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Trigonometric Functions — JEE Previous Year Questions

Every Trigonometric Functions question asked in JEE Main and JEE Advanced across the last 186 papers — 189 questions, each with its correct answer. Free to read, no account needed.

Questions

189

Papers it appeared in

144/186

Appearance rate

77%

All 189 Trigonometric Functions questions

Most recent papers first.

Q1·MathematicsIntegerJEE Advanced 2026
Let α=(1−2cos⁡(π11))(1−2cos⁡(3π11))(1−2cos⁡(9π11))(1−2cos⁡(27π11))(1−2cos⁡(81π11))\alpha = \left(1 - 2\cos\left(\frac{\pi}{11}\right)\right) \left(1 - 2\cos\left(\frac{3\pi}{11}\right)\right) \left(1 - 2\cos\left(\frac{9\pi}{11}\right)\right) \left(1 - 2\cos\left(\frac{27\pi}{11}\right)\right) \left(1 - 2\cos\left(\frac{81\pi}{11}\right)\right)α=(1−2cos(11π​))(1−2cos(113π​))(1−2cos(119π​))(1−2cos(1127π​))(1−2cos(1181π​)). Then the value of 5−α25 - \alpha^25−α2 is ______.

Correct answer: 4

Step-by-step solution →
Q2·MathematicsSingle correctJEE Advanced 2026
Match each entry in List-I to the correct entry in List-II and choose the correct option.
List-IList-II
P.The number of elements in the set {x∈[−π,π]:sin⁡6x+cos⁡4x=1}\{x \in [-\pi, \pi] : \sin^6 x + \cos^4 x = 1\}{x∈[−π,π]:sin6x+cos4x=1}1.is 1
Q.The number of elements in the set {x∈[−π2,π2]:sin⁡2x+cos⁡6x=1}\left\{x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] : \sin^2 x + \cos^6 x = 1\right\}{x∈[−2π​,2π​]:sin2x+cos6x=1}2.is 2
R.The number of elements in the set {x∈[−π,π]:cos⁡2(x2)−sin⁡2x=12}\left\{x \in [-\pi, \pi] : \cos^2\left(\frac{x}{2}\right) - \sin^2 x = \frac{1}{2}\right\}{x∈[−π,π]:cos2(2x​)−sin2x=21​}3.is 3
S.The number of elements in the set {x∈[−2π,2π]:6sin⁡2(x2)−cos⁡3x=3}\left\{x \in [-2\pi, 2\pi] : 6\sin^2\left(\frac{x}{2}\right) - \cos 3x = 3\right\}{x∈[−2π,2π]:6sin2(2x​)−cos3x=3}4.is 4
5.is 5
  1. (A)(P) → (2), (Q) → (5), (R) → (3), (S) → (4)
  2. (B)(P) → (5), (Q) → (3), (R) → (2), (S) → (4)
  3. (C)(P) → (5), (Q) → (4), (R) → (1), (S) → (3)
  4. (D)(P) → (4), (Q) → (3), (R) → (2), (S) → (5)

Correct answer: (B)

Step-by-step solution →
Q3·MathematicsNumericalJEE Advanced 2026
Passage: Consider the curve C1C_{1}C1​ given by y=e−xy = e^{-x}y=e−x for x∈[0,10π]x \in [0, 10\pi]x∈[0,10π], and the curve C2C_{2}C2​ given by y=e−x(sin⁡x+cos⁡x)y = e^{-x}(\sin x + \cos x)y=e−x(sinx+cosx) for x∈[0,10π]x \in [0, 10\pi]x∈[0,10π]. Let nnn be the total number of points of intersection of the curves C1C_{1}C1​ and C2C_{2}C2​. Suppose that α1,α2,…,αn∈[0,10π]\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} \in [0, 10\pi]α1​,α2​,…,αn​∈[0,10π] are the xxx-coordinates of the points of intersection of the curves C1C_{1}C1​ and C2C_{2}C2​ such that α1<α2<⋯<αn\alpha_{1} < \alpha_{2} < \cdots < \alpha_{n}α1​<α2​<⋯<αn​. Question: The value of nnn is _____.

Correct answer: 11

Step-by-step solution →
Q4·MathematicsSingle correctJEE Main 2026
Let S = {θ ∈ (−2π, 2π) : cos θ + 1 = 3\sqrt{3}3​ sin θ}. Then ∑θ∈Sθ\sum_{\theta \in S} \theta∑θ∈S​θ is equal to:
  1. (A)−2π3-\frac{2\pi}{3}−32π​
  2. (B)−4π3-\frac{4\pi}{3}−34π​
  3. (C)2π3\frac{2\pi}{3}32π​
  4. (D)4π3\frac{4\pi}{3}34π​

Correct answer: (B)

Step-by-step solution →
Q5·MathematicsSingle correctJEE Main 2026
Let tan⁡A\tan AtanA, tan⁡B\tan BtanB, where A,B∈(−π2,π2)A, B \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)A,B∈(−2π​,2π​), be the roots of the quadratic equation x2−2x−5=0x^2 - 2x - 5 = 0x2−2x−5=0. Then 20sin⁡2(A+B2)20\sin^2\left(\frac{A+B}{2}\right)20sin2(2A+B​) is equal to:
  1. (A)10+1010 + \sqrt{10}10+10​
  2. (B)10−21010 - 2\sqrt{10}10−210​
  3. (C)10−31010 - 3\sqrt{10}10−310​
  4. (D)10−1010 - \sqrt{10}10−10​

Correct answer: (C)

Step-by-step solution →
Q6·MathematicsSingle correctJEE Main 2026
The sum of all the integral values of ppp such that the equation 3sin⁡2x+12cos⁡x−3=p3\sin^2 x + 12\cos x - 3 = p3sin2x+12cosx−3=p, x∈Rx \in \mathbb{R}x∈R, has at least one solution, is:
  1. (A)−54-54−54
  2. (B)−60-60−60
  3. (C)−75-75−75
  4. (D)−84-84−84

Correct answer: (C)

Step-by-step solution →
Q7·MathematicsNumericalJEE Main 2026
If S={θ∈[−π,π]:cos⁡θcos⁡5θ2=cos⁡7θcos⁡7θ2}S = \left\{\theta \in [-\pi, \pi] : \cos\theta \cos\frac{5\theta}{2} = \cos 7\theta \cos\frac{7\theta}{2}\right\}S={θ∈[−π,π]:cosθcos25θ​=cos7θcos27θ​}, then n(S) is equal to _______.

Correct answer: 19

Step-by-step solution →
Q8·MathematicsNumericalJEE Main 2026
Let a⃗k=(tan⁡θk)i^+j^\vec{a}_k = (\tan\theta_k)\hat{i} + \hat{j}ak​=(tanθk​)i^+j^​ and b⃗k=i^−(cot⁡θk)j^\vec{b}_k = \hat{i} - (\cot\theta_k)\hat{j}bk​=i^−(cotθk​)j^​, where θk=2k−1π2n+1\theta_k = \frac{2^{k-1}\pi}{2^n + 1}θk​=2n+12k−1π​, for some n∈Nn \in \mathbb{N}n∈N, n>5n > 5n>5. Then the value of ∑k=1n∣a⃗k∣2∑k=1n∣b⃗k∣2\frac{\sum_{k=1}^{n} |\vec{a}_k|^2}{\sum_{k=1}^{n} |\vec{b}_k|^2}∑k=1n​∣bk​∣2∑k=1n​∣ak​∣2​ is _______.

Correct answer: 3

Step-by-step solution →
Q9·MathematicsNumericalJEE Main 2026
If A=sin⁡3∘cos⁡9∘+sin⁡9∘cos⁡27∘+sin⁡27∘cos⁡81∘A = \frac{\sin 3^\circ}{\cos 9^\circ} + \frac{\sin 9^\circ}{\cos 27^\circ} + \frac{\sin 27^\circ}{\cos 81^\circ}A=cos9∘sin3∘​+cos27∘sin9∘​+cos81∘sin27∘​ and B=tan⁡81∘−tan⁡3∘B = \tan 81^\circ - \tan 3^\circB=tan81∘−tan3∘, then BA\frac{B}{A}AB​ is equal to _______.

Correct answer: 2

Step-by-step solution →
Q10·MathematicsSingle correctJEE Main 2026
If sin⁡(π18)sin⁡(5π18)sin⁡(7π18)=K\sin\left(\frac{\pi}{18}\right)\sin\left(\frac{5\pi}{18}\right)\sin\left(\frac{7\pi}{18}\right)=Ksin(18π​)sin(185π​)sin(187π​)=K, then the value of sin⁡(10Kπ3)\sin\left(\frac{10K\pi}{3}\right)sin(310Kπ​) is :
  1. (A)3+122\frac{\sqrt{3}+1}{2\sqrt{2}}22​3​+1​
  2. (B)3−12\frac{\sqrt{3}-1}{\sqrt{2}}2​3​−1​
  3. (C)32\frac{\sqrt{3}}{2}23​​
  4. (D)12\frac{1}{2}21​

Correct answer: (A)

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2026
Let P={θ∈[0,4π]:tan⁡2θ≠1}P = \{\theta \in [0, 4\pi] : \tan^2\theta \neq 1\}P={θ∈[0,4π]:tan2θ=1} and S={a∈Z:2(cos⁡8θ−sin⁡8θ)sec⁡2θ=a2,θ∈P}S = \{a \in \mathbb{Z} : 2(\cos^8\theta - \sin^8\theta)\sec 2\theta = a^2, \theta \in P\}S={a∈Z:2(cos8θ−sin8θ)sec2θ=a2,θ∈P}. Then n(S)n(S)n(S) is:
  1. (A)000
  2. (B)111
  3. (C)222
  4. (D)333

Correct answer: (A)

Step-by-step solution →
Q12·MathematicsSingle correctJEE Main 2026
Let S={x∈[−π,π]:sin⁡x(sin⁡x+cos⁡x)=a,a∈Z}S=\{x\in[-\pi,\pi]:\sin x(\sin x+\cos x)=a,a\in\mathbb{Z}\}S={x∈[−π,π]:sinx(sinx+cosx)=a,a∈Z}. Then n(S)n(S)n(S) is equal to :
  1. (A)3
  2. (B)6
  3. (C)7
  4. (D)9

Correct answer: (D)

Step-by-step solution →
Q13·MathematicsSingle correctJEE Main 2026
If tan⁡(A−B)tan⁡A+sin⁡2Csin⁡2A=1\dfrac{\tan(A-B)}{\tan A} + \dfrac{\sin^2 C}{\sin^2 A} = 1tanAtan(A−B)​+sin2Asin2C​=1, AAA, BBB, C∈(0,π2)C \in \left(0, \dfrac{\pi}{2}\right)C∈(0,2π​), then
  1. (A)tan⁡A\tan AtanA, tan⁡C\tan CtanC, tan⁡B\tan BtanB are in G.P.
  2. (B)tan⁡A\tan AtanA, tan⁡B\tan BtanB, tan⁡C\tan CtanC are in G.P.
  3. (C)tan⁡A\tan AtanA, tan⁡C\tan CtanC, tan⁡B\tan BtanB are in A.P.
  4. (D)tan⁡A\tan AtanA, tan⁡B\tan BtanB, tan⁡C\tan CtanC are in A.P.

Correct answer: (A)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2026
If cot⁡x=512\cot x=\frac{5}{12}cotx=125​ for some x∈(π,3π2)x\in\left(\pi,\frac{3\pi}{2}\right)x∈(π,23π​), then sin⁡7x(cos⁡13x2+sin⁡13x2)+cos⁡7x(cos⁡13x2−sin⁡13x2)\sin 7x\left(\cos\frac{13x}{2}+\sin\frac{13x}{2}\right)+\cos 7x\left(\cos\frac{13x}{2}-\sin\frac{13x}{2}\right)sin7x(cos213x​+sin213x​)+cos7x(cos213x​−sin213x​) is equal to
  1. (A)426\frac{4}{\sqrt{26}}26​4​
  2. (B)626\frac{6}{\sqrt{26}}26​6​
  3. (C)113\frac{1}{\sqrt{13}}13​1​
  4. (D)513\frac{5}{\sqrt{13}}13​5​

Correct answer: (C)

Step-by-step solution →
Q15·MathematicsSingle correctJEE Main 2026
The value of 3csc⁡20∘−sec⁡20∘cos⁡20∘cos⁡40∘cos⁡60∘cos⁡80∘\frac{\sqrt{3}\csc 20^\circ - \sec 20^\circ}{\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ}cos20∘cos40∘cos60∘cos80∘3​csc20∘−sec20∘​ is equal to
  1. (A)32
  2. (B)16
  3. (C)64
  4. (D)12

Correct answer: (C)

Step-by-step solution →
Q16·MathematicsNumericalJEE Main 2026
The number of elements in the set {x∈[0,180∘]:tan⁡(x+100∘)=tan⁡(x+50∘)tan⁡xtan⁡(x−50∘)}\{x \in [0, 180^{\circ}] : \tan(x + 100^{\circ}) = \tan(x + 50^{\circ}) \tan x \tan(x - 50^{\circ})\}{x∈[0,180∘]:tan(x+100∘)=tan(x+50∘)tanxtan(x−50∘)} is ________.

Correct answer: 4

Step-by-step solution →
Q17·MathematicsSingle correctJEE Main 2026
Let α\alphaα and β\betaβ respectively be the maximum and the minimum values of the function f(θ)=4(sin⁡4(7π2−θ)+sin⁡4(11π+θ))−2(sin⁡6(3π2−θ)+sin⁡6(9π−θ))f(\theta) = 4\left(\sin^{4}\left(\frac{7\pi}{2} - \theta\right) + \sin^{4}(11\pi + \theta)\right) - 2\left(\sin^{6}\left(\frac{3\pi}{2} - \theta\right) + \sin^{6}(9\pi - \theta)\right)f(θ)=4(sin4(27π​−θ)+sin4(11π+θ))−2(sin6(23π​−θ)+sin6(9π−θ)), θ∈R\theta \in \mathbf{R}θ∈R. Then α+2β\alpha + 2\betaα+2β is equal to :
  1. (A)4
  2. (B)5
  3. (C)3
  4. (D)6

Correct answer: (B)

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2026
The least value of (cos⁡2θ−6sin⁡θ cos⁡θ+3sin⁡2θ+2)(\cos^2\theta - 6\sin\theta\ \cos\theta + 3\sin^2\theta + 2)(cos2θ−6sinθ cosθ+3sin2θ+2) is
  1. (A)−1
  2. (B)4+104 + \sqrt{10}4+10​
  3. (C)4−104 - \sqrt{10}4−10​
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q19·MathematicsSingle correctJEE Main 2026
Number of solutions of 3 cos⁡2θ+8cos⁡θ+33=0\sqrt{3}\ \cos 2\theta + 8 \cos\theta + 3\sqrt{3} = 03​ cos2θ+8cosθ+33​=0, θ∈[−3π,2π]\theta \in [-3\pi, 2\pi]θ∈[−3π,2π] is:
  1. (A)0
  2. (B)5
  3. (C)3
  4. (D)4

Correct answer: (B)

Step-by-step solution →
Q20·MathematicsNumericalJEE Main 2026
Let cos⁡(α+β)=−110\cos(\alpha + \beta) = -\frac{1}{10}cos(α+β)=−101​ and sin⁡(α−β)=38\sin(\alpha - \beta) = \frac{3}{8}sin(α−β)=83​, where 0<α<π30 < \alpha < \frac{\pi}{3}0<α<3π​ and 0<β<π40 < \beta < \frac{\pi}{4}0<β<4π​. If tan⁡2α=3(1−r5)11(s+5)\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}tan2α=11​(s+5​)3(1−r5​)​, r, s ∈ N\mathbb{N}N, then r + s is equal to______ .

Correct answer: 20

Step-by-step solution →
Q21·MathematicsNumericalJEE Main 2026
If cos⁡248∘−sin⁡212∘sin⁡224∘−sin⁡26∘=α+β52\frac{\cos^{2} 48^{\circ} - \sin^{2} 12^{\circ}}{\sin^{2} 24^{\circ} - \sin^{2} 6^{\circ}} = \frac{\alpha + \beta\sqrt{5}}{2}sin224∘−sin26∘cos248∘−sin212∘​=2α+β5​​, where α\alphaα, β∈N\beta \in \mathbb{N}β∈N, then α+β\alpha + \betaα+β is equal to _______.

