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JEE Main 4 April 2025 Shift 1 Question Paper with Answers

4 April 2025 · April session · 74 questions

74 of the 75 questions from the JEE Main 4 April 2025 Shift 1 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

1 question is held back from this paper while we re-check the transcription or the answer key. We would rather show you nothing than show you an answer we are not confident is right.

Physics
24
Chemistry
25
Mathematics
25

Physics — JEE Main 4 April 2025 Shift 1

Q1·Physics·Kinetic Theory of GasesSingle correct
The mean free path and the average speed of oxygen molecules at 300 K and 1 atm are 3×10−73\times10^{-7}3×10−7 m and 600 m/s, respectively. Find the frequency of its collisions.
  1. (A)2×10102\times10^{10}2×1010/s
  2. (B)9×1099\times10^{9}9×109/s
  3. (C)2×1092\times10^{9}2×109/s
  4. (D)5×1095\times10^{9}5×109/s

Correct answer: (C)

Step-by-step solution →
Q2·Physics·Electromagnetic WavesSingle correct
A small mirror of mass mmm is suspended by a massless thread of length lll. Then the small angle through which the thread will be deflected when a short pulse of laser of energy EEE falls normal on the mirror (ccc = speed of light in vacuum and ggg = acceleration due to gravity)
  1. (A)θ=3E4mcgl\theta=\dfrac{3E}{4mc\sqrt{gl}}θ=4mcgl​3E​
  2. (B)θ=Emcgl\theta=\dfrac{E}{mc\sqrt{gl}}θ=mcgl​E​
  3. (C)θ=E2mcgl\theta=\dfrac{E}{2mc\sqrt{gl}}θ=2mcgl​E​
  4. (D)θ=2Emcgl\theta=\dfrac{2E}{mc\sqrt{gl}}θ=mcgl​2E​

Correct answer: (D)

Step-by-step solution →
Q3·Physics·Properties of Solids and LiquidsSingle correct
Two liquids A and B have θA\theta_AθA​ and θB\theta_BθB​ as contact angles in a capillary tube. If K=cos⁡θAcos⁡θBK=\dfrac{\cos\theta_A}{\cos\theta_B}K=cosθB​cosθA​​, then identify the correct statement:
  1. (A)KKK is negative, then liquid A and liquid B have convex meniscus.
  2. (B)KKK is negative, then liquid A and liquid B have concave meniscus.
  3. (C)KKK is negative, then liquid A has concave meniscus and liquid B has convex meniscus.
  4. (D)KKK is zero, then liquid A has convex meniscus and liquid B has concave meniscus.

Correct answer: (C)

Step-by-step solution →
Q4·Physics·Rotational MotionSingle correct
Which of the following are correct expression for torque acting on a body? A. τ⃗=r⃗×L⃗\vec{\tau}=\vec{r}\times\vec{L}τ=r×L B. τ⃗=ddt(r⃗×p⃗)\vec{\tau}=\dfrac{d}{dt}(\vec{r}\times\vec{p})τ=dtd​(r×p​) C. τ⃗=r⃗×dp⃗dt\vec{\tau}=\vec{r}\times\dfrac{d\vec{p}}{dt}τ=r×dtdp​​ D. τ⃗=Iα⃗\vec{\tau}=I\vec{\alpha}τ=Iα E. τ⃗=r⃗×F⃗\vec{\tau}=\vec{r}\times\vec{F}τ=r×F (r⃗\vec{r}r = position vector; p⃗\vec{p}p​ = linear momentum; L⃗\vec{L}L = angular momentum; α⃗\vec{\alpha}α = angular acceleration; III = moment of inertia; F⃗\vec{F}F = force; ttt = time). Choose the correct answer from the options given below:
  1. (A)B, D and E Only
  2. (B)C and D Only
  3. (C)B, C, D and E Only
  4. (D)A, B, D and E Only

Correct answer: (C)

Step-by-step solution →
Q5·Physics·Wave OpticsSingle correct
In a Young's double slit experiment, the slits are separated by 0.2 mm. If the slits separation is increased to 0.4 mm, the percentage change of the fringe width is:
  1. (A)0%
  2. (B)100%
  3. (C)50%
  4. (D)25%

Correct answer: (C)

Step-by-step solution →
Q6·Physics·Alternating CurrentsSingle correct
An alternating current is represented by the equation, i=1002sin⁡(100πt)i=100\sqrt{2}\sin(100\pi t)i=1002​sin(100πt) ampere. The RMS value of current and the frequency of the given alternating current are
  1. (A)1002100\sqrt{2}1002​ A, 100 Hz
  2. (B)1002\dfrac{100}{\sqrt{2}}2​100​ A, 100 Hz
  3. (C)100 A, 50 Hz
  4. (D)50250\sqrt{2}502​ A, 50 Hz

Correct answer: (C)

Step-by-step solution →
Q7·Physics·Units and MeasurementsSingle correct
In an electromagnetic system, the quantity representing the ratio of electric flux and magnetic flux has dimension of MPLQTRASM^P L^Q T^R A^SMPLQTRAS, where value of QQQ and RRR are
  1. (A)(3,−5)(3,-5)(3,−5)
  2. (B)(−2,2)(-2,2)(−2,2)
  3. (C)(−2,1)(-2,1)(−2,1)
  4. (D)(1,−1)(1,-1)(1,−1)

Correct answer: (D)

Step-by-step solution →
Q8·Physics·Geometrical OpticsSingle correct
When an object is placed 40 cm away from a spherical mirror an image of magnification 12\dfrac{1}{2}21​ is produced. To obtain an image with magnification of 13\dfrac{1}{3}31​, the object is to be moved:
  1. (A)40 cm away from the mirror.
  2. (B)80 cm away from the mirror.
  3. (C)20 cm towards the mirror.
  4. (D)20 cm away from the mirror.

