−32​−7

Correct answer: (A)

Step-by-step solution →
Q21·MathematicsSingle correctJEE Main 2026
Let y=y(x)y = y(x)y=y(x) be the solution curve of the differential equation (1+x2)dy+(y−tan⁡−1x)dx=0(1 + x^{2})dy + (y - \tan^{-1}x)dx = 0(1+x2)dy+(y−tan−1x)dx=0, y(0)=1y(0) = 1y(0)=1. Then the value of y(1)y(1)y(1) is :
  1. (A)2eπ4+π4−1\frac{2}{e^{\frac{\pi}{4}}} + \frac{\pi}{4} - 1e4π​2​+4π​−1
  2. (B)2eπ4−π4−1\frac{2}{e^{\frac{\pi}{4}}} - \frac{\pi}{4} - 1e4π​2​−4π​−1
  3. (C)4eπ4+π2−1\frac{4}{e^{\frac{\pi}{4}}} + \frac{\pi}{2} - 1e4π​4​+2π​−1
  4. (D)4eπ4−π2−1\frac{4}{e^{\frac{\pi}{4}}} - \frac{\pi}{2} - 1e4π​4​−2π​−1

Correct answer: (A)

Step-by-step solution →
Q22·MathematicsNumericalJEE Advanced 2025
Let y(x) be the solution of the differential equation x2dydx+xy=x2+y2, x>1ex^2 \frac{dy}{dx} + xy = x^2 + y^2,\, x > \frac{1}{e}x2dxdy​+xy=x2+y2,x>e1​, satisfying y(1) = 0. Then the value of 2(y(e))2y(e2)2\frac{\left(y(e)\right)^2}{y\left(e^2\right)}2y(e2)(y(e))2​ is ______

Correct answer: 0.75

Step-by-step solution →
Q23·MathematicsNumericalJEE Advanced 2025
For all x>0x > 0x>0, let y1(x)y_1(x)y1​(x), y2(x)y_2(x)y2​(x), and y3(x)y_3(x)y3​(x) be the functions satisfying dy1dx−(sin⁡x)2y1=0\frac{dy_1}{dx} - \left(\sin x\right)^2 y_1 = 0dxdy1​​−(sinx)2y1​=0, y1(1)=5y_1(1) = 5y1​(1)=5, dy2dx−(cos⁡x)2y2=0\frac{dy_2}{dx} - \left(\cos x\right)^2 y_2 = 0dxdy2​​−(cosx)2y2​=0, y2(1)=13y_2(1) = \frac{1}{3}y2​(1)=31​, dy3dx−(2−x3x3)y3=0\frac{dy_3}{dx} - \left(\frac{2 - x^3}{x^3}\right) y_3 = 0dxdy3​​−(x32−x3​)y3​=0, y3(1)=35ey_3(1) = \frac{3}{5e}y3​(1)=5e3​, respectively. Then lim⁡x→0+y1(x)y2(x)y3(x)+2xe3xsin⁡x\lim_{x \to 0^{+}} \frac{y_1(x) y_2(x) y_3(x) + 2x}{e^{3x} \sin x}limx→0+​e3xsinxy1​(x)y2​(x)y3​(x)+2x​ is equal to __________ .

Correct answer: 2

Step-by-step solution →
Q24·MathematicsSingle correctJEE Main 2025
Let f(x)=x−1f(x)=x-1f(x)=x−1 and g(x)=exg(x)=e^xg(x)=ex for x∈Rx\in\mathbb{R}x∈R. If dydx=(e−2x g(f(f(x)))−yx)\dfrac{dy}{dx}=\left(e^{-2\sqrt x}\,g\big(f(f(x))\big)-\dfrac{y}{\sqrt x}\right)dxdy​=(e−2x​g(f(f(x)))−x​y​), y(0)=0y(0)=0y(0)=0, then y(1)y(1)y(1) is:
  1. (A)1−e2e4\dfrac{1-e^2}{e^4}e41−e2​
  2. (B)2e−1e3\dfrac{2e-1}{e^3}e32e−1​
  3. (C)e−1e4\dfrac{e-1}{e^4}e4e−1​
  4. (D)1−e3e4\dfrac{1-e^3}{e^4}e41−e3​

Correct answer: (C)

Step-by-step solution →
Q25·MathematicsSingle correctJEE Main 2025
Let y=y(x)y=y(x)y=y(x) be the solution curve of the differential equation x(x2+ex)dy+(ex(x−2)y−x3)dx=0x(x^2+e^x)dy+(e^x(x-2)y-x^3)dx=0x(x2+ex)dy+(ex(x−2)y−x3)dx=0, x>0x>0x>0, passing through the point (1,0)(1,0)(1,0). Then y(2)y(2)y(2) is equal to:
  1. (A)44−e2\dfrac{4}{4-e^2}4−e24​
  2. (B)22+e2\dfrac{2}{2+e^2}2+e22​
  3. (C)22−e2\dfrac{2}{2-e^2}2−e22​
  4. (D)44+e2\dfrac{4}{4+e^2}4+e24​

Correct answer: (D)

Step-by-step solution →
Q26·MathematicsSingle correctJEE Main 2025
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (x2+1)y′−2xy=(x4+2x2+1)cos⁡x(x^2+1)y'-2xy=(x^4+2x^2+1)\cos x(x2+1)y′−2xy=(x4+2x2+1)cosx, y(0)=1y(0)=1y(0)=1. Then ∫−33y(x) dx\displaystyle\int_{-3}^{3} y(x)\,dx∫−33​y(x)dx is:
  1. (A)24
  2. (B)36
  3. (C)30
  4. (D)18

Correct answer: (A)

Step-by-step solution →
Q27·MathematicsSingle correctJEE Main 2025
If a curve y=y(x)y=y(x)y=y(x) passes through the point (1,π2)\left(1,\dfrac{\pi}{2}\right)(1,2π​) and satisfies the differential equation (7x4cot⁡y−excosec⁡y)dxdy=x5(7x^4\cot y-e^x\operatorname{cosec}y)\dfrac{dx}{dy}=x^5(7x4coty−excosecy)dydx​=x5, x≥1x\ge1x≥1, then at x=2x=2x=2, the value of cos⁡y\cos ycosy is:
  1. (A)2e2−e64\dfrac{2e^2-e}{64}642e2−e​
  2. (B)2e2+e64\dfrac{2e^2+e}{64}642e2+e​
  3. (C)2e2−e128\dfrac{2e^2-e}{128}1282e2−e​
  4. (D)2e2+e128\dfrac{2e^2+e}{128}1282e2+e​

Correct answer: (C)

Step-by-step solution →
Q28·MathematicsSingle correctJEE Main 2025
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dydx+3(tan⁡2x) y+3y=sec⁡2x\dfrac{dy}{dx}+3(\tan^2 x)\,y+3y=\sec^2 xdxdy​+3(tan2x)y+3y=sec2x, y(0)=13+e3y(0)=\dfrac{1}{3}+e^3y(0)=31​+e3. Then y ⁣(π4)y\!\left(\dfrac{\pi}{4}\right)y(4π​) is equal to:
  1. (A)23\dfrac{2}{3}32​
  2. (B)43\dfrac{4}{3}34​
  3. (C)43+e3\dfrac{4}{3}+e^334​+e3
  4. (D)23+e3\dfrac{2}{3}+e^332​+e3

Correct answer: (B)

Step-by-step solution →
Q29·MathematicsSingle correctJEE Main 2025
Let ggg be a differentiable function such that ∫0xg(t) dt=x−∫0xt g(t) dt\displaystyle\int_0^x g(t)\,dt=x-\int_0^x t\,g(t)\,dt∫0x​g(t)dt=x−∫0x​tg(t)dt, x≥0x\ge 0x≥0 and let y=y(x)y=y(x)y=y(x) satisfy the differential equation dydx−ytan⁡x=2(x+1)sec⁡x g(x)\dfrac{dy}{dx}-y\tan x=2(x+1)\sec x\,g(x)dxdy​−ytanx=2(x+1)secxg(x), x∈[0,π2)x\in\left[0,\dfrac{\pi}{2}\right)x∈[0,2π​). If y(0)=0y(0)=0y(0)=0, then y ⁣(π3)y\!\left(\dfrac{\pi}{3}\right)y(3π​) is equal to:
  1. (A)2π33\dfrac{2\pi}{3\sqrt3}33​2π​
  2. (B)4π3\dfrac{4\pi}{3}34π​
  3. (C)2π3\dfrac{2\pi}{3}32π​
  4. (D)4π33\dfrac{4\pi}{3\sqrt3}33​4π​

Correct answer: (B)

Step-by-step solution →
Q30·MathematicsIntegerJEE Main 2025
Let f:R→Rf:R\to Rf:R→R be a thrice differentiable odd function satisfying f′′(x)≥0f''(x)\ge0f′′(x)≥0, f′′(x)=f(x)f''(x)=f(x)f′′(x)=f(x), f(0)=0f(0)=0f(0)=0, f′(0)=3f'(0)=3f′(0)=3. Then 9f(log⁡e3)9f(\log_e 3)9f(loge​3) is equal to ______.

Correct answer: 36

Step-by-step solution →
Q31·MathematicsIntegerJEE Main 2025
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dydx+2ysec⁡2x=2sec⁡2x+3tan⁡x⋅sec⁡2x\dfrac{dy}{dx}+2y\sec^2x=2\sec^2x+3\tan x\cdot\sec^2xdxdy​+2ysec2x=2sec2x+3tanx⋅sec2x such that y(0)=54y(0)=\dfrac{5}{4}y(0)=45​. Then 12(y(π4)−e−2)12\left(y\left(\dfrac{\pi}{4}\right)-e^{-2}\right)12(y(4π​)−e−2) is equal to ______.

Correct answer: 21

Step-by-step solution →
Q32·MathematicsSingle correctJEE Main 2025
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation cos⁡x (log⁡e(cos⁡x))2 dy+(sin⁡x−3ysin⁡x log⁡e(cos⁡x)) dx=0\cos x\,(\log_e(\cos x))^2\,dy+(\sin x-3y\sin x\,\log_e(\cos x))\,dx=0cosx(loge​(cosx))2dy+(sinx−3ysinxloge​(cosx))dx=0, x∈(0,π2)x\in\left(0,\dfrac{\pi}{2}\right)x∈(0,2π​). If y(π4)=−1log⁡e2y\left(\dfrac{\pi}{4}\right)=\dfrac{-1}{\log_e 2}y(4π​)=loge​2−1​, then y(π6)y\left(\dfrac{\pi}{6}\right)y(6π​) is:
  1. (A)2log⁡e3−log⁡e4\dfrac{2}{\log_e 3-\log_e 4}loge​3−loge​42​
  2. (B)1log⁡e4−log⁡e3\dfrac{1}{\log_e 4-\log_e 3}loge​4−loge​31​
  3. (C)−1log⁡e4\dfrac{-1}{\log_e 4}loge​4−1​
  4. (D)1log⁡e3−log⁡e4\dfrac{1}{\log_e 3-\log_e 4}loge​3−loge​41​

Correct answer: (D)

Step-by-step solution →
Q33·MathematicsSingle correctJEE Main 2025
If for the solution curve y=f(x)y=f(x)y=f(x) of the differential equation dydx+(tan⁡x)y=2+sec⁡x(1+2sec⁡x)2\dfrac{dy}{dx}+(\tan x)y=\dfrac{2+\sec x}{(1+2\sec x)^2}dxdy​+(tanx)y=(1+2secx)22+secx​, x∈(−π2,π2)x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)x∈(−2π​,2π​), f(π3)=310f\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{10}f(3π​)=103​​, then f(π4)f\left(\dfrac{\pi}{4}\right)f(4π​) is equal to:
  1. (A)93+310(4+3)\dfrac{9\sqrt{3}+3}{10(4+\sqrt{3})}10(4+3​)93​+3​
  2. (B)3+110(4+3)\dfrac{\sqrt{3}+1}{10(4+\sqrt{3})}10(4+3​)3​+1​
  3. (C)5−322\dfrac{5-\sqrt{3}}{2\sqrt{2}}22​5−3​​
  4. (D)4−214\dfrac{4-\sqrt{2}}{14}144−2​​

Correct answer: (D)

Step-by-step solution →
Q34·MathematicsIntegerJEE Main 2025
If y=y(x)y=y(x)y=y(x) is the solution of the differential equation 4−x2dydx=((sin⁡−1x2)2−y)sin⁡−1x2\sqrt{4-x^2}\frac{dy}{dx}=\left(\left(\sin^{-1}\frac{x}{2}\right)^2-y\right)\sin^{-1}\frac{x}{2}4−x2​dxdy​=((sin−12x​)2−y)sin−12x​, −2≤x≤2-2\le x\le2−2≤x≤2, y(2)=π2−84y(2)=\frac{\pi^2-8}{4}y(2)=4π2−8​, then y2(0)y^2(0)y2(0) is equal to ______.

Correct answer: 4

Step-by-step solution →
Q35·MathematicsSingle correctJEE Main 2025
Let for some function y=f(x)y=f(x)y=f(x), ∫0xt f(t)dt=x2f(x)\int_0^x t\,f(t)dt=x^2 f(x)∫0x​tf(t)dt=x2f(x), x>0x>0x>0 and f(2)=3f(2)=3f(2)=3. Then f(6)f(6)f(6) is equal to:
  1. (A)1
  2. (B)2
  3. (C)6
  4. (D)3

Correct answer: (A)

Step-by-step solution →
Q36·MathematicsSingle correctJEE Main 2025
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (xy−5x21+x2)dx+(1+x2)dy=0(xy-5x^2\sqrt{1+x^2})dx+(1+x^2)dy=0(xy−5x21+x2​)dx+(1+x2)dy=0, y(0)=0y(0)=0y(0)=0. Then y(3)y(\sqrt3)y(3​) is equal to
  1. (A)532\dfrac{5\sqrt3}{2}253​​
  2. (B)143\sqrt{\dfrac{14}{3}}314​​
  3. (C)222\sqrt222​
  4. (D)152\sqrt{\dfrac{15}{2}}215​​

Correct answer: (A)

Step-by-step solution →
Q37·MathematicsSingle correctJEE Main 2025
Let f:(0,∞)→Rf:(0,\infty)\to\mathbb{R}f:(0,∞)→R be a function differentiable at all points of its domain and satisfies the condition x2f′(x)=2xf(x)+3x^2 f'(x)=2x f(x)+3x2f′(x)=2xf(x)+3, with f(1)=4f(1)=4f(1)=4. Then 2f(2)2f(2)2f(2) is equal to:
  1. (A)29
  2. (B)19
  3. (C)39
  4. (D)23

Correct answer: (C)

Step-by-step solution →
Q38·MathematicsIntegerJEE Main 2025
Let f be a differentiable function such that 2(x+2)2f(x)−3(x+2)2=10∫0x(t+2)f(t) dt2(x+2)^2 f(x)-3(x+2)^2=10\int_0^x (t+2)f(t)\,dt2(x+2)2f(x)−3(x+2)2=10∫0x​(t+2)f(t)dt, x≥0x\ge 0x≥0. Then f(2)f(2)f(2) is equal to _______

Correct answer: 19

Step-by-step solution →
Q39·MathematicsIntegerJEE Main 2025
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation 2cos⁡xdydx=sin⁡2x−4ysin⁡x2\cos x\dfrac{dy}{dx}=\sin 2x-4y\sin x2cosxdxdy​=sin2x−4ysinx, x∈(0,π2)x\in\left(0,\dfrac{\pi}{2}\right)x∈(0,2π​). If y(π3)=0y\left(\dfrac{\pi}{3}\right)=0y(3π​)=0, then y(π4)+y′(π4)y\left(\dfrac{\pi}{4}\right)+y'\left(\dfrac{\pi}{4}\right)y(4π​)+y′(4π​) is equal to __________.

Correct answer: 1

Step-by-step solution →
Q40·MathematicsSingle correctJEE Main 2025
Let x=x(y)x=x(y)x=x(y) be the solution of the differential equation y=(x−ydxdy)sin⁡ ⁣(xy), y>0y=\left(x-y\dfrac{dx}{dy}\right)\sin\!\left(\dfrac{x}{y}\right),\ y>0y=(x−ydydx​)sin(yx​), y>0 and x(1)=π2x(1)=\dfrac{\pi}{2}x(1)=2π​. Then cos⁡(x(2))\cos\big(x(2)\big)cos(x(2)) is equal to :
  1. (A)1−2(log⁡e2)21-2(\log_e 2)^21−2(loge​2)2
  2. (B)2(log⁡e2)2−12(\log_e 2)^2-12(loge​2)2−1
  3. (C)2(log⁡e2)−12(\log_e 2)-12(loge​2)−1
  4. (D)1−2(log⁡e2)1-2(\log_e 2)1−2(loge​2)

Correct answer: (B)

Step-by-step solution →
Q41·MathematicsSingle correctJEE Main 2025
Let a curve y=f(x)y=f(x)y=f(x) pass through the points (0,5)(0,5)(0,5) and (log⁡e2,k)(\log_e 2, k)(loge​2,k). If the curve satisfies the differential equation 2(3+y)e2xdx−(7+e2x)dy=02(3+y)e^{2x}dx-(7+e^{2x})dy=02(3+y)e2xdx−(7+e2x)dy=0, then kkk is equal to
  1. (A)161616
  2. (B)888
  3. (C)323232
  4. (D)444

Correct answer: (B)

Step-by-step solution →
Q42·MathematicsIntegerJEE Main 2025
Let y=f(x)y=f(x)y=f(x) be the solution of the differential equation dydx+xyx2−1=x6+4x1−x2\dfrac{dy}{dx}+\dfrac{xy}{x^2-1}=\dfrac{x^6+4x}{\sqrt{1-x^2}}dxdy​+x2−1xy​=1−x2​x6+4x​, −1<x<1-1<x<1−1<x<1 such that f(0)=0f(0)=0f(0)=0. If 6∫−1/21/2f(x) dx=2π−α6\displaystyle\int_{-1/2}^{1/2}f(x)\,dx=2\pi-\alpha6∫−1/21/2​f(x)dx=2π−α then α2\alpha^2α2 is equal to ______.

Correct answer: 27

Step-by-step solution →
Q43·MathematicsSingle correctJEE Main 2025
Let x=x(y)x=x(y)x=x(y) be the solution of dxdy+xy2=1y3\dfrac{dx}{dy}+\dfrac{x}{y^{2}}=\dfrac{1}{y^{3}}dydx​+y2x​=y31​ with x(1)=1x(1)=1x(1)=1. Then x(12)x\big(\tfrac12\big)x(21​) is:
  1. (A)12+e\tfrac12+e21​+e
  2. (B)32+e\tfrac32+e23​+e
  3. (C)3-e
  4. (D)3+e

Correct answer: (C)

Step-by-step solution →
Q44·MathematicsSingle correctJEE Main 2025
If x=f(y)x=f(y)x=f(y) is the solution of the differential equation (1+y2)+(x−2etan⁡−1y)dydx=0(1+y^2)+\left(x-2e^{\tan^{-1}y}\right)\dfrac{dy}{dx}=0(1+y2)+(x−2etan−1y)dxdy​=0, y∈(−π2,π2)y\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)y∈(−2π​,2π​) with f(0)=1f(0)=1f(0)=1, then f(13)f\left(\dfrac{1}{\sqrt{3}}\right)f(3​1​) is equal to:
  1. (A)eπ/4e^{\pi/4}eπ/4
  2. (B)eπ/12e^{\pi/12}eπ/12
  3. (C)eπ/3e^{\pi/3}eπ/3
  4. (D)eπ/6e^{\pi/6}eπ/6

Correct answer: (D)

Step-by-step solution →
Q45·MathematicsSingle correctJEE Advanced 2024
Let f(x)f(x)f(x) be a continuously differentiable function on the interval (0,∞)(0, \infty)(0,∞) such that f(1)=2f(1) = 2f(1)=2 and lim⁡t→xt10f(x)−x10f(t)t9−x9=1\lim_{t \to x} \frac{t^{10} f(x) - x^{10} f(t)}{t^{9} - x^{9}} = 1limt→x​t9−x9t10f(x)−x10f(t)​=1 for each x>0x > 0x>0. Then, for all x>0x > 0x>0, f(x)f(x)f(x) is equal to
  1. (A)3111x−911x10\frac{31}{11x} - \frac{9}{11} x^{10}11x31​−119​x10
  2. (B)911x+1311x10\frac{9}{11x} + \frac{13}{11} x^{10}11x9​+1113​x10
  3. (C)−911x+3111x10\frac{-9}{11x} + \frac{31}{11} x^{10}11x−9​+1131​x10
  4. (D)1311x+911x10\frac{13}{11x} + \frac{9}{11} x^{10}11x13​+119​x10

Correct answer: (B)

Step-by-step solution →
Q46·MathematicsNumericalJEE Main 2024
For a differentiable function f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R, f′(x)=3f(x)+αf'(x)=3f(x)+\alphaf′(x)=3f(x)+α, where α∈R\alpha\in\mathbb{R}α∈R, f(0)=1f(0)=1f(0)=1 and lim⁡x→−∞f(x)=7\displaystyle\lim_{x\to-\infty}f(x)=7x→−∞lim​f(x)=7. Then 9f(−log⁡e3)9f(-\log_{e}3)9f(−loge​3) is equal to _______.