Correct answer: 4

Step-by-step solution →
Q22·MathematicsSingle correctJEE Main 2026
The value of cosec⁡10∘−3 sec⁡10∘\operatorname{cosec}10^{\circ}-\sqrt{3}\ \sec10^{\circ}cosec10∘−3​ sec10∘ is equal to:
  1. (A)4
  2. (B)2
  3. (C)8
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q23·MathematicsNumericalJEE Advanced 2025
Let α=1sin⁡60∘sin⁡61∘+1sin⁡62∘sin⁡63∘+……+1sin⁡118∘sin⁡119∘\alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \ldots\ldots + \frac{1}{\sin 118^\circ \sin 119^\circ}α=sin60∘sin61∘1​+sin62∘sin63∘1​+……+sin118∘sin119∘1​ Then the value of (cosec 1∘α)2\left( \frac{\mathrm{cosec}\, 1^\circ}{\alpha} \right)^2(αcosec1∘​)2 is ______

Correct answer: 3

Step-by-step solution →
Q24·MathematicsSingle correctJEE Main 2025
If for θ∈[−π3,0]\theta\in\left[-\dfrac{\pi}{3},0\right]θ∈[−3π​,0], the points (x,y)=(3tan⁡(θ+π3),2tan⁡(θ+π6))(x,y)=\left(3\tan\left(\theta+\dfrac{\pi}{3}\right),2\tan\left(\theta+\dfrac{\pi}{6}\right)\right)(x,y)=(3tan(θ+3π​),2tan(θ+6π​)) lie on xy+αx+βy+γ=0xy+\alpha x+\beta y+\gamma=0xy+αx+βy+γ=0, then α2+β2+γ2\alpha^2+\beta^2+\gamma^2α2+β2+γ2 is equal to:
  1. (A)80
  2. (B)72
  3. (C)96
  4. (D)75

Correct answer: (D)

Step-by-step solution →
Q25·MathematicsSingle correctJEE Main 2025
The number of solutions of the equation cos⁡2θ cos⁡θ2+cos⁡5θ2=2cos⁡35θ2\cos 2\theta\,\cos\dfrac{\theta}{2}+\cos\dfrac{5\theta}{2}=2\cos^3\dfrac{5\theta}{2}cos2θcos2θ​+cos25θ​=2cos325θ​ in [−π2,π2]\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right][−2π​,2π​] is:
  1. (A)7
  2. (B)5
  3. (C)6
  4. (D)9

Correct answer: (A)

Step-by-step solution →
Q26·MathematicsSingle correctJEE Main 2025
If 10sin⁡4θ+15cos⁡4θ=610\sin^4\theta+15\cos^4\theta=610sin4θ+15cos4θ=6, then the value of 27 cosec6θ+8sec⁡6θ16sec⁡8θ\dfrac{27\,\mathrm{cosec}^6\theta+8\sec^6\theta}{16\sec^8\theta}16sec8θ27cosec6θ+8sec6θ​ is:
  1. (A)25\dfrac{2}{5}52​
  2. (B)34\dfrac{3}{4}43​
  3. (C)35\dfrac{3}{5}53​
  4. (D)15\dfrac{1}{5}51​

Correct answer: (A)

Step-by-step solution →
Q27·MathematicsSingle correctJEE Main 2025
The number of solutions of the equation 2x+3tan⁡x=π2x+3\tan x=\pi2x+3tanx=π, x∈[−2π,2π]−{±π2,±3π2}x\in[-2\pi,2\pi]-\left\{\pm\dfrac{\pi}{2},\pm\dfrac{3\pi}{2}\right\}x∈[−2π,2π]−{±2π​,±23π​} is:
  1. (A)6
  2. (B)5
  3. (C)4
  4. (D)3

Correct answer: (B)

Step-by-step solution →
Q28·MathematicsSingle correctJEE Main 2025
The number of solutions of the equation (4−3)sin⁡x−23cos⁡2x=−41+3(4-\sqrt3)\sin x-2\sqrt3\cos^2 x=-\dfrac{4}{1+\sqrt3}(4−3​)sinx−23​cos2x=−1+3​4​, x∈[−2π,5π2]x\in\left[-2\pi,\dfrac{5\pi}{2}\right]x∈[−2π,25π​] is:
  1. (A)4
  2. (B)3
  3. (C)6
  4. (D)5

Correct answer: (D)

Step-by-step solution →
Q29·MathematicsSingle correctJEE Main 2025
If θ∈[−7π6,4π3]\theta\in\left[-\dfrac{7\pi}{6},\dfrac{4\pi}{3}\right]θ∈[−67π​,34π​], then the number of solutions of 3 cosec⁡2θ−2(3−1)cosec⁡θ−4=0\sqrt{3}\,\operatorname{cosec}^2\theta-2(\sqrt{3}-1)\operatorname{cosec}\theta-4=03​cosec2θ−2(3​−1)cosecθ−4=0 is equal to:
  1. (A)6
  2. (B)8
  3. (C)10
  4. (D)7

Correct answer: (A)

Step-by-step solution →
Q30·MathematicsSingle correctJEE Main 2025
If θ∈[−2π,2π]\theta\in[-2\pi,2\pi]θ∈[−2π,2π], then the number of solutions of 22cos⁡2θ+(2−6)cos⁡θ−3=02\sqrt{2}\cos^2\theta+(2-\sqrt{6})\cos\theta-\sqrt{3}=022​cos2θ+(2−6​)cosθ−3​=0, is equal to:
  1. (A)12
  2. (B)6
  3. (C)8
  4. (D)10

Correct answer: (C)

Step-by-step solution →
Q31·MathematicsSingle correctJEE Main 2025
If sin⁡x+sin⁡2x=1\sin x+\sin^2 x=1sinx+sin2x=1, x∈(0,π2)x\in\left(0,\dfrac{\pi}{2}\right)x∈(0,2π​), then (cos⁡12x+tan⁡12x)+3(cos⁡10x+tan⁡10x+cos⁡8x+tan⁡8x)+(cos⁡6x+tan⁡6x)(\cos^{12}x+\tan^{12}x)+3(\cos^{10}x+\tan^{10}x+\cos^{8}x+\tan^{8}x)+(\cos^{6}x+\tan^{6}x)(cos12x+tan12x)+3(cos10x+tan10x+cos8x+tan8x)+(cos6x+tan6x) is equal to
  1. (A)4
  2. (B)3
  3. (C)2
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q32·MathematicsSingle correctJEE Main 2025
If ∑r=113{1sin⁡(π4+(r−1)π6)sin⁡(π4+rπ6)}=a3+b\sum_{r=1}^{13}\left\{\frac{1}{\sin\left(\frac{\pi}{4}+(r-1)\frac{\pi}{6}\right)\sin\left(\frac{\pi}{4}+r\frac{\pi}{6}\right)}\right\}=a\sqrt{3}+b∑r=113​{sin(4π​+(r−1)6π​)sin(4π​+r6π​)1​}=a3​+b, a,b∈Za,b\in Za,b∈Z, then a2+b2a^2+b^2a2+b2 is equal to:
  1. (A)10
  2. (B)2
  3. (C)8
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q33·MathematicsSingle correctJEE Main 2025
Let the range of the function f(x)=6+16cos⁡x⋅cos⁡ ⁣(π3−x)⋅cos⁡ ⁣(π3+x)⋅sin⁡3x⋅cos⁡6x, x∈Rf(x)=6+16\cos x\cdot\cos\!\left(\dfrac{\pi}{3}-x\right)\cdot\cos\!\left(\dfrac{\pi}{3}+x\right)\cdot\sin 3x\cdot\cos 6x,\ x\in\mathbb{R}f(x)=6+16cosx⋅cos(3π​−x)⋅cos(3π​+x)⋅sin3x⋅cos6x, x∈R be [α,β][\alpha,\beta][α,β]. Then the distance of the point (α,β)(\alpha,\beta)(α,β) from the line 3x+4y+12=03x+4y+12=03x+4y+12=0 is :
  1. (A)11
  2. (B)8
  3. (C)10
  4. (D)9

Correct answer: (A)

Step-by-step solution →
Q34·MathematicsSingle correctJEE Main 2025
The value of (sin⁡70∘)(cot⁡10∘cot⁡70∘−1)(\sin 70^\circ)(\cot 10^\circ\cot 70^\circ-1)(sin70∘)(cot10∘cot70∘−1) is
  1. (A)111
  2. (B)000
  3. (C)32\dfrac{3}{2}23​
  4. (D)23\dfrac{2}{3}32​

Correct answer: (A)

Step-by-step solution →
Q35·MathematicsSingle correctJEE Main 2025
The sum of all values of θ∈[0,2π]\theta\in[0,2\pi]θ∈[0,2π] satisfying 2sin⁡2θ=cos⁡2θ2\sin^2\theta=\cos2\theta2sin2θ=cos2θ and 2cos⁡2θ=3sin⁡θ2\cos^2\theta=3\sin\theta2cos2θ=3sinθ is
  1. (A)π2\dfrac{\pi}{2}2π​
  2. (B)4\pi
  3. (C)5π6\dfrac{5\pi}{6}65π​
  4. (D)\pi

Correct answer: (D)

Step-by-step solution →
Q36·MathematicsSingle correctJEE Advanced 2024
Let π2<x<π\frac{\pi}{2} < x < \pi2π​<x<π be such that cot⁡x=−511\cot x = \frac{-5}{\sqrt{11}}cotx=11​−5​. Then (sin⁡11x2)(sin⁡6x−cos⁡6x)+(cos⁡11x2)(sin⁡6x+cos⁡6x)\left(\sin \frac{11x}{2}\right)(\sin 6x - \cos 6x) + \left(\cos \frac{11x}{2}\right)(\sin 6x + \cos 6x)(sin211x​)(sin6x−cos6x)+(cos211x​)(sin6x+cos6x) is equal to
  1. (A)11−123\frac{\sqrt{11} - 1}{2\sqrt{3}}23​11​−1​
  2. (B)11+123\frac{\sqrt{11} + 1}{2\sqrt{3}}23​11​+1​
  3. (C)11+132\frac{\sqrt{11} + 1}{3\sqrt{2}}32​11​+1​
  4. (D)11−132\frac{\sqrt{11} - 1}{3\sqrt{2}}32​11​−1​

Correct answer: (B)

Step-by-step solution →
Q37·MathematicsSingle correctJEE Main 2024
Let ∣cos⁡θcos⁡(60−θ)cos⁡(60+θ)∣≤18|\cos\theta \cos(60 - \theta)\cos(60 + \theta)| \le \frac{1}{8}∣cosθcos(60−θ)cos(60+θ)∣≤81​, θ∈[0,2π]\theta \in [0, 2\pi]θ∈[0,2π]. Then, the sum of all θ∈[0,2π]\theta \in [0, 2\pi]θ∈[0,2π], where cos⁡3θ\cos 3\thetacos3θ attains its maximum value, is:
  1. (A)9π9\pi9π
  2. (B)18π18\pi18π
  3. (C)6π6\pi6π
  4. (D)15π15\pi15π

Correct answer: (C)

Step-by-step solution →
Q38·MathematicsSingle correctJEE Main 2024
Let the range of the function f(x)=12+sin⁡3x+cos⁡3xf(x)=\dfrac{1}{2+\sin 3x+\cos 3x}f(x)=2+sin3x+cos3x1​, x∈Rx\in\mathbb{R}x∈R be [a,b][a, b][a,b]. If α\alphaα and β\betaβ are respectively the A.M. and the G.M. of aaa and bbb, then αβ\dfrac{\alpha}{\beta}βα​ is equal to:
  1. (A)2\sqrt{2}2​
  2. (B)222
  3. (C)π\sqrt{\pi}π​
  4. (D)π\piπ

Correct answer: (A)

Step-by-step solution →
Q39·MathematicsSingle correctJEE Main 2024
If sin⁡x=−35\sin x=-\dfrac35sinx=−53​, where π<x<3π2\pi<x<\dfrac{3\pi}{2}π<x<23π​, then 80(tan⁡2x−cos⁡x)80(\tan^2 x-\cos x)80(tan2x−cosx) is equal to:
  1. (A)109109109
  2. (B)108108108
  3. (C)181818
  4. (D)191919

Correct answer: (A)

Step-by-step solution →
Q40·MathematicsSingle correctJEE Main 2024
If the value of 3cos⁡36∘+5sin⁡18∘5cos⁡36∘−3sin⁡18∘\dfrac{3\cos 36^\circ+5\sin 18^\circ}{5\cos 36^\circ-3\sin 18^\circ}5cos36∘−3sin18∘3cos36∘+5sin18∘​ is a5−bc\dfrac{a\sqrt{5}-b}{c}ca5​−b​, where aaa, bbb, ccc are natural numbers and gcd⁡(a,c)=1\gcd(a,c)=1gcd(a,c)=1, then a+b+ca+b+ca+b+c is equal to
  1. (A)505050
  2. (B)404040
  3. (C)525252
  4. (D)545454

Correct answer: (C)

Step-by-step solution →
Q41·MathematicsSingle correctJEE Main 2024
A circle is inscribed in an equilateral triangle of side length 12. If the area and perimeter of any square inscribed in this circle are mmm and nnn, respectively, then m+n2m+n^{2}m+n2 is equal to
  1. (A)396396396
  2. (B)408408408
  3. (C)312312312
  4. (D)414414414

Correct answer: (B)

Step-by-step solution →
Q42·MathematicsNumericalJEE Main 2024
In a triangle ABCABCABC, BC=7BC=7BC=7, AC=8AC=8AC=8, AB=α∈NAB=\alpha\in NAB=α∈N and cos⁡A=23\cos A=\frac{2}{3}cosA=32​. If 49cos⁡(3C)+42=mn49\cos(3C)+42=\frac{m}{n}49cos(3C)+42=nm​, where gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1, then m+nm+nm+n is equal to ______.

Correct answer: 39

Step-by-step solution →
Q43·MathematicsNumericalJEE Main 2024
The number of solutions of sin⁡2x+(2+2x−x2)sin⁡x−3(x−1)2=0\sin^2 x+(2+2x-x^2)\sin x-3(x-1)^2=0sin2x+(2+2x−x2)sinx−3(x−1)2=0, where −π≤x≤π-\pi\le x\le\pi−π≤x≤π, is __________.

Correct answer: 2

Step-by-step solution →
Q44·MathematicsSingle correctJEE Main 2024
Suppose θ∈[0,π4]\theta \in \left[0, \dfrac{\pi}{4}\right]θ∈[0,4π​] is a solution of 4cos⁡θ−3sin⁡θ=14\cos\theta - 3\sin\theta = 14cosθ−3sinθ=1. Then cos⁡θ\cos\thetacosθ is equal to:
  1. (A)4(36−2)\dfrac{4}{\left(3\sqrt{6} - 2\right)}(36​−2)4​
  2. (B)6−6(36−2)\dfrac{6 - \sqrt{6}}{\left(3\sqrt{6} - 2\right)}(36​−2)6−6​​
  3. (C)6+6(36+2)\dfrac{6 + \sqrt{6}}{\left(3\sqrt{6} + 2\right)}(36​+2)6+6​​
  4. (D)4(36+2)\dfrac{4}{\left(3\sqrt{6} + 2\right)}(36​+2)4​

Correct answer: (A)

Step-by-step solution →
Q45·MathematicsSingle correctJEE Main 2024
The number of solutions of the equation 4sin⁡2x−4cos⁡3x+9−4cos⁡x=04\sin^2 x-4\cos^3 x+9-4\cos x=04sin2x−4cos3x+9−4cosx=0, x∈[−2π, 2π]x\in[-2\pi,\,2\pi]x∈[−2π,2π] is:
  1. (A)1
  2. (B)3
  3. (C)2
  4. (D)0

Correct answer: (D)

Step-by-step solution →
Q46·MathematicsNumericalJEE Main 2024
Let ABC be an isosceles triangle in which A is at (−1,0)(-1,0)(−1,0), ∠A=2π3\angle A=\dfrac{2\pi}{3}∠A=32π​, AB=ACAB=ACAB=AC and B is on the positive x-axis. If BC=43BC=4\sqrt 3BC=43​ and the line BC intersects the line y=x+3y=x+3y=x+3 at (α,β)(\alpha,\beta)(α,β), then β4α2\dfrac{\beta^4}{\alpha^2}α2β4​ is __________.