Correct answer: (A)

Step-by-step solution →
Q9·Physics·Dual Nature of Matter and RadiationSingle correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: In photoelectric effect, on increasing the intensity of incident light the stopping potential increases. Reason R: Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than threshold frequency. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both A and R are true but R is NOT the correct explanation of A
  2. (B)A is false but R is true
  3. (C)A is true but R is false
  4. (D)Both A and R are true and R is the correct explanation of A

Correct answer: (B)

Step-by-step solution →
Q10·Physics·KinematicsSingle correct
If L⃗\vec{L}L and P⃗\vec{P}P represent the angular momentum and linear momentum respectively of a particle of mass mmm having position vector r⃗=a(i^cos⁡ωt+j^sin⁡ωt)\vec{r}=a(\hat{i}\cos\omega t+\hat{j}\sin\omega t)r=a(i^cosωt+j^​sinωt). The direction of force is
  1. (A)Opposite to the direction of r⃗\vec{r}r
  2. (B)Opposite to the direction of L⃗\vec{L}L
  3. (C)Opposite to the direction of P⃗\vec{P}P
  4. (D)Opposite to the direction of L⃗×P⃗\vec{L}\times\vec{P}L×P

Correct answer: (A)

Step-by-step solution →
Q11·Physics·Laws of MotionSingle correct
A body of mass mmm is suspended by two strings making angles θ1\theta_1θ1​ and θ2\theta_2θ2​ with the horizontal ceiling with tensions T1T_1T1​ and T2T_2T2​ simultaneously. T1T_1T1​ and T2T_2T2​ are related by T1=3 T2T_1=\sqrt{3}\,T_2T1​=3​T2​. The angles θ1\theta_1θ1​ and θ2\theta_2θ2​ are
  1. (A)θ1=30∘\theta_1=30^\circθ1​=30∘, θ2=60∘\theta_2=60^\circθ2​=60∘ with T2=3mg4T_2=\dfrac{3mg}{4}T2​=43mg​
  2. (B)θ1=60∘\theta_1=60^\circθ1​=60∘, θ2=30∘\theta_2=30^\circθ2​=30∘ with T2=mg2T_2=\dfrac{mg}{2}T2​=2mg​
  3. (C)θ1=45∘\theta_1=45^\circθ1​=45∘, θ2=45∘\theta_2=45^\circθ2​=45∘ with T2=3mg4T_2=\dfrac{3mg}{4}T2​=43mg​
  4. (D)θ1=30∘\theta_1=30^\circθ1​=30∘, θ2=60∘\theta_2=60^\circθ2​=60∘ with T2=4mg5T_2=\dfrac{4mg}{5}T2​=54mg​

Correct answer: (B)

Step-by-step solution →
Q12·Physics·Current ElectricitySingle correct
Current passing through a wire as function of time is given as I(t)=0.02 t+0.01I(t)=0.02\,t+0.01I(t)=0.02t+0.01 A. The charge that will flow through the wire from t=1t=1t=1 s to t=2t=2t=2 s is:
  1. (A)0.06 C
  2. (B)0.02 C
  3. (C)0.07 C
  4. (D)0.04 C

Correct answer: (D)

Step-by-step solution →
Q13·Physics·GravitationSingle correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The kinetic energy needed to project a body of mass mmm from earth surface to infinity is 12mgR\dfrac{1}{2}mgR21​mgR, where RRR is the radius of earth. Reason R: The maximum potential energy of a body is zero when it is projected to infinity from earth surface. In the light of the above statements, choose the correct answer from the option given below:
  1. (A)A False but R is true
  2. (B)Both A and R are true and R is the correct explanation of A
  3. (C)A is true but R is false
  4. (D)Both A and R are true but R is NOT the correct explanation of A

Correct answer: (A)

Step-by-step solution →
Q14·Physics·Electronic DevicesSingle correct
The Boolean expression Y=ABˉC+AˉCˉY=A\bar{B}C+\bar{A}\bar{C}Y=ABˉC+AˉCˉ can be realised with which of the following gate configurations. A. One 3-input AND gate, 3 NOT gates and one 2-input OR gate, One 2-input AND gate, B. One 3-input AND gate, 1 NOT gate and one 2-input NOR gate and one 2-input OR gate C. 3-input OR gate, 3 NOT gates and one 2-input AND gate. Choose the correct answer from the options given below
  1. (A)B, C Only
  2. (B)A, B Only
  3. (C)A, B, C Only
  4. (D)A, C Only

Correct answer: (B)

Step-by-step solution →
Q15·Physics·WavesSingle correct
In an experiment with a closed organ pipe, it is filled with water by 15\dfrac{1}{5}51​th of its volume. The frequency of the fundamental note will change by
  1. (A)25%
  2. (B)20%
  3. (C)−20-20−20%
  4. (D)−25-25−25%

Correct answer: (A)

Step-by-step solution →
Q16·Physics·OscillationsSingle correct
Two simple pendulums having lengths l1l_1l1​ and l2l_2l2​ with negligible string mass undergo angular displacements θ1\theta_1θ1​ and θ2\theta_2θ2​ from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct?
  1. (A)θ1l12=θ2l22\theta_1 l_1^2=\theta_2 l_2^2θ1​l12​=θ2​l22​
  2. (B)θ1l1=θ2l2\theta_1 l_1=\theta_2 l_2θ1​l1​=θ2​l2​
  3. (C)θ1l1=θ2l22\theta_1 l_1=\theta_2 l_2^2θ1​l1​=θ2​l22​
  4. (D)θ1l2=θ2l1\theta_1 l_2=\theta_2 l_1θ1​l2​=θ2​l1​

Correct answer: (D)

Step-by-step solution →
Q17·Physics·Electric Field and Coulomb's LawSingle correct
Two infinite identical charged sheets and a charged spherical body of charge density ρ\rhoρ are arranged as shown in figure. Then the correct relation between the electrical fields at A, B, C and D points is:
  1. (A)E⃗A=E⃗B\vec{E}_A=\vec{E}_BEA​=EB​; E⃗C=E⃗D\vec{E}_C=\vec{E}_DEC​=ED​
  2. (B)E⃗A>E⃗B\vec{E}_A>\vec{E}_BEA​>EB​; E⃗C=E⃗D\vec{E}_C=\vec{E}_DEC​=ED​
  3. (C)E⃗C≠E⃗D\vec{E}_C\ne\vec{E}_DEC​=ED​; E⃗A>E⃗B\vec{E}_A>\vec{E}_BEA​>EB​
  4. (D)∣E⃗A∣=∣E⃗B∣|\vec{E}_A|=|\vec{E}_B|∣EA​∣=∣EB​∣; E⃗C>E⃗D\vec{E}_C>\vec{E}_DEC​>ED​