Correct answer: 61

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2024
The solution curve, of the differential equation 2ydydx+3=5dydx2y\frac{dy}{dx} + 3 = 5\frac{dy}{dx}2ydxdy​+3=5dxdy​, passing through the point (0,1)(0, 1)(0,1) is a conic, whose vertex lies on the line:
  1. (A)2x+3y=92x + 3y = 92x+3y=9
  2. (B)2x+3y=−92x + 3y = -92x+3y=−9
  3. (C)2x+3y=−62x + 3y = -62x+3y=−6
  4. (D)2x+3y=62x + 3y = 62x+3y=6

Correct answer: (A)

Step-by-step solution →
Q48·MathematicsSingle correctJEE Main 2024
The solution of the differential equation (x2+y2)dx−5xy dy=0(x^2 + y^2)dx - 5xy\, dy = 0(x2+y2)dx−5xydy=0, y(1)=0y(1) = 0y(1)=0, is :
  1. (A)∣x2−4y2∣5=x2\left|x^2 - 4y^2\right|^5 = x^2​x2−4y2​5=x2
  2. (B)∣x2−2y2∣6=x\left|x^2 - 2y^2\right|^6 = x​x2−2y2​6=x
  3. (C)∣x2−4y2∣6=x\left|x^2 - 4y^2\right|^6 = x​x2−4y2​6=x
  4. (D)∣x2−2y2∣5=x2\left|x^2 - 2y^2\right|^5 = x^2​x2−2y2​5=x2

Correct answer: (A)

Step-by-step solution →
Q49·MathematicsSingle correctJEE Main 2024
Let ∫0x1−(y′(t))2 dt=∫0xy(t) dt\displaystyle\int_{0}^{x}\sqrt{1-(y'(t))^{2}}\,dt=\int_{0}^{x}y(t)\,dt∫0x​1−(y′(t))2​dt=∫0x​y(t)dt, 0≤x≤30\le x\le 30≤x≤3, y≥0y\ge 0y≥0, y(0)=0y(0)=0y(0)=0. Then at x=2x=2x=2, y′′+y+1y''+y+1y′′+y+1 is equal to:
  1. (A)111
  2. (B)222
  3. (C)2\sqrt{2}2​
  4. (D)12\dfrac{1}{2}21​

Correct answer: (A)

Step-by-step solution →
Q50·MathematicsSingle correctJEE Main 2024
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (1+y2)etan⁡x dx+cos⁡2x(1+e2tan⁡x) dy=0(1+y^2)e^{\tan x}\,dx+\cos^2 x(1+e^{2\tan x})\,dy=0(1+y2)etanxdx+cos2x(1+e2tanx)dy=0, y(0)=1y(0)=1y(0)=1. Then y(π4)y\left(\dfrac\pi4\right)y(4π​) is equal to:
  1. (A)2e\dfrac2ee2​
  2. (B)1e2\dfrac{1}{e^2}e21​
  3. (C)1e\dfrac1ee1​
  4. (D)2e2\dfrac{2}{e^2}e22​

Correct answer: (C)

Step-by-step solution →
Q51·MathematicsSingle correctJEE Main 2024
Let y=y(x)y=y(x)y=y(x) be the solution curve of the differential equation sec⁡y dydx+2xsin⁡y=x3cos⁡y\sec y\,\dfrac{dy}{dx}+2x\sin y=x^3\cos ysecydxdy​+2xsiny=x3cosy, y(1)=0y(1)=0y(1)=0. Then y(3)y(\sqrt{3})y(3​) is equal to
  1. (A)π3\tfrac{\pi}{3}3π​
  2. (B)π6\tfrac{\pi}{6}6π​
  3. (C)π4\tfrac{\pi}{4}4π​
  4. (D)π12\tfrac{\pi}{12}12π​

Correct answer: (C)

Step-by-step solution →
Q52·MathematicsSingle correctJEE Main 2024
Let f(x)f(x)f(x) be a positive function such that the area bounded by y=f(x)y=f(x)y=f(x), y=0y=0y=0 from x=0x=0x=0 to x=a>0x=a>0x=a>0 is e−a+4a2+a−1e^{-a}+4a^2+a-1e−a+4a2+a−1. Then the differential equation, whose general solution is y=c1f(x)+c2y=c_1 f(x)+c_2y=c1​f(x)+c2​, where c1c_1c1​ and c2c_2c2​ are arbitrary constants, is:
  1. (A)(8ex−1)d2ydx2+dydx=0(8e^x-1)\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=0(8ex−1)dx2d2y​+dxdy​=0
  2. (B)(8ex−1)d2ydx2−dydx=0(8e^x-1)\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}=0(8ex−1)dx2d2y​−dxdy​=0
  3. (C)(8ex+1)d2ydx2+dydx=0(8e^x+1)\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=0(8ex+1)dx2d2y​+dxdy​=0
  4. (D)(8ex+1)d2ydx2−dydx=0(8e^x+1)\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}=0(8ex+1)dx2d2y​−dxdy​=0

Correct answer: (C)

Step-by-step solution →
Q53·MathematicsNumericalJEE Main 2024
Let α∣x∣=∣y∣exy−β\alpha|x|=|y|e^{xy-\beta}α∣x∣=∣y∣exy−β, α,β∈N\alpha,\beta\in\mathbb{N}α,β∈N be the solution of the differential equation x dy−y dx+xy(x dy+y dx)=0x\,dy-y\,dx+xy(x\,dy+y\,dx)=0xdy−ydx+xy(xdy+ydx)=0, y(1)=2y(1)=2y(1)=2. Then α+β\alpha+\betaα+β is equal to _______ .

Correct answer: 4

Step-by-step solution →
Q54·MathematicsSingle correctJEE Main 2024
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (1+x2)dydx+y=etan⁡−1x(1+x^{2})\dfrac{dy}{dx}+y=e^{\tan^{-1}x}(1+x2)dxdy​+y=etan−1x, y(1)=0y(1)=0y(1)=0. Then y(0)y(0)y(0) is
  1. (A)14(eπ/2−1)\tfrac{1}{4}\left(e^{\pi/2}-1\right)41​(eπ/2−1)
  2. (B)12(1−eπ/2)\tfrac{1}{2}\left(1-e^{\pi/2}\right)21​(1−eπ/2)
  3. (C)14(1−eπ/2)\tfrac{1}{4}\left(1-e^{\pi/2}\right)41​(1−eπ/2)
  4. (D)12(eπ/2−1)\tfrac{1}{2}\left(e^{\pi/2}-1\right)21​(eπ/2−1)

Correct answer: (B)

Step-by-step solution →
Q55·MathematicsSingle correctJEE Main 2024
Suppose the solution of the differential equation dydx=(2+α)x−βy+2βx−2αy−(βγ−4α)\frac{dy}{dx} = \frac{(2+\alpha)x - \beta y + 2}{\beta x - 2\alpha y - (\beta\gamma - 4\alpha)}dxdy​=βx−2αy−(βγ−4α)(2+α)x−βy+2​ represents a circle passing through origin. Then the radius of this circle is:
  1. (A)17\sqrt{17}17​
  2. (B)12\tfrac{1}{2}21​
  3. (C)172\tfrac{\sqrt{17}}{2}217​​
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q56·MathematicsNumericalJEE Main 2024
If the solution y(x)y(x)y(x) of the given differential equation (ey+1)cos⁡x dx+eysin⁡x dy=0(e^{y}+1)\cos x\,dx+e^{y}\sin x\,dy=0(ey+1)cosxdx+eysinxdy=0 passes through the point (π2,0)\left(\frac{\pi}{2},0\right)(2π​,0), then the value of ey(π6)e^{y\left(\frac{\pi}{6}\right)}ey(6π​) is equal to ______.

Correct answer: 3

Step-by-step solution →
Q57·MathematicsSingle correctJEE Main 2024
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (2xlog⁡ex)dydx+2y=3xlog⁡ex(2x\log_{e}x)\dfrac{dy}{dx}+2y=\dfrac{3}{x}\log_{e}x(2xloge​x)dxdy​+2y=x3​loge​x, x>0x>0x>0 and y(e−1)=0y(e^{-1})=0y(e−1)=0. Then y(e)y(e)y(e) is equal to
  1. (A)32e\tfrac{3}{2e}2e3​
  2. (B)23e\tfrac{2}{3e}3e2​
  3. (C)−3e-\tfrac{3}{e}−e3​
  4. (D)2e\tfrac{2}{e}e2​

Correct answer: (C)

Step-by-step solution →
Q58·MathematicsNumericalJEE Main 2024
Let fff be a differentiable function in the interval (0,∞)(0, \infty)(0,∞) such that f(1)=1f(1) = 1f(1)=1 and lim⁡t→xt2f(x)−x2f(t)t−x=1\displaystyle\lim_{t \to x} \dfrac{t^2 f(x) - x^2 f(t)}{t - x} = 1t→xlim​t−xt2f(x)−x2f(t)​=1 for each x>0x > 0x>0. Then 2f(2)+3f(3)2f(2) + 3f(3)2f(2)+3f(3) is equal to ___.

Correct answer: 24

Step-by-step solution →
Q59·MathematicsSingle correctJEE Main 2024
The differential equation of the family of circles passing through the origin and having their centre on the line y=xy=xy=x is:
  1. (A)(x2−y2+2xy) dx=(x2−y2+2xy) dy(x^2-y^2+2xy)\,dx=(x^2-y^2+2xy)\,dy(x2−y2+2xy)dx=(x2−y2+2xy)dy
  2. (B)(x2+y2+2xy) dx=(x2+y2−2xy) dy(x^2+y^2+2xy)\,dx=(x^2+y^2-2xy)\,dy(x2+y2+2xy)dx=(x2+y2−2xy)dy
  3. (C)(x2−y2+2xy) dx=(x2−y2−2xy) dy(x^2-y^2+2xy)\,dx=(x^2-y^2-2xy)\,dy(x2−y2+2xy)dx=(x2−y2−2xy)dy
  4. (D)(x2+y2−2xy) dx=(x2+y2+2xy) dy(x^2+y^2-2xy)\,dx=(x^2+y^2+2xy)\,dy(x2+y2−2xy)dx=(x2+y2+2xy)dy

Correct answer: (C)

Step-by-step solution →
Q60·MathematicsSingle correctJEE Main 2024
If y=y(x)y = y(x)y=y(x) is the solution of the differential equation dydx+2y=sin⁡(2x)\dfrac{dy}{dx} + 2y = \sin(2x)dxdy​+2y=sin(2x), y(0)=34y(0) = \dfrac{3}{4}y(0)=43​, then y(π8)y\left(\dfrac{\pi}{8}\right)y(8π​) is equal to:
  1. (A)e−π/8e^{-\pi/8}e−π/8
  2. (B)e−π/4e^{-\pi/4}e−π/4
  3. (C)eπ/4e^{\pi/4}eπ/4
  4. (D)eπ/8e^{\pi/8}eπ/8

Correct answer: (B)

Step-by-step solution →
Q61·MathematicsNumericalJEE Main 2024
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dydx+2x(1+x2)2 y=x e11+x2\dfrac{dy}{dx}+\dfrac{2x}{(1+x^2)^2}\,y=x\,e^{\frac{1}{1+x^2}}dxdy​+(1+x2)22x​y=xe1+x21​; y(0)=0y(0)=0y(0)=0. Then the area enclosed by the curve f(x)=y(x) e−11+x2f(x)=y(x)\,e^{-\frac{1}{1+x^2}}f(x)=y(x)e−1+x21​ and the line y−x=4y-x=4y−x=4 is __________.

Correct answer: 18

Step-by-step solution →
Q62·MathematicsNumericalJEE Main 2024
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (x+y+2)2 dx=dy(x+y+2)^2\,dx=dy(x+y+2)2dx=dy, y(0)=−2y(0)=-2y(0)=−2. Let the maximum and minimum values of the function y=y(x)y=y(x)y=y(x) in [0,π3]\left[0,\dfrac{\pi}{3}\right][0,3π​] be α\alphaα and β\betaβ, respectively. If (3α+π)2+β2=γ+δ3(3\alpha+\pi)^2+\beta^2=\gamma+\delta\sqrt{3}(3α+π)2+β2=γ+δ3​, γ,δ∈Z\gamma,\delta\in\mathbb{Z}γ,δ∈Z, then γ+δ\gamma+\deltaγ+δ equals

Correct answer: 31

Step-by-step solution →
Q63·MathematicsSingle correctJEE Main 2024
If the solution y=y(x)y=y(x)y=y(x) of the differential equation (x4+2x3+3x2+2x+2) dy=(2x2+2x+3) dx(x^4+2x^3+3x^2+2x+2)\,dy=(2x^2+2x+3)\,dx(x4+2x3+3x2+2x+2)dy=(2x2+2x+3)dx satisfies y(−1)=−π4y(-1)=-\dfrac\pi4y(−1)=−4π​, then y(0)y(0)y(0) is equal to:
  1. (A)−π12-\dfrac{\pi}{12}−12π​
  2. (B)000
  3. (C)π4\dfrac\pi44π​
  4. (D)π2\dfrac\pi22π​

Correct answer: (C)

Step-by-step solution →
Q64·MathematicsSingle correctJEE Main 2024
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (x2+4)2 dy+(2x3y+8xy−2) dx=0(x^2+4)^2\,dy+(2x^3y+8xy-2)\,dx=0(x2+4)2dy+(2x3y+8xy−2)dx=0. If y(0)=0y(0)=0y(0)=0, then y(2)y(2)y(2) is equal to
  1. (A)π8\tfrac{\pi}{8}8π​
  2. (B)π16\tfrac{\pi}{16}16π​
  3. (C)2π2\pi2π
  4. (D)π32\tfrac{\pi}{32}32π​

Correct answer: (D)

Step-by-step solution →
Q65·MathematicsNumericalJEE Main 2024
Let the solution y=y(x)y=y(x)y=y(x) of the differential equation dydx−y=1+4sin⁡x\dfrac{dy}{dx}-y=1+4\sin xdxdy​−y=1+4sinx satisfy y(π)=1y(\pi)=1y(π)=1. Then y(π2)+10y\left(\dfrac\pi2\right)+10y(2π​)+10 is equal to ___

Correct answer: 7

Step-by-step solution →
Q66·MathematicsNumericalJEE Main 2024
If dxdy=1+x−y2y\dfrac{dx}{dy}=\dfrac{1+x-y^2}{y}dydx​=y1+x−y2​, x(1)=1x(1)=1x(1)=1, then 5x(2)5x(2)5x(2) is equal to __________.

Correct answer: 5

Step-by-step solution →
Q67·MathematicsSingle correctJEE Main 2024
Let α\alphaα be a non-zero real number. Suppose f:R→Rf:\mathbb R\to\mathbb Rf:R→R is a differentiable function such that f(0)=2f(0)=2f(0)=2 and lim⁡x→−∞f(x)=1\displaystyle\lim_{x\to-\infty}f(x)=1x→−∞lim​f(x)=1. If f′(x)=αf(x)+3f'(x)=\alpha f(x)+3f′(x)=αf(x)+3, for all x∈Rx\in\mathbb Rx∈R, then f(−log⁡e2)f(-\log_e 2)f(−loge​2) is equal to:
  1. (A)3
  2. (B)5
  3. (C)9
  4. (D)7

Correct answer: (C)

Step-by-step solution →
Q68·MathematicsNumericalJEE Main 2024
If y=x(t)y=x(t)y=x(t) is the solution of the differential equation (t+1) dx=(2x+(t+1)4)dt(t+1)\,dx=\left(2x+(t+1)^4\right)dt(t+1)dx=(2x+(t+1)4)dt, x(0)=2x(0)=2x(0)=2, then x(1)x(1)x(1) equals ___

Correct answer: 14

Step-by-step solution →
Q69·MathematicsSingle correctJEE Main 2024
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dydx=2x(x+y)3−x(x+y)−1\dfrac{dy}{dx}=2x(x+y)^3-x(x+y)-1dxdy​=2x(x+y)3−x(x+y)−1, y(0)=1y(0)=1y(0)=1. Then (12+y(12))2\left(\dfrac{1}{\sqrt2}+y\left(\dfrac{1}{\sqrt2}\right)\right)^2(2​1​+y(2​1​))2 equals:
  1. (A)14+e\dfrac{1}{4+\sqrt e}4+e​1​
  2. (B)33−e\dfrac{3}{3-\sqrt e}3−e​3​
  3. (C)21+e\dfrac{2}{1+\sqrt e}1+e​2​
  4. (D)12−e\dfrac{1}{2-\sqrt e}2−e​1​

Correct answer: (D)

Step-by-step solution →
Q70·MathematicsSingle correctJEE Main 2024
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dydx=tan⁡x+ysin⁡x(sec⁡x−sin⁡xtan⁡x)\dfrac{dy}{dx}=\dfrac{\tan x+y}{\sin x(\sec x-\sin x\tan x)}dxdy​=sinx(secx−sinxtanx)tanx+y​, x∈(0,π2)x\in\left(0,\dfrac{\pi}{2}\right)x∈(0,2π​) satisfying the condition y(π4)=2y\left(\dfrac{\pi}{4}\right)=2y(4π​)=2. Then, y(π3)y\left(\dfrac{\pi}{3}\right)y(3π​) is
  1. (A)3(2+log⁡e3)\sqrt{3}\left(2+\log_e\sqrt{3}\right)3​(2+loge​3​)
  2. (B)32(2+log⁡e3)\dfrac{\sqrt{3}}{2}\left(2+\log_e 3\right)23​​(2+loge​3)
  3. (C)3(1+2log⁡e3)\sqrt{3}\left(1+2\log_e 3\right)3​(1+2loge​3)
  4. (D)3(2+log⁡e3)\sqrt{3}\left(2+\log_e 3\right)3​(2+loge​3)

Correct answer: (A)

Step-by-step solution →
Q71·MathematicsSingle correctJEE Main 2024
The temperature T(t)T(t)T(t) of a body at time t=0t=0t=0 is 160∘160^\circ160∘F and it decreases continuously as per the differential equation dTdt=−K(T−80)\dfrac{dT}{dt}=-K(T-80)dtdT​=−K(T−80), where K is a positive constant. If T(15)=120∘T(15)=120^\circT(15)=120∘F, then T(45)T(45)T(45) is equal to
  1. (A)85∘85^\circ85∘F
  2. (B)95∘95^\circ95∘F
  3. (C)90∘90^\circ90∘F
  4. (D)80∘80^\circ80∘F

Correct answer: (C)

Step-by-step solution →
Q72·MathematicsSingle correctJEE Main 2024
The solution curve of the differential equation ydxdy=x(log⁡ex−log⁡ey+1)y\dfrac{dx}{dy}=x\left(\log_e x-\log_e y+1\right)ydydx​=x(loge​x−loge​y+1), x>0, y>0x>0,\ y>0x>0, y>0 passing through the point (e,1)(e,1)(e,1) is
  1. (A)∣log⁡eyx∣=x\left|\log_e\dfrac{y}{x}\right|=x​loge​xy​​=x
  2. (B)∣log⁡eyx∣=y2\left|\log_e\dfrac{y}{x}\right|=y^2​loge​xy​​=y2
  3. (C)∣log⁡exy∣=y\left|\log_e\dfrac{x}{y}\right|=y​loge​yx​​=y
  4. (D)2∣log⁡exy∣=y+12\left|\log_e\dfrac{x}{y}\right|=y+12​loge​yx​​=y+1

Correct answer: (C)

Step-by-step solution →
Q73·MathematicsNumericalJEE Main 2024
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation sec⁡2x dx+(e2ytan⁡2x+tan⁡x)dy=0\sec^2 x\,dx+\left(e^{2y}\tan^2 x+\tan x\right)dy=0sec2xdx+(e2ytan2x+tanx)dy=0, 0<x<π20<x<\dfrac{\pi}{2}0<x<2π​, y(π4)=0y\left(\dfrac{\pi}{4}\right)=0y(4π​)=0. If y(π6)=αy\left(\dfrac{\pi}{6}\right)=\alphay(6π​)=α, Then e8αe^{8\alpha}e8α is equal to ______.

Correct answer: 9

Step-by-step solution →
Q74·MathematicsNumericalJEE Main 2024
Let Y=Y(X)Y=Y(X)Y=Y(X) be a curve lying in the first quadrant such that the area enclosed by the line Y−y=Y′(x)(X−x)Y-y=Y'(x)(X-x)Y−y=Y′(x)(X−x) and the co-ordinate axes, where (x,y)(x,y)(x,y) is any point on the curve, is always −y22Y′(x)+1\dfrac{-y^2}{2Y'(x)}+12Y′(x)−y2​+1, Y′(x)≠0Y'(x)\ne 0Y′(x)=0. If Y(1)=1Y(1)=1Y(1)=1, then 12 Y(2)12\,Y(2)12Y(2) equals ______.

Correct answer: 20

Step-by-step solution →
Q75·MathematicsSingle correctJEE Main 2024
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation sec⁡x dy+{2(1−x)tan⁡x+x(2−x)}dx=0\sec x\,dy+\{2(1-x)\tan x+x(2-x)\}dx=0secxdy+{2(1−x)tanx+x(2−x)}dx=0 such that y(0)=2y(0)=2y(0)=2. Then y(2)y(2)y(2) is equal to:
  1. (A)222
  2. (B)2(1−sin⁡(2))2(1-\sin(2))2(1−sin(2))
  3. (C)2sin⁡(2)+12\sin(2)+12sin(2)+1
  4. (D)111

Correct answer: (A)

Step-by-step solution →
Q76·MathematicsNumericalJEE Main 2024
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (1−x2) dy=[xy+(x3+2)3(1−x2)]dx(1-x^2)\,dy=\left[xy+(x^3+2)\sqrt{3(1-x^2)}\right]dx(1−x2)dy=[xy+(x3+2)3(1−x2)​]dx, −1<x<1-1<x<1−1<x<1, y(0)=0y(0)=0y(0)=0. If y(12)=mny\left(\dfrac12\right)=\dfrac{m}{n}y(21​)=nm​, mmm and nnn are co-prime numbers, then m+nm+nm+n is equal to ___

Correct answer: 97

Step-by-step solution →
Q77·MathematicsSingle correctJEE Main 2024
If sin⁡(yx)=log⁡e∣x∣+α2\sin\left(\dfrac{y}{x}\right)=\log_e|x|+\dfrac{\alpha}{2}sin(xy​)=loge​∣x∣+2α​ is the solution of the differential equation xcos⁡(yx)dydx=ycos⁡(yx)+xx\cos\left(\dfrac{y}{x}\right)\dfrac{dy}{dx}=y\cos\left(\dfrac{y}{x}\right)+xxcos(xy​)dxdy​=ycos(xy​)+x and y(1)=π3y(1)=\dfrac{\pi}{3}y(1)=3π​, then α2\alpha^2α2 is equal to:
  1. (A)3
  2. (B)12
  3. (C)4
  4. (D)9

Correct answer: (A)

Step-by-step solution →
Q78·MathematicsNumericalJEE Main 2024
If the solution curve y=y(x)y=y(x)y=y(x) of the differential equation (1+y2)(1+log⁡ex)dx+x dy=0\left(1+y^2\right)\left(1+\log_e x\right)dx+x\,dy=0(1+y2)(1+loge​x)dx+xdy=0, x>0x>0x>0 passes through the point (1,1)(1,1)(1,1) and y(e)=α−tan⁡(32)β+tan⁡(32)y(e)=\dfrac{\alpha-\tan\left(\frac{3}{2}\right)}{\beta+\tan\left(\frac{3}{2}\right)}y(e)=β+tan(23​)α−tan(23​)​, then α+2β\alpha+2\betaα+2β is ______.