Correct answer: 36

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2024
If tan⁡A=1x(x2+x+1)\tan A=\dfrac{1}{\sqrt{x(x^2+x+1)}}tanA=x(x2+x+1)​1​, tan⁡B=xx2+x+1\tan B=\dfrac{\sqrt x}{\sqrt{x^2+x+1}}tanB=x2+x+1​x​​ and tan⁡C=(x−3+x−2+x−1)1/2\tan C=\left(x^{-3}+x^{-2}+x^{-1}\right)^{1/2}tanC=(x−3+x−2+x−1)1/2, 0<A,B,C<π20<A,B,C<\dfrac\pi20<A,B,C<2π​, then A+BA+BA+B is equal to:
  1. (A)CCC
  2. (B)π−C\pi-Cπ−C
  3. (C)2π−C2\pi-C2π−C
  4. (D)π2−C\dfrac\pi2-C2π​−C

Correct answer: (A)

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Q48·MathematicsSingle correctJEE Main 2024
The number of solutions, of the equation esin⁡x−2e−sin⁡x=2e^{\sin x}-2e^{-\sin x}=2esinx−2e−sinx=2 is
  1. (A)2
  2. (B)more than 2
  3. (C)1
  4. (D)0

Correct answer: (D)

Step-by-step solution →
Q49·MathematicsSingle correctJEE Main 2024
For α,β∈(0,π2)\alpha,\beta\in\left(0,\dfrac{\pi}{2}\right)α,β∈(0,2π​), let 3sin⁡(α+β)=2sin⁡(α−β)3\sin(\alpha+\beta)=2\sin(\alpha-\beta)3sin(α+β)=2sin(α−β) and a real number kkk be such that tan⁡α=ktan⁡β\tan\alpha=k\tan\betatanα=ktanβ. Then the value of kkk is equal to:
  1. (A)−23-\dfrac{2}{3}−32​
  2. (B)−5-5−5
  3. (C)23\dfrac{2}{3}32​
  4. (D)555

Correct answer: (B)

Step-by-step solution →
Q50·MathematicsSingle correctJEE Main 2024
If 2sin⁡3x+sin⁡2xcos⁡x+4sin⁡x−4=02\sin^3 x+\sin 2x\cos x+4\sin x-4=02sin3x+sin2xcosx+4sinx−4=0 has exactly 333 solutions in the interval [0,nπ2]\left[0,\dfrac{n\pi}{2}\right][0,2nπ​], n∈Nn\in\mathbb{N}n∈N, then the roots of the equation x2+nx+(n−3)=0x^2+nx+(n-3)=0x2+nx+(n−3)=0 belong to:
  1. (A)(0,∞)(0,\infty)(0,∞)
  2. (B)(−∞,0)(-\infty,0)(−∞,0)
  3. (C)(−172,172)\left(-\dfrac{\sqrt{17}}{2},\dfrac{\sqrt{17}}{2}\right)(−217​​,217​​)
  4. (D)Z\mathbb{Z}Z

Correct answer: (B)

Step-by-step solution →
Q51·MathematicsSingle correctJEE Main 2024
The sum of the solutions x∈Rx\in\mathbb{R}x∈R of the equation 3cos⁡2x+cos⁡32xcos⁡6x−sin⁡6x=x3−x2+6\dfrac{3\cos 2x+\cos^3 2x}{\cos^6 x-\sin^6 x}=x^3-x^2+6cos6x−sin6x3cos2x+cos32x​=x3−x2+6 is:
  1. (A)0
  2. (B)1
  3. (C)−1-1−1
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q52·MathematicsSingle correctJEE Main 2024
If α\alphaα, −π2<α<π2-\dfrac{\pi}{2}<\alpha<\dfrac{\pi}{2}−2π​<α<2π​ is the solution of 4cos⁡θ+5sin⁡θ=14\cos\theta+5\sin\theta=14cosθ+5sinθ=1, then the value of tan⁡α\tan\alphatanα is
  1. (A)10−106\dfrac{10-\sqrt{10}}{6}610−10​​
  2. (B)10−1012\dfrac{10-\sqrt{10}}{12}1210−10​​
  3. (C)10−1012\dfrac{\sqrt{10}-10}{12}1210​−10​
  4. (D)10−106\dfrac{\sqrt{10}-10}{6}610​−10​

Correct answer: (C)

Step-by-step solution →
Q53·MathematicsNumericalJEE Main 2024
Let the set of all a∈Ra\in\mathbb Ra∈R such that the equation cos⁡2x+asin⁡x=2a−7\cos 2x+a\sin x=2a-7cos2x+asinx=2a−7 has a solution be [p,q][p,q][p,q] and r=tan⁡9∘−tan⁡27∘−1cot⁡63∘+tan⁡81∘r=\tan 9^{\circ}-\tan 27^{\circ}-\dfrac{1}{\cot 63^{\circ}}+\tan 81^{\circ}r=tan9∘−tan27∘−cot63∘1​+tan81∘, then pqrpqrpqr is equal to __________.

Correct answer: 48

Step-by-step solution →
Q54·MathematicsSingle correctJEE Main 2024
If 2tan⁡2θ−5sec⁡θ=12\tan^2\theta-5\sec\theta=12tan2θ−5secθ=1 has exactly 7 solutions in the interval [0,nπ2]\left[0,\dfrac{n\pi}{2}\right][0,2nπ​], for the least value of n∈Nn\in\mathbb{N}n∈N, then ∑k=1nk2k\displaystyle\sum_{k=1}^{n}\dfrac{k}{2^k}k=1∑n​2kk​ is equal to :
  1. (A)1215(214−14)\dfrac{1}{2^{15}}(2^{14}-14)2151​(214−14)
  2. (B)1214(215−15)\dfrac{1}{2^{14}}(2^{15}-15)2141​(215−15)
  3. (C)1−152131-\dfrac{15}{2^{13}}1−21315​
  4. (D)1213(214−15)\dfrac{1}{2^{13}}(2^{14}-15)2131​(214−15)

Correct answer: (D)

Step-by-step solution →
Q55·MathematicsNumericalJEE Advanced 2023
Consider an obtuse angled triangle ABC in which the difference between the largest and the smallest angle is π2\frac{\pi}{2}2π​ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1. Then the inradius of the triangle ABCABCABC is

Correct answer: 0.25

Step-by-step solution →
Q56·MathematicsNumericalJEE Advanced 2023
Consider an obtuse angled triangle ABC in which the difference between the largest and the smallest angle is π2\frac{\pi}{2}2π​ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1. Let aaa be the area of the triangle ABCABCABC. Then the value of (64a)2(64a)^{2}(64a)2 is

Correct answer: 1008

Step-by-step solution →
Q57·MathematicsSingle correctJEE Main 2023
In a triangle ABC, if cos⁡A+2cos⁡B+cos⁡C=2\cos A+2\cos B+\cos C=2cosA+2cosB+cosC=2 and the lengths of the sides opposite to the angles A and C are 3 and 7 respectively, then cos⁡A−cos⁡C\cos A-\cos CcosA−cosC is equal to
  1. (A)37\dfrac3773​
  2. (B)97\dfrac9779​
  3. (C)107\dfrac{10}{7}710​
  4. (D)57\dfrac5775​

Correct answer: (C)

Step-by-step solution →
Q58·MathematicsSingle correctJEE Main 2023
The number of elements in the set S={θ∈[0,2π]:3cos⁡4θ−5cos⁡2θ−2sin⁡2θ+2=0}S = \{\theta \in [0, 2\pi] : 3\cos^4\theta - 5\cos^2\theta - 2\sin^2\theta + 2 = 0\}S={θ∈[0,2π]:3cos4θ−5cos2θ−2sin2θ+2=0} is
  1. (A)101010
  2. (B)888
  3. (C)999
  4. (D)121212

Correct answer: (C)

Step-by-step solution →
Q59·MathematicsSingle correctJEE Main 2023
The angle of elevation of the top PPP of a tower from the feet of one person standing due South of the tower is 45∘45^\circ45∘ and from the feet of another person standing due West of the tower is 30∘30^\circ30∘. If the height of the tower is 555 metres, then the distance (in metres) between the two persons is equal to
  1. (A)101010
  2. (B)52\dfrac{5}{2}25​
  3. (C)555\sqrt{5}55​
  4. (D)525\dfrac{5}{2}\sqrt{5}25​5​

Correct answer: (A)

Step-by-step solution →
Q60·MathematicsSingle correctJEE Main 2023
96cos⁡π33cos⁡2π33cos⁡4π33cos⁡8π33cos⁡16π3396\cos\dfrac{\pi}{33}\cos\dfrac{2\pi}{33}\cos\dfrac{4\pi}{33}\cos\dfrac{8\pi}{33}\cos\dfrac{16\pi}{33}96cos33π​cos332π​cos334π​cos338π​cos3316π​ is equal to
  1. (A)333
  2. (B)222
  3. (C)444
  4. (D)111

Correct answer: (A)

Step-by-step solution →
Q61·MathematicsSingle correctJEE Main 2023
Let S={x∈(−π2,π2):91−tan⁡2x+9tan⁡2x=10}S=\left\{x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right):9^{1-\tan^2 x}+9^{\tan^2 x}=10\right\}S={x∈(−2π​,2π​):91−tan2x+9tan2x=10} and β=∑x∈Stan⁡2(x3)\beta=\sum_{x\in S}\tan^2\left(\frac{x}{3}\right)β=∑x∈S​tan2(3x​), then 16(β−14)2\frac{1}{6}(\beta-14)^261​(β−14)2 is equal to:
  1. (A)323232
  2. (B)888
  3. (C)646464
  4. (D)161616

Correct answer: (A)

Step-by-step solution →
Q62·MathematicsNumericalJEE Main 2023
In the figure, θ1+θ2=π2\theta_1+\theta_2=\frac{\pi}{2}θ1​+θ2​=2π​ and 3 (BE)=4(AB)\sqrt{3}\,(\mathrm{BE})=4(\mathrm{AB})3​(BE)=4(AB). If the area of △CAB\triangle CAB△CAB is 23−32\sqrt{3}-323​−3 unit2^22, when θ1θ2\frac{\theta_1}{\theta_2}θ2​θ1​​ is the largest, then the perimeter (in unit) of △CED\triangle CED△CED is equal to _______ .

Correct answer: 6

Step-by-step solution →
Q63·MathematicsSingle correctJEE Main 2023
Let f(x)=sin⁡x+cos⁡x−2sin⁡x−cos⁡x, x∈[0,π]−{π4}f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x},\ x\in[0,\pi]-\left\{\frac{\pi}{4}\right\}f(x)=sinx−cosxsinx+cosx−2​​, x∈[0,π]−{4π​}. Then f(7π12)f′′(7π12)f\left(\frac{7\pi}{12}\right)f''\left(\frac{7\pi}{12}\right)f(127π​)f′′(127π​) is equal to
  1. (A)−23\frac{-2}{3}3−2​
  2. (B)29\frac{2}{9}92​
  3. (C)−133-\frac{1}{3\sqrt{3}}−33​1​
  4. (D)−233\frac{-2}{3\sqrt{3}}33​−2​

Correct answer: (B)

Step-by-step solution →
Q64·MathematicsSingle correctJEE Main 2023
The value of 36(4cos⁡29∘−1)(4cos⁡227∘−1)(4cos⁡281∘−1)(4cos⁡2243∘−1)36(4\cos^{2}9^{\circ}-1)(4\cos^{2}27^{\circ}-1)(4\cos^{2}81^{\circ}-1)(4\cos^{2}243^{\circ}-1)36(4cos29∘−1)(4cos227∘−1)(4cos281∘−1)(4cos2243∘−1) is
  1. (A)54
  2. (B)18
  3. (C)27
  4. (D)36

Correct answer: (D)

Step-by-step solution →
Q65·MathematicsSingle correctJEE Main 2023
From the top AAA of a vertical wall ABABAB of height 30 m30\,m30m, the angles of depression of the top PPP and bottom QQQ of a vertical tower CPCPCP of height hhh are 15∘15^\circ15∘ and 60∘60^\circ60∘ respectively. BBB and QQQ are on the same horizontal level. If CCC is a point on ABABAB such that CB=PQCB=PQCB=PQ, then the area (in m2m^2m2) of the quadrilateral BCPQBCPQBCPQ is equal to:
  1. (A)600(3−1)600(\sqrt3-1)600(3​−1)
  2. (B)300(3+1)300(\sqrt3+1)300(3​+1)
  3. (C)200(3−3)200(3-\sqrt3)200(3−3​)
  4. (D)300(3−1)300(\sqrt3-1)300(3​−1)

Correct answer: (A)

Step-by-step solution →
Q66·MathematicsNumericalJEE Main 2023
The value of tan⁡9∘−tan⁡27∘−tan⁡63∘+tan⁡81∘\tan 9^\circ-\tan 27^\circ-\tan 63^\circ+\tan 81^\circtan9∘−tan27∘−tan63∘+tan81∘ is

Correct answer: 4

Step-by-step solution →
Q67·MathematicsSingle correctJEE Main 2023
For a triangle ABCABCABC, the value of cos⁡2A+cos⁡2B+cos⁡2C\cos 2A+\cos 2B+\cos 2Ccos2A+cos2B+cos2C is least. If its inradius is 333 and incentre is MMM, then which of the following is NOT correct?
  1. (A)perimeter of △ABC\triangle ABC△ABC is 18318\sqrt{3}183​
  2. (B)sin⁡2A+sin⁡2B+sin⁡2C=sin⁡A+sin⁡B+sin⁡C\sin 2A+\sin 2B+\sin 2C=\sin A+\sin B+\sin Csin2A+sin2B+sin2C=sinA+sinB+sinC
  3. (C)MA→⋅MB→=−18\overrightarrow{MA}\cdot\overrightarrow{MB}=-18MA⋅MB=−18
  4. (D)area of △ABC\triangle ABC△ABC is 2732\frac{27\sqrt{3}}{2}2273​​

Correct answer: (D)

Step-by-step solution →
Q68·MathematicsSingle correctJEE Main 2023
If tan⁡15°+1tan⁡75°+1tan⁡105°+tan⁡195°=2a\tan15°+\dfrac{1}{\tan75°}+\dfrac{1}{\tan105°}+\tan195°=2atan15°+tan75°1​+tan105°1​+tan195°=2a, then the value of (a+1a)\left(a+\dfrac1a\right)(a+a1​) is:
  1. (A)222
  2. (B)4−234-2\sqrt34−23​
  3. (C)5−3235-\dfrac32\sqrt35−23​3​
  4. (D)444

Correct answer: (D)

Step-by-step solution →
Q69·MathematicsSingle correctJEE Main 2023
If the solution of the equation log⁡cos⁡xcot⁡x+4log⁡sin⁡xtan⁡x=1\log_{\cos x}\cot x+4\log_{\sin x}\tan x=1logcosx​cotx+4logsinx​tanx=1, x∈(0,π2)x\in\left(0,\dfrac{\pi}{2}\right)x∈(0,2π​), is sin⁡−1(α+β2)\sin^{-1}\left(\dfrac{\alpha+\sqrt\beta}{2}\right)sin−1(2α+β​​), where α\alphaα and β\betaβ are integers, then α+β\alpha+\betaα+β is equal to:
  1. (A)555
  2. (B)666
  3. (C)444
  4. (D)333

Correct answer: (C)

Step-by-step solution →
Q70·MathematicsSingle correctJEE Main 2023
The set of all values of λ\lambdaλ for which the equation cos⁡22x−2sin⁡4x−2cos⁡2x=λ\cos^2 2x-2\sin^4 x-2\cos^2 x=\lambdacos22x−2sin4x−2cos2x=λ has a real solution xxx, is:
  1. (A)[−2,−1][-2,-1][−2,−1]
  2. (B)[−1,−12]\left[-1,-\dfrac12\right][−1,−21​]
  3. (C)[−32,−1]\left[-\dfrac32,-1\right][−23​,−1]
  4. (D)[−2,−32]\left[-2,-\dfrac32\right][−2,−23​]

Correct answer: (C)

Step-by-step solution →
Q71·MathematicsSingle correctJEE Main 2023
Let f(θ)=3(sin⁡4(3π2−θ)+sin⁡4(3π+θ))−2(1−sin⁡22θ)f(\theta) = 3\left(\sin^4\left(\dfrac{3\pi}{2} - \theta\right) + \sin^4(3\pi + \theta)\right) - 2(1 - \sin^2 2\theta)f(θ)=3(sin4(23π​−θ)+sin4(3π+θ))−2(1−sin22θ) and S={θ∈[0,π]:f′(θ)=−32}S = \left\{\theta \in [0, \pi] : f'(\theta) = -\dfrac{\sqrt{3}}{2}\right\}S={θ∈[0,π]:f′(θ)=−23​​}. If β∈S\beta \in Sβ∈S, then f(β)f(\beta)f(β) is equal to
  1. (A)54\dfrac{5}{4}45​
  2. (B)32\dfrac{3}{2}23​
  3. (C)98\dfrac{9}{8}89​
  4. (D)118\dfrac{11}{8}811​

Correct answer: (B)

Step-by-step solution →
Q72·MathematicsNumericalJEE Main 2023
If mmm and nnn respectively are the numbers of positive and negative values of θ\thetaθ in the interval [−π,π][-\pi,\pi][−π,π] that satisfy the equation cos⁡2θcos⁡θ2=cos⁡3θcos⁡9θ2\cos 2\theta\cos\dfrac{\theta}{2}=\cos 3\theta\cos\dfrac{9\theta}{2}cos2θcos2θ​=cos3θcos29θ​, then mnmnmn is equal to

Correct answer: 25

Step-by-step solution →
Q73·MathematicsNumericalJEE Main 2023
Let S={θ∈[0,2π):tan⁡(πcos⁡θ)+tan⁡(πsin⁡θ)=0}S=\{\theta\in[0,2\pi):\tan(\pi\cos\theta)+\tan(\pi\sin\theta)=0\}S={θ∈[0,2π):tan(πcosθ)+tan(πsinθ)=0}. Then ∑θ∈Ssin⁡2(θ+π4)\sum_{\theta\in S}\sin^2\left(\theta+\dfrac{\pi}{4}\right)∑θ∈S​sin2(θ+4π​) is equal to

Correct answer: 2

Step-by-step solution →
Q74·MathematicsIntegerJEE Advanced 2022
Let α and β be real numbers such that −π4<β<0<α<π4-\frac{\pi}{4} < \beta < 0 < \alpha < \frac{\pi}{4}−4π​<β<0<α<4π​. If sin⁡(α+β)=13\sin(\alpha + \beta) = \frac{1}{3}sin(α+β)=31​ and cos⁡(α−β)=23\cos\left(\alpha - \beta\right) = \frac{2}{3}cos(α−β)=32​, then the greatest integer less than or equal to (sin⁡αcos⁡β+cos⁡βsin⁡α+cos⁡αsin⁡β+sin⁡βcos⁡α)2\left(\frac{\sin\alpha}{\cos\beta} + \frac{\cos\beta}{\sin\alpha} + \frac{\cos\alpha}{\sin\beta} + \frac{\sin\beta}{\cos\alpha}\right)^{2}(cosβsinα​+sinαcosβ​+sinβcosα​+cosαsinβ​)2 is ________.