Correct answer: (C)

Step-by-step solution →
Q18·Physics·Electric Field and Coulomb's LawSingle correct
Two small spherical balls of mass 10 g each with charges −2 μC-2\,\mu C−2μC and 2 μC2\,\mu C2μC, are attached to two ends of very light rigid rod of length 20 cm. The arrangement is now placed near an infinite non-conducting charge sheet with uniform charge density of 100 μC/m2100\,\mu C/m^2100μC/m2 such that length of rod makes an angle of 30∘30^\circ30∘ with electric field generated by charge sheet. Net torque acting on the rod is: (Take ε0:8.85×10−12\varepsilon_0:8.85\times10^{-12}ε0​:8.85×10−12 C2^22/Nm2^22)
  1. (A)112 Nm
  2. (B)1.12 Nm
  3. (C)2.24 Nm
  4. (D)11.2 Nm

Correct answer: (B)

Step-by-step solution →
Q19·Physics·Atoms and NucleiSingle correct
Considering the Bohr model of hydrogen like atoms, the ratio of the ratio of the radius 5th5^{\text{th}}5th orbit of the electron in Li2+Li^{2+}Li2+ and He+He^{+}He+ is
  1. (A)32\dfrac{3}{2}23​
  2. (B)49\dfrac{4}{9}94​
  3. (C)94\dfrac{9}{4}49​
  4. (D)23\dfrac{2}{3}32​

Correct answer: (D)

Step-by-step solution →
Q20·Physics·Rotational MotionInteger
A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is x5\sqrt{\dfrac{x}{5}}5x​​ where x=x=x= ______.

Correct answer: 4

Step-by-step solution →
Q21·Physics·Properties of Solids and LiquidsInteger
Two slabs with square cross section of different materials (1, 2) with equal sides (lll) and thickness d1d_1d1​ and d2d_2d2​ such that d2=2d1d_2=2d_1d2​=2d1​ and l>d2l>d_2l>d2​. Considering lower edges of these slabs are fixed to the floor, we apply equal shearing force on the narrow faces. The angle of deformation is θ2=2θ1\theta_2=2\theta_1θ2​=2θ1​. If the shear moduli of material 1 is 4×1094\times10^94×109 N/m2^22, then shear moduli of material 2 is x×109x\times10^9x×109 N/m2^22, where value of xxx is ______.

Correct answer: 1

Step-by-step solution →
Q22·Physics·Geometrical OpticsInteger
Distance between object and its image (magnified by −13-\dfrac{1}{3}−31​) is 30 cm. The focal length of the mirror used is (x4)\left(\dfrac{x}{4}\right)(4x​) cm, where magnitude of value of xxx is ______.

Correct answer: 45

Step-by-step solution →
Q23·Physics·Capacitors and DielectricsInteger
Four capacitor each of capacitance 16 μ\muμF are connected as shown in the figure. The capacitance between points A and B is ______ (in μ\muμF).

Correct answer: 64

Step-by-step solution →
Q24·Physics·Electromagnetic InductionInteger
Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of 12\dfrac{1}{\sqrt{2}}2​1​ Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of 10 cm/s, induced emf between points A and E is ______ mV.

Correct answer: 10

Step-by-step solution →

Chemistry — JEE Main 4 April 2025 Shift 1

Q25·Chemistry·SolutionsSingle correct
XY is the membrane / partition between two chambers 1 and 2 containing sugar solutions of concentration c1c_1c1​ and c2c_2c2​ (c1>c2c_1>c_2c1​>c2​) mol L−1^{-1}−1 (shown in the figure). For the reverse osmosis to take place identify the correct condition. (Here p1p_1p1​ and p2p_2p2​ are pressures applied on chamber 1 and 2) (A) Membrane/Partition; Cellophane, p1>πp_1>\pip1​>π (B) Membrane/Partition; Porous, p2>πp_2>\pip2​>π (C) Membrane/Partition; Parchment paper, p1>πp_1>\pip1​>π (D) Membrane/Partition; Cellophane, p2>πp_2>\pip2​>π. Choose the correct answer from the option given below:
  1. (A)B and D only
  2. (B)A and D only
  3. (C)A and C only
  4. (D)C only

Correct answer: (C)

Step-by-step solution →
Q26·Chemistry·Chemical ThermodynamicsSingle correct
Let us consider a reversible reaction at temperature, T. In this reaction, both ΔH\Delta HΔH and ΔS\Delta SΔS were observed to have positive values. If the equilibrium temperature is TeT_eTe​, then the reaction becomes spontaneous at:
  1. (A)T=TeT=T_eT=Te​
  2. (B)Te>TT_e>TTe​>T
  3. (C)T>TeT>T_eT>Te​
  4. (D)Te=5TT_e=5TTe​=5T

Correct answer: (C)

Step-by-step solution →
Q27·Chemistry·Chemical Bonding and Molecular StructureSingle correct
Which of the following molecules(s) show/s paramagnetic behavior? (A) O2O_2O2​ (B) N2N_2N2​ (C) F2F_2F2​ (D) S2S_2S2​ (E) Cl2Cl_2Cl2​. Choose the correct answer from the options given below:
  1. (A)B only
  2. (B)A & C only
  3. (C)A & E only
  4. (D)A & D only

Correct answer: (D)