Correct answer: 3

Step-by-step solution →
Q79·MathematicsSingle correctJEE Main 2024
A function y=f(x)y=f(x)y=f(x) satisfies f(x)sin⁡2x+sin⁡x−(1+cos⁡2x)f′(x)=0f(x)\sin 2x+\sin x-(1+\cos^2 x)f'(x)=0f(x)sin2x+sinx−(1+cos2x)f′(x)=0 with condition f(0)=0f(0)=0f(0)=0. Then f(π2)f\left(\dfrac{\pi}{2}\right)f(2π​) is equal to
  1. (A)111
  2. (B)000
  3. (C)−1-1−1
  4. (D)222

Correct answer: (A)

Step-by-step solution →
Q80·MathematicsNumericalJEE Main 2024
If the solution curve, of the differential equation dydx=x+y−2x−y\dfrac{dy}{dx}=\dfrac{x+y-2}{x-y}dxdy​=x−yx+y−2​ passing through the point (2,1)(2,1)(2,1) is tan⁡−1(y−1x−1)−1βlog⁡e(α+(y−1x−1)2)=log⁡e∣x−1∣\tan^{-1}\left(\dfrac{y-1}{x-1}\right)-\dfrac{1}{\beta}\log_e\left(\alpha+\left(\dfrac{y-1}{x-1}\right)^2\right)=\log_e|x-1|tan−1(x−1y−1​)−β1​loge​(α+(x−1y−1​)2)=loge​∣x−1∣, then 5β+α5\beta+\alpha5β+α is equal to __________.

Correct answer: 11

Step-by-step solution →
Q81·MathematicsSingle correctJEE Main 2024
Let x=x(t)x=x(t)x=x(t) and y=y(t)y=y(t)y=y(t) be solutions of the differential equations dxdt+ax=0\dfrac{dx}{dt}+ax=0dtdx​+ax=0 and dydt+by=0\dfrac{dy}{dt}+by=0dtdy​+by=0 respectively, a,b∈Ra,b\in\mathbb Ra,b∈R. Given that x(0)=2x(0)=2x(0)=2, y(0)=1y(0)=1y(0)=1 and 3y(1)=2x(1)3y(1)=2x(1)3y(1)=2x(1), the value of ttt, for which x(t)=y(t)x(t)=y(t)x(t)=y(t), is:
  1. (A)log⁡2/32\log_{2/3} 2log2/3​2
  2. (B)log⁡43\log_4 3log4​3
  3. (C)log⁡34\log_3 4log3​4
  4. (D)log⁡4/32\log_{4/3} 2log4/3​2

Correct answer: (D)

Step-by-step solution →
Q82·MathematicsNumericalJEE Main 2024
If the solution of the differential equation (2x+3y−2) dx+(4x+6y−7) dy=0(2x+3y-2)\,dx+(4x+6y-7)\,dy=0(2x+3y−2)dx+(4x+6y−7)dy=0, y(0)=3y(0)=3y(0)=3, is αx+βy+3log⁡e∣2x+3y−γ∣=6\alpha x+\beta y+3\log_e|2x+3y-\gamma|=6αx+βy+3loge​∣2x+3y−γ∣=6, then α+2β+3γ\alpha+2\beta+3\gammaα+2β+3γ is equal to __________.

Correct answer: 29

Step-by-step solution →
Q83·MathematicsNumericalJEE Main 2024
Let for a differentiable function f:(0,∞)→Rf:(0,\infty)\to\mathbb Rf:(0,∞)→R, f(x)−f(y)≥log⁡e(x/y)+x−yf(x)-f(y)\ge\log_e(x/y)+x-yf(x)−f(y)≥loge​(x/y)+x−y, ∀ x,y∈(0,∞)\forall\,x,y\in(0,\infty)∀x,y∈(0,∞). Then ∑n=120f′ ⁣(1n2)\displaystyle\sum_{n=1}^{20}f'\!\left(\dfrac{1}{n^2}\right)n=1∑20​f′(n21​) is equal to __________.

Correct answer: 2890

Step-by-step solution →
Q84·MathematicsSingle correctJEE Main 2024
If y=y(x)y=y(x)y=y(x) is the solution curve of the differential equation (x2−4) dy−(y2−3y) dx=0(x^2-4)\,dy-(y^2-3y)\,dx=0(x2−4)dy−(y2−3y)dx=0, x>2x>2x>2, y(4)=32y(4)=\dfrac32y(4)=23​ and the slope of the curve is never zero, then the value of y(10)y(10)y(10) equals :
  1. (A)31+(8)1/4\dfrac{3}{1+(8)^{1/4}}1+(8)1/43​
  2. (B)31+22\dfrac{3}{1+2\sqrt2}1+22​3​
  3. (C)31−22\dfrac{3}{1-2\sqrt2}1−22​3​
  4. (D)31−(8)1/4\dfrac{3}{1-(8)^{1/4}}1−(8)1/43​

Correct answer: (A)

Step-by-step solution →
Q85·MathematicsIntegerJEE Advanced 2023
For x∈Rx \in \mathbb{R}x∈R, let y(x)y(x)y(x) be a solution of the differential equation (x2−5)dydx−2xy=−2x(x2−5)2(x^{2} - 5)\frac{dy}{dx} - 2xy = -2x(x^{2} - 5)^{2}(x2−5)dxdy​−2xy=−2x(x2−5)2 such that y(2)=7y(2) = 7y(2)=7. Then the maximum value of the function y(x)y(x)y(x) is

Correct answer: 16

Step-by-step solution →
Q86·MathematicsSingle correctJEE Advanced 2023
Let f:[1,∞)→Rf : [1, \infty) \to \mathbb{R}f:[1,∞)→R be a differentiable function such that f(1)=13f(1) = \frac{1}{3}f(1)=31​ and 3∫1xf(t) dt=xf(x)−x333\int_{1}^{x} f(t)\,dt = x f(x) - \frac{x^{3}}{3}3∫1x​f(t)dt=xf(x)−3x3​, x∈[1,∞)x \in [1, \infty)x∈[1,∞). Let e denote the base of the natural logarithm. Then the value of f(e)f(e)f(e) is
  1. (A)e2+43\frac{e^{2}+4}{3}3e2+4​
  2. (B)log⁡e4+e3\frac{\log_{e} 4 + e}{3}3loge​4+e​
  3. (C)4e23\frac{4e^{2}}{3}34e2​
  4. (D)e2−43\frac{e^{2}-4}{3}3e2−4​

Correct answer: (C)

Step-by-step solution →
Q87·MathematicsSingle correctJEE Main 2023
Let x=x(y)x=x(y)x=x(y) be the solution of the differential equation 2(y+2)log⁡e(y+2) dx+(x+4−2log⁡e(y+2)) dy=02(y+2)\log_e(y+2)\,dx+(x+4-2\log_e(y+2))\,dy=02(y+2)loge​(y+2)dx+(x+4−2loge​(y+2))dy=0, y>−1y>-1y>−1 with x(e4−2)=1x(e^4-2)=1x(e4−2)=1. Then x(e9−2)x(e^9-2)x(e9−2) is equal to
  1. (A)49\frac{4}{9}94​
  2. (B)103\frac{10}{3}310​
  3. (C)329\frac{32}{9}932​
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q88·MathematicsSingle correctJEE Main 2023
Let y=y1(x)y=y_{1}(x)y=y1​(x) and y=y2(x)y=y_{2}(x)y=y2​(x) be the solution curves of the differential equation dydx=y+7\dfrac{dy}{dx}=y+7dxdy​=y+7 with initial conditions y1(0)=0, y2(0)=1y_{1}(0)=0,\,y_{2}(0)=1y1​(0)=0,y2​(0)=1 respectively. Then the curves y=y1(x)y=y_{1}(x)y=y1​(x) and y=y2(x)y=y_{2}(x)y=y2​(x) intersect at:
  1. (A)Two points
  2. (B)no point
  3. (C)infinite number of points
  4. (D)one point

Correct answer: (B)

Step-by-step solution →
Q89·MathematicsNumericalJEE Main 2023
If y=y(x)y = y(x)y=y(x) is the solution of the differential equation dydx+4x(x2−1)y=x+2(x2−1)52\dfrac{dy}{dx} + \dfrac{4x}{(x^2-1)}y = \dfrac{x+2}{(x^2-1)^{\frac{5}{2}}}dxdy​+(x2−1)4x​y=(x2−1)25​x+2​, x>1x > 1x>1 such that y(2)=29log⁡e(2+3)y(2) = \dfrac{2}{9}\log_e\left(2 + \sqrt{3}\right)y(2)=92​loge​(2+3​) and y(2)=αlog⁡e(α+β)+β−γy\left(\sqrt{2}\right) = \alpha\log_e\left(\sqrt{\alpha} + \beta\right) + \beta - \sqrt{\gamma}y(2​)=αloge​(α​+β)+β−γ​, α,β,γ∈N\alpha, \beta, \gamma \in \mathbb{N}α,β,γ∈N, then αβγ\alpha\beta\gammaαβγ is equal to

Correct answer: 6

Step-by-step solution →
Q90·MathematicsSingle correctJEE Main 2023
Let y=y(x)y=y(x)y=y(x), y>0y>0y>0, be a solution curve of the differential equation (1+x2) dy=y(x−y) dx(1+x^2)\,dy=y(x-y)\,dx(1+x2)dy=y(x−y)dx. If y(0)=1y(0)=1y(0)=1 and y(22)=βy(2\sqrt2)=\betay(22​)=β, then
  1. (A)e1/β=e(3+22)e^{1/\beta}=e(3+2\sqrt2)e1/β=e(3+22​)
  2. (B)e1/β=e−2(5+42)e^{1/\beta}=e^{-2}(5+4\sqrt2)e1/β=e−2(5+42​)
  3. (C)e1/β=e−2(3+22)e^{1/\beta}=e^{-2}(3+2\sqrt2)e1/β=e−2(3+22​)
  4. (D)e1/β=e(5+42)e^{1/\beta}=e(5+4\sqrt2)e1/β=e(5+42​)

Correct answer: (A)

Step-by-step solution →
Q91·MathematicsSingle correctJEE Main 2023
Let y=y(x)y = y(x)y=y(x) be a solution curve of the differential equation, (1−x2y2)dx=y dx+x dy(1 - x^2 y^2) dx = y\, dx + x\, dy(1−x2y2)dx=ydx+xdy. If the line x=1x = 1x=1 intersects the curve y=y(x)y = y(x)y=y(x) at y=2y = 2y=2 and the line x=2x = 2x=2 intersects the curve y=y(x)y = y(x)y=y(x) at y=αy = \alphay=α, then a value of α\alphaα is
  1. (A)3e22(3e2−1)\frac{3e^2}{2(3e^2 - 1)}2(3e2−1)3e2​
  2. (B)3e22(3e2+1)\frac{3e^2}{2(3e^2 + 1)}2(3e2+1)3e2​
  3. (C)1−3e22(3e2+1)\frac{1 - 3e^2}{2(3e^2 + 1)}2(3e2+1)1−3e2​
  4. (D)1+3e22(3e2−1)\frac{1 + 3e^2}{2(3e^2 - 1)}2(3e2−1)1+3e2​

Correct answer: (D)

Step-by-step solution →
Q92·MathematicsSingle correctJEE Main 2023
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dydx+5x(x5+1)y=(x5+1)2x7\dfrac{dy}{dx}+\dfrac{5}{x\left(x^5+1\right)}y=\dfrac{\left(x^5+1\right)^2}{x^7}dxdy​+x(x5+1)5​y=x7(x5+1)2​, x>0x>0x>0. If y(1)=2y(1)=2y(1)=2, then y(2)y(2)y(2) is equal to
  1. (A)637128\dfrac{637}{128}128637​
  2. (B)679128\dfrac{679}{128}128679​
  3. (C)693128\dfrac{693}{128}128693​
  4. (D)697128\dfrac{697}{128}128697​

Correct answer: (C)

Step-by-step solution →
Q93·MathematicsSingle correctJEE Main 2023
The slope of tangent at any point (x,y)(x,y)(x,y) on a curve y=y(x)y=y(x)y=y(x) is x2+y22xy\dfrac{x^2+y^2}{2xy}2xyx2+y2​, x>0x>0x>0. If y(2)=0y(2)=0y(2)=0, then a value of y(8)y(8)y(8) is
  1. (A)−23-2\sqrt3−23​
  2. (B)434\sqrt343​
  3. (C)232\sqrt323​
  4. (D)−42-4\sqrt2−42​

Correct answer: (B)

Step-by-step solution →
Q94·MathematicsNumericalJEE Main 2023
Let the tangent at any point P on a curve passing through the points (1,1)(1,1)(1,1) and (110,100)\left(\frac{1}{10},100\right)(101​,100), intersect the positive x-axis and y-axis at the points A and B respectively. If PA:PB=1:kPA:PB=1:kPA:PB=1:k and y=y(x)y=y(x)y=y(x) is the solution of the differential equation edydx=kx+k2e^{\frac{dy}{dx}}=kx+\frac{k}{2}edxdy​=kx+2k​, y(0)=ky(0)=ky(0)=k, then 4y(1)−5log⁡e34y(1)-5\log_e 34y(1)−5loge​3 is equal to _______ .

Correct answer: 6

Step-by-step solution →
Q95·MathematicsNumericalJEE Main 2023
If the solution curve of the differential equation (y−2log⁡ex)dx+(xlog⁡ex2)dy=0(y-2\log_{e}x)dx+(x\log_{e}x^{2})dy=0(y−2loge​x)dx+(xloge​x2)dy=0, x>1x>1x>1 passes through the points (e,43)\left(e,\frac{4}{3}\right)(e,34​) and (e4,α)(e^{4},\alpha)(e4,α), then α\alphaα is equal to

Correct answer: 3

Step-by-step solution →
Q96·MathematicsNumericalJEE Main 2023
Let the solution curve x=x(y)x=x(y)x=x(y), 0<y<π20<y<\dfrac{\pi}{2}0<y<2π​, of the differential equation (log⁡e(cos⁡y))2cos⁡y dx−(1+3xlog⁡e(cos⁡y))sin⁡y dy=0(\log_e(\cos y))^{2}\cos y\,dx-(1+3x\log_e(\cos y))\sin y\,dy=0(loge​(cosy))2cosydx−(1+3xloge​(cosy))sinydy=0 satisfy x(π3)=12log⁡e2x\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2\log_e 2}x(3π​)=2loge​21​. If x(π6)=1log⁡em−log⁡enx\left(\dfrac{\pi}{6}\right)=\dfrac{1}{\log_e m-\log_e n}x(6π​)=loge​m−loge​n1​, where m and n are co-prime, then mn is equal to

Correct answer: 12

Step-by-step solution →
Q97·MathematicsNumericalJEE Main 2023
Let a curve y=f(x)y=f(x)y=f(x), x∈(0,∞)x\in(0,\infty)x∈(0,∞) pass through the points P(1,32)P\left(1,\frac{3}{2}\right)P(1,23​) and Q(a,12)Q\left(a,\frac{1}{2}\right)Q(a,21​). If the tangent at any point R(b,f(b))R(b,f(b))R(b,f(b)) to the given curve cuts the y-axis at the point S(0,c)S(0,c)S(0,c) such that bc=3bc=3bc=3, then (PQ)2(PQ)^2(PQ)2 is equal to

Correct answer: 5

Step-by-step solution →
Q98·MathematicsSingle correctJEE Main 2023
If the solution curve f(x,y)=0f(x, y) = 0f(x,y)=0 of the differential equation (1+log⁡ex)dxdy−xlog⁡ex=ey(1+\log_e x)\frac{dx}{dy}-x\log_e x=e^y(1+loge​x)dydx​−xloge​x=ey, x>0x>0x>0, passes through the points (1,0)(1,0)(1,0) and (α,2)(\alpha,2)(α,2) then αα\alpha^\alphaαα is equal to
  1. (A)e2e2e^{2e^{\sqrt{2}}}e2e2​
  2. (B)e2e2e^{\sqrt{2}e^2}e2​e2
  3. (C)ee2e^{e^2}ee2
  4. (D)e2e2e^{2e^2}e2e2

Correct answer: (D)

Step-by-step solution →
Q99·MathematicsNumericalJEE Main 2023
Let y=y(x)y=y(x)y=y(x) be a solution of the differential equation (xcos⁡x)dy+(xysin⁡x+ycos⁡x−1)dx=0, 0<x<π2(x\cos x)dy+(xy\sin x+y\cos x-1)dx=0,\,0<x<\dfrac{\pi}{2}(xcosx)dy+(xysinx+ycosx−1)dx=0,0<x<2π​. If y(π3)=3πy\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt3}{\pi}y(3π​)=π3​​, then ∣π6 y(π6)+2y(π6)∣\left|\dfrac{\pi}{6}\,y\left(\dfrac{\pi}{6}\right)+2y\left(\dfrac{\pi}{6}\right)\right|​6π​y(6π​)+2y(6π​)​ is equal to _____.

Correct answer: 2

Step-by-step solution →
Q100·MathematicsSingle correctJEE Main 2023
If y=y(x)y=y(x)y=y(x) is the solution curve of the differential equation dydx+ytan⁡x=xsec⁡x\frac{dy}{dx}+y\tan x=x\sec xdxdy​+ytanx=xsecx, 0≤x≤π30\le x\le\frac{\pi}{3}0≤x≤3π​, y(0)=1y(0)=1y(0)=1, then y ⁣(π6)y\!\left(\frac{\pi}{6}\right)y(6π​) is equal to
  1. (A)π12−32log⁡e ⁣(23e)\frac{\pi}{12}-\frac{\sqrt{3}}{2}\log_e\!\left(\frac{2\sqrt{3}}{e}\right)12π​−23​​loge​(e23​​)
  2. (B)π12−32log⁡e ⁣(2e3)\frac{\pi}{12}-\frac{\sqrt{3}}{2}\log_e\!\left(\frac{2}{e\sqrt{3}}\right)12π​−23​​loge​(e3​2​)
  3. (C)π12+32log⁡e ⁣(2e3)\frac{\pi}{12}+\frac{\sqrt{3}}{2}\log_e\!\left(\frac{2}{e\sqrt{3}}\right)12π​+23​​loge​(e3​2​)
  4. (D)π12+32log⁡e ⁣(23e)\frac{\pi}{12}+\frac{\sqrt{3}}{2}\log_e\!\left(\frac{2\sqrt{3}}{e}\right)12π​+23​​loge​(e23​​)

Correct answer: (B)

Step-by-step solution →
Q101·MathematicsSingle correctJEE Main 2023
Let αx=exp⁡(xβyγ)\alpha x=\exp(x^{\beta}y^{\gamma})αx=exp(xβyγ) be the solution of the differential equation 2x2y dy−(1−xy2)dx=02x^2y\,dy-(1-xy^2)dx=02x2ydy−(1−xy2)dx=0, x>0x>0x>0, y(2)=log⁡e2y(2)=\sqrt{\log_e 2}y(2)=loge​2​. Then α+β+γ\alpha+\beta+\gammaα+β+γ equals:
  1. (A)111
  2. (B)−1-1−1
  3. (C)333
  4. (D)000

Correct answer: (A)

Step-by-step solution →
Q102·MathematicsSingle correctJEE Main 2023
The area enclosed by the closed curve CCC given by the differential equation dydx+x+ay−2=0\frac{dy}{dx}+\frac{x+a}{y-2}=0dxdy​+y−2x+a​=0, y(1)=0y(1)=0y(1)=0 is 4π4\pi4π. Let PPP and QQQ be the points of intersection of the curve CCC and the yyy-axis. If normals at PPP and QQQ on the curve CCC intersect xxx-axis at points RRR and SSS respectively, then the length of the line segment RSRSRS is
  1. (A)222
  2. (B)433\frac{4\sqrt{3}}{3}343​​
  3. (C)232\sqrt{3}23​
  4. (D)233\frac{2\sqrt{3}}{3}323​​

Correct answer: (B)

Step-by-step solution →
Q103·MathematicsNumericalJEE Main 2023
Let f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R be a differentiable function such that f′(x)+f(x)=∫02f(t) dtf'(x)+f(x)=\int_0^2 f(t)\,dtf′(x)+f(x)=∫02​f(t)dt. If f(0)=e−2f(0)=e^{-2}f(0)=e−2, then 2f(0)−f(2)2f(0)-f(2)2f(0)−f(2) is equal to _______ .

