Download JEE Main 2025 Question Paper (08 Apr - Shift 2) Mathematics
Important Instructions
- Section-A : Attempt all questions.
- Section-B : Attempt all questions.
- Section-A (01 – 20) contains 20 multiple choice questions which have only one correct answer. Each question carries +4 marks for correct answer and –1 mark for wrong answer.
- Section-B (01 – 05) contains 5 Numerical value based questions. The answer to each question is rounded off to the nearest integer. Each question carries +4 marks for correct answer and –1 mark for wrong answer.
SECTION - A
(One Option Correct Type)
This section contains 20 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE option is correct
- Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2}$=$\frac{y-2}{3}$=$\frac{z-3}{4}$ and $\frac{x-\lambda}{3}$=$\frac{y-4}{4}$=$\frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be $\lambda_1$ and $\lambda_2$. Then the radius of the circle passing through the points (0, 0), $(\lambda_1, \lambda_2)$ and $(\lambda_2, \lambda_1)$ is
- $\frac{5\sqrt{2}}{3}$
- 4
- $\frac{\sqrt{2}}{3}$
- 3
- Let $\alpha$ be a solution of $x^2$
+ $x$ + 1 = 0, and for some $a$ and $b$ in $R$
$[4 a b]$$\begin{equation*} \begin{bmatrix} 1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8 \end{bmatrix} \end{equation*}$=[0 0 0]. If $\frac{4}{\alpha^4}$+$\frac{m}{\alpha^a}$+$\frac{n}{\alpha^b}$=3, then $m+n$ is equal to .......- 3
- 11
- 7
- 8
- Let the function $f(x)$=$\frac{x}{3}$+$\frac{3}{x}$+3, $x \neq 0$ be strictly increasing in $(–\infty, \alpha_1) \cup (\alpha_2, \infty)$ and strictly decreasing in $(\alpha_3, \alpha_4) \cup (\alpha_4, \alpha_5)$. Then $\sum \limits_{i=1}^{5}\alpha_i^2$ is equal to:
- 48
- 28
- 40
- 36
- If $A$ and $B$ are two events such that $P(A)$ = 0.7, $P(B)$ = 0.4 and $P(A \cap \bar{B})$=0.5, where $\bar{B}$ denotes
the complement of $B$, then $P(B| A \cup \bar{B})$is equal:
- $\frac{1}{4}$
- $\frac{1}{2}$
- $\frac{1}{6}$
- $\frac{1}{3}$
- If $\frac{1}{1^4}$+$\frac{1}{2^4}$+$\frac{1}{3^4}$+...$\infty$=$\frac{\pi^4}{90}$,
$\frac{1}{1^4}$+$\frac{1}{3^4}$+$\frac{1}{5^4}$+...$\infty$=$\alpha$,
$\frac{1}{2^4}$+$\frac{1}{4^4}$+$\frac{1}{6^4}$+...$\infty$=$\beta$,
then $\frac{\alpha}{\beta}$ is equal to- 23
- 18
- 15
- 14
- The sum of the squares of the roots of $|x + 2|^2$+ $|x – 2|$ – 2 = 0 and the squares of the roots of $x^2$ – $2|x – 3|$ – 5 = 0, is
- 26
- 36
- 30
- 24
- Let $a$ be the length of a side of a square $OABC$ with $O$ being the origin. Its side $OA$ makes an acute angle $\alpha$ with the positive $x-$axis and the equations of its diagonals are$(\sqrt{3}+1)x$ + $(\sqrt{3}-1)y$= 0 and $(\sqrt{3}-1)x$$-(\sqrt{3}+1)y$+$8\sqrt{3}$=0 Then $a^2$ is equal to
- 48
- 32
- 16
- 24
- Let $f(x)$ be a positive function and $I_1$=$\int \limits_{\frac{1}{2}}^1 2xf(2x(1-2x))dx$ and $I_2$=$\int \limits_{-1}^2f(x(1-x))dx$. Then the value of $\frac{I_2}{I_1}$ is equal to......
