Download JEE Main 2024 Question Paper (27 Jan - Shift 2) Mathematics
Important Instructions
- Section-A : Attempt all questions.
- Section-B : Do any 5 questions out of 10 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 (1 – 10) contains 10 Numerical based questions. The answer to each question is rounded off to the nearest integer value. 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
- Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $\tan^{-1}(x)$+$\tan^{-1}(2x)$ is :
- more than 2
- 1
- 2
- 0
- Consider the function $f :(0,2) \to R$
defined by $f (x)$=$\frac{x}{2}$+$\frac{2}{x}$ and the function g(x) defined by
$g(x)$=$\left\{\begin{array}{cc}min{f(t)} &, 0 < t \leq x \text{ and } 0 < x \leq 1\\ \frac{3}{2}+x &, 1 < x < 2 \end{array} \right.$. Then- $g$ is continuous but not differentiable at $x$ = 1
- $g$ is not continuous for all $x \in (0,2)$
- $g$ is neither continuous nor differentiable at $x$ = 1
- $g$ is continuous and differentiable for all $x \in (0,2)$
- Let the image of the point (1, 0, 7) in the line $\frac{x}{1}$=$\frac{y-1}{2}$=$\frac{z-2}{3}$be the point $(\alpha, \beta, \gamma)$. Then which one of the following points lies on the line passing through $(\alpha, \beta, \gamma)$ and making angles with $y-$axis and $z-$axis respectively and an acute angle with $x-$axis ?
- $(1, -2, 1+\sqrt{2})$
- $(1, 2, 1-\sqrt{2})$
- $(3, 4, 3-2\sqrt{2})$
- $(3, -4, 3+2\sqrt{2})$
- Let $R$ be the interior region between the lines $3x-y$+1=0 and $x+2y-5$=0 containing the origin. The set of all values of $a$, for which the points $(a^2, a + 1)$ lie in $R$, is :
- $(-3, -1) \cup \left(-\frac{1}{3}, 1\right)$
- $(-3, 0) \cup \left(\frac{1}{3}, 1\right)$
- $(-3, 0) \cup \left(\frac{2}{3}, 1\right)$
- $(-3, -1) \cup \left(\frac{1}{3}, 1\right)$
- The $20^{th}$ term from the end of the progression 20, $19\frac{1}{4}$, $18\frac{1}{2}$, $17\frac{3}{4}$, ... ,$-129\frac{1}{4}$ is:-
- -118
- -110
- -115
- -100
- Let $f:R-\left\{\frac{-1}{2}\right\}\to R$ and $g:R-\left\{\frac{-5}{2}\right\}\to R$ be defined as $f(x)$=$\frac{2x+3}{2x+1}$ and $g(x)$=$\frac{|x|+1}{2x+5}$. Then the domain of the function $fog$ is :
- $R-\left\{-\frac{5}{2}\right\}$
- $R$
- $R-\left\{-\frac{7}{4}\right\}$
- $R-\left\{-\frac{5}{2}, -\frac{7}{4}\right\}$
- For $0 < a < 1$, the value of the integral $\int \limits_{0}^{\pi}\frac{dx}{1-2a\cos x+a^2}$ is
- $\frac{\pi^2}{\pi+a^2}$
- $\frac{\pi^2}{\pi-a^2}$
- $\frac{\pi}{1-a^2}$
- $\frac{\pi}{1+a^2}$
- Let $g(x)$=$3f \left(\frac{x}{3}\right)$+$f(3-x)$ and $f"(x) > 0$ for all $x \in (0,3)$. If $g$ is decreasing in $(0, \alpha)$ and increasing in $(\alpha, 3)$, then $8\alpha$ is
- 24
- 0
- 18
- 20
- If $\lim \limits_{x \to 0}\frac{3+\alpha \sin x+\beta \cos x+\log_e(1-x)}{3\tan^2x}$=$\frac{1}{3}$, then $2\alpha-\beta$ is equal to:
- 2
- 7
- 5
- 1
- If $\alpha$, $\beta$ are the roots of the equation, $x^2-x-1$=0 and $S_n$=$2023\alpha^n+2024\beta^n$, then
- $2S_{12}$=$S_{11}$+$S_{10}$
- $S_{12}$=$S_{11}$+$S_{10}$
- $2S_{11}$=$S_{12}$+$S_{10}$
- $S_{11}$=$S_{10}$+$S_{12}$
- Let $A$ and $B$ be two finite sets with $m$ and $n$ elements respectively. The total number of subsets of the set $A$ is 56 more than the total number of subsets of $B$. Then the distance of the point $P(m, n)$ from the point $Q(–2, –3)$ is
- 10
- 6
- 4
- 8
- The values of $\alpha$, for which
$\begin{equation*}\begin{vmatrix} 1 & \frac{3}{2} & \alpha + \frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2\alpha+3 & 3\alpha+1 & 0\end{vmatrix}\end{equation*}$=0, lie in the interval- (-2, 1)
- (-3, 0)
- $\left(-\frac{3}{2}, \frac{3}{2}\right)$
- (0, 3)
- An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability, that the first draw gives all white balls and the second draw gives all black balls, is :
- $\frac{5}{256}$
- $\frac{5}{715}$
- $\frac{3}{715}$
- $\frac{3}{256}$