Correct answer: 1

Step-by-step solution →
Q104·MathematicsSingle correctJEE Main 2023
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (3y2−5x2)y dx+2x(x2−y2)dy=0(3y^2-5x^2)y\,dx+2x(x^2-y^2)dy=0(3y2−5x2)ydx+2x(x2−y2)dy=0 such that y(1)=1y(1)=1y(1)=1. Then ∣(y(2))3−12y(2)∣\left|(y(2))^3-12y(2)\right|​(y(2))3−12y(2)​ is equal to:
  1. (A)16216\sqrt2162​
  2. (B)32232\sqrt2322​
  3. (C)323232
  4. (D)646464

Correct answer: (B)

Step-by-step solution →
Q105·MathematicsSingle correctJEE Main 2023
Let a differentiable function fff satisfy f(x)+∫3xf(t)t dt=x+1, x≥3f(x)+\int_{3}^{x}\frac{f(t)}{t}\,dt=\sqrt{x+1},\ x\ge 3f(x)+∫3x​tf(t)​dt=x+1​, x≥3. Then 12f(8)12f(8)12f(8) is equal to :
  1. (A)34
  2. (B)1
  3. (C)17
  4. (D)19

Correct answer: (C)

Step-by-step solution →
Q106·MathematicsSingle correctJEE Main 2023
Let the solution curve y=y(x)y=y(x)y=y(x) of the differential equation dydx−3x5tan⁡−1(x3)(1+x6)3/2y=2xexp⁡{x3−tan⁡−1x31+x6}\dfrac{dy}{dx}-\dfrac{3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}}y=2x\exp\left\{\dfrac{x^3-\tan^{-1}x^3}{\sqrt{1+x^6}}\right\}dxdy​−(1+x6)3/23x5tan−1(x3)​y=2xexp{1+x6​x3−tan−1x3​} pass through the origin. Then y(1)y(1)y(1) is equal to:
  1. (A)exp⁡(4+π42)\exp\left(\dfrac{4+\pi}{4\sqrt2}\right)exp(42​4+π​)
  2. (B)exp⁡(1−π42)\exp\left(\dfrac{1-\pi}{4\sqrt2}\right)exp(42​1−π​)
  3. (C)exp⁡(π−442)\exp\left(\dfrac{\pi-4}{4\sqrt2}\right)exp(42​π−4​)
  4. (D)exp⁡(4−π42)\exp\left(\dfrac{4-\pi}{4\sqrt2}\right)exp(42​4−π​)

Correct answer: (D)

Step-by-step solution →
Q107·MathematicsSingle correctJEE Main 2023
The solution of the differential equation dydx=−(x2+3y23x2+y2)\dfrac{dy}{dx}=-\left(\dfrac{x^2 + 3y^2}{3x^2 + y^2}\right)dxdy​=−(3x2+y2x2+3y2​), y(1)=0y(1)=0y(1)=0, is:
  1. (A)log⁡e∣x+y∣−xy(x+y)2=0\log_e|x + y| - \dfrac{xy}{(x + y)^2}=0loge​∣x+y∣−(x+y)2xy​=0
  2. (B)log⁡e∣x+y∣+2xy(x+y)2=0\log_e|x + y| + \dfrac{2xy}{(x + y)^2}=0loge​∣x+y∣+(x+y)22xy​=0
  3. (C)log⁡e∣x+y∣−2xy(x+y)2=0\log_e|x + y| - \dfrac{2xy}{(x + y)^2}=0loge​∣x+y∣−(x+y)22xy​=0
  4. (D)log⁡e∣x+y∣+xy(x+y)2=0\log_e|x + y| + \dfrac{xy}{(x + y)^2}=0loge​∣x+y∣+(x+y)2xy​=0

Correct answer: (B)

Step-by-step solution →
Q108·MathematicsSingle correctJEE Main 2023
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation xlog⁡exdydx+y=x2log⁡ex, (x>1)x\log_e x\dfrac{dy}{dx}+y=x^2\log_e x,\,(x>1)xloge​xdxdy​+y=x2loge​x,(x>1). If y(2)=2y(2)=2y(2)=2, then y(e)y(e)y(e) is equal to:
  1. (A)1+e22\dfrac{1+e^2}{2}21+e2​
  2. (B)4+e24\dfrac{4+e^2}{4}44+e2​
  3. (C)2+e22\dfrac{2+e^2}{2}22+e2​
  4. (D)1+e24\dfrac{1+e^2}{4}41+e2​

Correct answer: (B)

Step-by-step solution →
Q109·MathematicsSingle correctJEE Main 2023
Let y=f(x)y = f(x)y=f(x) be the solution of the differential equation y(x+1)dx−x2dy=0y(x + 1)dx - x^2 dy = 0y(x+1)dx−x2dy=0, y(1)=ey(1) = ey(1)=e. Then lim⁡x→0+f(x)\lim_{x \to 0^+} f(x)limx→0+​f(x) is equal to
  1. (A)1e2\dfrac{1}{e^2}e21​
  2. (B)e2e^2e2
  3. (C)0
  4. (D)1e\dfrac{1}{e}e1​

Correct answer: (C)

Step-by-step solution →
Q110·MathematicsSingle correctJEE Main 2023
Let y=y(x)y=y(x)y=y(x) be the solution curve of the differential equation dydx=yx(1+xy2(1+log⁡ex))\dfrac{dy}{dx}=\dfrac{y}{x}\big(1+xy^2(1+\log_e x)\big)dxdy​=xy​(1+xy2(1+loge​x)), x>0x>0x>0, y(1)=3y(1)=3y(1)=3. Then y2(x)9\dfrac{y^2(x)}{9}9y2(x)​ is equal to:
  1. (A)x22x3(2+log⁡ex3)−3\dfrac{x^2}{2x^3(2+\log_e x^3)-3}2x3(2+loge​x3)−3x2​
  2. (B)x23x3(1+log⁡ex2)−2\dfrac{x^2}{3x^3(1+\log_e x^2)-2}3x3(1+loge​x2)−2x2​
  3. (C)x27−3x3(2+log⁡ex2)\dfrac{x^2}{7-3x^3(2+\log_e x^2)}7−3x3(2+loge​x2)x2​
  4. (D)x25−2x3(2+log⁡ex3)\dfrac{x^2}{5-2x^3(2+\log_e x^3)}5−2x3(2+loge​x3)x2​

Correct answer: (D)

Step-by-step solution →
Q111·MathematicsSingle correctJEE Main 2023
Let y=y(t)y = y(t)y=y(t) be a solution of the differential equation dydt+αy=γe−βt\dfrac{dy}{dt} + \alpha y = \gamma e^{-\beta t}dtdy​+αy=γe−βt where α>0\alpha > 0α>0, β>0\beta > 0β>0 and γ>0\gamma > 0γ>0. Then lim⁡t→∞y(t)\displaystyle\lim_{t \to \infty} y(t)t→∞lim​y(t)
  1. (A)is −1-1−1
  2. (B)is 1
  3. (C)does not exist
  4. (D)is 0

Correct answer: (D)

Step-by-step solution →
Q112·MathematicsSingle correctJEE Main 2023
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation x3 dy+(xy−1) dx=0x^3\,dy+(xy-1)\,dx=0x3dy+(xy−1)dx=0, x>0x>0x>0, y ⁣(12)=3−ey\!\left(\tfrac{1}{2}\right)=3-ey(21​)=3−e. Then y(1)y(1)y(1) is equal to
  1. (A)111
  2. (B)eee
  3. (C)333
  4. (D)2−e2-e2−e

Correct answer: (A)

Step-by-step solution →
Q113·MathematicsSingle correctJEE Main 2023
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (x2−3y2)dx+3xy dy=0, y(1)=1(x^2-3y^2)dx+3xy\,dy=0,\ y(1)=1(x2−3y2)dx+3xydy=0, y(1)=1. Then 6y2(e)6y^2(e)6y2(e) is equal to
  1. (A)2e22e^22e2
  2. (B)3e23e^23e2
  3. (C)e2e^2e2
  4. (D)32e2\dfrac32 e^223​e2

Correct answer: (A)

Step-by-step solution →
Q114·MathematicsNumericalJEE Main 2023
Let fff be a differentiable function defined on [0,π2]\left[0,\dfrac{\pi}{2}\right][0,2π​] such that f(x)>0f(x)>0f(x)>0 and f(x)+∫0xf(t)1−(log⁡ef(t))2 dt=e, ∀x∈[0,π2]f(x)+\int_0^x f(t)\sqrt{1-(\log_e f(t))^2}\,dt=e,\ \forall x\in\left[0,\dfrac{\pi}{2}\right]f(x)+∫0x​f(t)1−(loge​f(t))2​dt=e, ∀x∈[0,2π​]. Then (6log⁡ef(π6))2\left(6\log_e f\left(\dfrac{\pi}{6}\right)\right)^2(6loge​f(6π​))2 is equal to

Correct answer: 27

Step-by-step solution →
Q115·MathematicsIntegerJEE Advanced 2022
If y(x) is the solution of the differential equation xdy−(y2−4y)dx=0xdy - \left(y^{2} - 4y\right)dx = 0xdy−(y2−4y)dx=0 for x > 0, y(1) = 2, and the slope of the curve y = y(x) is never zero, then the value of 10y(2)10y\left(\sqrt{2}\right)10y(2​) is ________.

Correct answer: 8

Step-by-step solution →
Q116·MathematicsMultiple correctJEE Advanced 2022
For x ∈ R, let the function y(x) be the solution of the differential equation dydx+12y=cos⁡(π12x)\frac{dy}{dx} + 12y = \cos\left(\frac{\pi}{12}x\right)dxdy​+12y=cos(12π​x), y(0) = 0 . Then, which of the following statements is/are TRUE?
  1. (A)y(x) is an increasing function
  2. (B)y(x) is a decreasing function
  3. (C)There exists a real number β such that the line y = β intersects the curve y = y(x) at infinitely many points
  4. (D)y(x) is a periodic function

Correct answer: (C)

Step-by-step solution →
Q117·MathematicsSingle correctJEE Main 2022
Let y=y(x)y = y(x)y=y(x) be the solution curve of the differential equation dydx+(2x2+11x+13x3+6x2+11x+6)y=(x+3)x+1\frac{dy}{dx} + \left( \frac{2x^2 + 11x + 13}{x^3 + 6x^2 + 11x + 6} \right) y = \frac{(x + 3)}{x + 1}dxdy​+(x3+6x2+11x+62x2+11x+13​)y=x+1(x+3)​, x>−1x > -1x>−1, which passes through the point (0,1)(0, 1)(0,1). Then y(1)y(1)y(1) is equal to:
  1. (A)12\frac{1}{2}21​
  2. (B)32\frac{3}{2}23​
  3. (C)52\frac{5}{2}25​
  4. (D)72\frac{7}{2}27​

Correct answer: (B)

Step-by-step solution →
Q118·MathematicsSingle correctJEE Main 2022
Let the solution curve y=y(x)y = y(x)y=y(x) of the differential equation (1+e2x)(dydx+y)=1(1 + e^{2x})\left(\frac{dy}{dx} + y\right) = 1(1+e2x)(dxdy​+y)=1 pass through the point (0,π2)\left(0, \frac{\pi}{2}\right)(0,2π​). Then, lim⁡x→∞exy(x)\lim_{x \to \infty} e^{x} y(x)limx→∞​exy(x) is equal to :
  1. (A)π4\frac{\pi}{4}4π​
  2. (B)3π4\frac{3\pi}{4}43π​
  3. (C)π2\frac{\pi}{2}2π​
  4. (D)3π2\frac{3\pi}{2}23π​

Correct answer: (B)

Step-by-step solution →
Q119·MathematicsSingle correctJEE Main 2022
If the solution curve of the differential equation dydx=x+y−2x−y\frac{dy}{dx} = \frac{x + y - 2}{x - y}dxdy​=x−yx+y−2​ passes through the point (2,1)(2, 1)(2,1) and (k+1,2)(k + 1, 2)(k+1,2), k>0k > 0k>0, then
  1. (A)2tan⁡−1(1k)=log⁡e(k2+1)2\tan^{-1}\left( \frac{1}{k} \right) = \log_e \left( k^2 + 1 \right)2tan−1(k1​)=loge​(k2+1)
  2. (B)tan⁡−1(1k)=log⁡e(k2+1)\tan^{-1}\left( \frac{1}{k} \right) = \log_e \left( k^2 + 1 \right)tan−1(k1​)=loge​(k2+1)
  3. (C)2tan⁡−1(1k+1)=log⁡e(k2+2k+2)2\tan^{-1}\left( \frac{1}{k + 1} \right) = \log_e \left( k^2 + 2k + 2 \right)2tan−1(k+11​)=loge​(k2+2k+2)
  4. (D)2tan⁡−1(1k)=log⁡e(k2+1k2)2\tan^{-1}\left( \frac{1}{k} \right) = \log_e \left( \frac{k^2 + 1}{k^2} \right)2tan−1(k1​)=loge​(k2k2+1​)

Correct answer: (A)

Step-by-step solution →
Q120·MathematicsSingle correctJEE Main 2022
Let the solution curve of the differential equation xdy=(x2+y2+y)dxxdy = \left(\sqrt{x^{2}+y^{2}} + y\right)dxxdy=(x2+y2​+y)dx, x>0x > 0x>0, intersect the line x=1x = 1x=1 at y=0y = 0y=0 and the line x=2x = 2x=2 at y=αy = \alphay=α. Then the value of α\alphaα is :
  1. (A)12\frac{1}{2}21​
  2. (B)32\frac{3}{2}23​
  3. (C)−32-\frac{3}{2}−23​
  4. (D)52\frac{5}{2}25​

Correct answer: (B)

Step-by-step solution →
Q121·MathematicsSingle correctJEE Main 2022
If y=y(x)y = y(x)y=y(x), x∈(0,π2)x \in \left(0, \frac{\pi}{2}\right)x∈(0,2π​) be the solution curve of the differential equation (sin⁡22x)dydx+(8sin⁡22x+2sin⁡4x)y=2e−4x(2sin⁡2x+cos⁡2x)\left(\sin^{2}2x\right)\frac{dy}{dx} + \left(8\sin^{2}2x + 2\sin 4x\right)y = 2e^{-4x}\left(2\sin 2x + \cos 2x\right)(sin22x)dxdy​+(8sin22x+2sin4x)y=2e−4x(2sin2x+cos2x), with y(π4)=e−πy\left(\frac{\pi}{4}\right) = e^{-\pi}y(4π​)=e−π, then y(π6)y\left(\frac{\pi}{6}\right)y(6π​) is equal to :
  1. (A)23e−2π/3\frac{2}{\sqrt{3}}e^{-2\pi/3}3​2​e−2π/3
  2. (B)23e2π/3\frac{2}{\sqrt{3}}e^{2\pi/3}3​2​e2π/3
  3. (C)13e−2π/3\frac{1}{\sqrt{3}}e^{-2\pi/3}3​1​e−2π/3
  4. (D)13e2π/3\frac{1}{\sqrt{3}}e^{2\pi/3}3​1​e2π/3

Correct answer: (A)

Step-by-step solution →
Q122·MathematicsSingle correctJEE Main 2022
The differential equation of the family of circles passing through the points (0,2)(0,2)(0,2) and (0,−2)(0,-2)(0,−2) is
  1. (A)2xydydx+(x2−y2+4)=02xy\frac{dy}{dx}+\left(x^{2}-y^{2}+4\right)=02xydxdy​+(x2−y2+4)=0
  2. (B)2xydydx+(x2+y2−4)=02xy\frac{dy}{dx}+\left(x^{2}+y^{2}-4\right)=02xydxdy​+(x2+y2−4)=0
  3. (C)2xydydx+(y2−x2+4)=02xy\frac{dy}{dx}+\left(y^{2}-x^{2}+4\right)=02xydxdy​+(y2−x2+4)=0
  4. (D)2xydydx−(x2−y2+4)=02xy\frac{dy}{dx}-\left(x^{2}-y^{2}+4\right)=02xydxdy​−(x2−y2+4)=0

Correct answer: (A)

Step-by-step solution →
Q123·MathematicsSingle correctJEE Main 2022
Let y=y(x)y=y(x)y=y(x) be the solution curve of the differential equation dydx+1x2−1y=(x−1x+1)12\frac{dy}{dx}+\frac{1}{x^{2}-1}y=\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}}dxdy​+x2−11​y=(x+1x−1​)21​, x>1x>1x>1 passing through the point (2,13)\left(2,\sqrt{\frac{1}{3}}\right)(2,31​​). Then 7y(8)\sqrt{7}y(8)7​y(8) is equal to
  1. (A)11+6log⁡e311+6\log_{e}311+6loge​3
  2. (B)19
  3. (C)12−2log⁡e312-2\log_{e}312−2loge​3
  4. (D)19−6log⁡e319-6\log_{e}319−6loge​3

Correct answer: (D)

Step-by-step solution →
Q124·MathematicsSingle correctJEE Main 2022
Consider a curve y=y(x)y = y(x)y=y(x) in the first quadrant as shown in the figure. Let the area A1A_{1}A1​ is twice the area A2A_{2}A2​. Then the normal to the curve perpendicular to the line 2x−12y=152x - 12y = 152x−12y=15 does NOT pass through the point.
  1. (A)(6,21)(6, 21)(6,21)
  2. (B)(8,9)(8, 9)(8,9)
  3. (C)(10,−4)(10, -4)(10,−4)
  4. (D)(12,−15)(12, -15)(12,−15)

Correct answer: (C)

Step-by-step solution →
Q125·MathematicsSingle correctJEE Main 2022
Let y=y1(x)y = y_1(x)y=y1​(x) and y=y2(x)y = y_2(x)y=y2​(x) be two distinct solutions of the differential equation dydx=x+y\frac{dy}{dx} = x + ydxdy​=x+y, with y1(0)=0y_1(0) = 0y1​(0)=0 and y2(0)=1y_2(0) = 1y2​(0)=1 respectively. Then, the number of points of intersection of y=y1(x)y = y_1(x)y=y1​(x) and y=y2(x)y = y_2(x)y=y2​(x) is
  1. (A)0
  2. (B)1
  3. (C)2
  4. (D)3

Correct answer: (A)

Step-by-step solution →
Q126·MathematicsNumericalJEE Main 2022
Let y = y(x) be the solution curve of the differential equation sin⁡(2x2)log⁡e(tan⁡x2)dy+(4xy−42xsin⁡(x2−π4))dx=0\sin\left(2x^{2}\right)\log_{e}\left(\tan x^{2}\right)dy+\left(4xy-4\sqrt{2}x\sin\left(x^{2}-\frac{\pi}{4}\right)\right)dx=0sin(2x2)loge​(tanx2)dy+(4xy−42​xsin(x2−4π​))dx=0, 0<x<π20<x<\sqrt{\frac{\pi}{2}}0<x<2π​​, which passes through the point (π6,1)\left(\sqrt{\frac{\pi}{6}},1\right)(6π​​,1). Then ∣y(π3)∣\left|y\left(\sqrt{\frac{\pi}{3}}\right)\right|​y(3π​​)​ is equal to ______.

Correct answer: 1

Step-by-step solution →
Q127·MathematicsNumericalJEE Main 2022
Let a curve y=y(x)y = y\left(x\right)y=y(x) pass through the point (3, 3) and the area of the region under this curve, above the x-axis and between the abscissae 3 and x(>3) be (yx)3\left(\frac{y}{x}\right)^{3}(xy​)3. If this curve also passes through the point (α,610)\left(\alpha, 6\sqrt{10}\right)(α,610​) in the first quadrant, then α is equal to _______

Correct answer: 6

Step-by-step solution →
Q128·MathematicsNumericalJEE Main 2022
Suppose y=y(x)y=y(x)y=y(x) be the solution curve to the differential equation dydx−y=2−e−x\frac{dy}{dx}-y=2-e^{-x}dxdy​−y=2−e−x such that lim⁡x→∞y(x)\lim_{x\to\infty}y(x)limx→∞​y(x) is finite. If a and b are respectively the x- and y- intercepts of the tangent to the curve at x=0x=0x=0, then the value of a−4ba-4ba−4b is equal to __________.

Correct answer: 3

Step-by-step solution →
Q129·MathematicsSingle correctJEE Main 2022
Let the solution curve y=f(x)y = f(x)y=f(x) of the differential equation dydx+xyx2−1=x4+2x1−x2,x∈(−1,1)\frac{dy}{dx}+\frac{xy}{x^{2}-1}=\frac{x^{4}+2x}{\sqrt{1-x^{2}}}, x\in\left(-1,1\right)dxdy​+x2−1xy​=1−x2​x4+2x​,x∈(−1,1) pass through the origin. Then ∫−3232f(x)dx\int\limits_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}f\left(x\right)dx−23​​∫23​​​f(x)dx is equal to
  1. (A)π3−14\frac{\pi}{3}-\frac{1}{4}3π​−41​
  2. (B)π3−34\frac{\pi}{3}-\frac{\sqrt{3}}{4}3π​−43​​
  3. (C)π6−34\frac{\pi}{6}-\frac{\sqrt{3}}{4}6π​−43​​
  4. (D)π6−32\frac{\pi}{6}-\frac{\sqrt{3}}{2}6π​−23​​

Correct answer: (B)

Step-by-step solution →
Q130·MathematicsSingle correctJEE Main 2022
If dydx\dfrac{dy}{dx}dxdy​ + 2y tan x = sin x, 0 < x < π2\dfrac{\pi}{2}2π​ and y(π3)y\left(\dfrac{\pi}{3}\right)y(3π​) = 0, then the maximum value of y(x) is
  1. (A)18\dfrac{1}{8}81​
  2. (B)34\dfrac{3}{4}43​
  3. (C)14\dfrac{1}{4}41​
  4. (D)38\dfrac{3}{8}83​

Correct answer: (A)

Step-by-step solution →
Q131·MathematicsSingle correctJEE Main 2022
The slope of the tangent to a curve C : y = y(x) at any point [x,y)[x, y)[x,y) on it is 2e2x−6e−x+92+9e−2x\frac{2e^{2x} - 6e^{-x} + 9}{2 + 9e^{-2x}}2+9e−2x2e2x−6e−x+9​. If C passes through the points (0,12+π22)\left( 0, \frac{1}{2} + \frac{\pi}{2\sqrt{2}} \right)(0,21​+22​π​) and (α,12e2α)\left( \alpha, \frac{1}{2} e^{2\alpha} \right)(α,21​e2α) then eαe^{\alpha}eα is equal to :
  1. (A)3+23−2\frac{3 + \sqrt{2}}{3 - \sqrt{2}}3−2​3+2​​
  2. (B)32(3+23−2)\frac{3}{\sqrt{2}} \left( \frac{3 + \sqrt{2}}{3 - \sqrt{2}} \right)2​3​(3−2​3+2​​)
  3. (C)12(2+12−1)\frac{1}{\sqrt{2}} \left( \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right)2​1​(2​−12​+1​)
  4. (D)2+12−1\frac{\sqrt{2} + 1}{\sqrt{2} - 1}2​−12​+1​

Correct answer: (B)

Step-by-step solution →
Q132·MathematicsSingle correctJEE Main 2022
The general solution of the differential equation (x−y2)dx+y(5x+y2)dy=0(x - y^{2})dx + y(5x + y^{2})dy = 0(x−y2)dx+y(5x+y2)dy=0 is :
  1. (A)(y2+x)4=C∣(y2+2x)3∣(y^{2} + x)^{4} = C|(y^{2} + 2x)^{3}|(y2+x)4=C∣(y2+2x)3∣
  2. (B)(y2+2x)4=C∣(y2+x)3∣(y^{2} + 2x)^{4} = C|(y^{2} + x)^{3}|(y2+2x)4=C∣(y2+x)3∣
  3. (C)∣(y2+x)3∣=C(2y2+x)4|(y^{2} + x)^{3}| = C(2y^{2} + x)^{4}∣(y2+x)3∣=C(2y2+x)4
  4. (D)∣(y2+2x)3∣=C(2y2+x)4|(y^{2} + 2x)^{3}| = C(2y^{2} + x)^{4}∣(y2+2x)3∣=C(2y2+x)4

Correct answer: (A)

Step-by-step solution →
Q133·MathematicsSingle correctJEE Main 2022
Let the solution curve of the differential equation xdydx−y=y2+16x2x\frac{dy}{dx} - y = \sqrt{y^{2} + 16x^{2}}xdxdy​−y=y2+16x2​, y(1)=3y(1) = 3y(1)=3 be y=y(x)y = y(x)y=y(x). Then y(2)y(2)y(2) is equal to :
  1. (A)15
  2. (B)11
  3. (C)13
  4. (D)17

Correct answer: (A)

Step-by-step solution →
Q134·MathematicsSingle correctJEE Main 2022
If y=y(x)y = y(x)y=y(x) is the solution of the differential equation (1+e2x)dydx+2(1+y2)ex=0\left(1 + e^{2x}\right)\frac{dy}{dx} + 2\left(1 + y^{2}\right)e^{x} = 0(1+e2x)dxdy​+2(1+y2)ex=0 and y(0)=0y(0) = 0y(0)=0, then 6(y′(0)+(y(log⁡e3))2)6\left(y'(0) + \left(y\left(\log_{e}\sqrt{3}\right)\right)^{2}\right)6(y′(0)+(y(loge​3​))2) is equal to:
  1. (A)222
  2. (B)−2-2−2
  3. (C)−4-4−4
  4. (D)−1-1−1

Correct answer: (C)

Step-by-step solution →
Q135·MathematicsNumericalJEE Main 2022
Let y=y(x)y = y(x)y=y(x), x>1x > 1x>1, be the solution of the differential equation (x−1)dydx+2xy=1x−1(x - 1)\frac{dy}{dx} + 2xy = \frac{1}{x - 1}(x−1)dxdy​+2xy=x−11​, with y(2)=1+e42e4y(2) = \frac{1 + e^4}{2e^4}y(2)=2e41+e4​. If y(3)=eα+1βeαy(3) = \frac{e^\alpha + 1}{\beta e^\alpha}y(3)=βeαeα+1​. then the value of α+β\alpha + \betaα+β is equal to______.