Correct answer: 1

Step-by-step solution →
Q75·MathematicsSingle correctJEE Advanced 2022
Consider the following lists: The correct option is:
List-IList-II
I.{x∈[−2π3,2π3]:cos⁡x+sin⁡x=1}\left\{x \in \left[-\frac{2\pi}{3}, \frac{2\pi}{3}\right] : \cos x + \sin x = 1\right\}{x∈[−32π​,32π​]:cosx+sinx=1}P.has two elements
II.{x∈[−5π18,5π18]:3tan⁡3x=1}\left\{x \in \left[-\frac{5\pi}{18}, \frac{5\pi}{18}\right] : \sqrt{3}\tan 3x = 1\right\}{x∈[−185π​,185π​]:3​tan3x=1}Q.has three elements
III.{x∈[−6π5,6π5]:2cos⁡(2x)=3}\left\{x \in \left[-\frac{6\pi}{5}, \frac{6\pi}{5}\right] : 2\cos\left(2x\right) = \sqrt{3}\right\}{x∈[−56π​,56π​]:2cos(2x)=3​}R.has four elements
IV.{x∈[−7π4,7π4]:sin⁡x−cos⁡x=1}\left\{x \in \left[-\frac{7\pi}{4}, \frac{7\pi}{4}\right] : \sin x - \cos x = 1\right\}{x∈[−47π​,47π​]:sinx−cosx=1}S.has five elements
T.has six elements
  1. (A)(I) → (P); (II) → (S); (III) → (P); (IV) → (S)
  2. (B)(I) → (P); (II) → (P); (III) → (T); (IV) → (R)
  3. (C)(I) → (Q); (II) → (P); (III) → (T); (IV) → (S)
  4. (D)(I) → (Q); (II) → (S); (III) → (P); (IV) → (R)

Correct answer: (B)

Step-by-step solution →
Q76·MathematicsMultiple correctJEE Advanced 2022
Let PQRS be a quadrilateral in a plane, where QR=1QR = 1QR=1, ∠PQR=∠QRS=70∘\angle PQR = \angle QRS = 70^\circ∠PQR=∠QRS=70∘, ∠PQS=15∘\angle PQS = 15^\circ∠PQS=15∘ and ∠PRS=40∘\angle PRS = 40^\circ∠PRS=40∘. If ∠RPS=θ∘\angle RPS = \theta^\circ∠RPS=θ∘, PQ=αPQ = \alphaPQ=α and PS=βPS = \betaPS=β, then the interval(s) that contain(s) the value of 4αβsin⁡θ∘4\alpha\beta \sin\theta^\circ4αβsinθ∘ is/are
  1. (A)(0,2)\left(0, \sqrt{2}\right)(0,2​)
  2. (B)(1,2)(1, 2)(1,2)
  3. (C)(2,3)\left(\sqrt{2}, 3\right)(2​,3)
  4. (D)(22,32)\left(2\sqrt{2}, 3\sqrt{2}\right)(22​,32​)

Correct answer: (A), (B)

Step-by-step solution →
Q77·MathematicsSingle correctJEE Main 2022
The angle of elevation of the top of a tower from a point A due north of it is α\alphaα and from a point B at a distance of 9 units due west of A is cos⁡−1(313)\cos^{-1}\left(\frac{3}{\sqrt{13}}\right)cos−1(13​3​). If the distance of the point B from the tower is 15 units, then cot⁡α\cot \alphacotα is equal to :
  1. (A)65\frac{6}{5}56​
  2. (B)95\frac{9}{5}59​
  3. (C)43\frac{4}{3}34​
  4. (D)73\frac{7}{3}37​

Correct answer: (A)

Step-by-step solution →
Q78·MathematicsNumericalJEE Main 2022
Let S={θ∈(0, 2π):7cos⁡2θ−3sin⁡2θ−2cos⁡22θ=2}S = \left\{\theta \in (0,\,2\pi) : 7\cos^{2}\theta - 3\sin^{2}\theta - 2\cos^{2}2\theta = 2\right\}S={θ∈(0,2π):7cos2θ−3sin2θ−2cos22θ=2}. Then, the sum of roots of all the equations x2−2(tan⁡2θ+cot⁡2θ)x+6sin⁡2θ=0x^{2} - 2\left(\tan^{2}\theta + \cot^{2}\theta\right)x + 6\sin^{2}\theta = 0x2−2(tan2θ+cot2θ)x+6sin2θ=0, θ∈S\theta \in Sθ∈S, is __________.

Correct answer: 16

Step-by-step solution →
Q79·MathematicsSingle correctJEE Main 2022
The number of elements in the set S={x∈R:2cos⁡(x2+x6)=4x+4−x}S = \left\{ x \in \mathbb{R} : 2\cos\left( \frac{x^2 + x}{6} \right) = 4^x + 4^{-x} \right\}S={x∈R:2cos(6x2+x​)=4x+4−x} is:
  1. (A)111
  2. (B)333
  3. (C)000
  4. (D)infinite

Correct answer: (A)

Step-by-step solution →
Q80·MathematicsSingle correctJEE Main 2022
A horizontal park is in the shape of a triangle OAB with AB=16AB=16AB=16. A vertical lamp post OP is erected at the point O such that ∠PAO=∠PBO=15∘\angle PAO = \angle PBO = 15^{\circ}∠PAO=∠PBO=15∘ and ∠PCO=45∘\angle PCO = 45^{\circ}∠PCO=45∘, where C is the midpoint of AB. Then (OP)2(OP)^{2}(OP)2 is equal to
  1. (A)323(3−1)\frac{32}{\sqrt{3}}\left(\sqrt{3}-1\right)3​32​(3​−1)
  2. (B)323(2−3)\frac{32}{\sqrt{3}}\left(2-\sqrt{3}\right)3​32​(2−3​)
  3. (C)163(3−1)\frac{16}{\sqrt{3}}\left(\sqrt{3}-1\right)3​16​(3​−1)
  4. (D)163(2−3)\frac{16}{\sqrt{3}}\left(2-\sqrt{3}\right)3​16​(2−3​)

Correct answer: (B)

Step-by-step solution →
Q81·MathematicsNumericalJEE Main 2022
Let S=[−π,π2)−{−π2,−π4,−3π4,π4}S=\left[-\pi,\frac{\pi}{2}\right)-\left\{-\frac{\pi}{2},-\frac{\pi}{4},-\frac{3\pi}{4},\frac{\pi}{4}\right\}S=[−π,2π​)−{−2π​,−4π​,−43π​,4π​}. Then the number of elements in the set A={θ∈S:tan⁡θ(1+5tan⁡(2θ))=5−tan⁡(2θ)}A=\left\{\theta\in S:\tan\theta\left(1+\sqrt{5}\tan(2\theta)\right)=\sqrt{5}-\tan(2\theta)\right\}A={θ∈S:tanθ(1+5​tan(2θ))=5​−tan(2θ)} is _____

Correct answer: 5

Step-by-step solution →
Q82·MathematicsSingle correctJEE Main 2022
Let a vertical tower AB of height 2h2h2h stands on a horizontal ground. Let from a point P on the ground a man can see upto height hhh of the tower with an angle of elevation 2α2\alpha2α. When from PPP, he moves a distance d in the direction of AP→\overrightarrow{AP}AP, he can see the top B of the tower with an angle of elevation α\alphaα. If d=7hd = \sqrt{7}hd=7​h, then tan⁡α\tan\alphatanα is equal to
  1. (A)5−2\sqrt{5} - 25​−2
  2. (B)3−1\sqrt{3} - 13​−1
  3. (C)7−2\sqrt{7} - 27​−2
  4. (D)7−3\sqrt{7} - \sqrt{3}7​−3​

Correct answer: (C)

Step-by-step solution →
Q83·MathematicsSingle correctJEE Main 2022
The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is 45∘45^\circ45∘. Let R be a point on AQ and from a point B, vertically above R, the angle of elevation of P is 60∘60^\circ60∘. If ∠BAQ=30∘\angle BAQ = 30^\circ∠BAQ=30∘, AB=dAB = dAB=d and the area of the trapezium PQRB is α\alphaα, then the ordered pair (d,α)(d, \alpha)(d,α) is :
  1. (A)(10(3−1),25)\left(10\left(\sqrt{3}-1\right), 25\right)(10(3​−1),25)
  2. (B)(10(3−1),252)\left(10\left(\sqrt{3}-1\right), \frac{25}{2}\right)(10(3​−1),225​)
  3. (C)(10(3+1),25)\left(10\left(\sqrt{3}+1\right), 25\right)(10(3​+1),25)
  4. (D)(10(3+1),252)\left(10\left(\sqrt{3}+1\right), \frac{25}{2}\right)(10(3​+1),225​)

Correct answer: (A)

Step-by-step solution →
Q84·MathematicsSingle correctJEE Main 2022
Let S={θ∈(0,π2):∑m=19sec⁡(θ+(m−1)π6)sec⁡(θ+mπ6)=−83}S = \left\{\theta \in \left(0, \frac{\pi}{2}\right) : \sum_{m=1}^{9} \sec\left(\theta + (m-1)\frac{\pi}{6}\right)\sec\left(\theta + \frac{m\pi}{6}\right) = -\frac{8}{\sqrt{3}}\right\}S={θ∈(0,2π​):∑m=19​sec(θ+(m−1)6π​)sec(θ+6mπ​)=−3​8​} Then
  1. (A)S={π12}S = \left\{\frac{\pi}{12}\right\}S={12π​}
  2. (B)S={2π3}S = \left\{\frac{2\pi}{3}\right\}S={32π​}
  3. (C)∑θ∈Sθ=π2\sum_{\theta \in S} \theta = \frac{\pi}{2}∑θ∈S​θ=2π​
  4. (D)∑θ∈Sθ=3π4\sum_{\theta \in S} \theta = \frac{3\pi}{4}∑θ∈S​θ=43π​

Correct answer: (C)

Step-by-step solution →
Q85·MathematicsSingle correctJEE Main 2022
Let S={θ∈[0,2π]:82sin⁡2θ+82cos⁡2θ=16}S = \{\theta \in [0, 2\pi] : 8^{2\sin^{2}\theta} + 8^{2\cos^{2}\theta} = 16\}S={θ∈[0,2π]:82sin2θ+82cos2θ=16}. Then n(S)+∑θ∈S(sec⁡(π4+2θ)cosec⁡(π4+2θ))n(S) + \sum_{\theta \in S}\left( \sec\left(\frac{\pi}{4} + 2\theta\right)\operatorname{cosec}\left(\frac{\pi}{4} + 2\theta\right) \right)n(S)+∑θ∈S​(sec(4π​+2θ)cosec(4π​+2θ)) is equal to :
  1. (A)0
  2. (B)−2
  3. (C)−4
  4. (D)12

Correct answer: (C)

Step-by-step solution →
Q86·MathematicsNumericalJEE Main 2022
If the sum of solutions of the system of equations 2sin⁡2θ−cos⁡2θ=02\sin^{2}\theta-\cos 2\theta=02sin2θ−cos2θ=0 and 2cos⁡2θ+3sin⁡θ=02\cos^{2}\theta+3\sin\theta=02cos2θ+3sinθ=0 in the interval [0,2π][0,2\pi][0,2π] is kπk\pikπ, then k is equal to ________.

Correct answer: 3

Step-by-step solution →
Q87·MathematicsSingle correctJEE Main 2022
The number of solutions of ∣cos⁡x∣=sin⁡x|\cos x| = \sin x∣cosx∣=sinx, such that −4π≤x≤4π-4\pi \le x \le 4\pi−4π≤x≤4π is :
  1. (A)4
  2. (B)6
  3. (C)8
  4. (D)12

Correct answer: (C)

Step-by-step solution →
Q88·MathematicsSingle correctJEE Main 2022
A tower PQ stands on a horizontal ground with base Q on the ground. The point R divides the tower in two parts such that QR = 15 m. If from a point A on the ground the angle of elevation of R is 60∘60^\circ60∘ and the part PR of the tower subtends an angle of 15∘15^\circ15∘ at A, then the height of the tower is :
  1. (A)5(23+3)5(2\sqrt{3}+3)5(23​+3) m
  2. (B)5(3+3)5(\sqrt{3}+3)5(3​+3) m
  3. (C)10(3+1)10(\sqrt{3}+1)10(3​+1) m
  4. (D)10(23+1)10(2\sqrt{3}+1)10(23​+1) m

Correct answer: (A)

Step-by-step solution →
Q89·MathematicsNumericalJEE Main 2022
The number of solutions of the equation sin⁡x=cos⁡2x\sin x = \cos^2 xsinx=cos2x in the interval (0,10)(0,10)(0,10) is____.

Correct answer: 4

Step-by-step solution →
Q90·MathematicsNumericalJEE Main 2022
The number of elements in the set S={θ∈[−4π,4π]:3cos⁡22θ+6cos⁡2θ−10cos⁡2θ+5=0}S = \{\theta \in [-4\pi, 4\pi] : 3\cos^2 2\theta + 6\cos 2\theta - 10\cos^2\theta + 5 = 0\}S={θ∈[−4π,4π]:3cos22θ+6cos2θ−10cos2θ+5=0} is ________.

Correct answer: 32

Step-by-step solution →
Q91·MathematicsSingle correctJEE Main 2022
From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60∘60^\circ60∘. The pole subtends an angle 30∘30^\circ30∘ at the top of the tower. Then the height of the tower is:
  1. (A)15315\sqrt{3}153​
  2. (B)20320\sqrt{3}203​
  3. (C)20+10320 + 10\sqrt{3}20+103​
  4. (D)30

Correct answer: (D)

Step-by-step solution →
Q92·MathematicsNumericalJEE Main 2022
The number of solutions of the equation 2θ−cos⁡2θ+2=02\theta - \cos^2\theta + \sqrt{2} = 02θ−cos2θ+2​=0 is R is equal to ______.

Correct answer: 1

Step-by-step solution →
Q93·MathematicsSingle correctJEE Main 2022
Let AB and PQ be two vertical poles, 160 m apart from each other. Let C be the middle point of B and Q, which are feet of these two poles. Let π8\frac{\pi}{8}8π​ and θ be the angles of elevation from C to P and A, respectively. If the height of pole PQ is twice the height of pole AB, then tan⁡2θ\tan^2 \thetatan2θ is equal to
  1. (A)3−222\frac{3-2\sqrt{2}}{2}23−22​​
  2. (B)3+22\frac{3+\sqrt{2}}{2}23+2​​
  3. (C)3−224\frac{3-2\sqrt{2}}{4}43−22​​
  4. (D)3−24\frac{3-\sqrt{2}}{4}43−2​​

Correct answer: (C)

Step-by-step solution →
Q94·MathematicsSingle correctJEE Main 2022
If cot⁡α=1\cot\alpha = 1cotα=1 and sec⁡β=−53\sec\beta = -\frac{5}{3}secβ=−35​, where π<α<3π2\pi < \alpha < \frac{3\pi}{2}π<α<23π​ and π2<β<π\frac{\pi}{2} < \beta < \pi2π​<β<π, then the value of tan⁡(α+β)\tan(\alpha + \beta)tan(α+β) and the quadrant in which α+β\alpha + \betaα+β lies, respectively are
  1. (A)−17-\frac{1}{7}−71​ and IVth\mathrm{IV}^{\mathrm{th}}IVth quadrant
  2. (B)777 and Ist\mathrm{I}^{\mathrm{st}}Ist quadrant
  3. (C)−7-7−7 and IVth\mathrm{IV}^{\mathrm{th}}IVth quadrant
  4. (D)17\frac{1}{7}71​ and Ist\mathrm{I}^{\mathrm{st}}Ist quadrant

Correct answer: (A)

Step-by-step solution →
Q95·MathematicsSingle correctJEE Main 2022
α=sin⁡36∘\alpha=\sin 36^{\circ}α=sin36∘ is a root of which of the following equation
  1. (A)10x4−10x2−5=010x^{4}-10x^{2}-5=010x4−10x2−5=0
  2. (B)16x4+20x2−5=016x^{4}+20x^{2}-5=016x4+20x2−5=0
  3. (C)16x4−20x2+5=016x^{4}-20x^{2}+5=016x4−20x2+5=0
  4. (D)16x4−10x2+5=016x^{4}-10x^{2}+5=016x4−10x2+5=0

Correct answer: (C)

Step-by-step solution →
Q96·MathematicsSingle correctJEE Main 2022
The value of cos⁡(2π7)+cos⁡(4π7)+cos⁡(6π7)\cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{6\pi}{7}\right)cos(72π​)+cos(74π​)+cos(76π​) is equal to :
  1. (A)-1
  2. (B)−12-\frac{1}{2}−21​
  3. (C)−13-\frac{1}{3}−31​
  4. (D)−14-\frac{1}{4}−41​

Correct answer: (B)

Step-by-step solution →
Q97·MathematicsNumericalJEE Main 2022
If sin⁡2(10∘)sin⁡(20∘)sin⁡(40∘)sin⁡(50∘)sin⁡(70∘)=α−116sin⁡(10∘)\sin^{2}(10^{\circ})\sin(20^{\circ})\sin(40^{\circ})\sin(50^{\circ})\sin(70^{\circ}) = \alpha - \frac{1}{16}\sin(10^{\circ})sin2(10∘)sin(20∘)sin(40∘)sin(50∘)sin(70∘)=α−161​sin(10∘), then 16+α−116 + \alpha^{-1}16+α−1 is equal to ___________.