Step-by-step solution →
Q28·Chemistry·Aldehydes and KetonesSingle correct
Aldol condensation is a popular and classical method to prepare α,β\alpha,\betaα,β-unsaturated carbonyl compounds. This reaction can be both intermolecular and intramolecular. Predict which one of the following is not a product of intramolecular aldol condensation?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q29·Chemistry·Chemical ThermodynamicsSingle correct
One mole of an ideal gas expands isothermally and reversibly from 10 dm3^33 to 20 dm3^33 at 300 K. ΔU\Delta UΔU, qqq and work done in the process respectively are: (Given: R=8.3R=8.3R=8.3 JK−1^{-1}−1 mol−1^{-1}−1, ln⁡10=2.3\ln 10=2.3ln10=2.3, log⁡2=0.30\log 2=0.30log2=0.30, log⁡3=0.48\log 3=0.48log3=0.48)
  1. (A)0, 21.84 kJ, −1.26-1.26−1.26 kJ
  2. (B)0, −17.18-17.18−17.18 kJ, 1.718 J
  3. (C)0, 21.84 kJ, 21.84 kJ
  4. (D)0, 1.718 kJ, −1.718-1.718−1.718 kJ

Correct answer: (D)

Step-by-step solution →
Q30·Chemistry·Coordination CompoundsSingle correct
Which one of the following complexes will have Δ0=0\Delta_0=0Δ0​=0 and μ=5.96\mu=5.96μ=5.96 B.M.?
  1. (A)[Fe(CN)6]4−[Fe(CN)_6]^{4-}[Fe(CN)6​]4−
  2. (B)[Co(NH3)6]3+[Co(NH_3)_6]^{3+}[Co(NH3​)6​]3+
  3. (C)[FeF6]4−[FeF_6]^{4-}[FeF6​]4−
  4. (D)[Mn(SCN)6]4−[Mn(SCN)_6]^{4-}[Mn(SCN)6​]4−

Correct answer: (D)

Step-by-step solution →
Q31·Chemistry·Chemical KineticsSingle correct
For A2+B2⇌2ABA_2+B_2\rightleftharpoons 2ABA2​+B2​⇌2AB, EaE_aEa​ for forward and backward reaction are 180 and 200 kJ mol−1^{-1}−1 respectively. If catalyst lowers EaE_aEa​ for both reaction by 100 kJ mol−1^{-1}−1. Which of the following statement is correct?
  1. (A)Catalyst does not alter the Gibbs energy change of a reaction.
  2. (B)Catalyst can cause non-spontaneous reactions to occur.
  3. (C)The enthalpy change for the reaction is +20+20+20 kJ mol−1^{-1}−1.
  4. (D)The enthalpy change for the catalysed reaction is different from that of uncatalysed reaction.

Correct answer: (A)

Step-by-step solution →
Q32·Chemistry·Chemical KineticsSingle correct
Rate law for a reaction between A and B is given by R=k[A]n[B]mR=k[A]^n[B]^mR=k[A]n[B]m. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction (r2r1)\left(\dfrac{r_2}{r_1}\right)(r1​r2​​) is:
  1. (A)2(n−m)2^{(n-m)}2(n−m)
  2. (B)(n−m)(n-m)(n−m)
  3. (C)(m+n)(m+n)(m+n)
  4. (D)12m+n\dfrac{1}{2^{m+n}}2m+n1​

Correct answer: (A)

Step-by-step solution →
Q33·Chemistry·Coordination CompoundsSingle correct
Number of stereoisomers possible for the complexes, [CrCl3(py)3][CrCl_3(py)_3][CrCl3​(py)3​] and [CrCl2(ox)2]3−[CrCl_2(ox)_2]^{3-}[CrCl2​(ox)2​]3− are respectively (py = pyridine, ox = oxalate)
  1. (A)3 & 3
  2. (B)2 & 2
  3. (C)2 & 3
  4. (D)1 & 2

Correct answer: (C)

Step-by-step solution →
Q34·Chemistry·AminesSingle correct
The major product (A) formed when nitrobenzene is treated with the following reagents in sequence — (i) Sn, HCl; (ii) Ac2_22​O, Pyridine; (iii) Br2_22​, AcOH; (iv) NaOH(aq) — is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q35·Chemistry·Redox Reactions and ElectrochemistrySingle correct
On charging the lead storage battery, the oxidation state of lead changes from x1x_1x1​ to y1y_1y1​ at the anode and from x2x_2x2​ to y2y_2y2​ at the cathode. The values of x1,y1,x2,y2x_1, y_1, x_2, y_2x1​,y1​,x2​,y2​ are respectively:
  1. (A)+4,+2,0,+2+4, +2, 0, +2+4,+2,0,+2
  2. (B)+2,0,+2,+4+2, 0, +2, +4+2,0,+2,+4
  3. (C)0,+2,+4,+20, +2, +4, +20,+2,+4,+2
  4. (D)+2,0,0,+4+2, 0, 0, +4+2,0,0,+4

Correct answer: (B)

Step-by-step solution →
Q36·Chemistry·p-Block ElementsSingle correct
Given below are two statements: Statement I: Nitrogen forms oxides with +1+1+1 to +5+5+5 oxidation states due to the formation of pπ−pπp\pi-p\pipπ−pπ bond with oxygen. Statement II: Nitrogen does not form halides with +5+5+5 oxidation state due to the absence of d-orbital in it. In the light of given statements, choose the correct answer from the options below:
  1. (A)Statement I is true but Statement II is false
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is false but Statement II is true
  4. (D)Both Statement I and Statement II are true

Correct answer: (D)

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Q37·Chemistry·HydrocarbonsSingle correct
Benzene is treated with oleum to produce compound (X) which when further heated with molten sodium hydroxide followed by acidification produces compound (Y). The compound Y is treated with zinc metal to produce compound (Z). Identify the structure of compound (Z) from the following option.
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q38·Chemistry·BiomoleculesSingle correct
Identify the pair of reactants that upon reaction, with elimination of HCl will give rise to the dipeptide Gly-Ala.
  1. (A)NH2−CH2−COClNH_2-CH_2-COClNH2​−CH2​−COCl and NH2−CH(CH3)−COOHNH_2-CH(CH_3)-COOHNH2​−CH(CH3​)−COOH
  2. (B)NH2−CH2−COClNH_2-CH_2-COClNH2​−CH2​−COCl and N⊕H3−CH(CH3)−COCl\overset{\oplus}{N}H_3-CH(CH_3)-COClN⊕H3​−CH(CH3​)−COCl
  3. (C)NH2−CH2−COOHNH_2-CH_2-COOHNH2​−CH2​−COOH and NH2−CH(CH3)−COClNH_2-CH(CH_3)-COClNH2​−CH(CH3​)−COCl
  4. (D)NH2−CH2−COOHNH_2-CH_2-COOHNH2​−CH2​−COOH and NH2−CH(CH3)−COOHNH_2-CH(CH_3)-COOHNH2​−CH(CH3​)−COOH