Correct answer: 14

Step-by-step solution →
Q136·MathematicsNumericalJEE Main 2022
Let y = y(x) be the solution of the differential equation dydx+2 y2cos⁡4x−cos⁡2x=xetan⁡−1(2cot⁡2x),0<x<π/2\frac{dy}{dx} + \frac{\sqrt{2}\,y}{2\cos^4 x - \cos 2x} = x e^{\tan^{-1}(\sqrt{2}\cot 2x)}, 0 < x < \pi/2dxdy​+2cos4x−cos2x2​y​=xetan−1(2​cot2x),0<x<π/2 with y(π4)=π232y\left(\frac{\pi}{4}\right) = \frac{\pi^2}{32}y(4π​)=32π2​. If y(π3)=π218e−tan⁡−1(α)y\left(\frac{\pi}{3}\right) = \frac{\pi^2}{18} e^{-\tan^{-1}(\alpha)}y(3π​)=18π2​e−tan−1(α), then the value of 3α23\alpha^23α2 is equal to _____.

Correct answer: 2

Step-by-step solution →
Q137·MathematicsSingle correctJEE Main 2022
Let x=x(y)x = x(y)x=x(y) be the solution of the differential equation 2y ex/y2dx+(y2−4xex/y2)dy=02y\,e^{x/y^{2}}dx + \left(y^{2} - 4xe^{x/y^{2}}\right)dy = 02yex/y2dx+(y2−4xex/y2)dy=0 such that x(1)=0x(1) = 0x(1)=0. Then, x(e)x(e)x(e) is equal to
  1. (A)elog⁡e(2)e\log_{e}(2)eloge​(2)
  2. (B)−elog⁡e(2)-e\log_{e}(2)−eloge​(2)
  3. (C)e2log⁡e(2)e^{2}\log_{e}(2)e2loge​(2)
  4. (D)−e2log⁡e(2)-e^{2}\log_{e}(2)−e2loge​(2)

Correct answer: (D)

Step-by-step solution →
Q138·MathematicsSingle correctJEE Main 2022
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation x(1−x2)dydx+(3x2y−y−4x3)=0,x>1,x\left(1-x^{2}\right)\frac{dy}{dx}+\left(3x^{2}y-y-4x^{3}\right)=0, x>1,x(1−x2)dxdy​+(3x2y−y−4x3)=0,x>1, with y(2)=−2.y(2) = -2.y(2)=−2. Then y(3)y(3)y(3) is equal to
  1. (A)−18-18−18
  2. (B)−12-12−12
  3. (C)−6-6−6
  4. (D)−3-3−3

Correct answer: (A)

Step-by-step solution →
Q139·MathematicsSingle correctJEE Main 2022
Let the slope of the tangent to a curve y=f(x)y = f(x)y=f(x) at (x,y)(x, y)(x,y) be given by 2tan⁡x(cos⁡x−y)2\tan x(\cos x - y)2tanx(cosx−y). if the curve passes through the point (π4,0)\left(\frac{\pi}{4}, 0\right)(4π​,0), then the value of ∫0π/2y dx\int_{0}^{\pi/2} y\,dx∫0π/2​ydx is equal to
  1. (A)(2−2)+π2\left(2 - \sqrt{2}\right) + \frac{\pi}{\sqrt{2}}(2−2​)+2​π​
  2. (B)2−π22 - \frac{\pi}{\sqrt{2}}2−2​π​
  3. (C)(2+2)+π2\left(2 + \sqrt{2}\right) + \frac{\pi}{\sqrt{2}}(2+2​)+2​π​
  4. (D)2+π22 + \frac{\pi}{\sqrt{2}}2+2​π​

Correct answer: (B)

Step-by-step solution →
Q140·MathematicsSingle correctJEE Main 2022
Let the solution curve y=y(x)y = y(x)y=y(x) of the differential equation, [xx2−y2+eyx]xdydx=x+[xx2−y2+eyx]y\left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right]x\frac{dy}{dx}=x+\left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right]y[x2−y2​x​+exy​]xdxdy​=x+[x2−y2​x​+exy​]y pass through the points (1, 0) and (2α,α),α>0.(2\alpha, \alpha), \alpha > 0.(2α,α),α>0. Then α\alphaα is equal to
  1. (A)12exp⁡(π6+e−1)\frac{1}{2}\exp\left(\frac{\pi}{6}+\sqrt{e}-1\right)21​exp(6π​+e​−1)
  2. (B)12exp⁡(π3+e−1)\frac{1}{2}\exp\left(\frac{\pi}{3}+\sqrt{e}-1\right)21​exp(3π​+e​−1)
  3. (C)exp⁡(π6+e+1)\exp\left(\frac{\pi}{6}+\sqrt{e}+1\right)exp(6π​+e​+1)
  4. (D)2exp⁡(π3+e−1)2\exp\left(\frac{\pi}{3}+\sqrt{e}-1\right)2exp(3π​+e​−1)

Correct answer: (A)

Step-by-step solution →
Q141·MathematicsNumericalJEE Main 2022
Let v be the solution of the differential equation (1−x2)dy=(xy+(x3+2)1−x2)dx, −1<x<1(1-x^{2})dy =\left(xy+(x^{3}+2)\sqrt{1-x^{2}}\right)dx,\ -1<x<1(1−x2)dy=(xy+(x3+2)1−x2​)dx, −1<x<1 and y (0) = 0 if ∫−12121−x2  y(x)dx=k\int_{-\frac{1}{2}}^{\frac{1}{2}}\sqrt{1-x^{2}}\;y(x)dx=k∫−21​21​​1−x2​y(x)dx=k then k−1k^{-1}k−1 is equal to :

Correct answer: 320

Step-by-step solution →
Q142·MathematicsSingle correctJEE Main 2022
Let dydx=ax−by+abx+cy+a\frac{dy}{dx} = \frac{ax - by + a}{bx + cy + a}dxdy​=bx+cy+aax−by+a​, where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is :
  1. (A)10
  2. (B)8
  3. (C)7
  4. (D)5

Correct answer: (B)

Step-by-step solution →
Q143·MathematicsSingle correctJEE Main 2022
If the solution curve of the differential equation ((tan⁡−1y)−x)dy=(1+y2)dx\left((\tan^{-1}y)-x\right)dy=(1+y^{2})dx((tan−1y)−x)dy=(1+y2)dx passes through the point (1, 0) then the abscissa of the point on the curve whose ordinate is tan⁡(1)\tan(1)tan(1) is :
  1. (A)2e2e2e
  2. (B)2e\dfrac{2}{e}e2​
  3. (C)222
  4. (D)1e\dfrac{1}{e}e1​

Correct answer: (B)

Step-by-step solution →
Q144·MathematicsNumericalJEE Main 2022
Let S=(0,2π)−{π2,3π4,3π2,7π4}S = (0, 2\pi) - \left\{\frac{\pi}{2}, \frac{3\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}\right\}S=(0,2π)−{2π​,43π​,23π​,47π​}. Let y=y(x)y = y(x)y=y(x), x∈Sx \in Sx∈S, be the solution curve of the differential equation dydx=11+sin⁡2x\frac{dy}{dx} = \frac{1}{1 + \sin 2x}dxdy​=1+sin2x1​, y(π4)=12y\left(\frac{\pi}{4}\right) = \frac{1}{2}y(4π​)=21​. if the sum of abscissas of all the points of intersection of the curve y=y(x)y = y(x)y=y(x) with the curve y=2sin⁡xy = \sqrt{2}\sin xy=2​sinx is kπ12\frac{k\pi}{12}12kπ​, then k is equal to __________.

Correct answer: 42

Step-by-step solution →
Q145·MathematicsSingle correctJEE Main 2022
If y = y(x) is the solution of the differential equation xdydx+2y=xex, y(1)=0x\frac{dy}{dx}+2y=xe^x,\ y(1)=0xdxdy​+2y=xex, y(1)=0, then the local maximum value of the function z(x)=x2y(x)−exz(x)=x^2y(x)-e^xz(x)=x2y(x)−ex, x∈R is :
  1. (A)1 – e
  2. (B)0
  3. (C)12\frac{1}{2}21​
  4. (D)4e−e\frac{4}{e}-ee4​−e

Correct answer: (D)

Step-by-step solution →
Q146·MathematicsSingle correctJEE Main 2022
If the solution of the differential equation dydx+ex(x2−2)y=(x2−2x)(x2−2)e2x\frac{dy}{dx}+e^x\left(x^2-2\right)y=\left(x^2-2x\right)\left(x^2-2\right)e^{2x}dxdy​+ex(x2−2)y=(x2−2x)(x2−2)e2x satisfies y(0) = 0, then the value of y(2) is ________ .
  1. (A)–1
  2. (B)1
  3. (C)0
  4. (D)e

Correct answer: (C)

Step-by-step solution →
Q147·MathematicsNumericalJEE Main 2022
Let the solution curve y=y(x)y = y(x)y=y(x) of the differential equation (4+x2)dy−2x(x2+3y+4)dx=0(4 + x^{2})dy - 2x(x^{2} + 3y + 4)dx = 0(4+x2)dy−2x(x2+3y+4)dx=0 pass through the origin. Then y(2)y(2)y(2) is equal to_________.

Correct answer: 12

Step-by-step solution →
Q148·MathematicsSingle correctJEE Main 2022
If y=y(x)y = y(x)y=y(x) is the solution of the differential equation 2x2dydx−2xy+3y2=02x^2 \frac{dy}{dx} - 2xy + 3y^2 = 02x2dxdy​−2xy+3y2=0 such that y(e)=e3y(e) = \frac{e}{3}y(e)=3e​, then y(1)y(1)y(1) is equal to
  1. (A)13\frac{1}{3}31​
  2. (B)23\frac{2}{3}32​
  3. (C)32\frac{3}{2}23​
  4. (D)3

Correct answer: (B)

Step-by-step solution →
Q149·MathematicsSingle correctJEE Main 2022
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation (x+1)y′−y=e3x(x+1)2(x + 1)y' - y = e^{3x}(x + 1)^2(x+1)y′−y=e3x(x+1)2, with y(0)=13y(0) = \frac{1}{3}y(0)=31​. Then, the point x=−43x = -\frac{4}{3}x=−34​ for the curve y=y(x)y = y(x)y=y(x) is:
  1. (A)not a critical point
  2. (B)a point of local minima
  3. (C)a point of local maxima
  4. (D)a point of inflection

Correct answer: (B)

Step-by-step solution →
Q150·MathematicsSingle correctJEE Main 2022
If the solution curve y=y(x)y = y(x)y=y(x) of the differential equation y2dx+(x2−xy+y2)dy=0y^2dx + (x^2 - xy + y^2)dy = 0y2dx+(x2−xy+y2)dy=0, which passes through the point (1, 1) and intersects the line y=3 xy = \sqrt{3}\,xy=3​x at the point (α,3 α)(\alpha, \sqrt{3}\,\alpha)(α,3​α), then value of log⁡e(3 α)\log_e(\sqrt{3}\,\alpha)loge​(3​α) is equal to
  1. (A)π3\frac{\pi}{3}3π​
  2. (B)π2\frac{\pi}{2}2π​
  3. (C)π12\frac{\pi}{12}12π​
  4. (D)π6\frac{\pi}{6}6π​

Correct answer: (C)

Step-by-step solution →
Q151·MathematicsSingle correctJEE Main 2022
The slope of normal at any point (x,y)(x, y)(x,y), x>0x > 0x>0, y>0y > 0y>0 on the curve y=y(x)y = y(x)y=y(x) is given by x2xy−x2y2−1\dfrac{x^2}{xy - x^2 y^2 - 1}xy−x2y2−1x2​. If the curve passes through the point (1,1)(1, 1)(1,1), then e⋅y(e)e \cdot y(e)e⋅y(e) is equal to
  1. (A)1−tan⁡(1)1+tan⁡(1)\dfrac{1-\tan(1)}{1+\tan(1)}1+tan(1)1−tan(1)​
  2. (B)tan⁡(1)\tan(1)tan(1)
  3. (C)111
  4. (D)1+tan⁡(1)1−tan⁡(1)\dfrac{1+\tan(1)}{1-\tan(1)}1−tan(1)1+tan(1)​

Correct answer: (D)

Step-by-step solution →
Q152·MathematicsSingle correctJEE Main 2022
If x=x(y)x=x(y)x=x(y) is the solution of the differential equation ydxdy=2x+y3(y+1)eyy\dfrac{dx}{dy}=2x+y^{3}(y+1)e^{y}ydydx​=2x+y3(y+1)ey, x(1)=0x(1)=0x(1)=0 ; then x(e)x(e)x(e) is equal to :
  1. (A)e3(ee−1)e^{3}(e^{e}-1)e3(ee−1)
  2. (B)ee(e3−1)e^{e}(e^{3}-1)ee(e3−1)
  3. (C)e2(ee+1)e^{2}(e^{e}+1)e2(ee+1)
  4. (D)ee(e2−1)e^{e}(e^{2}-1)ee(e2−1)

Correct answer: (A)

Step-by-step solution →
Q153·MathematicsSingle correctJEE Main 2022
A particle is moving in the xy-plane along a curve C passing through the point (3, 3). The tangent to the curve C at the point P meets the x-axis at Q. If the y-axis bisects the segment PQ, then C is a parabola with
  1. (A)length of latus rectum 3
  2. (B)length of latus rectum 6
  3. (C)focus (43,0)\left(\dfrac{4}{3},0\right)(34​,0)
  4. (D)focus (0,34)\left(0,\dfrac{3}{4}\right)(0,43​)

Correct answer: (A)

Step-by-step solution →
Q154·MathematicsMultiple correctJEE Advanced 2021
For any real numbers α and β, let yα,β(x)y_{\alpha,\beta}(x)yα,β​(x), x ∈ ℝ, be the solution of the differential equation dydx+αy=xeβx\frac{dy}{dx} + \alpha y = xe^{\beta x}dxdy​+αy=xeβx, y(1)=1y(1) = 1y(1)=1 Let S={yα,β(x):α,β∈R}S = \{y_{\alpha,\beta}(x) : \alpha, \beta \in \mathbb{R}\}S={yα,β​(x):α,β∈R}. Then which of the following functions belong(s) to the set S?
  1. (A)f(x)=x22e−x+(e−12)e−xf(x) = \frac{x^2}{2}e^{-x} + \left(e - \frac{1}{2}\right)e^{-x}f(x)=2x2​e−x+(e−21​)e−x
  2. (B)f(x)=−x22e−x+(e+12)e−xf(x) = -\frac{x^2}{2}e^{-x} + \left(e + \frac{1}{2}\right)e^{-x}f(x)=−2x2​e−x+(e+21​)e−x
  3. (C)f(x)=ex2(x−12)+(e−e24)e−xf(x) = \frac{e^x}{2}\left(x - \frac{1}{2}\right) + \left(e - \frac{e^2}{4}\right)e^{-x}f(x)=2ex​(x−21​)+(e−4e2​)e−x
  4. (D)f(x)=ex2(12−x)+(e+e24)e−xf(x) = \frac{e^x}{2}\left(\frac{1}{2} - x\right) + \left(e + \frac{e^2}{4}\right)e^{-x}f(x)=2ex​(21​−x)+(e+4e2​)e−x

Correct answer: (A), (C)

Step-by-step solution →
Q155·MathematicsSingle correctJEE Main 2021
If y = y (x) is the solution curve of the differential equation x2dy+(y−1x)dx=0x^{2}dy + \left(y - \frac{1}{x}\right)dx = 0x2dy+(y−x1​)dx=0 ; x > 0 and y(1) = 1, then y(12)y\left(\frac{1}{2}\right)y(21​) is equal to :
  1. (A)32−1e\frac{3}{2} - \frac{1}{\sqrt{e}}23​−e​1​
  2. (B)3+1e3 + \frac{1}{\sqrt{e}}3+e​1​
  3. (C)3 + e
  4. (D)3 − e

Correct answer: (D)

Step-by-step solution →
Q156·MathematicsSingle correctJEE Main 2021
If dydx=2x+y−2x2y\frac{dy}{dx} = \frac{2^{x+y} - 2^{x}}{2^{y}}dxdy​=2y2x+y−2x​, y(0)=1y(0) = 1y(0)=1, then y(1)y(1)y(1) is equal to :
  1. (A)log⁡2(2+e)\log_{2}(2 + e)log2​(2+e)
  2. (B)log⁡2(1+e)\log_{2}(1 + e)log2​(1+e)
  3. (C)log⁡2(2e)\log_{2}(2e)log2​(2e)
  4. (D)log⁡2(1+e2)\log_{2}(1 + e^{2})log2​(1+e2)

Correct answer: (B)

Step-by-step solution →
Q157·MathematicsSingle correctJEE Main 2021
If ydydx=x[y2x2+ϕ(y2x2)ϕ′(y2x2)]y \frac{dy}{dx} = x\left[\frac{y^2}{x^2} + \frac{\phi\left(\frac{y^2}{x^2}\right)}{\phi'\left(\frac{y^2}{x^2}\right)}\right]ydxdy​=x​x2y2​+ϕ′(x2y2​)ϕ(x2y2​)​​, x>0x > 0x>0, ϕ>0\phi > 0ϕ>0, and y(1)=−1y(1) = -1y(1)=−1, then ϕ(y24)\phi\left(\frac{y^2}{4}\right)ϕ(4y2​) is equal to :
  1. (A)4 ϕ(2)4\,\phi(2)4ϕ(2)
  2. (B)4 ϕ(1)4\,\phi(1)4ϕ(1)
  3. (C)2 ϕ(1)2\,\phi(1)2ϕ(1)
  4. (D)ϕ(1)\phi(1)ϕ(1)

Correct answer: (B)

Step-by-step solution →
Q158·MathematicsSingle correctJEE Main 2021
If dydx=2xy+2y⋅2x2x+2x+ylog⁡e2\frac{dy}{dx} = \frac{2^x y + 2^y \cdot 2^x}{2^x + 2^{x+y} \log_e 2}dxdy​=2x+2x+yloge​22xy+2y⋅2x​, y(0)=0y(0) = 0y(0)=0, then for y=1y = 1y=1, the value of xxx lies in the interval:
  1. (A)(1,2)(1, 2)(1,2)
  2. (B)(12,1]\left(\frac{1}{2}, 1\right](21​,1]
  3. (C)(2,3)(2, 3)(2,3)
  4. (D)(0,12]\left(0, \frac{1}{2}\right](0,21​]

Correct answer: (A)

Step-by-step solution →
Q159·MathematicsSingle correctJEE Main 2021
Let us consider a curve, y=f(x)y = f(x)y=f(x) passing through the point (−2,2)(-2, 2)(−2,2) and the slope of the tangent to the curve at any point (x,f(x))(x, f(x))(x,f(x)) is given by f(x)+xf′(x)=x2f(x) + xf'(x) = x^2f(x)+xf′(x)=x2. Then :
  1. (A)x2+2xf(x)−12=0x^2 + 2xf(x) - 12 = 0x2+2xf(x)−12=0
  2. (B)x3+xf(x)+12=0x^3 + xf(x) + 12 = 0x3+xf(x)+12=0
  3. (C)x3−3xf(x)−4=0x^3 - 3xf(x) - 4 = 0x3−3xf(x)−4=0
  4. (D)x2+2xf(x)+4=0x^2 + 2xf(x) + 4 = 0x2+2xf(x)+4=0

Correct answer: (C)

Step-by-step solution →
Q160·MathematicsSingle correctJEE Main 2021
If the solution curve of the differential equation (2x−10y3) dy+y dx=0(2x - 10y^{3})\,dy + y\,dx = 0(2x−10y3)dy+ydx=0, passes through the points (0, 1) and (2, β), then β is a root of the equation:
  1. (A)y5−2y−2=0y^{5} - 2y - 2 = 0y5−2y−2=0
  2. (B)2y5−2y−1=02y^{5} - 2y - 1 = 02y5−2y−1=0
  3. (C)2y5−y2−2=02y^{5} - y^{2} - 2 = 02y5−y2−2=0
  4. (D)y5−y2−1=0y^{5} - y^{2} - 1 = 0y5−y2−1=0