- 9
- 6
- 12
- 4
- Let $\vec{a}$=$\hat{i}$+$2\hat{j}$+$\hat{k}$ and $\vec{b}$=$2\hat{i}$+$\hat{j}$$-\hat{k}$. Let $\hat{c}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$. Then such a vector $\hat{c}$ is:
- $\frac{1}{\sqrt{5}}(\hat{j}-2\hat{k})$
- $\frac{1}{\sqrt{3}}(-\hat{i}+\hat{j}-\hat{k})$
- $\frac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})$
- $\frac{1}{\sqrt{2}}(-\hat{i}+\hat{k})$
- Let the ellipse $3x^2$+ $py^2$= 4 pass through the centre $C$ of the circle $x^2$ + $y^2$– $2x$ – $4y$ – 11 = 0 of radius $r$. Let $f_1$, $f_2$ be the focal distances of the point
$C$ on the ellipse. Then $6f_1f_2 – r$ is equal to
- 74
- 68
- 70
- 78
- The integral $\int \limits_{-1}^{\frac{3}{2}}(|\pi^2x \sin (\pi x)|)dx$ is equal to:
- $3+2\pi$
- $4+\pi$
- $1+3\pi$
- $2+3\pi$
- A line passing through the point $P(a, \theta)$ makes an acute angle $\alpha$ with the positive $x-$axis. Let this line be rotated about the point $P$ through an angle $\frac{\alpha}{2}$ in the clock-wise direction. If in the new position, the slope of the line is $2 – \sqrt{3}$ and its distance from the origin is $\frac{1}{\sqrt{2}}$, then the value of $3a^2\tan^2 \alpha – 2\sqrt{3}$ is
- 4
- 6
- 5
- 8
- There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is
- 230
- 220
- 200
- 210
- Let $A$=$\left\{ \theta \in [0, 2\pi]:1+10Re\left(\frac{2\cos \theta+i\sin \theta}{\cos \theta - 3i\sin \theta}\right)=0\right\}$. Then $\sum \limits_{\theta \in A}\theta^2$ is equal to:
- $\frac{21}{4}\pi^2$
- $8\pi^2$
- $\frac{27}{4}\pi^2$
- $6\pi^2$
- Let $A$ = {0, 1, 2, 3, 4, 5}. Let $R$ be a relation on $A$ defined by $(x, y) \in R$ if and only if max ${x, y} \in {3, 4}$. Then among the statements
$(S1)$ : The number of elements in $R$ is 18, and
$(S2)$ : The relation $R$ is symmetric but neither reflexive nor transitive- both are true
- both are false
- only (S2) is true
- only (S1) is true
- The number of integral terms in the expansion of
$\left(5^{\frac{1}{2}}+7^{\frac{1}{8}}\right)^{1016}$ is- 127
- 130
- 129
- 128
- Let $f(x)$ = $x – 1$ and $g(x)$ = $e^x$ for $x \in R$. If $\frac{dy}{dx}$=$\left(e^{-2\sqrt{x}}g(f(f(x)))-\frac{y}{\sqrt{x}}\right)$, $y(0)$=0, then $y(1)$ is:
- $\frac{1-e^2}{e^4}$
- $\frac{2e-1}{e^3}$
- $\frac{e-1}{e^4}$
- $\frac{1-e^3}{e^4}$
- The value of $\cot^{-1}\left(\frac{\sqrt{1+\tan^2(2)}-1}{\tan(2)}\right)$$-\cot^{-1}\left(\frac{\sqrt{1+\tan^2\left(\frac{1}{2}\right)}+1}{\tan\left(\frac{1}{2}\right)}\right)$ is equal to
- $\pi-\frac{5}{4}$
- $\pi-\frac{3}{2}$
- $\pi+\frac{3}{2}$
- $\pi+\frac{5}{2}$
- Let $A$=$\begin{equation*} \begin{bmatrix} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{bmatrix} \end{equation*}$. If $det (adj (adj(3A)))$ = $2^m• 3^n$, $m, n \in N$ , then $m + n$ is equal to
- 22
- 24
- 26
- 20
- Given below are two statements :
Statement I : $\lim \limits_{x \to 0}\left(\frac{\tan^{-1}x+\log_e\sqrt{\frac{1+x}{1-x}}-2x}{x^5}\right)$=$\frac{2}{5}$
Statement II : $\lim \limits_{x \to 1}\left(x^{\frac{2}{1-x}}\right)$=$\frac{1}{e^2}$
In the light of the above statements, choose the correct answer from the options given below :- Statement I is false but Statement II is true
- Statement I is true but Statement II is false
- Both Statement I and Statement II are false
- Both Statement I and Statement II are true
SECTION - B
(Numerical Value Type Questions)
This section contains 05 Numerical based questions. The answer to each question is rounded off to the nearest integer.
- Let the area of the bounded region {$(x, y) :0 \leq 9x \leq y^2, y \geq 3x – 6$} be $A$. Then $6A$ is equal to ______
- Let the domain of the function
$f(x)$=$\cos^{-1}\left(\frac{4x+5}{3x-7}\right)$ be $[\alpha, \beta]$ and the domain of $g(x)$=$\log_2(2-6\log_{27}(2x+5))$ be $(\gamma, \delta)$. Then $|7(\alpha+\beta)+4(\gamma+\delta)|$ is equal to ....... - Let the area of the triangle formed by the lines $x+2$=$y-1$=$z$, $\frac{x-3}{5}$=$\frac{y}{-1}$=$\frac{z-1}{1}$ and $\frac{x}{-3}$=$\frac{y-3}{3}$=$\frac{z-2}{1}$ be $A$. Then $A^2$ is equal to........
- The product of the last two digits of $(1919)^{1919}$ is ______
- Let $r$ be the radius of the circle, which touches $x-$axis at point $(a, 0)$, $a < 0$ and the parabola $y^2$= $9x$at the point (4, 6). Then $r$ is equal to ______
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