Correct answer: (A)

Step-by-step solution →
Q39·Chemistry·p-Block ElementsSingle correct
Given below are the pairs of group 13 elements showing their relation in terms of atomic radius. (B<Al)(B<Al)(B<Al), (Al<Ga)(Al<Ga)(Al<Ga), (Ga<In)(Ga<In)(Ga<In) and (In<Tl)(In<Tl)(In<Tl). Identify the elements present in the incorrect pair and in that pair find out the element (X) that has higher ionic radius (M3+M^{3+}M3+) than the other one. The atomic number of the element (X) is
  1. (A)31
  2. (B)49
  3. (C)13
  4. (D)81

Correct answer: (A)

Step-by-step solution →
Q40·Chemistry·Aldehydes and KetonesSingle correct
An organic compound (X) with molecular formula C3H6OC_3H_6OC3​H6​O is not readily oxidised. On reduction it gives (C3H8O)(C_3H_8O)(C3​H8​O) (Y) which reacts with HBr to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis give 2,3-dimethylbutan-2-ol. Compounds (X), (Y) and (Z) respectively are:
  1. (A)CH3COCH3CH_3COCH_3CH3​COCH3​, CH3CH2CH2OHCH_3CH_2CH_2OHCH3​CH2​CH2​OH, CH3CH(Br)CH3CH_3CH(Br)CH_3CH3​CH(Br)CH3​
  2. (B)CH3COCH3CH_3COCH_3CH3​COCH3​, CH3CH(OH)CH3CH_3CH(OH)CH_3CH3​CH(OH)CH3​, CH3CH(Br)CH3CH_3CH(Br)CH_3CH3​CH(Br)CH3​
  3. (C)CH3CH2CHOCH_3CH_2CHOCH3​CH2​CHO, CH3CH2CH2OHCH_3CH_2CH_2OHCH3​CH2​CH2​OH, CH3CH2CH2BrCH_3CH_2CH_2BrCH3​CH2​CH2​Br
  4. (D)CH3CH2CHOCH_3CH_2CHOCH3​CH2​CHO, CH3CH=CH2CH_3CH=CH_2CH3​CH=CH2​, CH3CH(Br)CH3CH_3CH(Br)CH_3CH3​CH(Br)CH3​

Correct answer: (B)

Step-by-step solution →
Q41·Chemistry·HydrocarbonsSingle correct
Predict the major product when methylcyclohexane is subjected to the following reagents in sequence: (1) Br2_22​/hν\nuν; (2) Alcoholic/KOH, Δ\DeltaΔ; (3) H−-−Br/R−-−O−-−O−-−R, hν\nuν.
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q42·Chemistry·Electronic Effects and StabilitySingle correct
Given below are two statements. Statement I: The dipole moment of C4H3−C3H=C2H−C1H=O\overset{4}{C}H_3-\overset{3}{C}H=\overset{2}{C}H-\overset{1}{C}H=OC4H3​−C3H=C2H−C1H=O is greater than C4H3−C3H2−C2H2−C1H=O\overset{4}{C}H_3-\overset{3}{C}H_2-\overset{2}{C}H_2-\overset{1}{C}H=OC4H3​−C3H2​−C2H2​−C1H=O. Statement II: C1−C2C_1-C_2C1​−C2​ bond length of C4H3−C3H=C2H−C1H=O\overset{4}{C}H_3-\overset{3}{C}H=\overset{2}{C}H-\overset{1}{C}H=OC4H3​−C3H=C2H−C1H=O is greater than C1−C2C_1-C_2C1​−C2​ bond length of C4H3−C3H2−C2H2−C1H=O\overset{4}{C}H_3-\overset{3}{C}H_2-\overset{2}{C}H_2-\overset{1}{C}H=OC4H3​−C3H2​−C2H2​−C1H=O. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Statement I is false but Statement II is true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Both Statement I and Statement II are true

Correct answer: (C)

Step-by-step solution →
Q43·Chemistry·d- and f-Block ElementsSingle correct
Pair of transition metal ions having the same number of unpaired electrons is:
  1. (A)V2+,Co2+V^{2+}, Co^{2+}V2+,Co2+
  2. (B)Ti2+,Co2+Ti^{2+}, Co^{2+}Ti2+,Co2+
  3. (C)Fe3+,Cr2+Fe^{3+}, Cr^{2+}Fe3+,Cr2+
  4. (D)Ti3+,Mn2+Ti^{3+}, Mn^{2+}Ti3+,Mn2+

Correct answer: (A)

Step-by-step solution →
Q44·Chemistry·Atomic StructureSingle correct
Which one of the following about an electron occupying the 1s orbital in a hydrogen atom is incorrect? (Bohr's radius is represented by a0a_0a0​)
  1. (A)The probability density of finding the electron is maximum at the nucleus
  2. (B)The electron can be found at a distance 2a02a_02a0​ from the nucleus
  3. (C)The 1s orbital is spherically symmetrical
  4. (D)The total energy of the electron is maximum when it is at a distance a0a_0a0​ from the nucleus

Correct answer: (D)

Step-by-step solution →
Q45·Chemistry·Purification and Characterisation of Organic CompoundsInteger
In Dumas' method for estimation of nitrogen 1 g of an organic compound gave 150 mL of nitrogen collected at 300 K temperature and 900 mm Hg pressure. The percentage composition of nitrogen in the compound is ______ % (nearest integer). (Aqueous tension at 300 K = 15 mm Hg)

Correct answer: 20

Step-by-step solution →
Q46·Chemistry·Redox Reactions and ElectrochemistryInteger
KMnO4KMnO_4KMnO4​ acts as an oxidising agent in acidic medium. 'X' is the difference between the oxidation states of Mn in reactant and product. 'Y' is the number of 'd' electrons present in the brown red precipitate formed at the end of the action test with neutral ferric chloride. The value of X+YX+YX+Y is ______.