Correct answer: (D)

Step-by-step solution →
Q161·MathematicsSingle correctJEE Main 2021
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation dydx=2(y+2sin⁡x−5) x−2cos⁡x\frac{dy}{dx} = 2(y + 2\sin x - 5)\, x - 2\cos xdxdy​=2(y+2sinx−5)x−2cosx such that y(0)=7y(0) = 7y(0)=7. Then y(π)y(\pi)y(π) is equal to :
  1. (A)2eπ2+52e^{\pi^2} + 52eπ2+5
  2. (B)eπ2+5e^{\pi^2} + 5eπ2+5
  3. (C)3eπ2+53e^{\pi^2} + 53eπ2+5
  4. (D)7eπ2+57e^{\pi^2} + 57eπ2+5

Correct answer: (A)

Step-by-step solution →
Q162·MathematicsSingle correctJEE Main 2021
A differential equation representing the family of parabolas with axis parallel to y-axis and whose length of latus rectum is the distance of the point (2, −3) form the line 3x + 4y = 5, is given by :
  1. (A)10d2ydx2=1110\frac{d^{2}y}{dx^{2}} = 1110dx2d2y​=11
  2. (B)11d2xdy2=1011\frac{d^{2}x}{dy^{2}} = 1011dy2d2x​=10
  3. (C)10d2xdy2=1110\frac{d^{2}x}{dy^{2}} = 1110dy2d2x​=11
  4. (D)11d2ydx2=1011\frac{d^{2}y}{dx^{2}} = 1011dx2d2y​=10

Correct answer: (D)

Step-by-step solution →
Q163·MathematicsSingle correctJEE Main 2021
Let y=y(x)y = y(x)y=y(x) be a solution curve of the differential equation (y+1)tan⁡2x dx+tan⁡x dy+y dx=0(y + 1)\tan^2 x\, dx + \tan x\, dy + y\, dx = 0(y+1)tan2xdx+tanxdy+ydx=0, x∈(0,π2)x \in \left(0, \frac{\pi}{2}\right)x∈(0,2π​). If lim⁡x→0+xy(x)=1\lim_{x \to 0+} xy(x) = 1limx→0+​xy(x)=1, then the value of y(π4)y\left(\frac{\pi}{4}\right)y(4π​) is :
  1. (A)−π4-\frac{\pi}{4}−4π​
  2. (B)π4−1\frac{\pi}{4} - 14π​−1
  3. (C)π4+1\frac{\pi}{4} + 14π​+1
  4. (D)π4\frac{\pi}{4}4π​

Correct answer: (D)

Step-by-step solution →
Q164·MathematicsSingle correctJEE Main 2021
Let y(x)y(x)y(x) be the solution of the differential equation 2x2dy+(ey−2x)dx=02x^{2}dy + (e^{y} - 2x)dx = 02x2dy+(ey−2x)dx=0, x>0x > 0x>0. If y(e)=1y(e) = 1y(e)=1, then y(1)y(1)y(1) is equal to :
  1. (A)000
  2. (B)222
  3. (C)log⁡e2\log_{e} 2loge​2
  4. (D)log⁡e(2e)\log_{e}(2e)loge​(2e)

Correct answer: (C)

Step-by-step solution →
Q165·MathematicsNumericalJEE Main 2021
If y=y(x),y∈[0,π2]y=y\left(x\right), y \in \left[0,\frac{\pi}{2}\right]y=y(x),y∈[0,2π​] is the solution of the differential equation sec⁡ydydx−sin⁡(x+y)−sin⁡(x−y)=0\sec y\frac{dy}{dx}-\sin\left(x+y\right)-\sin\left(x-y\right)=0secydxdy​−sin(x+y)−sin(x−y)=0, with y(0)=0y\left(0\right)=0y(0)=0, then 5y′(π2)5y'\left(\frac{\pi}{2}\right)5y′(2π​) is equal to ____.

Correct answer: 2

Step-by-step solution →
Q166·MathematicsSingle correctJEE Main 2021
Let y=y(x)y = y\left(x\right)y=y(x) be solution of the differential equation log⁡e(dydx)=3x+4y\log_e\left(\frac{dy}{dx}\right) = 3x + 4yloge​(dxdy​)=3x+4y, with y(0)=0y\left(0\right) = 0y(0)=0. If y(−23log⁡e2)=αlog⁡e2y\left(-\frac{2}{3}\log_e 2\right) = \alpha \log_e 2y(−32​loge​2)=αloge​2, then the value of α\alphaα is equal to :
  1. (A)14\frac{1}{4}41​
  2. (B)−14-\frac{1}{4}−41​
  3. (C)2
  4. (D)−12-\frac{1}{2}−21​

Correct answer: (B)

Step-by-step solution →
Q167·MathematicsSingle correctJEE Main 2021
Let y = y(x) be the solution of the differential equation (x−x3)dy=(y+yx2−3x4)dx\left( x - x^3 \right) dy = \left( y + yx^2 - 3x^4 \right) dx(x−x3)dy=(y+yx2−3x4)dx, x > 2. If y(3) = 3 then y(4) is equal to
  1. (A)12
  2. (B)4
  3. (C)8
  4. (D)16

Correct answer: (A)

Step-by-step solution →
Q168·MathematicsNumericalJEE Main 2021
Let y = y(x)\left(x\right)(x) be the solution of the differential equation dy = eαx+ye^{\alpha x+y}eαx+ydx; α ∈ N . If y(log⁡e2)\left(\log_{e}2\right)(loge​2) = log⁡e2\log_{e}2loge​2 and y(0) = log⁡e(12)\log_{e}\left(\frac{1}{2}\right)loge​(21​), then the value of α is equal to………….

Correct answer: 2

Step-by-step solution →
Q169·MathematicsSingle correctJEE Main 2021
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dydx=1+xey−x\frac{dy}{dx}=1+xe^{y-x}dxdy​=1+xey−x, −2<x<2-\sqrt{2}<x<\sqrt{2}−2​<x<2​, y(0)=0y(0)=0y(0)=0 then, the minimum value of y(x)y(x)y(x), x∈(−2,2)x \in \left(-\sqrt{2},\sqrt{2}\right)x∈(−2​,2​) is equal to :
  1. (A)(1+3)−log⁡e(3−1)\left(1+\sqrt{3}\right)-\log_{e}\left(\sqrt{3}-1\right)(1+3​)−loge​(3​−1)
  2. (B)(2+3)+log⁡e2\left(2+\sqrt{3}\right)+\log_{e}2(2+3​)+loge​2
  3. (C)(1−3)−log⁡e(3−1)\left(1-\sqrt{3}\right)-\log_{e}\left(\sqrt{3}-1\right)(1−3​)−loge​(3​−1)
  4. (D)(2−3)−log⁡e2\left(2-\sqrt{3}\right)-\log_{e}2(2−3​)−loge​2

Correct answer: (C)

Step-by-step solution →
Q170·MathematicsNumericalJEE Main 2021
Let y=y(x)y = y(x)y=y(x) be solution of the following differential equation eydydx−2eysin⁡x+sin⁡xcos⁡2x=0e^y \frac{dy}{dx} - 2e^y \sin x + \sin x \cos^2 x = 0eydxdy​−2eysinx+sinxcos2x=0, y(π2)=0y\left( \frac{\pi}{2} \right) = 0y(2π​)=0. If y(0)=log⁡e(α+βe−2)y(0) = \log_e \left( \alpha + \beta e^{-2} \right)y(0)=loge​(α+βe−2), then 4(α+β)4\left( \alpha + \beta \right)4(α+β) is equal to.......

Correct answer: 4

Step-by-step solution →
Q171·MathematicsSingle correctJEE Main 2021
Let y=y(x)y = y\left(x\right)y=y(x) be the solution of the differential equation xdy=(y+x3cos⁡x)dxxdy = \left(y + x^3\cos x\right)dxxdy=(y+x3cosx)dx with y(π)=0y\left(\pi\right) = 0y(π)=0, then y(π2)y\left(\frac{\pi}{2}\right)y(2π​) is equal to :
  1. (A)π24−π2\frac{\pi^2}{4} - \frac{\pi}{2}4π2​−2π​
  2. (B)π24+π2\frac{\pi^2}{4} + \frac{\pi}{2}4π2​+2π​
  3. (C)π22−π4\frac{\pi^2}{2} - \frac{\pi}{4}2π2​−4π​
  4. (D)π22+π4\frac{\pi^2}{2} + \frac{\pi}{4}2π2​+4π​

Correct answer: (B)

Step-by-step solution →
Q172·MathematicsNumericalJEE Main 2021
Let a curve y=f(x)y = f(x)y=f(x) pass through the point (2,(log⁡e2)2)\left(2, \left(\log_{e}2\right)^{2}\right)(2,(loge​2)2) and have slope 2yxlog⁡ex\dfrac{2y}{x\log_{e}x}xloge​x2y​ for all positive real value of x. Then the value of f(e)f(e)f(e) is equal to..........

Correct answer: 1

Step-by-step solution →
Q173·MathematicsNumericalJEE Main 2021
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation ((x+2)ey+1x+2+(y+1))dx=(x+2)dy\left((x + 2)e^{\frac{y+1}{x+2}} + (y + 1)\right)dx = (x + 2)dy((x+2)ex+2y+1​+(y+1))dx=(x+2)dy, y(1)=1y(1) = 1y(1)=1. If the domain y=y(x)y = y(x)y=y(x) is an open interval (α,β)(\alpha, \beta)(α,β), then ∣α+β∣|\alpha + \beta|∣α+β∣ is equal to..........

Correct answer: 4

Step-by-step solution →
Q174·MathematicsSingle correctJEE Main 2021
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation cosec⁡2x dy+2 dx=(1+ycos⁡2x)cosec⁡2x dx\cosec^2 x \, dy + 2 \, dx = (1 + y \cos 2x) \cosec^2 x \, dxcosec2xdy+2dx=(1+ycos2x)cosec2xdx, with y(π4)=0y\left(\dfrac{\pi}{4}\right) = 0y(4π​)=0. Then, the value of (y(0)+1)2\left(y(0) + 1\right)^2(y(0)+1)2 is equal to :
  1. (A)e1/2e^{1/2}e1/2
  2. (B)e−1/2e^{-1/2}e−1/2
  3. (C)e−1e^{-1}e−1
  4. (D)eee

Correct answer: (C)

Step-by-step solution →
Q175·MathematicsSingle correctJEE Main 2021
Let y=y(x)y=y(x)y=y(x) satisfies the equation dydx−∣A∣=0,\frac{dy}{dx}-|A|=0,dxdy​−∣A∣=0, for all x>0,x>0,x>0, where A=[ysin⁡x10−11201x]A=\begin{bmatrix} y & \sin x & 1 \\ 0 & -1 & 1 \\ 2 & 0 & \frac{1}{x} \end{bmatrix}A=​y02​sinx−10​11x1​​​. If y(π)=π+2,y(\pi)=\pi+2,y(π)=π+2, then the value of y(π2)y\left(\frac{\pi}{2}\right)y(2π​) is :
  1. (A)π2+4π\frac{\pi}{2}+\frac{4}{\pi}2π​+π4​
  2. (B)3π2−1π\frac{3\pi}{2}-\frac{1}{\pi}23π​−π1​
  3. (C)π2−4π\frac{\pi}{2}-\frac{4}{\pi}2π​−π4​
  4. (D)π2−1π\frac{\pi}{2}-\frac{1}{\pi}2π​−π1​

Correct answer: (A)

Step-by-step solution →
Q176·MathematicsSingle correctJEE Main 2021
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation xtan⁡(yx)dy=(ytan⁡(yx)−x)dxx\tan\left(\frac{y}{x}\right)dy = \left(y\tan\left(\frac{y}{x}\right)-x\right)dxxtan(xy​)dy=(ytan(xy​)−x)dx, −1≤x≤1-1 \leq x \leq 1−1≤x≤1, y(12)=π6y\left(\frac{1}{2}\right) = \frac{\pi}{6}y(21​)=6π​ Then the area of the region bounded by the curves x=0x = 0x=0, x=12x = \frac{1}{\sqrt{2}}x=2​1​ and y=y(x)y = y(x)y=y(x) in the upper half plane is :
  1. (A)16(π−1)\frac{1}{6}(\pi-1)61​(π−1)
  2. (B)18(π−1)\frac{1}{8}(\pi-1)81​(π−1)
  3. (C)14(π−2)\frac{1}{4}(\pi-2)41​(π−2)
  4. (D)112(π−3)\frac{1}{12}(\pi-3)121​(π−3)

Correct answer: (B)

Step-by-step solution →
Q177·MathematicsNumericalJEE Main 2021
Let a curve y=y(x)y=y(x)y=y(x) be given by the solution of the differential equation cos⁡(12cos⁡−1(e−x))dx=e2x−1 dy\cos\left(\frac{1}{2}\cos^{-1}\left(e^{-x}\right)\right)dx=\sqrt{e^{2x}-1}\,dycos(21​cos−1(e−x))dx=e2x−1​dy If it intersects y-axis at y=−1y=-1y=−1, and the intersection point of the curve with x-axis is (α,0)(\alpha,0)(α,0), then eαe^{\alpha}eα is equal to……..

Correct answer: 2

Step-by-step solution →
Q178·MathematicsSingle correctJEE Main 2021
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation ex1−y2 dx+(yx)dy=0e^x\sqrt{1-y^2}\,dx+\left(\frac{y}{x}\right)dy = 0ex1−y2​dx+(xy​)dy=0, y(1)=−1y\left(1\right) = -1y(1)=−1 Then the value of (y(3))2\left(y(3)\right)^2(y(3))2 is equal to :
  1. (A)1+4e61+4e^61+4e6
  2. (B)1−4e31-4e^31−4e3
  3. (C)1+4e31+4e^31+4e3
  4. (D)1−4e61-4e^61−4e6

Correct answer: (D)

Step-by-step solution →
Q179·MathematicsSingle correctJEE Main 2021
The differential equation satisfied by the system of parabolas y2=4a(x+a)y^{2} = 4a(x + a)y2=4a(x+a) is :
  1. (A)y(dydx)2−2x(dydx)−y=0y\left(\dfrac{dy}{dx}\right)^{2} - 2x\left(\dfrac{dy}{dx}\right) - y = 0y(dxdy​)2−2x(dxdy​)−y=0
  2. (B)y(dydx)2−2x(dydx)+y=0y\left(\dfrac{dy}{dx}\right)^{2} - 2x\left(\dfrac{dy}{dx}\right) + y = 0y(dxdy​)2−2x(dxdy​)+y=0
  3. (C)y(dydx)2+2x(dydx)−y=0y\left(\dfrac{dy}{dx}\right)^{2} + 2x\left(\dfrac{dy}{dx}\right) - y = 0y(dxdy​)2+2x(dxdy​)−y=0
  4. (D)y(dydx)+2x(dydx)−y=0y\left(\dfrac{dy}{dx}\right) + 2x\left(\dfrac{dy}{dx}\right) - y = 0y(dxdy​)+2x(dxdy​)−y=0

Correct answer: (C)

Step-by-step solution →
Q180·MathematicsNumericalJEE Main 2021
Let y = y(x) be the solution of the differential equation xdy−ydx=(x2−y2) dxxdy - ydx = \sqrt{\left(x^{2} - y^{2}\right)}\,dxxdy−ydx=(x2−y2)​dx , x ≥ 1, with y(1) = 0. If the area bounded by the line x = 1, x=eπx = e^{\pi}x=eπ, y = 0 and y = y(x) is αe2π+β\alpha e^{2\pi} + \betaαe2π+β, then the value of 10(α + β) is equal to ________ .

Correct answer: 4

Step-by-step solution →
Q181·MathematicsSingle correctJEE Main 2021
Let y = y(x) be the solution of the differential equation dydx=(y+1)((y+1)ex2/2−x)\frac{dy}{dx} = (y+1)\left((y+1)e^{x^{2}/2} - x\right)dxdy​=(y+1)((y+1)ex2/2−x), 0 < x < 2.1, with y(2) = 0. Then the value of dydx\frac{dy}{dx}dxdy​ at x = 1 is equal to :
  1. (A)−e3/2(e2+1)2\frac{-e^{3/2}}{\left(e^{2}+1\right)^{2}}(e2+1)2−e3/2​
  2. (B)−2e2(1+e2)2-\frac{2e^{2}}{\left(1+e^{2}\right)^{2}}−(1+e2)22e2​
  3. (C)e5/2(1+e2)2\frac{e^{5/2}}{\left(1+e^{2}\right)^{2}}(1+e2)2e5/2​
  4. (D)5e1/2(e2+1)2\frac{5e^{1/2}}{\left(e^{2}+1\right)^{2}}(e2+1)25e1/2​

Correct answer: (A)

Step-by-step solution →
Q182·MathematicsSingle correctJEE Main 2021
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation cos⁡x (3sin⁡x+cos⁡x+3)dy=(1+ysin⁡x (3sin⁡x+cos⁡x+3))dx\cos x\,(3\sin x + \cos x + 3)dy = (1 + y\sin x\,(3\sin x + \cos x + 3))dxcosx(3sinx+cosx+3)dy=(1+ysinx(3sinx+cosx+3))dx, 0≤x≤π20 \le x \le \frac{\pi}{2}0≤x≤2π​, y(0)=0y(0) = 0y(0)=0. Then, y(π3)y\left(\frac{\pi}{3}\right)y(3π​) is equal to:
  1. (A)2log⁡e(23+96)2\log_{e}\left(\frac{2\sqrt{3}+9}{6}\right)2loge​(623​+9​)
  2. (B)2log⁡e(23+1011)2\log_{e}\left(\frac{2\sqrt{3}+10}{11}\right)2loge​(1123​+10​)
  3. (C)2log⁡e(3+72)2\log_{e}\left(\frac{\sqrt{3}+7}{2}\right)2loge​(23​+7​)
  4. (D)2log⁡e(33−84)2\log_{e}\left(\frac{3\sqrt{3}-8}{4}\right)2loge​(433​−8​)

Correct answer: (B)

Step-by-step solution →
Q183·MathematicsSingle correctJEE Main 2021
If the curve y=y(x)y = y(x)y=y(x) is the solution of the differential equation 2(x2+x5/4)dy−y(x+x1/4)dx=2x9/4dx2(x^{2} + x^{5/4})dy - y(x + x^{1/4})dx = 2x^{9/4}dx2(x2+x5/4)dy−y(x+x1/4)dx=2x9/4dx , x>0x > 0x>0 which passes through the point (1,1−43log⁡e2)\left(1, 1-\frac{4}{3}\log_{e}2\right)(1,1−34​loge​2), then the value of y(16)y(16)y(16) is equal to :
  1. (A)4(313+83log⁡e3)4\left(\frac{31}{3}+\frac{8}{3}\log_{e}3\right)4(331​+38​loge​3)
  2. (B)(313+83log⁡e3)\left(\frac{31}{3}+\frac{8}{3}\log_{e}3\right)(331​+38​loge​3)
  3. (C)4(313−83log⁡e3)4\left(\frac{31}{3}-\frac{8}{3}\log_{e}3\right)4(331​−38​loge​3)
  4. (D)(313−83log⁡e3)\left(\frac{31}{3}-\frac{8}{3}\log_{e}3\right)(331​−38​loge​3)

Correct answer: (C)

Step-by-step solution →
Q184·MathematicsSingle correctJEE Main 2021
Which of the following is true for y(x)y(x)y(x) that satisfies the differential equation dydx=xy−1+x−y\frac{dy}{dx} = xy - 1 + x - ydxdy​=xy−1+x−y ; y(0)=0y(0) = 0y(0)=0 :
  1. (A)y(1)=e−12−1y(1) = e^{-\frac{1}{2}} - 1y(1)=e−21​−1
  2. (B)y(1)=e12−e−12y(1) = e^{\frac{1}{2}} - e^{-\frac{1}{2}}y(1)=e21​−e−21​
  3. (C)y(1)=1y(1) = 1y(1)=1
  4. (D)y(1)=e12−1y(1) = e^{\frac{1}{2}} - 1y(1)=e21​−1

Correct answer: (A)

Step-by-step solution →
Q185·MathematicsSingle correctJEE Main 2021
If y=y(x)y = y(x)y=y(x) is the solution of the differential equation dydx+(tan⁡x) y=sin⁡x, 0≤x≤π3\frac{dy}{dx}+(\tan x)\,y=\sin x,\ 0 \le x \le \frac{\pi}{3}dxdy​+(tanx)y=sinx, 0≤x≤3π​, with y(0)=0y(0) = 0y(0)=0, then y(π4)y\left(\frac{\pi}{4}\right)y(4π​) equal to :
  1. (A)14log⁡e2\frac{1}{4}\log_{e}241​loge​2
  2. (B)(122)log⁡e2\left(\frac{1}{2\sqrt{2}}\right)\log_{e}2(22​1​)loge​2
  3. (C)log⁡e2\log_{e}2loge​2
  4. (D)12log⁡e2\frac{1}{2}\log_{e}221​loge​2

Correct answer: (B)

Step-by-step solution →
Q186·MathematicsSingle correctJEE Main 2021
Let C1C_{1}C1​ be the curve obtained by the solution of differential equation 2xydydx=y2−x22xy\frac{dy}{dx}=y^{2}-x^{2}2xydxdy​=y2−x2, x>0x > 0x>0. Let the curve C2C_{2}C2​ be the solution of 2xyx2−y2=dydx\frac{2xy}{x^{2}-y^{2}}=\frac{dy}{dx}x2−y22xy​=dxdy​. If both the curves pass through (1,1), then the area enclosed by the curves C1C_{1}C1​ and C2C_{2}C2​ is equal to :
  1. (A)π−1\pi-1π−1
  2. (B)π2−1\frac{\pi}{2}-12π​−1
  3. (C)π+1\pi+1π+1
  4. (D)π4+1\frac{\pi}{4}+14π​+1

Correct answer: (B)

Step-by-step solution →
Q187·MathematicsSingle correctJEE Main 2021
If y = y(x) is the solution of the differential equation, dydx+2ytan⁡x=sin⁡x\frac{dy}{dx} + 2y\tan x = \sin xdxdy​+2ytanx=sinx, y(π3)=0y\left(\frac{\pi}{3}\right) = 0y(3π​)=0, then the maximum value of the function y(x) over ℝ is equal to :
  1. (A)8
  2. (B)12\frac{1}{2}21​
  3. (C)−154-\frac{15}{4}−415​
  4. (D)18\frac{1}{8}81​

Correct answer: (D)

Step-by-step solution →
Q188·MathematicsNumericalJEE Main 2021
Let the normals at all the points on a given curve pass through a fixed point (a, b). If the curve passes through (3, -3) and (4,−22)\left(4, -2\sqrt{2}\right)(4,−22​), and given that a−22 b=3a - 2\sqrt{2}\ b = 3a−22​ b=3, then (a2+b2+ab)(a^{2}+b^{2}+ab)(a2+b2+ab) is equal to________________.