Correct answer: 80

Step-by-step solution →
Q98·MathematicsSingle correctJEE Main 2022
16sin⁡(20∘) sin⁡(40∘) sin⁡(80∘)16\sin(20^\circ)\,\sin(40^\circ)\,\sin(80^\circ)16sin(20∘)sin(40∘)sin(80∘) is equal to :
  1. (A)3\sqrt{3}3​
  2. (B)232\sqrt{3}23​
  3. (C)3
  4. (D)434\sqrt{3}43​

Correct answer: (B)

Step-by-step solution →
Q99·MathematicsNumericalJEE Main 2022
The number of values of x in the interval (π4,7π4)\left(\frac{\pi}{4}, \frac{7\pi}{4}\right)(4π​,47π​) for which 14 cosec2x−2sin⁡2x=21−4cos⁡2x14\,\mathrm{cosec}^2x - 2\sin^2x = 21 - 4\cos^2x14cosec2x−2sin2x=21−4cos2x holds, is ________

Correct answer: 4

Step-by-step solution →
Q100·MathematicsSingle correctJEE Main 2022
The value of 2sin⁡(12∘)−sin⁡(72∘)2\sin(12^\circ) - \sin(72^\circ)2sin(12∘)−sin(72∘) is :
  1. (A)5(1−3)4\frac{\sqrt{5}\left(1 - \sqrt{3}\right)}{4}45​(1−3​)​
  2. (B)1−58\frac{1 - \sqrt{5}}{8}81−5​​
  3. (C)3(1−5)2\frac{\sqrt{3}\left(1 - \sqrt{5}\right)}{2}23​(1−5​)​
  4. (D)3(1−5)4\frac{\sqrt{3}\left(1 - \sqrt{5}\right)}{4}43​(1−5​)​

Correct answer: (D)

Step-by-step solution →
Q101·MathematicsSingle correctJEE Main 2022
Let a, b and c be the length of sides of a triangle ABC such that a+b7=b+c8=c+a9\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9}7a+b​=8b+c​=9c+a​. If r and R are the radius of incircle and radius of circumcircle of the triangle ABC, respectively, then the value of Rr\frac{R}{r}rR​ is equal to
  1. (A)52\frac{5}{2}25​
  2. (B)222
  3. (C)32\frac{3}{2}23​
  4. (D)111

Correct answer: (A)

Step-by-step solution →
Q102·MathematicsSingle correctJEE Main 2022
Let S={θ∈[−π,π]−{±π2}:sin⁡θtan⁡θ+tan⁡θ=sin⁡2θ}S=\left\{\theta\in[-\pi,\pi]-\left\{\pm\dfrac{\pi}{2}\right\}:\sin\theta\tan\theta+\tan\theta=\sin 2\theta\right\}S={θ∈[−π,π]−{±2π​}:sinθtanθ+tanθ=sin2θ}. If T=∑θ∈Scos⁡2θT=\displaystyle\sum_{\theta\in S}\cos 2\thetaT=θ∈S∑​cos2θ, then T+n(S)T+n(S)T+n(S) is equal
  1. (A)7+37+\sqrt{3}7+3​
  2. (B)999
  3. (C)8+38+\sqrt{3}8+3​
  4. (D)101010

Correct answer: (B)

Step-by-step solution →
Q103·MathematicsSingle correctJEE Main 2022
The number of solutions of the equation cos⁡(x+π3)cos⁡(π3−x)=14cos⁡22x\cos\left(x+\dfrac{\pi}{3}\right)\cos\left(\dfrac{\pi}{3}-x\right)=\dfrac{1}{4}\cos^2 2xcos(x+3π​)cos(3π​−x)=41​cos22x, x∈[−3π,3π]x \in [-3\pi, 3\pi]x∈[−3π,3π] is :
  1. (A)8
  2. (B)5
  3. (C)6
  4. (D)7

Correct answer: (D)

Step-by-step solution →
Q104·MathematicsMultiple correctJEE Advanced 2021
Consider a triangle PQR having sides of lengths p, q and r opposite to the angles P, Q and R, respectively. Then which of the following statements is (are) TRUE?
  1. (A)cos⁡P≥1−p22qr\cos P \ge 1 - \frac{p^2}{2qr}cosP≥1−2qrp2​
  2. (B)cos⁡R≥(q−rp+q)cos⁡P+(p−rp+q)cos⁡Q\cos R \ge \left(\frac{q-r}{p+q}\right)\cos P + \left(\frac{p-r}{p+q}\right)\cos QcosR≥(p+qq−r​)cosP+(p+qp−r​)cosQ
  3. (C)q+rp<2sin⁡Qsin⁡Rsin⁡P\frac{q+r}{p} < 2\frac{\sqrt{\sin Q \sin R}}{\sin P}pq+r​<2sinPsinQsinR​​
  4. (D)If p<qp < qp<q and p<rp < rp<r, then cos⁡Q>pr\cos Q > \frac{p}{r}cosQ>rp​ and cos⁡R>pq\cos R > \frac{p}{q}cosR>qp​

Correct answer: (A), (B)

Step-by-step solution →
Q105·MathematicsIntegerJEE Advanced 2021
In a triangle ABC, let AB=23AB = \sqrt{23}AB=23​, BC=3BC = 3BC=3 and CA=4CA = 4CA=4. Then the value of cot⁡A+cot⁡Ccot⁡B\frac{\cot A + \cot C}{\cot B}cotBcotA+cotC​ is____.

Correct answer: 2

Step-by-step solution →
Q106·MathematicsSingle correctJEE Main 2021
If n is the number of solutions of the equation 2cos⁡x(4sin⁡(π4+x)sin⁡(π4−x)−1)=12\cos x\left(4\sin\left(\frac{\pi}{4} + x\right)\sin\left(\frac{\pi}{4} - x\right) - 1\right) = 12cosx(4sin(4π​+x)sin(4π​−x)−1)=1, x ∈ [0, π] and S is the sum of all these solutions, then the ordered pair (n, S) is :
  1. (A)(3, 13π / 9)
  2. (B)(2, 2π / 3)
  3. (C)(2, 8π / 9)
  4. (D)(3, 5π / 3)

Correct answer: (A)

Step-by-step solution →
Q107·MathematicsSingle correctJEE Main 2021
The range of the function, f(x)=log⁡5(3+cos⁡(3π4+x)+cos⁡(π4+x)+cos⁡(π4−x)−cos⁡(3π4−x))f(x) = \log_{\sqrt{5}}\left(3 + \cos\left(\frac{3\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} - x\right) - \cos\left(\frac{3\pi}{4} - x\right)\right)f(x)=log5​​(3+cos(43π​+x)+cos(4π​+x)+cos(4π​−x)−cos(43π​−x)) is :
  1. (A)(0,5)\left(0, \sqrt{5}\right)(0,5​)
  2. (B)[–2, 2]
  3. (C)[15,5]\left[\frac{1}{\sqrt{5}}, \sqrt{5}\right][5​1​,5​]
  4. (D)[0, 2]

Correct answer: (D)

Step-by-step solution →
Q108·MathematicsSingle correctJEE Main 2021
cosec⁡18∘\operatorname{cosec} 18^{\circ}cosec18∘ is a root of the equation :
  1. (A)x2+2x−4=0x^{2} + 2x - 4 = 0x2+2x−4=0
  2. (B)4x2+2x−1=04x^{2} + 2x - 1 = 04x2+2x−1=0
  3. (C)x2−2x+4=0x^{2} - 2x + 4 = 0x2−2x+4=0
  4. (D)x2−2x−4=0x^{2} - 2x - 4 = 0x2−2x−4=0

Correct answer: (D)

Step-by-step solution →
Q109·MathematicsSingle correctJEE Main 2021
The number of solutions of the equation 32tan⁡2x+32sec⁡2x=8132^{\tan^2 x} + 32^{\sec^2 x} = 8132tan2x+32sec2x=81, 0≤x≤π40 \leq x \leq \frac{\pi}{4}0≤x≤4π​ is :
  1. (A)3
  2. (B)1
  3. (C)0
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q110·MathematicsSingle correctJEE Main 2021
A vertical pole fixed to the horizontal ground is divided in the ratio 3:73 : 73:7 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground 18 m away from the base of the pole, then the height of the pole (in meters) is :
  1. (A)121512\sqrt{15}1215​
  2. (B)121012\sqrt{10}1210​
  3. (C)8108\sqrt{10}810​
  4. (D)6106\sqrt{10}610​

Correct answer: (B)

Step-by-step solution →
Q111·MathematicsNumericalJEE Main 2021
Let S be the sum of all solutions (in radians) of the equation sin⁡4θ+cos⁡4θ−sin⁡θ cos⁡θ=0\sin^{4}\theta + \cos^{4}\theta - \sin\theta\,\cos\theta = 0sin4θ+cos4θ−sinθcosθ=0 in [0, 4π][0,\ 4\pi][0, 4π]. Then 8Sπ\frac{8S}{\pi}π8S​ is equal to _________ .

Correct answer: 56

Step-by-step solution →
Q112·MathematicsSingle correctJEE Main 2021
Two poles, AB of length a metres and CD of length a + b (b ≠ a) metres are erected at the same horizontal level with bases at B and D. If BD = x and tan⁡∠ACB=12\tan \angle ACB = \frac{1}{2}tan∠ACB=21​, then:
  1. (A)x2+2(a+2b)x−b(a+b)=0x^{2} + 2(a + 2b)x - b(a + b) = 0x2+2(a+2b)x−b(a+b)=0
  2. (B)x2+2(a+2b)x+a(a+b)=0x^{2} + 2(a + 2b)x + a(a + b) = 0x2+2(a+2b)x+a(a+b)=0
  3. (C)x2−2ax+b(a+b)=0x^{2} - 2ax + b(a + b) = 0x2−2ax+b(a+b)=0
  4. (D)x2−2ax+a(a+b)=0x^{2} - 2ax + a(a + b) = 0x2−2ax+a(a+b)=0

Correct answer: (C)

Step-by-step solution →
Q113·MathematicsSingle correctJEE Main 2021
Let sin⁡Asin⁡B=sin⁡(A−C)sin⁡(C−B)\frac{\sin A}{\sin B} = \frac{\sin(A-C)}{\sin(C-B)}sinBsinA​=sin(C−B)sin(A−C)​ , where A, B, C are angles of a triangle ABC. If the lengths of the sides opposite these angles are a, b, c respectively, then :
  1. (A)b2−a2=a2+c2b^2 - a^2 = a^2 + c^2b2−a2=a2+c2
  2. (B)b2, c2,a2b^2,\ c^2, a^2b2, c2,a2 are in A.P.
  3. (C)c2,a2,b2c^2, a^2, b^2c2,a2,b2 are in A.P.
  4. (D)a2,b2,c2a^2, b^2, c^2a2,b2,c2 are in A.P.

Correct answer: (B)

Step-by-step solution →
Q114·MathematicsSingle correctJEE Main 2021
The sum of solutions of the equation cos⁡x1+sin⁡x=∣tan⁡2x∣\frac{\cos x}{1+\sin x} = |\tan 2x|1+sinxcosx​=∣tan2x∣, x∈(−π2,π2)−{π4,−π4}x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) - \left\{\frac{\pi}{4}, -\frac{\pi}{4}\right\}x∈(−2π​,2π​)−{4π​,−4π​} is :
  1. (A)−11π30-\frac{11\pi}{30}−3011π​
  2. (B)π10\frac{\pi}{10}10π​
  3. (C)−7π30-\frac{7\pi}{30}−307π​
  4. (D)−π15-\frac{\pi}{15}−15π​

Correct answer: (A)

Step-by-step solution →
Q115·MathematicsSingle correctJEE Main 2021
The value of 2sin⁡(π8)sin⁡(2π8)sin⁡(3π8)sin⁡(5π8)sin⁡(6π8)sin⁡(7π8)2\sin\left(\frac{\pi}{8}\right)\sin\left(\frac{2\pi}{8}\right)\sin\left(\frac{3\pi}{8}\right)\sin\left(\frac{5\pi}{8}\right)\sin\left(\frac{6\pi}{8}\right)\sin\left(\frac{7\pi}{8}\right)2sin(8π​)sin(82π​)sin(83π​)sin(85π​)sin(86π​)sin(87π​) is :
  1. (A)142\frac{1}{4\sqrt{2}}42​1​
  2. (B)14\frac{1}{4}41​
  3. (C)18\frac{1}{8}81​
  4. (D)182\frac{1}{8\sqrt{2}}82​1​

Correct answer: (C)

Step-by-step solution →
Q116·MathematicsSingle correctJEE Main 2021
Let f : ℝ → ℝ be defined as f(x + y) + f(x − y) = 2f(x) f(y), f(12)f\left( \frac{1}{2} \right)f(21​) = −1. Then, the value of ∑k=1201sin⁡(k)sin⁡(k+f(k))\sum_{k=1}^{20} \frac{1}{\sin\left( k \right) \sin\left( k + f\left( k \right) \right)}∑k=120​sin(k)sin(k+f(k))1​ is equal to :
  1. (A)cosec⁡2(21)cos⁡(20)cos⁡(2)\cosec^2\left( 21 \right) \cos\left( 20 \right) \cos\left( 2 \right)cosec2(21)cos(20)cos(2)
  2. (B)sec⁡2(1)sec⁡(21)cos⁡(20)\sec^2\left( 1 \right) \sec\left( 21 \right) \cos\left( 20 \right)sec2(1)sec(21)cos(20)
  3. (C)sec⁡2(21)sin⁡(20)sin⁡(2)\sec^2\left( 21 \right) \sin\left( 20 \right) \sin\left( 2 \right)sec2(21)sin(20)sin(2)
  4. (D)cosec⁡2(1)cosec⁡(21)sin⁡(20)\cosec^2\left( 1 \right) \cosec\left( 21 \right) \sin\left( 20 \right)cosec2(1)cosec(21)sin(20)

Correct answer: (D)

Step-by-step solution →
Q117·MathematicsSingle correctJEE Main 2021
If sin⁡θ+cos⁡θ=12\sin\theta + \cos\theta = \frac{1}{2}sinθ+cosθ=21​, then 16(sin⁡(2θ)+cos⁡(4θ)+sin⁡(6θ))16\left(\sin(2\theta) + \cos(4\theta) + \sin(6\theta)\right)16(sin(2θ)+cos(4θ)+sin(6θ)) is equal to :
  1. (A)27
  2. (B)−23-23−23
  3. (C)−27-27−27
  4. (D)23

Correct answer: (B)

Step-by-step solution →
Q118·MathematicsSingle correctJEE Main 2021
If tan⁡(π9)\tan\left( \frac{\pi}{9} \right)tan(9π​), x, tan⁡(7π18)\tan\left( \frac{7\pi}{18} \right)tan(187π​) are in arithmetic progression and tan⁡(π9)\tan\left( \frac{\pi}{9} \right)tan(9π​), y, tan⁡(5π18)\tan\left( \frac{5\pi}{18} \right)tan(185π​) are also in arithmetic progression, then ∣x−2y∣\left| x - 2y \right|∣x−2y∣ is equal to :
  1. (A)1
  2. (B)0
  3. (C)4
  4. (D)3

Correct answer: (B)

Step-by-step solution →
Q119·MathematicsSingle correctJEE Main 2021
The value of cot⁡π24\cot\frac{\pi}{24}cot24π​ is :
  1. (A)2+3+2+6\sqrt{2} + \sqrt{3} + 2 + \sqrt{6}2​+3​+2+6​
  2. (B)32−3−63\sqrt{2} - \sqrt{3} - \sqrt{6}32​−3​−6​
  3. (C)2−3−2+6\sqrt{2} - \sqrt{3} - 2 + \sqrt{6}2​−3​−2+6​
  4. (D)2+3+2−6\sqrt{2} + \sqrt{3} + 2 - \sqrt{6}2​+3​+2−6​

Correct answer: (A)

Step-by-step solution →
Q120·MathematicsSingle correctJEE Main 2021
A spherical gas balloon of radius 16 meter subtends an angle 60° at the eye of the observer A while the angle of elevation of its center from the eye of A is 75°. Then the height (in meter) of the top most point of the balloon from the level of the observer's eye is :
  1. (A)8(2+23+2)8\left(2+2\sqrt{3}+\sqrt{2}\right)8(2+23​+2​)
  2. (B)8(6+2+2)8\left(\sqrt{6}+\sqrt{2}+2\right)8(6​+2​+2)
  3. (C)8(6−2+2)8\left(\sqrt{6}-\sqrt{2}+2\right)8(6​−2​+2)
  4. (D)8(2+2+3)8\left(\sqrt{2}+2+\sqrt{3}\right)8(2​+2+3​)