Correct answer: 10

Step-by-step solution →
Q47·Chemistry·Some Basic Concepts in ChemistryInteger
Fortification of food with iron is done using FeSO4⋅7H2OFeSO_4\cdot7H_2OFeSO4​⋅7H2​O. The mass in grams of the FeSO4⋅7H2OFeSO_4\cdot7H_2OFeSO4​⋅7H2​O required to achieve 12 ppm of iron in 150 kg of wheat is ______ (Nearest integer). [Given: Molar mass of Fe, S and O respectively are 56, 32 and 16 g mol−1^{-1}−1]

Correct answer: 9

Step-by-step solution →
Q48·Chemistry·EquilibriumInteger
The pH of a 0.01 M weak acid HX (Ka=4×10−10K_a=4\times10^{-10}Ka​=4×10−10) is found to be 5. Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6. The new concentration of the diluted weak acid is given as x×10−4x\times10^{-4}x×10−4 M. The value of xxx is ______ (nearest integer).

Correct answer: 25

Step-by-step solution →
Q49·Chemistry·BiomoleculesInteger
The total number of hydrogen bonds of a DNA-double Helix strand whose one strand has the following sequence of bases is ______. 5′−G−G−C−A−A−A−T−C−G−G−C−T−A−3′5'-G-G-C-A-A-A-T-C-G-G-C-T-A-3'5′−G−G−C−A−A−A−T−C−G−G−C−T−A−3′

Correct answer: 33

Step-by-step solution →

Mathematics — JEE Main 4 April 2025 Shift 1

Q50·Mathematics·Sets, Relations and FunctionsSingle correct
Let f,g:(1,∞)→Rf, g:(1,\infty)\to\mathbb{R}f,g:(1,∞)→R be defined as f(x)=2x+35x+2f(x)=\dfrac{2x+3}{5x+2}f(x)=5x+22x+3​ and g(x)=2−3x1−xg(x)=\dfrac{2-3x}{1-x}g(x)=1−x2−3x​. If the range of the function f∘g:[2,4]→Rf\circ g:[2,4]\to\mathbb{R}f∘g:[2,4]→R is [α,β][\alpha,\beta][α,β], then 1β−α\dfrac{1}{\beta-\alpha}β−α1​ is equal to
  1. (A)68
  2. (B)29
  3. (C)2
  4. (D)56

Correct answer: (D)

Step-by-step solution →
Q51·Mathematics·Permutations and CombinationsSingle correct
Consider the sets A={(x,y)∈R×R:x2+y2=25}A=\{(x,y)\in\mathbb{R}\times\mathbb{R}:x^2+y^2=25\}A={(x,y)∈R×R:x2+y2=25}, B={(x,y)∈R×R:x2+9y2=144}B=\{(x,y)\in\mathbb{R}\times\mathbb{R}:x^2+9y^2=144\}B={(x,y)∈R×R:x2+9y2=144}, C={(x,y)∈Z×Z:x2+y2≤4}C=\{(x,y)\in\mathbb{Z}\times\mathbb{Z}:x^2+y^2\le4\}C={(x,y)∈Z×Z:x2+y2≤4}, and D=A∩BD=A\cap BD=A∩B. The total number of one-one functions from the set DDD to the set CCC is:
  1. (A)15120
  2. (B)19320
  3. (C)17160
  4. (D)18290

Correct answer: (C)

Step-by-step solution →
Q52·Mathematics·Sequence and SeriesSingle correct
Let A={1,6,11,16,…}A=\{1,6,11,16,\ldots\}A={1,6,11,16,…} and B={9,16,23,30,…}B=\{9,16,23,30,\ldots\}B={9,16,23,30,…} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A∪B)n(A\cup B)n(A∪B) is
  1. (A)3814
  2. (B)4027
  3. (C)3761
  4. (D)4003

Correct answer: (C)

Step-by-step solution →
Q53·Mathematics·Binomial Theorem and Its Simple ApplicationsSingle correct
For an integer n≥2n\ge2n≥2, if the arithmetic mean of all coefficients in the binomial expansion of (x+y)2n−3(x+y)^{2n-3}(x+y)2n−3 is 16, then the distance of the point P(2n−1,n2−4n)P(2n-1, n^2-4n)P(2n−1,n2−4n) from the line x+y=8x+y=8x+y=8 is:
  1. (A)2\sqrt{2}2​
  2. (B)222\sqrt{2}22​
  3. (C)525\sqrt{2}52​
  4. (D)323\sqrt{2}32​

Correct answer: (D)

Step-by-step solution →
Q54·Mathematics·Statistics and ProbabilitySingle correct
The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing at least 3 engineers and at least 1 doctor, is:
  1. (A)129182\dfrac{129}{182}182129​
  2. (B)103182\dfrac{103}{182}182103​
  3. (C)1726\dfrac{17}{26}2617​
  4. (D)1926\dfrac{19}{26}2619​

Correct answer: (A)

Step-by-step solution →
Q55·Mathematics·Three Dimensional GeometrySingle correct
Let the shortest distance between the lines x−33=y−α−1=z−31\dfrac{x-3}{3}=\dfrac{y-\alpha}{-1}=\dfrac{z-3}{1}3x−3​=−1y−α​=1z−3​ and x+3−3=y+72=z−β4\dfrac{x+3}{-3}=\dfrac{y+7}{2}=\dfrac{z-\beta}{4}−3x+3​=2y+7​=4z−β​ be 3303\sqrt{30}330​. Then the positive value of 5α+β5\alpha+\beta5α+β is:
  1. (A)42
  2. (B)46
  3. (C)48
  4. (D)40

Correct answer: (B)

Step-by-step solution →
Q56·Mathematics·Limits and ContinuitySingle correct
If lim⁡x→1(x−1)(6+λcos⁡(x−1))+μsin⁡(1−x)(x−1)3=−1\displaystyle\lim_{x\to1}\dfrac{(x-1)(6+\lambda\cos(x-1))+\mu\sin(1-x)}{(x-1)^3}=-1x→1lim​(x−1)3(x−1)(6+λcos(x−1))+μsin(1−x)​=−1, where λ,μ∈R\lambda,\mu\in\mathbb{R}λ,μ∈R, then λ+μ\lambda+\muλ+μ is equal to
  1. (A)18
  2. (B)20
  3. (C)19
  4. (D)17