Correct answer: 9

Step-by-step solution →
Q189·MathematicsSingle correctJEE Main 2021
The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time t = 0. The number of bacteria is increased by 20% in 2 hours. If the population of bacteria is 2000 after klog⁡e(65)\frac{k}{\log_{e}\left(\frac{6}{5}\right)}loge​(56​)k​ hours, then (klog⁡e2)2\left(\frac{k}{\log_{e} 2}\right)^{2}(loge​2k​)2 is equal to
  1. (A)4
  2. (B)2
  3. (C)16
  4. (D)8

Correct answer: (A)

Step-by-step solution →
Q190·MathematicsNumericalJEE Main 2021
The difference betweeen degree and order of differential equation that represents the family of curves given by y2=a(x+a2)y^{2}=a\left(x+\frac{\sqrt{a}}{2}\right)y2=a(x+2a​​), a > 0 is _______.

Correct answer: 2

Step-by-step solution →
Q191·MathematicsSingle correctJEE Main 2021
Let f(x)=∫0xetf(t)dt+exf(x) = \int_{0}^{x} e^{t} f(t)dt + e^{x}f(x)=∫0x​etf(t)dt+ex be a differentiable function for all x∈Rx \in Rx∈R. Then f(x) equals.
  1. (A)2e(ex−1)−12e^{(e^{x}-1)} - 12e(ex−1)−1
  2. (B)e(ex−1)e^{(e^{x}-1)}e(ex−1)
  3. (C)2eex−12e^{e^{x}} - 12eex−1
  4. (D)eex−1e^{e^{x}} - 1eex−1

Correct answer: (A)

Step-by-step solution →
Q192·MathematicsSingle correctJEE Main 2021
Let slope of the tangent line to a curve at any point P(x, y) be given by xy2+yx\frac{xy^{2} + y}{x}xxy2+y​. If the curve intersects the line x + 2y = 4 at x = − 2, then the value of y, for which the point (3, y) lies on the curve, is :
  1. (A)−1811-\frac{18}{11}−1118​
  2. (B)−1819-\frac{18}{19}−1918​
  3. (C)−43-\frac{4}{3}−34​
  4. (D)1835\frac{18}{35}3518​

Correct answer: (B)

Step-by-step solution →
Q193·MathematicsNumericalJEE Main 2021
If y = y(x) is the solution of the equation esin⁡ycos⁡y dydx+esin⁡ycos⁡x=cos⁡xe^{\sin y}\cos y\,\frac{dy}{dx}+e^{\sin y}\cos x=\cos xesinycosydxdy​+esinycosx=cosx, y(0) = 0 ; then 1+y(π6)+32y(π3)+12y(π4)1+y\left(\frac{\pi}{6}\right)+\frac{\sqrt{3}}{2}y\left(\frac{\pi}{3}\right)+\frac{1}{\sqrt{2}}y\left(\frac{\pi}{4}\right)1+y(6π​)+23​​y(3π​)+2​1​y(4π​) is equal to _______.

Correct answer: 1

Step-by-step solution →
Q194·MathematicsSingle correctJEE Main 2021
If a curve passes through the origin and the slope of the tangent to it at any point (x,y)(x, y)(x,y) is x2−4x+y+8x−2\dfrac{x^2 - 4x + y + 8}{x - 2}x−2x2−4x+y+8​, then this curve also passes through the point :
  1. (A)(4,5)(4, 5)(4,5)
  2. (B)(5,4)(5, 4)(5,4)
  3. (C)(4,4)(4, 4)(4,4)
  4. (D)(5,5)(5, 5)(5,5)

Correct answer: (D)

Step-by-step solution →
Q195·MathematicsSingle correctJEE Main 2021
Let fff be a twice differentiable function defined on R such that f(0)=1, f′(0)=2f(0)=1,\ f'(0)=2f(0)=1, f′(0)=2 and f′(x)≠0f'(x)\neq 0f′(x)=0 for all x∈Rx\in Rx∈R. If ∣f(x)f′(x)f′(x)f′′(x)∣=0,\begin{vmatrix} f(x) & f'(x) \\ f'(x) & f''(x) \end{vmatrix}=0,​f(x)f′(x)​f′(x)f′′(x)​​=0, for all x∈Rx\in Rx∈R then the value of f(1)f(1)f(1) lies in the interval :
  1. (A)(9, 12)
  2. (B)(6, 9)
  3. (C)(3, 6)
  4. (D)(0, 3)

Correct answer: (B)

Step-by-step solution →
Q196·MathematicsSingle correctJEE Main 2021
If a curve y=f(x)y=f(x)y=f(x) passes through the point (1,2) and satisfies xdydx+y=bx4x\frac{dy}{dx}+y=bx^{4}xdxdy​+y=bx4, then for what value of b, ∫12f(x)dx=625\int_{1}^{2}f(x)dx=\frac{62}{5}∫12​f(x)dx=562​ ?
  1. (A)5
  2. (B)625\frac{62}{5}562​
  3. (C)315\frac{31}{5}531​
  4. (D)10

Correct answer: (D)

Step-by-step solution →
Q197·MathematicsSingle correctJEE Main 2021
The population P=P(t)P = P(t)P=P(t) at time 't' of a certain species follows the differential equation dPdt=0.5P−450\frac{dP}{dt} = 0.5P - 450dtdP​=0.5P−450. If P(0)=850P(0) = 850P(0)=850, then the time at which population becomes zero is :
  1. (A)12log⁡e18\frac{1}{2}\log_e 1821​loge​18
  2. (B)2log⁡e182\log_e 182loge​18
  3. (C)log⁡e9\log_e 9loge​9
  4. (D)log⁡e18\log_e 18loge​18

Correct answer: (B)

Step-by-step solution →
Q198·MathematicsMultiple correctJEE Advanced 2020
Let bbb be a nonzero real number. Suppose f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R is a differentiable function such that f(0)=1f(0) = 1f(0)=1. If the derivative f′f'f′ of fff satisfies the equation f′(x)=f(x)b2+x2f'(x) = \frac{f(x)}{b^{2} + x^{2}}f′(x)=b2+x2f(x)​ for all x∈Rx \in \mathbb{R}x∈R, then which of the following statements is/are TRUE?
  1. (A)If b>0b > 0b>0, then fff is an increasing function
  2. (B)If b<0b < 0b<0, then fff is a decreasing function
  3. (C)f(x)f(−x)=1f(x) f(-x) = 1f(x)f(−x)=1 for all x∈Rx \in \mathbb{R}x∈R
  4. (D)f(x)−f(−x)=0f(x) - f(-x) = 0f(x)−f(−x)=0 for all x∈Rx \in \mathbb{R}x∈R

Correct answer: (A), (C)

Step-by-step solution →
Q199·MathematicsSingle correctJEE Main 2020
If y=(2πx−1)cosec⁡xy = \left(\frac{2}{\pi}x - 1\right)\operatorname{cosec} xy=(π2​x−1)cosecx is the solution of the differential equation, dydx+p(x)y=2πcosec⁡ x,0<x<π2\frac{dy}{dx} + p(x)y = \frac{2}{\pi}\operatorname{cosec}\ x, 0 < x < \frac{\pi}{2}dxdy​+p(x)y=π2​cosec x,0<x<2π​, then the function p(x) is equal to:
  1. (A)cot⁡x\cot xcotx
  2. (B)cosec⁡x\operatorname{cosec} xcosecx
  3. (C)sec⁡x\sec xsecx
  4. (D)tan⁡x\tan xtanx

Correct answer: (A)

Step-by-step solution →
Q200·MathematicsSingle correctJEE Main 2020
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation cos⁡xdydx+2ysin⁡x=sin⁡2x,x∈(0,π2)\cos x\dfrac{dy}{dx}+2y\sin x=\sin 2x, x\in\left(0,\dfrac{\pi}{2}\right)cosxdxdy​+2ysinx=sin2x,x∈(0,2π​). If y(π/3)=0y(\pi/3)=0y(π/3)=0, then y(π/4)y(\pi/4)y(π/4) is equal to:
  1. (A)2+22+\sqrt{2}2+2​
  2. (B)12−1\dfrac{1}{\sqrt{2}}-12​1​−1
  3. (C)2−22-\sqrt{2}2−2​
  4. (D)2−2\sqrt{2}-22​−2

Correct answer: (D)

Step-by-step solution →
Q201·MathematicsSingle correctJEE Main 2020
If y = y (x) is the solution of the differential equation 5+ex2+y.dydx+ex=0\dfrac{5+e^{x}}{2+y}.\dfrac{dy}{dx} + e^{x} = 02+y5+ex​.dxdy​+ex=0 satisfying y(0) = 1, then a value of y(log⁡e13)y(\log_{e} 13)y(loge​13) is:
  1. (A)0
  2. (B)1
  3. (C)-1
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q202·MathematicsSingle correctJEE Main 2020
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation, xy′−y=x2(xcos⁡x+sin⁡x)xy' - y = x^{2}(x \cos x + \sin x)xy′−y=x2(xcosx+sinx), x>0x > 0x>0. If y(π)=πy(\pi) = \piy(π)=π, then y′′(π2)+y(π2)y''\left(\frac{\pi}{2}\right) + y\left(\frac{\pi}{2}\right)y′′(2π​)+y(2π​) is equal to:
  1. (A)1+π2+π241 + \frac{\pi}{2} + \frac{\pi^{2}}{4}1+2π​+4π2​
  2. (B)2+π2+π242 + \frac{\pi}{2} + \frac{\pi^{2}}{4}2+2π​+4π2​
  3. (C)2+π22 + \frac{\pi}{2}2+2π​
  4. (D)1+π21 + \frac{\pi}{2}1+2π​

Correct answer: (C)

Step-by-step solution →
Q203·MathematicsSingle correctJEE Main 2020
The solution of differential equation dydx−y+3xlog⁡e(y+3x)+3=0\frac{dy}{dx} - \frac{y + 3x}{\log_{e}\left(y + 3x\right)} + 3 = 0dxdy​−loge​(y+3x)y+3x​+3=0 is: (where C is a constant of integration.)
  1. (A)y+3x−12(log⁡ex)2=Cy + 3x - \frac{1}{2}\left(\log_{e} x\right)^{2} = Cy+3x−21​(loge​x)2=C
  2. (B)x−12(log⁡e(y+3x))2=Cx - \frac{1}{2}\left(\log_{e}\left(y + 3x\right)\right)^{2} = Cx−21​(loge​(y+3x))2=C
  3. (C)x−2log⁡e(y+3x)=Cx - 2\log_{e}\left(y + 3x\right) = Cx−2loge​(y+3x)=C
  4. (D)x−log⁡e(y+3x)=Cx - \log_{e}\left(y + 3x\right) = Cx−loge​(y+3x)=C

Correct answer: (B)

Step-by-step solution →
Q204·MathematicsSingle correctJEE Main 2020
Let f : (0, ∞) → (0, ∞) be a differentiable function such that f(1) = e and lim⁡t→xt2f2(x)−x2f2(t)t−x=0\lim_{t \to x} \frac{t^{2}f^{2}(x) - x^{2}f^{2}(t)}{t - x} = 0limt→x​t−xt2f2(x)−x2f2(t)​=0. If f(x) = 1, then x is equal to:
  1. (A)e
  2. (B)1e\frac{1}{e}e1​
  3. (C)12e\frac{1}{2e}2e1​
  4. (D)2e

Correct answer: (B)

Step-by-step solution →
Q205·MathematicsSingle correctJEE Main 2020
The solution curve of the differential equation, (1+e−x)(1+y2)dydx=y2\left(1+e^{-x}\right)\left(1+y^{2}\right)\frac{dy}{dx}=y^{2}(1+e−x)(1+y2)dxdy​=y2, which passes through the point (0, 1), is
  1. (A)y2+1=y(log⁡e(1+ex2)+2)y^{2}+1=y\left(\log_{e}\left(\frac{1+e^{x}}{2}\right)+2\right)y2+1=y(loge​(21+ex​)+2)
  2. (B)y2+1=ylog⁡e(1+ex2)y^{2}+1=y\log_{e}\left(\frac{1+e^{x}}{2}\right)y2+1=yloge​(21+ex​)
  3. (C)y2+1=y(log⁡e(1+e−x2)+2)y^{2}+1=y\left(\log_{e}\left(\frac{1+e^{-x}}{2}\right)+2\right)y2+1=y(loge​(21+e−x​)+2)
  4. (D)y2=1+ylog⁡e(1+ex2)y^{2}=1+y\log_{e}\left(\frac{1+e^{x}}{2}\right)y2=1+yloge​(21+ex​)

Correct answer: (D)

Step-by-step solution →
Q206·MathematicsSingle correctJEE Main 2020
If x3dy+xy dx=x2dy+2y dx;y(2)=ex^3 dy + xy\, dx = x^2 dy + 2y\, dx; y(2) = ex3dy+xydx=x2dy+2ydx;y(2)=e and x>1x > 1x>1, then y(4)y(4)y(4) is equal to:
  1. (A)e2\frac{\sqrt{e}}{2}2e​​
  2. (B)12+e\frac{1}{2} + \sqrt{e}21​+e​
  3. (C)32e\frac{3}{2}\sqrt{e}23​e​
  4. (D)32+e\frac{3}{2} + \sqrt{e}23​+e​

Correct answer: (C)

Step-by-step solution →
Q207·MathematicsNumericalJEE Main 2020
If for x≥0x \geq 0x≥0, y=y(x)y=y(x)y=y(x) is the solution of the differential equation, (x+1)dy=((x+1)2+y−3)dx,y(2)=0(x+1)dy=\left((x+1)^{2}+y-3\right)dx, y(2)=0(x+1)dy=((x+1)2+y−3)dx,y(2)=0, then y(3)y(3)y(3) is equal to ______

Correct answer: 3

Step-by-step solution →
Q208·MathematicsSingle correctJEE Main 2020
If dydx=xyx2+y2\dfrac{dy}{dx} = \dfrac{xy}{x^{2} + y^{2}}dxdy​=x2+y2xy​; y(1)=1y(1) = 1y(1)=1; then a value of x satisfying y(x)=ey(x) = ey(x)=e is:
  1. (A)2 e\sqrt{2}\,e2​e
  2. (B)123 e\dfrac{1}{2}\sqrt{3}\,e21​3​e
  3. (C)3 e\sqrt{3}\,e3​e
  4. (D)e2\dfrac{e}{\sqrt{2}}2​e​

Correct answer: (C)

Step-by-step solution →
Q209·MathematicsSingle correctJEE Main 2020
The differential equation of the family of curves, x2=4b(y+b),b∈Rx^{2}=4b\left(y+b\right), b \in Rx2=4b(y+b),b∈R, is:
  1. (A)xy′′=y′xy''=y'xy′′=y′
  2. (B)x(y′)2=x−2yy′x\left(y'\right)^{2}=x-2yy'x(y′)2=x−2yy′
  3. (C)x(y′)2=x+2yy′x\left(y'\right)^{2}=x+2yy'x(y′)2=x+2yy′
  4. (D)x(y′)2=2yy′−xx\left(y'\right)^{2}=2yy'-xx(y′)2=2yy′−x

Correct answer: (C)

Step-by-step solution →
Q210·MathematicsSingle correctJEE Main 2020
Let y=y(x)y=y(x)y=y(x) be a solution of the differential equation, 1−x2dydx+1−y2=0,∣x∣<1\sqrt{1-x^{2}}\dfrac{dy}{dx}+\sqrt{1-y^{2}}=0, |x|<11−x2​dxdy​+1−y2​=0,∣x∣<1. If y(12)=32y\left(\dfrac{1}{2}\right)=\dfrac{\sqrt{3}}{2}y(21​)=23​​, then y(12)y\left(\dfrac{1}{\sqrt{2}}\right)y(2​1​) is equal to:
  1. (A)−32-\dfrac{\sqrt{3}}{2}−23​​
  2. (B)−12-\dfrac{1}{\sqrt{2}}−2​1​
  3. (C)12\dfrac{1}{\sqrt{2}}2​1​
  4. (D)32\dfrac{\sqrt{3}}{2}23​​

Correct answer: (C)

Step-by-step solution →
Q211·MathematicsSingle correctJEE Main 2020
Let xk+yk=ak,(a,k>0)x^{k} + y^{k} = a^{k}, (a,k > 0)xk+yk=ak,(a,k>0) and dydx+(yx)13=0\frac{dy}{dx} + \left(\frac{y}{x}\right)^{\frac{1}{3}} = 0dxdy​+(xy​)31​=0, then k is:
  1. (A)32\frac{3}{2}23​
  2. (B)43\frac{4}{3}34​
  3. (C)13\frac{1}{3}31​
  4. (D)23\frac{2}{3}32​

Correct answer: (D)

Step-by-step solution →
Q212·MathematicsSingle correctJEE Main 2020
If y=y(x)y = y(x)y=y(x) is the solution of the differential equation, ey(dydx−1)=exe^{y}\left(\frac{dy}{dx} - 1\right) = e^{x}ey(dxdy​−1)=ex such that y(0)=0y(0) = 0y(0)=0, then y(1)y(1)y(1) is equal to:
  1. (A)1+log⁡e21 + \log_{e} 21+loge​2
  2. (B)2e2e2e
  3. (C)log⁡e2\log_{e} 2loge​2
  4. (D)2+log⁡e22 + \log_{e} 22+loge​2

Correct answer: (A)

Step-by-step solution →
Q213·MathematicsSingle correctJEE Main 2020
Let y=y(x)y=y(x)y=y(x) be the solution curve of the differential equation, (y2−x)dydx=1(y^{2}-x)\dfrac{dy}{dx}=1(y2−x)dxdy​=1, satisfying y(0)=1y(0)=1y(0)=1. This curve intersects the x-axis at a point whose abscissa is:
  1. (A)2+e2+e2+e
  2. (B)−e-e−e
  3. (C)2
  4. (D)2−e2-e2−e

Correct answer: (D)

Step-by-step solution →
Q214·MathematicsMultiple correctJEE Advanced 2019
Let Γ\GammaΓ denote a curve y=y(x)y = y(x)y=y(x) which is in the first quadrant and let the point (1, 0) lie on it. Let the tangent to Γ\GammaΓ at a point P intersect the y-axis at YPY_PYP​. If PYPPY_PPYP​ has length 1 for each point P on Γ\GammaΓ, then which of the following option is/are correct?
  1. (A)y=log⁡e(1+1−x2x)−1−x2y = \log_e\left(\frac{1 + \sqrt{1 - x^2}}{x}\right) - \sqrt{1 - x^2}y=loge​(x1+1−x2​​)−1−x2​
  2. (B)xy′+1−x2=0xy' + \sqrt{1 - x^2} = 0xy′+1−x2​=0
  3. (C)y=−log⁡e(1+1−x2x)+1−x2y = -\log_e\left(\frac{1 + \sqrt{1 - x^2}}{x}\right) + \sqrt{1 - x^2}y=−loge​(x1+1−x2​​)+1−x2​
  4. (D)xy′−1−x2=0xy' - \sqrt{1 - x^2} = 0xy′−1−x2​=0

Correct answer: (A), (B)

Step-by-step solution →
Q215·MathematicsSingle correctJEE Main 2019
Consider the differential equation, y2dx+(x−1y)dy=0y^{2}dx+\left(x-\frac{1}{y}\right)dy=0y2dx+(x−y1​)dy=0. If value of y is 1 when x = 1, then the value of x for which y = 2, is :
  1. (A)32−e\frac{3}{2}-\sqrt{e}23​−e​
  2. (B)12+1e\frac{1}{2}+\frac{1}{\sqrt{e}}21​+e​1​
  3. (C)32−1e\frac{3}{2}-\frac{1}{\sqrt{e}}23​−e​1​
  4. (D)52+1e\frac{5}{2}+\frac{1}{\sqrt{e}}25​+e​1​

Correct answer: (C)

Step-by-step solution →
Q216·MathematicsSingle correctJEE Main 2019
The general solution of the differential equation (y2−x3)dx−xy dy=0(y^{2}-x^{3})dx - xy\,dy = 0(y2−x3)dx−xydy=0 (x≠0)(x \neq 0)(x=0) is : (where c is a constant of integration)
  1. (A)y2+2x3+cx2=0y^{2}+2x^{3}+cx^{2}=0y2+2x3+cx2=0
  2. (B)y2−2x3+cx2=0y^{2}-2x^{3}+cx^{2}=0y2−2x3+cx2=0
  3. (C)y2+2x2+cx3=0y^{2}+2x^{2}+cx^{3}=0y2+2x2+cx3=0
  4. (D)y2−2x2+cx3=0y^{2}-2x^{2}+cx^{3}=0y2−2x2+cx3=0

Correct answer: (A)

Step-by-step solution →
Q217·MathematicsSingle correctJEE Main 2019
If y = 1f(x) is the solution of the differential equation dydx=(tan⁡x−y)sec⁡2x\frac{dy}{dx}=(\tan x - y)\sec^{2}xdxdy​=(tanx−y)sec2x, x ∈ (−π2,π2-\frac{\pi}{2}, \frac{\pi}{2}−2π​,2π​), such that y(0) = 0, then y(−π4-\frac{\pi}{4}−4π​) is equal to
  1. (A)12−e\frac{1}{2}-e21​−e
  2. (B)1e−2\frac{1}{e}-2e1​−2
  3. (C)e−2e-2e−2
  4. (D)2+1e2+\frac{1}{e}2+e1​

Correct answer: (C)