Correct answer: (B)

Step-by-step solution →
Q121·MathematicsSingle correctJEE Main 2021
The sum of all values of x in [0,2π][0,2\pi][0,2π], for which sin⁡x+sin⁡2x+sin⁡3x+sin⁡4x=0\sin x+\sin 2x+\sin 3x+\sin 4x=0sinx+sin2x+sin3x+sin4x=0, is equal to :
  1. (A)11π11\pi11π
  2. (B)9π9\pi9π
  3. (C)8π8\pi8π
  4. (D)12π12\pi12π

Correct answer: (B)

Step-by-step solution →
Q122·MathematicsSingle correctJEE Main 2021
The number of solutions of sin⁡7x+cos⁡7x=1\sin^7 x + \cos^7 x = 1sin7x+cos7x=1, x∈[0,4π]x \in \left[0, 4\pi\right]x∈[0,4π] is equal to :
  1. (A)777
  2. (B)111111
  3. (C)555
  4. (D)999

Correct answer: (C)

Step-by-step solution →
Q123·MathematicsSingle correctJEE Main 2021
Let in a right angled triangle, the smallest angle be θ\thetaθ. If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then sin⁡θ\sin\thetasinθ is equal to :
  1. (A)2−12\frac{\sqrt{2}-1}{2}22​−1​
  2. (B)5+14\frac{\sqrt{5}+1}{4}45​+1​
  3. (C)5−12\frac{\sqrt{5}-1}{2}25​−1​
  4. (D)5−14\frac{\sqrt{5}-1}{4}45​−1​

Correct answer: (C)

Step-by-step solution →
Q124·MathematicsSingle correctJEE Main 2021
If in a triangle ABC, AB = 5 units, ∠B=cos⁡−1(35)\angle B = \cos^{-1}\left(\frac{3}{5}\right)∠B=cos−1(53​) and radius of circumcircle of ΔABC is 5 units, then the area (in sq. units) of ΔABC is :
  1. (A)6+836 + 8\sqrt{3}6+83​
  2. (B)8+228 + 2\sqrt{2}8+22​
  3. (C)10+6210 + 6\sqrt{2}10+62​
  4. (D)4+234 + 2\sqrt{3}4+23​

Correct answer: (A)

Step-by-step solution →
Q125·MathematicsNumericalJEE Main 2021
The number of solutions of the equation ∣cot⁡x∣=cot⁡x+1sin⁡x\left|\cot x\right| = \cot x + \dfrac{1}{\sin x}∣cotx∣=cotx+sinx1​ in the interval [0,2π][0, 2\pi][0,2π] is

Correct answer: 1

Step-by-step solution →
Q126·MathematicsSingle correctJEE Main 2021
If 15sin⁡4α+10cos⁡4α=615\sin^{4}\alpha + 10\cos^{4}\alpha = 615sin4α+10cos4α=6, for some α ∈ R, then the value of 27sec⁡6α+8 cosec6α27\sec^{6}\alpha + 8\,\mathrm{cosec}^{6}\alpha27sec6α+8cosec6α is equal to :
  1. (A)350
  2. (B)500
  3. (C)400
  4. (D)250

Correct answer: (D)

Step-by-step solution →
Q127·MathematicsSingle correctJEE Main 2021
A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from each corner of the park be π3\frac{\pi}{3}3π​. If the radius of the circumcircle ot ΔABC is 2, then the height of the pole is equal to :
  1. (A)233\frac{2\sqrt{3}}{3}323​​
  2. (B)232\sqrt{3}23​
  3. (C)3\sqrt{3}3​
  4. (D)13\frac{1}{\sqrt{3}}3​1​

Correct answer: (B)

Step-by-step solution →
Q128·MathematicsSingle correctJEE Main 2021
The number of solutions of the equation x+2tan⁡x=π2x + 2\tan x = \frac{\pi}{2}x+2tanx=2π​ in the interval [0,2π][0, 2\pi][0,2π] is :
  1. (A)333
  2. (B)444
  3. (C)222
  4. (D)555

Correct answer: (A)

Step-by-step solution →
Q129·MathematicsNumericalJEE Main 2021
Let tan⁡α\tan\alphatanα, tan⁡β\tan\betatanβ and tan⁡γ\tan\gammatanγ; α,β,γ≠(2n−1)π2\alpha, \beta, \gamma \neq \frac{(2n-1)\pi}{2}α,β,γ=2(2n−1)π​, n∈Nn \in Nn∈N be the slopes of three line segments OA, OB and OC, respectively, where O is origin. If circumcentre of ΔABC\Delta ABCΔABC coincides with origin and its orthocentre lies on y-axis, then the value of (cos⁡3α+cos⁡3β+cos⁡3γcos⁡αcos⁡βcos⁡γ)2\left(\frac{\cos 3\alpha + \cos 3\beta + \cos 3\gamma}{\cos\alpha\cos\beta\cos\gamma}\right)^{2}(cosαcosβcosγcos3α+cos3β+cos3γ​)2 is equal to :

Correct answer: 144

Step-by-step solution →
Q130·MathematicsNumericalJEE Main 2021
In ΔABC\Delta ABCΔABC, the lengths of sides AC and AB are 12 cm and 5 cm, respectively. If the area of ΔABC\Delta ABCΔABC is 30 cm2^{2}2 and R and r are respectively the radii of circumcircle and incircle of ΔABC\Delta ABCΔABC, then the value of 2R+r2R + r2R+r (in cm) is equal to _______.

Correct answer: 15

Step-by-step solution →
Q131·MathematicsSingle correctJEE Main 2021
The number of roots of the equation, (81)sin⁡2x+(81)cos⁡2x=30(81)^{\sin^{2} x} + (81)^{\cos^{2} x} = 30(81)sin2x+(81)cos2x=30 in the interval [0, π] is equal to :
  1. (A)3
  2. (B)4
  3. (C)8
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q132·MathematicsSingle correctJEE Main 2021
If for x∈(0,π2)x \in \left(0, \frac{\pi}{2}\right)x∈(0,2π​), log⁡10sin⁡x+log⁡10cos⁡x=−1\log_{10}\sin x + \log_{10}\cos x = -1log10​sinx+log10​cosx=−1 and log⁡10(sin⁡x+cos⁡x)=12(log⁡10n−1)\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - 1)log10​(sinx+cosx)=21​(log10​n−1), n > 0, then the value of n is equal to :
  1. (A)20
  2. (B)12
  3. (C)9
  4. (D)16

Correct answer: (B)

Step-by-step solution →
Q133·MathematicsNumericalJEE Main 2021
The number of integral values of 'k' for which the equation 3sinx + 4cos x = k + 1 has a solution, k ∈ R is _______.

Correct answer: 11

Step-by-step solution →
Q134·MathematicsNumericalJEE Main 2021
If 3(cos⁡2x)=(3−1)cos⁡x+1\sqrt{3}(\cos^{2}x)=(\sqrt{3}-1)\cos x+13​(cos2x)=(3​−1)cosx+1, the number of solutions of the given equation when x∈[0,π2]x \in \left[0,\frac{\pi}{2}\right]x∈[0,2π​] is _______.

Correct answer: 1

Step-by-step solution →
Q135·MathematicsSingle correctJEE Main 2021
All possible values of θ∈[0,2π]\theta \in [0, 2\pi]θ∈[0,2π] for which sin⁡2θ+tan⁡2θ>0\sin 2\theta + \tan 2\theta > 0sin2θ+tan2θ>0 lie in:
  1. (A)(0,π2)∪(π,3π2)\left(0, \frac{\pi}{2}\right) \cup \left(\pi, \frac{3\pi}{2}\right)(0,2π​)∪(π,23π​)
  2. (B)(0,π4)∪(π2,3π4)∪(π,5π4)∪(3π2,7π4)\left(0, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{2}, \frac{3\pi}{4}\right) \cup \left(\pi, \frac{5\pi}{4}\right) \cup \left(\frac{3\pi}{2}, \frac{7\pi}{4}\right)(0,4π​)∪(2π​,43π​)∪(π,45π​)∪(23π​,47π​)
  3. (C)(0,π2)∪(π2,3π4)∪(π,7π6)\left(0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \frac{3\pi}{4}\right) \cup \left(\pi, \frac{7\pi}{6}\right)(0,2π​)∪(2π​,43π​)∪(π,67π​)
  4. (D)(0,π4)∪(π2,3π4)∪(3π2,11π6)\left(0, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{2}, \frac{3\pi}{4}\right) \cup \left(\frac{3\pi}{2}, \frac{11\pi}{6}\right)(0,4π​)∪(2π​,43π​)∪(23π​,611π​)

Correct answer: (B)

Step-by-step solution →
Q136·MathematicsSingle correctJEE Main 2021
If 0 < x, y <π and cosx + cosy – cos(x + y) = 32 , then sinx + cosy is equal to:
  1. (A)1+3 2
  2. (B)1–3 2
  3. (C)23
  4. (D)12

Correct answer: (A)

Step-by-step solution →
Q137·MathematicsSingle correctJEE Main 2021
A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man's eye is 30∘30^{\circ}30∘ (Ignore man's height). After sailing for 20 seconds towards the base of the tower (which is at the level of water), the boat has reached a point B, where the angle of depression is 45∘45^{\circ}45∘. Then the time taken (in seconds) by the boat from B to reach the base of the tower is :
  1. (A)10(3−1)10(\sqrt{3} - 1)10(3​−1)
  2. (B)10310\sqrt{3}103​
  3. (C)101010
  4. (D)10(3+1)10(\sqrt{3} + 1)10(3​+1)

Correct answer: (D)

Step-by-step solution →
Q138·MathematicsSingle correctJEE Main 2021
If e(cos⁡2x+cos⁡4x+cos⁡6x+...∞)log⁡e2e^{\left(\cos^2 x + \cos^4 x + \cos^6 x + ...\infty\right)\log_e 2}e(cos2x+cos4x+cos6x+...∞)loge​2 satisfies the equation t2−9t+8=0t^2 - 9t + 8 = 0t2−9t+8=0, then the value of 2sin⁡xsin⁡x+3cos⁡x(0<x<π2)\frac{2\sin x}{\sin x + \sqrt{3}\cos x}\left(0 < x < \frac{\pi}{2}\right)sinx+3​cosx2sinx​(0<x<2π​) is :
  1. (A)32\frac{3}{2}23​
  2. (B)232\sqrt{3}23​
  3. (C)12\frac{1}{2}21​
  4. (D)3\sqrt{3}3​

Correct answer: (C)

Step-by-step solution →
Q139·MathematicsSingle correctJEE Main 2021
The angle of elevation of a jet plane from a point AAA on the ground is 60°. After a flight of 20 seconds at the speed of 432 km/ hour, the angle of elevation changes to 30°. If the jet plane is flying at a constant height, then its height is:
  1. (A)120031200\sqrt{3}12003​m
  2. (B)180031800\sqrt{3}18003​m
  3. (C)360033600\sqrt{3}36003​m
  4. (D)240032400\sqrt{3}24003​m

Correct answer: (A)

Step-by-step solution →
Q140·MathematicsSingle correctJEE Main 2021
Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:
  1. (A)25
  2. (B)20320\sqrt{3}203​
  3. (C)30
  4. (D)25325\sqrt{3}253​

Correct answer: (D)

Step-by-step solution →
Q141·MathematicsNumericalJEE Advanced 2020
Let f:[0,2]→Rf : [0, 2] \to \mathbb{R}f:[0,2]→R be the function defined by f(x)=(3−sin⁡(2πx))sin⁡(πx−π4)−sin⁡(3πx+π4)f(x) = (3 - \sin(2\pi x))\sin\left(\pi x - \frac{\pi}{4}\right) - \sin\left(3\pi x + \frac{\pi}{4}\right)f(x)=(3−sin(2πx))sin(πx−4π​)−sin(3πx+4π​) If α,β∈[0,2]\alpha, \beta \in [0, 2]α,β∈[0,2] are such that {x∈[0,2]:f(x)≥0}=[α,β]\{x \in [0, 2] : f(x) \geq 0\} = [\alpha, \beta]{x∈[0,2]:f(x)≥0}=[α,β], then the value of β−α\beta - \alphaβ−α is ______

Correct answer: 1.00

Step-by-step solution →
Q142·MathematicsMultiple correctJEE Advanced 2020
Let x, y and z be positive real numbers. Suppose x, y and z are lengths of the sides of a triangle opposite to its angles X, Y and Z, respectively. If tan⁡X2+tan⁡Z2=2yx+y+z\tan\frac{X}{2} + \tan\frac{Z}{2} = \frac{2y}{x + y + z}tan2X​+tan2Z​=x+y+z2y​, then which of the following statements is/are TRUE?
  1. (A)2Y=X+Z2Y = X + Z2Y=X+Z
  2. (B)Y=X+ZY = X + ZY=X+Z
  3. (C)tan⁡X2=xy+z\tan\frac{X}{2} = \frac{x}{y + z}tan2X​=y+zx​
  4. (D)x2+z2−y2=xzx^{2} + z^{2} - y^{2} = xzx2+z2−y2=xz

Correct answer: (B), (C)

Step-by-step solution →
Q143·MathematicsNumericalJEE Main 2020
The angle of elevation of the top of a hill from a point on the horizontal plane passing through the foot of the hill is found to be 45∘45^{\circ}45∘. After walking a distance of 80 meters towards the top, up a slope inclined at an angle of 30∘30^{\circ}30∘ to the horizontal plane, the angle of elevation of the top of the hill becomes 75∘75^{\circ}75∘. Then the height of the hill (in metres) is ____.

Correct answer: 80

Step-by-step solution →
Q144·MathematicsSingle correctJEE Main 2020
The angle of elevation of the summit of a mountain from a point on the ground is 45°. After climbing up one km towards the summit at an inclination of 30° from the ground, the angle of elevation of the summit is found to be 60°. Then the height (in km) of the summit from the ground is:
  1. (A)3−13+1\frac{\sqrt{3}-1}{\sqrt{3}+1}3​+13​−1​
  2. (B)3+13−1\frac{\sqrt{3}+1}{\sqrt{3}-1}3​−13​+1​
  3. (C)13−1\frac{1}{\sqrt{3}-1}3​−11​
  4. (D)13+1\frac{1}{\sqrt{3}+1}3​+11​

Correct answer: (C)

Step-by-step solution →
Q145·MathematicsSingle correctJEE Main 2020
If L=sin⁡2(π16)−sin⁡2(π8)L=\sin^2\left(\dfrac{\pi}{16}\right)-\sin^2\left(\dfrac{\pi}{8}\right)L=sin2(16π​)−sin2(8π​) and M=cos⁡2(π16)−sin⁡2(π8)M=\cos^2\left(\dfrac{\pi}{16}\right)-\sin^2\left(\dfrac{\pi}{8}\right)M=cos2(16π​)−sin2(8π​), then:
  1. (A)M=142+14cos⁡π8M=\dfrac{1}{4\sqrt{2}}+\dfrac{1}{4}\cos\dfrac{\pi}{8}M=42​1​+41​cos8π​
  2. (B)L=−122+12cos⁡π8L=-\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2}\cos\dfrac{\pi}{8}L=−22​1​+21​cos8π​
  3. (C)M=122+12cos⁡π8M=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2}\cos\dfrac{\pi}{8}M=22​1​+21​cos8π​
  4. (D)L=142−14cos⁡π8L=\dfrac{1}{4\sqrt{2}}-\dfrac{1}{4}\cos\dfrac{\pi}{8}L=42​1​−41​cos8π​

Correct answer: (C)

Step-by-step solution →
Q146·MathematicsSingle correctJEE Main 2020
Two vertical poles AB = 15 m and CD = 10 m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m) above the line AC is:
  1. (A)103\frac{10}{3}310​
  2. (B)666
  3. (C)555
  4. (D)203\frac{20}{3}320​

Correct answer: (B)

Step-by-step solution →
Q147·MathematicsSingle correctJEE Main 2020
The angle of elevation of a cloud C from a point P, 200 m above a still take is 30°. If the angle of depression of the image of C in the lake from the point P is 60°, then PC (in m) is equal to
  1. (A)2003200\sqrt{3}2003​
  2. (B)100
  3. (C)400
  4. (D)4003400\sqrt{3}4003​

Correct answer: (C)

Step-by-step solution →
Q148·MathematicsSingle correctJEE Main 2020
The minimum value of 2sin⁡x+2cos⁡x2^{\sin x} + 2^{\cos x}2sinx+2cosx is:
  1. (A)21−122^{1 - \frac{1}{\sqrt{2}}}21−2​1​
  2. (B)21−22^{1 - \sqrt{2}}21−2​
  3. (C)2−1+22^{-1 + \sqrt{2}}2−1+2​
  4. (D)2−1+122^{-1 + \frac{1}{\sqrt{2}}}2−1+2​1​

Correct answer: (A)

Step-by-step solution →
Q149·MathematicsNumericalJEE Main 2020
The number of distinct solutions of the equation, log⁡12∣sin⁡x∣=2−log⁡12∣cos⁡x∣\log_{\frac{1}{2}}|\sin x|=2-\log_{\frac{1}{2}}|\cos x|log21​​∣sinx∣=2−log21​​∣cosx∣ in the interval [0,2π][0,2\pi][0,2π], is ______.