Correct answer: (A)

Step-by-step solution →
Q57·Mathematics·Area Under CurvesSingle correct
Let f:[0,∞)→Rf:[0,\infty)\to\mathbb{R}f:[0,∞)→R be differentiable function such that f(x)=1−2x+∫0xex−tf(t) dtf(x)=1-2x+\displaystyle\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 the area of the region bounded by y=f(x)y=f(x)y=f(x) and the coordinate axes is
  1. (A)5\sqrt{5}5​
  2. (B)12\dfrac{1}{2}21​
  3. (C)2\sqrt{2}2​
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q58·Mathematics·Three Dimensional GeometrySingle correct
Let AAA and BBB be two distinct points on the line L:x−63=y−72=z−7−2L:\dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{z-7}{-2}L:3x−6​=2y−7​=−2z−7​. Both AAA and BBB are at a distance 2172\sqrt{17}217​ from the foot of perpendicular drawn from the point (1,2,3)(1,2,3)(1,2,3) on the line LLL. If OOO is the origin, then OA→⋅OB→\overrightarrow{OA}\cdot\overrightarrow{OB}OA⋅OB is equal to:
  1. (A)49
  2. (B)47
  3. (C)21
  4. (D)62

Correct answer: (B)

Step-by-step solution →
Q59·Mathematics·Sets, Relations and FunctionsSingle correct
Let f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R be a continuous function satisfying f(0)=1f(0)=1f(0)=1 and f(2x)−f(x)=xf(2x)-f(x)=xf(2x)−f(x)=x for all x∈Rx\in\mathbb{R}x∈R. If lim⁡n→∞{f(x)−f(x2n)}=G(x)\displaystyle\lim_{n\to\infty}\left\{f(x)-f\left(\dfrac{x}{2^n}\right)\right\}=G(x)n→∞lim​{f(x)−f(2nx​)}=G(x), then ∑r=110G(r2)\displaystyle\sum_{r=1}^{10}G(r^2)r=1∑10​G(r2) is equal to:
  1. (A)540
  2. (B)385
  3. (C)420
  4. (D)215

Correct answer: (B)

Step-by-step solution →
Q60·Mathematics·Sequence and SeriesSingle correct
1+3+52+7+92+…1+3+5^2+7+9^2+\ldots1+3+52+7+92+… upto 40 terms is equal to
  1. (A)43890
  2. (B)41880
  3. (C)33980
  4. (D)40870

Correct answer: (B)

Step-by-step solution →
Q61·Mathematics·Binomial Theorem and Its Simple ApplicationsSingle correct
In the expansion of (23+133)n\left(\sqrt[3]{2}+\dfrac{1}{\sqrt[3]{3}}\right)^n(32​+33​1​)n, n∈Nn\in\mathbb{N}n∈N, if the ratio of 15th15^{\text{th}}15th term from the beginning to the 15th15^{\text{th}}15th term from the end is 16\dfrac{1}{6}61​, then the value of nC3{}^nC_3nC3​ is:
  1. (A)4060
  2. (B)1040
  3. (C)2300
  4. (D)4960

Correct answer: (C)

Step-by-step solution →
Q62·Mathematics·Inverse Trigonometric FunctionsSingle correct
Considering the principal values of the inverse trigonometric functions, sin⁡−1(32x+121−x2)\sin^{-1}\left(\dfrac{\sqrt{3}}{2}x+\dfrac{1}{2}\sqrt{1-x^2}\right)sin−1(23​​x+21​1−x2​), −12<x<12-\dfrac{1}{2}<x<\dfrac{1}{\sqrt{2}}−21​<x<2​1​, is equal to:
  1. (A)π4+sin⁡−1x\dfrac{\pi}{4}+\sin^{-1}x4π​+sin−1x
  2. (B)π6+sin⁡−1x\dfrac{\pi}{6}+\sin^{-1}x6π​+sin−1x
  3. (C)−5π6−sin⁡−1x-\dfrac{5\pi}{6}-\sin^{-1}x−65π​−sin−1x
  4. (D)5π6−sin⁡−1x\dfrac{5\pi}{6}-\sin^{-1}x65π​−sin−1x

Correct answer: (B)

Step-by-step solution →
Q63·Mathematics·Vector AlgebraSingle correct
Consider two vectors u⃗=3i^−j^\vec{u}=3\hat{i}-\hat{j}u=3i^−j^​ and v⃗=2i^+j^−λk^\vec{v}=2\hat{i}+\hat{j}-\lambda\hat{k}v=2i^+j^​−λk^, λ>0\lambda>0λ>0. The angle between them is given by cos⁡−1(527)\cos^{-1}\left(\dfrac{\sqrt{5}}{2\sqrt{7}}\right)cos−1(27​5​​). Let v⃗=v1⃗+v2⃗\vec{v}=\vec{v_1}+\vec{v_2}v=v1​​+v2​​, where v1⃗\vec{v_1}v1​​ is parallel to u⃗\vec{u}u and v2⃗\vec{v_2}v2​​ is perpendicular to u⃗\vec{u}u. Then the value of ∣v1⃗∣2+∣v2⃗∣2|\vec{v_1}|^2+|\vec{v_2}|^2∣v1​​∣2+∣v2​​∣2 is equal to:
  1. (A)232\dfrac{23}{2}223​
  2. (B)14
  3. (C)252\dfrac{25}{2}225​
  4. (D)10

Correct answer: (B)

Step-by-step solution →
Q64·Mathematics·CirclesSingle correct
Let the three sides of a triangle are on the lines 4x−7y+10=04x-7y+10=04x−7y+10=0, x+y=5x+y=5x+y=5 and 7x+4y=157x+4y=157x+4y=15. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines x=0x=0x=0, y=0y=0y=0 and x+y=1x+y=1x+y=1 is
  1. (A)5
  2. (B)5\sqrt{5}5​
  3. (C)20\sqrt{20}20​
  4. (D)20