Step-by-step solution →
Q218·MathematicsSingle correctJEE Main 2019
The solution of the differential equation xdydx+2y=x2 (x≠0)x\dfrac{dy}{dx}+2y=x^{2}\ (x\neq 0)xdxdy​+2y=x2 (x=0) with y(1)=1y(1)=1y(1)=1, is:
  1. (A)y=x35+15x2y=\dfrac{x^{3}}{5}+\dfrac{1}{5x^{2}}y=5x3​+5x21​
  2. (B)y=x24+34x2y=\dfrac{x^{2}}{4}+\dfrac{3}{4x^{2}}y=4x2​+4x23​
  3. (C)y=45x3+15x2y=\dfrac{4}{5}x^{3}+\dfrac{1}{5x^{2}}y=54​x3+5x21​
  4. (D)y=34x2+14x2y=\dfrac{3}{4}x^{2}+\dfrac{1}{4x^{2}}y=43​x2+4x21​

Correct answer: (B)

Step-by-step solution →
Q219·MathematicsSingle correctJEE Main 2019
If cos⁡xdydx−ysin⁡x=6x,(0<x<π2)\cos x\dfrac{dy}{dx} - y\sin x = 6x, \left(0 < x < \dfrac{\pi}{2}\right)cosxdxdy​−ysinx=6x,(0<x<2π​) and y(π3)=0y\left(\dfrac{\pi}{3}\right) = 0y(3π​)=0, then y(π6)y\left(\dfrac{\pi}{6}\right)y(6π​) is equal to:
  1. (A)−π243-\dfrac{\pi^{2}}{4\sqrt{3}}−43​π2​
  2. (B)−π22-\dfrac{\pi^{2}}{2}−2π2​
  3. (C)π223\dfrac{\pi^{2}}{2\sqrt{3}}23​π2​
  4. (D)−π223-\dfrac{\pi^{2}}{2\sqrt{3}}−23​π2​

Correct answer: (D)

Step-by-step solution →
Q220·MathematicsSingle correctJEE Main 2019
Given that the slope of the tangent to a curve y=y(x)y=y(x)y=y(x) at any point (x,y)(x,y)(x,y) is 2yx2\frac{2y}{x^{2}}x22y​. If the curve passes through the centre of the circle x2+y2−2x−2y=0x^{2}+y^{2}-2x-2y=0x2+y2−2x−2y=0, then its equation is:
  1. (A)xlog⁡e∣y∣=x−1x\log_{e}|y|=x-1xloge​∣y∣=x−1
  2. (B)xlog⁡e∣y∣=−2(x−1)x\log_{e}|y|=-2(x-1)xloge​∣y∣=−2(x−1)
  3. (C)x2log⁡e∣y∣=−2(x−1)x^{2}\log_{e}|y|=-2(x-1)x2loge​∣y∣=−2(x−1)
  4. (D)xlog⁡e∣y∣=2(x−1)x\log_{e}|y|=2(x-1)xloge​∣y∣=2(x−1)

Correct answer: (D)

Step-by-step solution →
Q221·MathematicsSingle correctJEE Main 2019
Let y=y(x)y=y(x)y=y(x) be the solutions of the differential equation, (x2+1)2dydx+2x(x2+1)y=1\left(x^{2}+1\right)^{2}\frac{dy}{dx}+2x\left(x^{2}+1\right)y=1(x2+1)2dxdy​+2x(x2+1)y=1 such that y(0)=0y(0)=0y(0)=0. If a y(1)=π32\sqrt{a}\,y(1)=\frac{\pi}{32}a​y(1)=32π​, then the value of 'a' is
  1. (A)12\frac{1}{2}21​
  2. (B)111
  3. (C)116\frac{1}{16}161​
  4. (D)14\frac{1}{4}41​

Correct answer: (C)

Step-by-step solution →
Q222·MathematicsSingle correctJEE Main 2019
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation, xdydx+y=xlog⁡ex,(x>1)x\dfrac{dy}{dx} + y = x\log_{e} x, (x > 1)xdxdy​+y=xloge​x,(x>1). If 2y(2)=log⁡e4−12y(2) = \log_{e} 4 - 12y(2)=loge​4−1, then y(e)y(e)y(e) is equal to:
  1. (A)−e2-\dfrac{e}{2}−2e​
  2. (B)−e22-\dfrac{e^{2}}{2}−2e2​
  3. (C)e4\dfrac{e}{4}4e​
  4. (D)e24\dfrac{e^{2}}{4}4e2​

Correct answer: (C)

Step-by-step solution →
Q223·MathematicsSingle correctJEE Main 2019
If a curve passes through the point (1, −2) and has slope of the tangent at any point (x, y) on it as x2−2yx\dfrac{x^{2} - 2y}{x}xx2−2y​, then the curve also passes through the point :
  1. (A)(3,0)(3, 0)(3,0)
  2. (B)(3,0)(\sqrt{3}, 0)(3​,0)
  3. (C)(−1,2)(-1, 2)(−1,2)
  4. (D)(−2,1)(-\sqrt{2}, 1)(−2​,1)

Correct answer: (B)

Step-by-step solution →
Q224·MathematicsSingle correctJEE Main 2019
If y (x) is the solution of the differential equation dydx+(2x+1x)y=e−2x,x>0\frac{dy}{dx}+\left(\frac{2x+1}{x}\right)y=e^{-2x}, x>0dxdy​+(x2x+1​)y=e−2x,x>0 where y(1)=12e−2y(1)=\frac{1}{2}e^{-2}y(1)=21​e−2, then:
  1. (A)y(log⁡e2)=log⁡e4y\left(\log_{e}2\right)=\log_{e}4y(loge​2)=loge​4
  2. (B)y(log⁡e2)=log⁡e24y\left(\log_{e}2\right)=\frac{\log_{e}2}{4}y(loge​2)=4loge​2​
  3. (C)y(x) is decreasing in (12,1)\left(\frac{1}{2},1\right)(21​,1)
  4. (D)y(x) is decreasing in (0, 1)

Correct answer: (C)

Step-by-step solution →
Q225·MathematicsSingle correctJEE Main 2019
The solution of the differential equation dydx=(x−y)2\frac{dy}{dx}=(x-y)^{2}dxdy​=(x−y)2 when y(1) = 1, is:
  1. (A)log⁡e∣2−x2−y∣=x−y\log_e\left|\frac{2-x}{2-y}\right|=x-yloge​​2−y2−x​​=x−y
  2. (B)−log⁡e∣1−x+y1+x−y∣=2(x−1)-\log_e\left|\frac{1-x+y}{1+x-y}\right|=2(x-1)−loge​​1+x−y1−x+y​​=2(x−1)
  3. (C)−log⁡e∣1+x−y1−x+y∣=x+y−2-\log_e\left|\frac{1+x-y}{1-x+y}\right|=x+y-2−loge​​1−x+y1+x−y​​=x+y−2
  4. (D)log⁡e∣2−y2−x∣=2(y−1)\log_e\left|\frac{2-y}{2-x}\right|=2(y-1)loge​​2−x2−y​​=2(y−1)

Correct answer: (B)

Step-by-step solution →
Q226·MathematicsSingle correctJEE Main 2019
If dydx+3cos⁡2xy=1cos⁡2x,x∈(−π3,π3)\frac{dy}{dx}+\frac{3}{\cos^{2}x}y=\frac{1}{\cos^{2}x},x\in\left(\frac{-\pi}{3},\frac{\pi}{3}\right)dxdy​+cos2x3​y=cos2x1​,x∈(3−π​,3π​) and y(π4)=43y\left(\frac{\pi}{4}\right)=\frac{4}{3}y(4π​)=34​, then y(−π4)y\left(-\frac{\pi}{4}\right)y(−4π​) equals:
  1. (A)13+e6\frac{1}{3}+e^{6}31​+e6
  2. (B)13\frac{1}{3}31​
  3. (C)−43-\frac{4}{3}−34​
  4. (D)13+e3\frac{1}{3}+e^{3}31​+e3

Correct answer: (A)

Step-by-step solution →
Q227·MathematicsSingle correctJEE Main 2019
If y=y(x)y = y(x)y=y(x) is solution of the differential equation xdydx+2y=x2x\dfrac{dy}{dx} + 2y = x^2xdxdy​+2y=x2 satisfying y(1)=1y(1) = 1y(1)=1, then y(12)y\left(\dfrac{1}{2}\right)y(21​) is equal to
  1. (A)764\dfrac{7}{64}647​
  2. (B)14\dfrac{1}{4}41​
  3. (C)4916\dfrac{49}{16}1649​
  4. (D)1316\dfrac{13}{16}1613​

Correct answer: (C)

Step-by-step solution →
Q228·MathematicsSingle correctJEE Main 2019
Let f:[0,1]→Rf:[0,1]\rightarrow Rf:[0,1]→R be such that f(xy)=f(x)f(y)f(xy)=f(x)f(y)f(xy)=f(x)f(y) for all x,y∈[0,1]x,y\in[0,1]x,y∈[0,1], and f(0)≠0f(0)\neq 0f(0)=0. If y=y(x)y=y(x)y=y(x) satisfies the differential equation, dydx=f(x)\frac{dy}{dx}=f(x)dxdy​=f(x) with y(0)=1y(0)=1y(0)=1, then y(14)+y(34)y\left(\frac{1}{4}\right)+y\left(\frac{3}{4}\right)y(41​)+y(43​) is equal to
  1. (A)4
  2. (B)3
  3. (C)5
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q229·MathematicsNumericalJEE Advanced 2018
Let f:R→Rf : R \to Rf:R→R be a differentiable function with f(0)=0f(0) = 0f(0)=0. If y=f(x)y = f(x)y=f(x) satisfies the differential equation dydx=(2+5y)(5y−2)\frac{dy}{dx} = (2+5y)(5y-2)dxdy​=(2+5y)(5y−2), then the value of lim⁡x→−∞f(x)\lim_{x \to -\infty} f(x)limx→−∞​f(x) is ______ .

Correct answer: 0.4

Step-by-step solution →
Q230·MathematicsMultiple correctJEE Advanced 2018
Let f:[0,∞)→Rf : [0, \infty) \to Rf:[0,∞)→R be a continuous function such that f(x)=1−2x+∫0xex−tf(t)dtf(x) = 1 - 2x + \int_{0}^{x} e^{x-t} f(t) dtf(x)=1−2x+∫0x​ex−tf(t)dt for all x∈[0,∞)x \in [0, \infty)x∈[0,∞). Then, which of the following statement(s) is (are) TRUE ?
  1. (A)The curve y=f(x)y = f(x)y=f(x) passes through the point (1,2)(1, 2)(1,2)
  2. (B)The curve y=f(x)y = f(x)y=f(x) passes through the point (2,−1)(2, -1)(2,−1)
  3. (C)The area of the region {(x,y)∈[0,1]×R:f(x)≤y≤1−x2}\{(x, y) \in [0, 1] \times R : f(x) \le y \le \sqrt{1-x^{2}}\}{(x,y)∈[0,1]×R:f(x)≤y≤1−x2​} is π−24\frac{\pi-2}{4}4π−2​
  4. (D)The area of the region {(x,y)∈[0,1]×R:f(x)≤y≤1−x2}\{(x, y) \in [0, 1] \times R : f(x) \le y \le \sqrt{1-x^{2}}\}{(x,y)∈[0,1]×R:f(x)≤y≤1−x2​} is π−14\frac{\pi-1}{4}4π−1​

Correct answer: (B), (C)

Step-by-step solution →
Q231·MathematicsNumericalJEE Advanced 2018
Let f:R→Rf : R \to Rf:R→R be a differentiable function with f(0)=1f(0) = 1f(0)=1 and satisfying the equation f(x+y)=f(x)f′(y)+f′(x)f(y)f(x + y) = f(x)f'(y) + f'(x)f(y)f(x+y)=f(x)f′(y)+f′(x)f(y) for all xxx, y∈Ry \in Ry∈R. Then, the value of log⁡e(f(4))\log_{e}(f(4))loge​(f(4)) is ______ .

Correct answer: 2

Step-by-step solution →
Q232·MathematicsMultiple correctJEE Advanced 2018
Let f:(0,π)→Rf : (0, \pi) \to Rf:(0,π)→R be a twice differentiable function such that lim⁡t→xf(x)sin⁡t−f(t)sin⁡xt−x=sin⁡2x\lim_{t \to x} \frac{f(x)\sin t - f(t)\sin x}{t - x} = \sin^{2} xlimt→x​t−xf(x)sint−f(t)sinx​=sin2x for all x∈(0,π)x \in (0, \pi)x∈(0,π) If f(π6)=−π12f\left(\frac{\pi}{6}\right) = -\frac{\pi}{12}f(6π​)=−12π​, then which of the following statement(s) is (are) TRUE ?
  1. (A)f(π4)=π42f\left(\frac{\pi}{4}\right) = \frac{\pi}{4\sqrt{2}}f(4π​)=42​π​
  2. (B)f(x)<x46−x2f(x) < \frac{x^{4}}{6} - x^{2}f(x)<6x4​−x2 for all x∈(0,π)x \in (0, \pi)x∈(0,π)
  3. (C)There exists α∈(0,π)\alpha \in (0, \pi)α∈(0,π) such that f′(α)=0f'(\alpha) = 0f′(α)=0
  4. (D)f′′(π2)+f(π2)=0f''\left(\frac{\pi}{2}\right) + f\left(\frac{\pi}{2}\right) = 0f′′(2π​)+f(2π​)=0

Correct answer: (B), (C), (D)

Step-by-step solution →
Q233·MathematicsMultiple correctJEE Advanced 2018
Let f:R→Rf : R \to Rf:R→R and g:R→Rg : R \to Rg:R→R be two non-constant differentiable functions. If f′(x)=(e(f(x)−g(x)))g′(x)f'(x) = \left(e^{(f(x)-g(x))}\right) g'(x)f′(x)=(e(f(x)−g(x)))g′(x) for all x∈Rx \in Rx∈R, and f(1)=g(2)=1f(1) = g(2) = 1f(1)=g(2)=1, then which of the following statement(s) is (are) TRUE ?
  1. (A)f(2)<1−log⁡e2f(2) < 1 - \log_{e} 2f(2)<1−loge​2
  2. (B)f(2)>1−log⁡e2f(2) > 1 - \log_{e} 2f(2)>1−loge​2
  3. (C)g(1)>1−log⁡e2g(1) > 1 - \log_{e} 2g(1)>1−loge​2
  4. (D)g(1)<1−log⁡e2g(1) < 1 - \log_{e} 2g(1)<1−loge​2

Correct answer: (B), (C)

Step-by-step solution →
Q234·MathematicsSingle correctJEE Advanced 2017
If y=y(x)y = y(x)y=y(x) satisfies the differential equation 8x(9+x)dy=(4+9+x)−1dx8\sqrt{x}\left(\sqrt{9+\sqrt{x}}\right)dy = \left(\sqrt{4+\sqrt{9+\sqrt{x}}}\right)^{-1}dx8x​(9+x​​)dy=(4+9+x​​​)−1dx, x>0\quad x > 0x>0 and y(0)=7y(0) = \sqrt{7}y(0)=7​, then y(256)=y(256) =y(256)=
  1. (A)333
  2. (B)999
  3. (C)161616
  4. (D)808080

Correct answer: (A)

Step-by-step solution →
Q235·MathematicsMultiple correctJEE Advanced 2016
A solution curve of the differential equation (x2+xy+4x+2y+4)dydx−y2=0\left(x^{2} + xy + 4x + 2y + 4\right)\frac{dy}{dx} - y^{2} = 0(x2+xy+4x+2y+4)dxdy​−y2=0, x>0x > 0x>0, passes through the point (1, 3). Then the solution curve
  1. (A)intersects y=x+2y = x + 2y=x+2 exactly at one point
  2. (B)intersects y=x+2y = x + 2y=x+2 exactly at two points
  3. (C)intersects y=(x+2)2y = (x + 2)^{2}y=(x+2)2
  4. (D)does NOT intersect y=(x+3)2y = (x + 3)^{2}y=(x+3)2

Correct answer: (A), (D)

Step-by-step solution →
Q236·MathematicsMultiple correctJEE Advanced 2016
Let f:(0,∞)→Rf : (0, \infty) \to \mathbb{R}f:(0,∞)→R be a differentiable function such that f′(x)=2−f(x)xf'(x) = 2 - \frac{f(x)}{x}f′(x)=2−xf(x)​ for all x∈(0,∞)x \in (0, \infty)x∈(0,∞) and f(1)≠1f(1) \neq 1f(1)=1. Then
  1. (A)lim⁡x→0+f′(1x)=1\lim_{x \to 0+} f'\left(\frac{1}{x}\right) = 1limx→0+​f′(x1​)=1
  2. (B)lim⁡x→0+xf(1x)=2\lim_{x \to 0+} x f\left(\frac{1}{x}\right) = 2limx→0+​xf(x1​)=2
  3. (C)lim⁡x→0+x2f′(x)=0\lim_{x \to 0+} x^{2} f'(x) = 0limx→0+​x2f′(x)=0
  4. (D)∣f(x)∣≤2|f(x)| \leq 2∣f(x)∣≤2 for all x∈(0,2)x \in (0, 2)x∈(0,2)

Correct answer: (A)

Step-by-step solution →
Q237·MathematicsMultiple correctJEE Advanced 2015
Consider the family of all circles whose centers lie on the straight line y=xy = xy=x. If this family of circles is represented by the differential equation Py′′+Qy′+1=0Py'' + Qy' + 1 = 0Py′′+Qy′+1=0, where PPP, QQQ are functions of xxx, yyy and y′y'y′ (here y′=dydxy' = \dfrac{dy}{dx}y′=dxdy​, y′′=d2ydx2y'' = \dfrac{d^{2}y}{dx^{2}}y′′=dx2d2y​), then which of the following statements is (are) true ?
  1. (A)P=y+xP = y + xP=y+x
  2. (B)P=y−xP = y - xP=y−x
  3. (C)P+Q=1−x+y+y′+(y′)2P + Q = 1 - x + y + y' + (y')^{2}P+Q=1−x+y+y′+(y′)2
  4. (D)P−Q=x+y−y′−(y′)2P - Q = x + y - y' - (y')^{2}P−Q=x+y−y′−(y′)2

Correct answer: (B), (C)

Step-by-step solution →
Q238·MathematicsMultiple correctJEE Advanced 2015
Let y(x)y(x)y(x) be a solution of the differential equation (1+ex)y′+yex=1(1 + e^{x})y' + ye^{x} = 1(1+ex)y′+yex=1. If y(0)=2y(0) = 2y(0)=2, then which of the following statements is (are) true ?
  1. (A)y(−4)=0y(-4) = 0y(−4)=0
  2. (B)y(−2)=0y(-2) = 0y(−2)=0
  3. (C)y(x)y(x)y(x) has a critical point in the interval (−1,0)(-1, 0)(−1,0)
  4. (D)y(x)y(x)y(x) has no critical point in the interval (−1,0)(-1, 0)(−1,0)

Correct answer: (A), (C)

Step-by-step solution →
Q239·MathematicsSingle correctJEE Advanced 2014
The function y=f(x)y = f(x)y=f(x) is the solution of the differential equation dydx+xyx2−1=x4+2x1−x2\frac{dy}{dx} + \frac{xy}{x^{2}-1} = \frac{x^{4}+2x}{\sqrt{1-x^{2}}}dxdy​+x2−1xy​=1−x2​x4+2x​ in (−1,1)(-1, 1)(−1,1) satisfying f(0)=0f(0) = 0f(0)=0. Then ∫−3232f(x) dx\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x)\, dx∫−23​​23​​​f(x)dx is
  1. (A)π3−32\frac{\pi}{3} - \frac{\sqrt{3}}{2}3π​−23​​
  2. (B)π3−34\frac{\pi}{3} - \frac{\sqrt{3}}{4}3π​−43​​
  3. (C)π6−34\frac{\pi}{6} - \frac{\sqrt{3}}{4}6π​−43​​
  4. (D)π6−32\frac{\pi}{6} - \frac{\sqrt{3}}{2}6π​−23​​

Correct answer: (B)

Step-by-step solution →
Q240·MathematicsSingle correctJEE Advanced 2013
A curve passes through the point (1,π6)(1,\frac{\pi}{6})(1,6π​). Let the slope of the curve at each point (x,y)(x,y)(x,y) be yx+sec⁡(yx)\frac{y}{x}+\sec\left(\frac{y}{x}\right)xy​+sec(xy​), x>0x>0x>0. Then the equation of the curve is
  1. (A)sin⁡(yx)=log⁡x+12\sin\left(\frac{y}{x}\right)=\log x+\frac{1}{2}sin(xy​)=logx+21​
  2. (B)cosec⁡(yx)=log⁡x+2\operatorname{cosec}\left(\frac{y}{x}\right)=\log x+2cosec(xy​)=logx+2
  3. (C)sec⁡(2yx)=log⁡x+2\sec\left(\frac{2y}{x}\right)=\log x+2sec(x2y​)=logx+2
  4. (D)cos⁡(2yx)=log⁡x+12\cos\left(\frac{2y}{x}\right)=\log x+\frac{1}{2}cos(x2y​)=logx+21​

Correct answer: (A)

Step-by-step solution →

Differential Equations — frequently asked

How many questions from Differential Equations appear in JEE?

Differential Equations has appeared in 167 of the last 186 JEE Main and JEE Advanced papers — about 90% of them — contributing 240 questions in total across those papers.

Is Differential Equations an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 90% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Differential Equations questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Mathematics chapters

  • Three Dimensional Geometry 344
  • Matrices and Determinants 342
  • Sets, Relations and Functions 313
  • Sequence and Series 287
  • Definite Integration 267
  • Vector Algebra 245
  • Probability 226
  • Permutations and Combinations 220

All 26 Mathematics chapters →

Practise Differential Equations until it stops costing you marks.

Build a timed test from these 240 questions in one click. Jarvis marks it, names the specific misconception behind each wrong answer, and brings the ones you failed back at the right interval.

Practise Differential Equations freeSee pricing

Free plan, no time limit · No credit card needed

Jarvis OS

AI-powered JEE preparation.

Question papers

JEE Main PYQsJEE Advanced PYQsPhysics PYQsChemistry PYQsMaths PYQs

Product

TourPricingSign inCreate account

Legal

Privacy PolicyTerms of ServiceRefund & CancellationShipping & DeliveryContact Us

Operated by

Venyou Craft Private Limited

info@jarvisos.net

© 2026 Jarvis OS