Correct answer: 8

Step-by-step solution →
Q150·MathematicsSingle correctJEE Main 2020
The value of cos⁡3(π8).cos⁡(3π8)+sin⁡3(π8).sin⁡(3π8)\cos^{3}\left(\dfrac{\pi}{8}\right).\cos\left(\dfrac{3\pi}{8}\right)+\sin^{3}\left(\dfrac{\pi}{8}\right).\sin\left(\dfrac{3\pi}{8}\right)cos3(8π​).cos(83π​)+sin3(8π​).sin(83π​) is:
  1. (A)12\dfrac{1}{\sqrt{2}}2​1​
  2. (B)14\dfrac{1}{4}41​
  3. (C)12\dfrac{1}{2}21​
  4. (D)122\dfrac{1}{2\sqrt{2}}22​1​

Correct answer: (D)

Step-by-step solution →
Q151·MathematicsNumericalJEE Main 2020
If 2sin⁡α1+cos⁡2α=17\dfrac{\sqrt{2}\sin\alpha}{\sqrt{1+\cos2\alpha}}=\dfrac{1}{7}1+cos2α​2​sinα​=71​ and 1−cos⁡2β2=110\sqrt{\dfrac{1-\cos2\beta}{2}}=\dfrac{1}{\sqrt{10}}21−cos2β​​=10​1​, α,β∈(0,π2)\alpha,\beta\in\left(0,\dfrac{\pi}{2}\right)α,β∈(0,2π​), then tan⁡(α+2β)\tan\left(\alpha+2\beta\right)tan(α+2β) is equal to __________.

Correct answer: 1

Step-by-step solution →
Q152·MathematicsSingle correctJEE Main 2020
If θ1\theta_{1}θ1​ and θ2\theta_{2}θ2​ be respectively the smallest and the largest values of θ\thetaθ in (0,2π)−{π}(0,2\pi)-\{\pi\}(0,2π)−{π} which satisfy the equation, 2cot⁡2θ−5sin⁡θ+4=02\cot^{2}\theta-\dfrac{5}{\sin\theta}+4=02cot2θ−sinθ5​+4=0, then ∫θ1θ2cos⁡23θ dθ\displaystyle\int_{\theta_{1}}^{\theta_{2}}\cos^{2}3\theta \, d\theta∫θ1​θ2​​cos23θdθ is equal to:
  1. (A)π3+16\dfrac{\pi}{3}+\dfrac{1}{6}3π​+61​
  2. (B)π3\dfrac{\pi}{3}3π​
  3. (C)π9\dfrac{\pi}{9}9π​
  4. (D)2π3\dfrac{2\pi}{3}32π​

Correct answer: (B)

Step-by-step solution →
Q153·MathematicsMultiple correctJEE Advanced 2019
In a non-right-angled triangle ΔPQR\Delta PQRΔPQR, let p, q, r denote the lengths of the sides opposite to the angles at P, Q, R respectively. The median from R meets the side PQ at S, the perpendicular from P meets the side QR at E, and RS and PE intersect at O. If p=3p = \sqrt{3}p=3​, q=1q = 1q=1, and the radius of the circumcircle of the ΔPQR\Delta PQRΔPQR equals 1, then which of the following options is/are correct?
  1. (A)length of OE=16OE = \frac{1}{6}OE=61​
  2. (B)Radius of incircle of ΔPQR=32(2−3)\Delta PQR = \frac{\sqrt{3}}{2}\left(2 - \sqrt{3}\right)ΔPQR=23​​(2−3​)
  3. (C)Length of RS=72RS = \frac{\sqrt{7}}{2}RS=27​​
  4. (D)Are of ΔSOE=312\Delta SOE = \frac{\sqrt{3}}{12}ΔSOE=123​​

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

Step-by-step solution →
Q154·MathematicsSingle correctJEE Advanced 2019
Answer the following by appropriately matching the lists based on the information given in the paragraph Let f(x)=sin⁡(πcos⁡x)f(x) = \sin(\pi \cos x)f(x)=sin(πcosx) and g(x)=cos⁡(2πsin⁡x)g(x) = \cos(2\pi \sin x)g(x)=cos(2πsinx) be two functions defined for x>0x > 0x>0. Define the following sets whose elements are written in the increasing order: X={x:f(x)=0}X = \{x : f(x) = 0\}X={x:f(x)=0}, Y={x:f′(x)=0}Y = \{x : f'(x) = 0\}Y={x:f′(x)=0}, Z={x:g(x)=0}Z = \{x : g(x) = 0\}Z={x:g(x)=0}, W={x:g′(x)=0}W = \{x : g'(x) = 0\}W={x:g′(x)=0} List – I contains the set X, Y, Z and W. List – II contains some information regarding these sets. Which of the following is the only CORRECT combination?
List – IList – II
I.XP.⊇{π2, 3π2, 4π, 7π}\supseteq\left\{\dfrac{\pi}{2},\ \dfrac{3\pi}{2},\ 4\pi,\ 7\pi\right\}⊇{2π​, 23π​, 4π, 7π}
II.YQ.an arithmetic progression
III.ZR.NOT an arithmetic progression
IV.WS.⊇{π6, 7π6, 13π6}\supseteq\left\{\dfrac{\pi}{6},\ \dfrac{7\pi}{6},\ \dfrac{13\pi}{6}\right\}⊇{6π​, 67π​, 613π​}
T.⊇{π3, 2π3, π}\supseteq\left\{\dfrac{\pi}{3},\ \dfrac{2\pi}{3},\ \pi\right\}⊇{3π​, 32π​, π}
U.⊇{π6, 3π4}\supseteq\left\{\dfrac{\pi}{6},\ \dfrac{3\pi}{4}\right\}⊇{6π​, 43π​}
  1. (A)(IV), (P), (R), (S)
  2. (B)(III), (P), (Q), (U)
  3. (C)(IV), (Q), (T)
  4. (D)(III), (R), (U)

Correct answer: (A)

Step-by-step solution →
Q155·MathematicsMultiple correctJEE Advanced 2019
For non–negative integers n, let f(n)=∑k=0nsin⁡(k+1n+2π)sin⁡(k+2n+2π)∑k=0nsin⁡2(k+1n+2π)f(n) = \dfrac{\displaystyle\sum_{k=0}^{n} \sin\left(\dfrac{k+1}{n+2}\pi\right)\sin\left(\dfrac{k+2}{n+2}\pi\right)}{\displaystyle\sum_{k=0}^{n} \sin^{2}\left(\dfrac{k+1}{n+2}\pi\right)}f(n)=k=0∑n​sin2(n+2k+1​π)k=0∑n​sin(n+2k+1​π)sin(n+2k+2​π)​ Assuming cos⁡−1x\cos^{-1}xcos−1x takes values in [0,π][0, \pi][0,π], which of the following options is/are correct?
  1. (A)f(4)=32f(4) = \dfrac{\sqrt{3}}{2}f(4)=23​​
  2. (B)If α=tan⁡(cos⁡−1f(6))\alpha = \tan(\cos^{-1}f(6))α=tan(cos−1f(6)), then α2+2α−1=0\alpha^{2} + 2\alpha - 1 = 0α2+2α−1=0
  3. (C)sin⁡(7 cos⁡−1f(5))=0\sin(7\,\cos^{-1}f(5)) = 0sin(7cos−1f(5))=0
  4. (D)lim⁡n→∞f(n)=12\lim\limits_{n \to \infty} f(n) = \dfrac{1}{2}n→∞lim​f(n)=21​

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

Step-by-step solution →
Q156·MathematicsSingle correctJEE Advanced 2019
Answer the following by appropriately matching the lists based on the information given in the paragraph Let f(x)=sin⁡(πcos⁡x)f(x) = \sin(\pi \cos x)f(x)=sin(πcosx) and g(x)=cos⁡(2πsin⁡x)g(x) = \cos(2\pi \sin x)g(x)=cos(2πsinx) be two functions defined for x>0x > 0x>0. Define the following sets whose elements are written in the increasing order: X={x:f(x)=0}X = \{x : f(x) = 0\}X={x:f(x)=0}, Y={x:f′(x)=0}Y = \{x : f'(x) = 0\}Y={x:f′(x)=0}, Z={x:g(x)=0}Z = \{x : g(x) = 0\}Z={x:g(x)=0}, W={x:g′(x)=0}W = \{x : g'(x) = 0\}W={x:g′(x)=0} List – I contains the set X, Y, Z and W. List – II contains some information regarding these sets. Which of the following is the only CORRECT combination?
List – IList – II
I.XP.⊇{π2, 3π2, 4π, 7π}\supseteq\left\{\dfrac{\pi}{2},\ \dfrac{3\pi}{2},\ 4\pi,\ 7\pi\right\}⊇{2π​, 23π​, 4π, 7π}
II.YQ.an arithmetic progression
III.ZR.NOT an arithmetic progression
IV.WS.⊇{π6, 7π6, 13π6}\supseteq\left\{\dfrac{\pi}{6},\ \dfrac{7\pi}{6},\ \dfrac{13\pi}{6}\right\}⊇{6π​, 67π​, 613π​}
T.⊇{π3, 2π3, π}\supseteq\left\{\dfrac{\pi}{3},\ \dfrac{2\pi}{3},\ \pi\right\}⊇{3π​, 32π​, π}
U.⊇{π6, 3π4}\supseteq\left\{\dfrac{\pi}{6},\ \dfrac{3\pi}{4}\right\}⊇{6π​, 43π​}
  1. (A)(II), (R), (S)
  2. (B)(I), (Q), (U)
  3. (C)(II), (Q), (T)
  4. (D)(I), (P), (R)

Correct answer: (C)

Step-by-step solution →
Q157·MathematicsSingle correctJEE Main 2019
The angle of elevation of the top of vertical tower standing on a horizontal plane is observed to be 45°45°45° from a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be 30°30°30°, then the distance (in m) of the foot of the tower from the point A is:
  1. (A)15(1+3)15\left(1+\sqrt{3}\right)15(1+3​)
  2. (B)15(3−3)15\left(3-\sqrt{3}\right)15(3−3​)
  3. (C)15(3+3)15\left(3+\sqrt{3}\right)15(3+3​)
  4. (D)15(5−3)15\left(5-\sqrt{3}\right)15(5−3​)

Correct answer: (C)

Step-by-step solution →
Q158·MathematicsSingle correctJEE Main 2019
Let S be the set of all α∈R\alpha \in Rα∈R such that the equation, cos⁡2x+αsin⁡x=2α−7\cos 2x + \alpha \sin x = 2\alpha - 7cos2x+αsinx=2α−7 has a solution. Then S is equal to :
  1. (A)[3,7][3,7][3,7]
  2. (B)RRR
  3. (C)[2,6][2,6][2,6]
  4. (D)[1,4][1,4][1,4]

Correct answer: (C)

Step-by-step solution →
Q159·MathematicsSingle correctJEE Main 2019
The number of solutions of the equation 1+sin⁡4x=cos⁡23x,x∈[−5π2,5π2]1+\sin^{4}x=\cos^{2}3x, x \in \left[-\frac{5\pi}{2}, \frac{5\pi}{2}\right]1+sin4x=cos23x,x∈[−25π​,25π​] is :
  1. (A)3
  2. (B)4
  3. (C)5
  4. (D)7

Correct answer: (C)

Step-by-step solution →
Q160·MathematicsSingle correctJEE Main 2019
All the pairs (x, y) that satisfy the inequality 2sin⁡2x−2sin⁡x+5⋅14sin⁡2y≤12^{\sqrt{\sin^{2}x-2\sin x+5}}\cdot\dfrac{1}{4^{\sin^{2}y}}\le 12sin2x−2sinx+5​⋅4sin2y1​≤1 also Satisfy the equation:
  1. (A)2∣sin⁡x∣=3sin⁡y2|\sin x|=3\sin y2∣sinx∣=3siny
  2. (B)sin⁡x=∣sin⁡y∣\sin x=|\sin y|sinx=∣siny∣
  3. (C)2sin⁡x=sin⁡y2\sin x=\sin y2sinx=siny
  4. (D)sin⁡x=2sin⁡y\sin x=2\sin ysinx=2siny

Correct answer: (B)

Step-by-step solution →
Q161·MathematicsSingle correctJEE Main 2019
ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cot−1^{-1}−1(32\sqrt{2}2​) and cosec−1^{-1}−1(22\sqrt{2}2​) respectively, then the height of the tower (in metres) is:
  1. (A)25
  2. (B)10510\sqrt{5}105​
  3. (C)10033\frac{100}{3\sqrt{3}}33​100​
  4. (D)20

Correct answer: (D)

Step-by-step solution →
Q162·MathematicsSingle correctJEE Main 2019
The angles A, B and C of a triangle ABC are in A.P and a:b=1:3a : b = 1 : \sqrt{3}a:b=1:3​. If c=4c = 4c=4 cm, then the area (in sq. cm) of this triangle is
  1. (A)232\sqrt{3}23​
  2. (B)43\dfrac{4}{\sqrt{3}}3​4​
  3. (C)434\sqrt{3}43​
  4. (D)23\dfrac{2}{\sqrt{3}}3​2​

Correct answer: (A)

Step-by-step solution →
Q163·MathematicsSingle correctJEE Main 2019
The value of sin⁡10∘sin⁡30∘sin⁡50∘sin⁡70∘\sin10^{\circ}\sin30^{\circ}\sin50^{\circ}\sin70^{\circ}sin10∘sin30∘sin50∘sin70∘ is
  1. (A)136\dfrac{1}{36}361​
  2. (B)132\dfrac{1}{32}321​
  3. (C)118\dfrac{1}{18}181​
  4. (D)116\dfrac{1}{16}161​

Correct answer: (D)

Step-by-step solution →
Q164·MathematicsSingle correctJEE Main 2019
The value of cos⁡210∘−cos⁡10∘cos⁡50∘+cos⁡250∘\cos^{2}10^{\circ}-\cos 10^{\circ}\cos 50^{\circ}+\cos^{2}50^{\circ}cos210∘−cos10∘cos50∘+cos250∘ is:
  1. (A)32(1+cos⁡20∘)\dfrac{3}{2}(1+\cos 20^{\circ})23​(1+cos20∘)
  2. (B)34\dfrac{3}{4}43​
  3. (C)32\dfrac{3}{2}23​
  4. (D)34+cos⁡20∘\dfrac{3}{4}+\cos 20^{\circ}43​+cos20∘

Correct answer: (B)

Step-by-step solution →
Q165·MathematicsSingle correctJEE Main 2019
Let S={θ∈[−2π,2π]:2cos⁡2θ+3sin⁡θ=0}S=\{\theta\in[-2\pi, 2\pi] : 2\cos^{2}\theta+3\sin\theta=0\}S={θ∈[−2π,2π]:2cos2θ+3sinθ=0}. Then the sum of the elements of S is:
  1. (A)13π6\dfrac{13\pi}{6}613π​
  2. (B)2π2\pi2π
  3. (C)π\piπ
  4. (D)5π3\dfrac{5\pi}{3}35π​

Correct answer: (B)

Step-by-step solution →
Q166·MathematicsSingle correctJEE Main 2019
Two poles standing on a horizontal ground are of heights 5m and 10 m respectively. The line joining their tops makes an angle of 15∘15^{\circ}15∘ with ground. Then the distance (in m) between the poles, is
  1. (A)52(2+3)\dfrac{5}{2}\left(2+\sqrt{3}\right)25​(2+3​)
  2. (B)5(3+1)5\left(\sqrt{3}+1\right)5(3​+1)
  3. (C)5(2+3)5\left(2+\sqrt{3}\right)5(2+3​)
  4. (D)10(3−1)10\left(\sqrt{3}-1\right)10(3​−1)

Correct answer: (C)

Step-by-step solution →
Q167·MathematicsSingle correctJEE Main 2019
If cos⁡(α+β)=35\cos(\alpha+\beta)=\frac{3}{5}cos(α+β)=53​, sin⁡(α−β)=513\sin(\alpha-\beta)=\frac{5}{13}sin(α−β)=135​ and 0<α,β<π40<\alpha,\beta<\frac{\pi}{4}0<α,β<4π​, then tan⁡(2α)\tan(2\alpha)tan(2α) is equal to:
  1. (A)6352\frac{63}{52}5263​
  2. (B)3352\frac{33}{52}5233​
  3. (C)6316\frac{63}{16}1663​
  4. (D)2116\frac{21}{16}1621​

Correct answer: (C)

Step-by-step solution →
Q168·MathematicsSingle correctJEE Main 2019
If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle:
  1. (A)4:5:6
  2. (B)5:6:7
  3. (C)3:4:5
  4. (D)5:9:13

Correct answer: (A)