Correct answer: (B)

Step-by-step solution →
Q65·Mathematics·Definite IntegrationSingle correct
The value of ∫−11(1+∣x∣−x)ex+(∣x∣−x)e−xex+e−x dx\displaystyle\int_{-1}^{1}\dfrac{\left(1+\sqrt{|x|-x}\right)e^x+\left(\sqrt{|x|-x}\right)e^{-x}}{e^x+e^{-x}}\,dx∫−11​ex+e−x(1+∣x∣−x​)ex+(∣x∣−x​)e−x​dx is equal to:
  1. (A)3−2233-\dfrac{2\sqrt{2}}{3}3−322​​
  2. (B)2+2232+\dfrac{2\sqrt{2}}{3}2+322​​
  3. (C)1−2231-\dfrac{2\sqrt{2}}{3}1−322​​
  4. (D)1+2231+\dfrac{2\sqrt{2}}{3}1+322​​

Correct answer: (D)

Step-by-step solution →
Q66·Mathematics·EllipseSingle correct
The length of the latus-rectum of the ellipse, whose foci are (2,5)(2,5)(2,5) and (2,−3)(2,-3)(2,−3) and eccentricity is 45\dfrac{4}{5}54​, is
  1. (A)65\dfrac{6}{5}56​
  2. (B)503\dfrac{50}{3}350​
  3. (C)103\dfrac{10}{3}310​
  4. (D)185\dfrac{18}{5}518​

Correct answer: (D)

Step-by-step solution →
Q67·Mathematics·Quadratic EquationsSingle correct
Consider the equation x2+4x−n=0x^2+4x-n=0x2+4x−n=0, where n∈[20,100]n\in[20,100]n∈[20,100] is a natural number. Then the number of all distinct values of nnn, for which the given equation has integral roots, is equal to
  1. (A)7
  2. (B)8
  3. (C)6
  4. (D)5

Correct answer: (C)

Step-by-step solution →
Q68·Mathematics·Statistics and ProbabilitySingle correct
A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let XXX denote the number of defective pens. Then the variance of XXX is
  1. (A)1115\dfrac{11}{15}1511​
  2. (B)2875\dfrac{28}{75}7528​
  3. (C)215\dfrac{2}{15}152​
  4. (D)35\dfrac{3}{5}53​

Correct answer: (B)

Step-by-step solution →
Q69·Mathematics·Trigonometric FunctionsSingle correct
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 →
Q70·Mathematics·Area Under CurvesInteger
If the area of the region {(x,y):∣x−5∣≤y≤4x}\{(x,y):|x-5|\le y\le4\sqrt{x}\}{(x,y):∣x−5∣≤y≤4x​} is AAA, then 3A3A3A is equal to ______.

Correct answer: 368

Step-by-step solution →
Q71·Mathematics·Matrices and DeterminantsInteger
Let A=[cos⁡θ0−sin⁡θ010sin⁡θ0cos⁡θ]A=\begin{bmatrix}\cos\theta & 0 & -\sin\theta\\0 & 1 & 0\\\sin\theta & 0 & \cos\theta\end{bmatrix}A=​cosθ0sinθ​010​−sinθ0cosθ​​. If for some θ∈(0,π)\theta\in(0,\pi)θ∈(0,π), A2=ATA^2=A^TA2=AT, then the sum of the diagonal elements of the matrix (A+I)3+(A−I)3−6A(A+I)^3+(A-I)^3-6A(A+I)3+(A−I)3−6A is equal to ______.

Correct answer: 6

Step-by-step solution →
Q72·Mathematics·Complex NumbersInteger
Let A={z∈C:∣z−2−i∣=3}A=\{z\in\mathbb{C}:|z-2-i|=3\}A={z∈C:∣z−2−i∣=3}, B={z∈C:Re(z−iz)=2}B=\{z\in\mathbb{C}:\mathrm{Re}(z-iz)=2\}B={z∈C:Re(z−iz)=2} and S=A∩BS=A\cap BS=A∩B. Then ∑z∈S∣z∣2\displaystyle\sum_{z\in S}|z|^2z∈S∑​∣z∣2 is equal to ______.

Correct answer: 22

Step-by-step solution →
Q73·Mathematics·CirclesInteger
Let CCC be the circle x2+(y−1)2=2x^2+(y-1)^2=2x2+(y−1)2=2, E1E_1E1​ and E2E_2E2​ be two ellipses whose centres lie at the origin and major axes lie on xxx-axis and yyy-axis respectively. Let the straight line x+y=3x+y=3x+y=3 touch the curves CCC, E1E_1E1​ and E2E_2E2​ at P(x1,y1)P(x_1,y_1)P(x1​,y1​), Q(x2,y2)Q(x_2,y_2)Q(x2​,y2​) and R(x3,y3)R(x_3,y_3)R(x3​,y3​) respectively. Given that PPP is the mid-point of the line segment QRQRQR and PQ=223PQ=\dfrac{2\sqrt{2}}{3}PQ=322​​, the value of 9(x1y1+x2y2+x3y3)9(x_1y_1+x_2y_2+x_3y_3)9(x1​y1​+x2​y2​+x3​y3​) is equal to ______.

Correct answer: 46

Step-by-step solution →
Q74·Mathematics·Limits and ContinuityInteger
Let mmm and nnn be the number of points at which the function f(x)=max⁡{x,x3,x5,…,x21}f(x)=\max\{x,x^3,x^5,\ldots,x^{21}\}f(x)=max{x,x3,x5,…,x21}, x∈Rx\in\mathbb{R}x∈R, is not differentiable and not continuous, respectively. Then m+nm+nm+n is equal to ______.

Correct answer: 3

Step-by-step solution →

Chapters tested in this paper

Open any chapter to practise every past-year question on that topic across all papers, and see how often it has actually appeared.

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • Chemical Bonding and Molecular Structure 151/186
  • Limits and Continuity 149/186
  • d- and f-Block Elements 126/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Ellipse 103/186
  • Inverse Trigonometric Functions 93/186
  • Electronic Effects and Stability 74/186
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