Step-by-step solution →
Q169·MathematicsSingle correctJEE Main 2019
The maximum value of 3cos⁡θ+5sin⁡(θ−π6)3\cos\theta + 5\sin\left(\theta - \dfrac{\pi}{6}\right)3cosθ+5sin(θ−6π​) for any real value of θ\thetaθ is:
  1. (A)19\sqrt{19}19​
  2. (B)792\dfrac{\sqrt{79}}{2}279​​
  3. (C)34\sqrt{34}34​
  4. (D)31\sqrt{31}31​

Correct answer: (A)

Step-by-step solution →
Q170·MathematicsSingle correctJEE Main 2019
If the angle of elevation of a cloud from a point P which is 25 m above a lake be 30°30°30° and the angle of depression of reflection of the cloud in the lake from P be 60°60°60°, then the height of the cloud (in meters) from the surface of the lake is :
  1. (A)60
  2. (B)50
  3. (C)45
  4. (D)42

Correct answer: (B)

Step-by-step solution →
Q171·MathematicsSingle correctJEE Main 2019
In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y. If x2−c2=yx^{2}-c^{2}=yx2−c2=y, where c is the length of the third side of the triangle, then the circumradius of the triangle is:
  1. (A)32y\frac{3}{2}y23​y
  2. (B)c3\frac{c}{\sqrt{3}}3​c​
  3. (C)c3\frac{c}{3}3c​
  4. (D)y3\frac{y}{\sqrt{3}}3​y​

Correct answer: (B)

Step-by-step solution →
Q172·MathematicsSingle correctJEE Main 2019
Given b+c11=c+a12=a+b13\dfrac{b+c}{11}=\dfrac{c+a}{12}=\dfrac{a+b}{13}11b+c​=12c+a​=13a+b​ for a ΔABC with usual nation. If cos⁡Aα=cos⁡ββ=cos⁡Cγ\dfrac{\cos A}{\alpha}=\dfrac{\cos\beta}{\beta}=\dfrac{\cos C}{\gamma}αcosA​=βcosβ​=γcosC​, then the ordered tried (α, β, γ) has a value:
  1. (A)(7, 19, 25)
  2. (B)(3, 4, 5)
  3. (C)(5, 12, 13)
  4. (D)(19, 7, 25)

Correct answer: (A)

Step-by-step solution →
Q173·MathematicsSingle correctJEE Main 2019
Let fk(x)=1k(sin⁡kx+cos⁡kx)f_{k}(x)=\frac{1}{k}(\sin^{k}x+\cos^{k}x)fk​(x)=k1​(sinkx+coskx) for k = 1, 2, 3, … Then for all x ∈ R, the value of f4(x)−f6(x)f_{4}(x)-f_{6}(x)f4​(x)−f6​(x) is equal to:
  1. (A)112\frac{1}{12}121​
  2. (B)14\frac{1}{4}41​
  3. (C)−112\frac{-1}{12}12−1​
  4. (D)512\frac{5}{12}125​

Correct answer: (A)

Step-by-step solution →
Q174·MathematicsSingle correctJEE Main 2019
The sum of all values of θ∈(0,π2)\theta\in\left(0,\frac{\pi}{2}\right)θ∈(0,2π​) satisfying sin⁡22θ+cos⁡42θ=34\sin^{2}2\theta+\cos^{4}2\theta=\frac{3}{4}sin22θ+cos42θ=43​ is:
  1. (A)π\piπ
  2. (B)5π4\frac{5\pi}{4}45π​
  3. (C)π2\frac{\pi}{2}2π​
  4. (D)3π8\frac{3\pi}{8}83π​

Correct answer: (C)

Step-by-step solution →
Q175·MathematicsSingle correctJEE Main 2019
Consider a triangular plot ABC with sides AB = 7 m, BC = 5 m and CA = 6 m. A vertical lamp-post at the mid point D of AC subtends an angle 30° at B. The height (in m) of the lamp-post is:
  1. (A)3221\dfrac{3}{2}\sqrt{21}23​21​
  2. (B)2321\dfrac{2}{3}\sqrt{21}32​21​
  3. (C)2212\sqrt{21}221​
  4. (D)7217\sqrt{21}721​

Correct answer: (B)

Step-by-step solution →
Q176·MathematicsSingle correctJEE Main 2019
If 0≤x<π20 \le x < \frac{\pi}{2}0≤x<2π​, then the number of values of x for which sin⁡x−sin⁡2x+sin⁡3x=0\sin x - \sin 2x + \sin 3x = 0sinx−sin2x+sin3x=0, is
  1. (A)222
  2. (B)111
  3. (C)333
  4. (D)444

Correct answer: (A)

Step-by-step solution →
Q177·MathematicsSingle correctJEE Main 2019
For any θ∈(π4,π2)\theta\in\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right)θ∈(4π​,2π​), the expression 3(sin⁡θ−cos⁡θ)4+6(sin⁡θ+cos⁡θ)2+4sin⁡6θ3(\sin\theta-\cos\theta)^{4}+6(\sin\theta+\cos\theta)^{2}+4\sin^{6}\theta3(sinθ−cosθ)4+6(sinθ+cosθ)2+4sin6θ equals:
  1. (A)13−4cos⁡2θ+6sin⁡2θcos⁡2θ13-4\cos^{2}\theta+6\sin^{2}\theta\cos^{2}\theta13−4cos2θ+6sin2θcos2θ
  2. (B)13−4cos⁡6θ13-4\cos^{6}\theta13−4cos6θ
  3. (C)13−4cos⁡2θ+6cos⁡4θ13-4\cos^{2}\theta+6\cos^{4}\theta13−4cos2θ+6cos4θ
  4. (D)13−4cos⁡4θ+2sin⁡2θcos⁡2θ13-4\cos^{4}\theta+2\sin^{2}\theta\cos^{2}\theta13−4cos4θ+2sin2θcos2θ

Correct answer: (B)

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Q178·MathematicsMultiple correctJEE Advanced 2018
In a triangle PQR, let ∠PQR=30∘\angle PQR = 30^{\circ}∠PQR=30∘ and the sides PQ and QR have lengths 10310\sqrt{3}103​ and 10, respectively. Then, which of the following statement(s) is (are) TRUE ?
  1. (A)∠QPR=45∘\angle QPR = 45^{\circ}∠QPR=45∘
  2. (B)The area of the triangle PQR is 25325\sqrt{3}253​ and ∠QRP=120∘\angle QRP = 120^{\circ}∠QRP=120∘
  3. (C)The radius of the incircle of the triangle PQR is 103−1510\sqrt{3} - 15103​−15
  4. (D)The area of the circumcircle of the triangle PQR is 100π100\pi100π

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

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Q179·MathematicsNumericalJEE Advanced 2018
Let a, b, c be three non-zero real numbers such that the equation 3acos⁡x+2bsin⁡x=c\sqrt{3}a \cos x + 2b \sin x = c3​acosx+2bsinx=c, x∈[−π2,π2]x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]x∈[−2π​,2π​], has two distinct real roots α\alphaα and β\betaβ with α+β=π3\alpha + \beta = \frac{\pi}{3}α+β=3π​. Then, the value of ba\frac{b}{a}ab​ is ______ .

Correct answer: 0.5

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Q180·MathematicsMultiple correctJEE Advanced 2017
Let α\alphaα and β\betaβ be nonzero real numbers such that 2(cos⁡β−cos⁡α)+cos⁡αcos⁡β=12(\cos\beta - \cos\alpha) + \cos\alpha\cos\beta = 12(cosβ−cosα)+cosαcosβ=1. Then which of the following is/are true ?
  1. (A)tan⁡(α2)+3tan⁡(β2)=0\tan\left(\dfrac{\alpha}{2}\right) + \sqrt{3}\tan\left(\dfrac{\beta}{2}\right) = 0tan(2α​)+3​tan(2β​)=0
  2. (B)3tan⁡(α2)+tan⁡(β2)=0\sqrt{3}\tan\left(\dfrac{\alpha}{2}\right) + \tan\left(\dfrac{\beta}{2}\right) = 03​tan(2α​)+tan(2β​)=0
  3. (C)tan⁡(α2)−3tan⁡(β2)=0\tan\left(\dfrac{\alpha}{2}\right) - \sqrt{3}\tan\left(\dfrac{\beta}{2}\right) = 0tan(2α​)−3​tan(2β​)=0
  4. (D)3tan⁡(α2)−tan⁡(β2)=0\sqrt{3}\tan\left(\dfrac{\alpha}{2}\right) - \tan\left(\dfrac{\beta}{2}\right) = 03​tan(2α​)−tan(2β​)=0

Correct answer: (A), (C)

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Q181·MathematicsSingle correctJEE Advanced 2017
PARAGRAPH 1 Let OOO be the origin, and OX→\overrightarrow{OX}OX, OY→\overrightarrow{OY}OY, OZ→\overrightarrow{OZ}OZ be three unit vectors in the directions of the sides QR→\overrightarrow{QR}QR​, RP→\overrightarrow{RP}RP, PQ→\overrightarrow{PQ}PQ​, respectively, of a triangle PQRPQRPQR. If the triangle PQRPQRPQR varies, then the minimum value of cos⁡(P+Q)+cos⁡(Q+R)+cos⁡(R+P)\cos(P+Q) + \cos(Q+R) + \cos(R+P)cos(P+Q)+cos(Q+R)+cos(R+P) is
  1. (A)−53-\dfrac{5}{3}−35​
  2. (B)−32-\dfrac{3}{2}−23​
  3. (C)32\dfrac{3}{2}23​
  4. (D)53\dfrac{5}{3}35​

Correct answer: (B)

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Q182·MathematicsSingle correctJEE Advanced 2016
The value of ∑k=1131sin⁡(π4+(k−1)π6)sin⁡(π4+kπ6)\sum_{k=1}^{13} \frac{1}{\sin\left( \frac{\pi}{4} + \frac{(k-1)\pi}{6} \right) \sin\left( \frac{\pi}{4} + \frac{k\pi}{6} \right)}∑k=113​sin(4π​+6(k−1)π​)sin(4π​+6kπ​)1​ is equal to
  1. (A)3−33 - \sqrt{3}3−3​
  2. (B)2(3−3)2\left( 3 - \sqrt{3} \right)2(3−3​)
  3. (C)2(3−1)2\left( \sqrt{3} - 1 \right)2(3​−1)
  4. (D)2(2+3)2\left( 2 + \sqrt{3} \right)2(2+3​)

Correct answer: (C)

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Q183·MathematicsMultiple correctJEE Advanced 2016
In a triangle XYZ, let x,y,zx, y, zx,y,z be the lengths of sides opposite to the angles X, Y, Z, respectively, and 2s=x+y+z2s = x + y + z2s=x+y+z. If s−x4=s−y3=s−z2\frac{s - x}{4} = \frac{s - y}{3} = \frac{s - z}{2}4s−x​=3s−y​=2s−z​ and area of incircle of the triangle XYZ is 8π3\frac{8\pi}{3}38π​, then
  1. (A)area of the triangle XYZ is 666\sqrt{6}66​
  2. (B)the radius of circumcircle of the triangle XYZ is 3566\frac{35}{6}\sqrt{6}635​6​
  3. (C)sin⁡X2sin⁡Y2sin⁡Z2=435\sin\frac{X}{2}\sin\frac{Y}{2}\sin\frac{Z}{2} = \frac{4}{35}sin2X​sin2Y​sin2Z​=354​
  4. (D)sin⁡2(X+Y2)=35\sin^{2}\left(\frac{X + Y}{2}\right) = \frac{3}{5}sin2(2X+Y​)=53​

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

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Q184·MathematicsSingle correctJEE Advanced 2016
Let S={x∈(−π,π):x≠0,±π2}S = \left\{ x \in (-\pi, \pi) : x \neq 0, \pm\frac{\pi}{2} \right\}S={x∈(−π,π):x=0,±2π​}. The sum of all distinct solutions of the equation 3sec⁡x+cosec⁡x+2(tan⁡x−cot⁡x)=0\sqrt{3}\sec x + \operatorname{cosec} x + 2(\tan x - \cot x) = 03​secx+cosecx+2(tanx−cotx)=0 in the set SSS is equal to
  1. (A)−7π9-\frac{7\pi}{9}−97π​
  2. (B)−2π9-\frac{2\pi}{9}−92π​
  3. (C)000
  4. (D)5π9\frac{5\pi}{9}95π​

Correct answer: (C)

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Q185·MathematicsMatrix matchJEE Advanced 2015
Match the entries in Column I with the entries in Column II.
Column – IColumn – II
A.In a triangle ΔXYZ\Delta XYZΔXYZ, let aaa, bbb and ccc be the lengths of the sides opposite to the angles XXX, YYY and ZZZ, respectively. If 2(a2−b2)=c22(a^{2} - b^{2}) = c^{2}2(a2−b2)=c2 and λ=sin⁡(X−Y)sin⁡Z\lambda = \dfrac{\sin(X - Y)}{\sin Z}λ=sinZsin(X−Y)​, then possible values of nnn for which cos⁡(nπλ)=0\cos(n\pi\lambda) = 0cos(nπλ)=0 is (are)P.1
B.In a triangle ΔXYZ\Delta XYZΔXYZ, let aaa, bbb and ccc be the lengths of the sides opposite to the angles XXX, YYY and ZZZ, respectively. If 1+cos⁡2X−2cos⁡2Y=2sin⁡Xsin⁡Y1 + \cos 2X - 2\cos 2Y = 2\sin X \sin Y1+cos2X−2cos2Y=2sinXsinY, then possible value(s) of ab\dfrac{a}{b}ba​ is (are)Q.2
C.In R2\mathbb{R}^{2}R2, let 3i^+j^\sqrt{3}\hat{i} + \hat{j}3​i^+j^​, i^+3j^\hat{i} + \sqrt{3}\hat{j}i^+3​j^​ and βi^+(1−β)j^\beta\hat{i} + (1-\beta)\hat{j}βi^+(1−β)j^​ be the position vectors of XXX, YYY and ZZZ with respect of the origin O, respectively. If the distance of ZZZ from the bisector of the acute angle of OX→\overrightarrow{OX}OX with OY→\overrightarrow{OY}OY is 32\dfrac{3}{\sqrt{2}}2​3​, then possible value(s) of ∣β∣|\beta|∣β∣ is (are)R.3
D.Suppose that F(α)F(\alpha)F(α) denotes the area of the region bounded by x=0x = 0x=0, x=2x = 2x=2, y2=4xy^{2} = 4xy2=4x and y=∣αx−1∣+∣αx−2∣+αxy = |\alpha x - 1| + |\alpha x - 2| + \alpha xy=∣αx−1∣+∣αx−2∣+αx, where α∈{0,1}\alpha \in \{0, 1\}α∈{0,1}. Then the value(s) of F(α)+832F(\alpha) + \dfrac{8}{3}\sqrt{2}F(α)+38​2​, when α=0\alpha = 0α=0 and α=1\alpha = 1α=1, is (are)S.5
T.6

Correct answer: A-(P,R,S); B-(P); C-(P,Q); D-(S,T)

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Q186·MathematicsIntegerJEE Advanced 2015
The number of distinct solutions of the equation 54cos⁡22x+cos⁡4x+sin⁡4x+cos⁡6x+sin⁡6x=2\dfrac{5}{4}\cos^{2}2x + \cos^{4}x + \sin^{4}x + \cos^{6}x + \sin^{6}x = 245​cos22x+cos4x+sin4x+cos6x+sin6x=2 in the interval [0,2π][0, 2\pi][0,2π] is

Correct answer: 8

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Q187·MathematicsSingle correctJEE Advanced 2014
In a triangle the sum of two sides is xxx and the product of the same two sides is yyy. If x2−c2=yx^{2} - c^{2} = yx2−c2=y, where ccc is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is
  1. (A)3y2x(x+c)\frac{3y}{2x(x+c)}2x(x+c)3y​
  2. (B)3y2c(x+c)\frac{3y}{2c(x+c)}2c(x+c)3y​
  3. (C)3y4x(x+c)\frac{3y}{4x(x+c)}4x(x+c)3y​
  4. (D)3y4c(x+c)\frac{3y}{4c(x+c)}4c(x+c)3y​

Correct answer: (B)

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Q188·MathematicsSingle correctJEE Advanced 2014
For x∈(0,π)x \in (0, \pi)x∈(0,π), the equation sin⁡x+2sin⁡2x−sin⁡3x=3\sin x + 2\sin 2x - \sin 3x = 3sinx+2sin2x−sin3x=3 has
  1. (A)infinitely many solutions
  2. (B)three solutions
  3. (C)one solution
  4. (D)no solution

Correct answer: (D)

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Q189·MathematicsMultiple correctJEE Advanced 2013
In a triangle PQR, P is the largest angle and cos⁡P=13\cos P = \frac{1}{3}cosP=31​. Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)
  1. (A)161616
  2. (B)181818
  3. (C)242424
  4. (D)222222

Correct answer: (B), (D)

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Trigonometric Functions — frequently asked

How many questions from Trigonometric Functions appear in JEE?

Trigonometric Functions has appeared in 144 of the last 186 JEE Main and JEE Advanced papers — about 77% of them — contributing 189 questions in total across those papers.

Is Trigonometric Functions an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 77% 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 Trigonometric Functions 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
  • Differential Equations 240
  • Probability 226

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