- The integral $\int \frac{(x^8-x^2)dx}{(x^{12}+3x^6+1)\tan^{-1}\left(x^3+\frac{1}{x^3}\right)}$ equal to:
- $\log_e\left(\left|\tan^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)^{1/3}+C$
- $\log_e\left(\left|\tan^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)^{1/2}+C$
- $\log_e\left(\left|\tan^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)+C$
- $\log_e\left(\left|\tan^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)^{3}+C$
- If $2 \tan^2 \theta$$-5 \sec \theta$=1 has exactly 7 solutions in the interval $\left[0, \frac{n\pi}{2}\right]$,for the least value of $n \in N$,then $\sum \limits_{k=1}^{n}\frac{k}{2^{k}}$ is equal to :
- $\frac{1}{2^{15}}(2^{14}-14)$
- $\frac{1}{2^{14}}(2^{15}-14)$
- $1-\frac{15}{2^{13}}$
- $\frac{1}{2^{13}}(2^{14}-15)$
- The position vectors of the vertices $A$, $B$ and $C$ of a triangle are $2\hat{i}$$-3\hat{j}$+$3\hat{k}$, $2\hat{i}$+$2\hat{j}$+$3\hat{k}$ and $-\hat{i}$+$\hat{j}$+$3\hat{k}$ respectively. Let $l$ denotes the length of the angle bisector $AD$ of $\angle{BAC}$ where $D$ is on the line segment $BC$, then $2l^2$ equals :
- 49
- 42
- 50
- 45
- If $y = y(x)$ is the solution curve of the differential equation $(x^2-4)dy-$$(y^2-3y)dx$=0, and the slope of the curve is never zero, then the value of $y(10)$ equals :
- $\frac{3}{1+(8)^{1/4}}$
- $\frac{3}{1+2\sqrt{2}}$
- $\frac{3}{1-2\sqrt{2}}$
- $\frac{3}{1-(8)^{1/4}}$
- Let $e_1$ be the eccentricity of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}$=1 and $e_2$ be the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}$=1, $a > b$, which passes through the foci of the hyperbola. If $e_1e_2$ = 1, then the length of the chord of the ellipse parallel to the x-axis and passing through (0, 2) is :
- $4\sqrt{5}$
- $\frac{8\sqrt{5}}{3}$
- $\frac{10\sqrt{5}}{3}$
- $3\sqrt{5}$
- Let $\alpha$=$\frac{(4!)!}{(4!)^{3!}}$ and $\beta$=$\frac{(5!)!}{(5!)^{4!}}$
- $\alpha \in N and $\beta \notin N$
- $\alpha \notin N and $\beta \in N$
- $\alpha \in N and $\beta \in N$
- $\alpha \notin N and $\beta \notin N$
- Let the position vectors of the vertices $A$, $B$ and $C$ of a triangle be $2\hat{i}$+$2\hat{j}$+$\hat{k}$, $\hat{i}$+$2\hat{j}$+$2\hat{k}$ and $2\hat{i}$+$\hat{j}$+$2\hat{k}$ respectively. Let $l_1$, $l_2$ and $l_3$ be the lengths of perpendiculars drawn from the ortho center of the triangle on the sides $AB$, $BC$ and $CA$ respectively, then $l_1^2$+$l_2^2$+$l_3^2$ equals :
- $\frac{1}{5}$
- $\frac{1}{2}$
- $\frac{1}{4}$
- $\frac{1}{3}$
SECTION - B
(Numerical Answer Type)
This section contains 10 Numerical based questions. The answer to each question is rounded off to the nearest integer value.
- The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12. If $µ$ and $\sigma^2$ denote the mean and variance of the correct observations respectively, then $15(\mu+\mu^2+\sigma^2)$ is equal to ……………………..
- If the area of the region
{$(x,y):0 \leq y \leq min{2x,6x - x^2 }$} is $A$, then $12 A$ is equal to……………… - Let $A$ be a 2 × 2 real matrix and $I$ be the identity matrix of order 2. If the roots of the equation $| A - xI|=0$ be –1 and 3, then the sum of the diagonal elements of the matrix $A^2$ is……………
- If the sum of squares of all real values of $\alpha$, for which the lines $2x – y$ + 3 = 0, $6x + 3y$ + 1 = 0 and $\alpha x + 2y$ –2 = 0 do not form a triangle is $p$, then the greatest integer less than or equal to $p$ is ………
- The coefficient of $x^{2012}$ in the expansion of $(1-x)^{2008}(1+x+x^2)^{2007}$ is equal to
- If the solution curve, of the differential equation $\frac{dy}{dx}$=$\frac{x+y-2}{x-y}$ passing through the point (2, 1) is $\tan^{-1}\left(\frac{y-1}{x-1}\right)$$-\frac{1}{\beta}\log_e\left(\alpha+\left(\frac{y-1}{x-1}\right)^2\right)$=$\log_e|x-1|$, then $5\beta+\alpha$ is equal to:
- Let $f(x)$=$\int \limits_{0}^{x}g(t)\log_e\left(\frac{1-t}{1+t}\right)dt$, where $g$ is a continuous odd function. If $\int \limits_{-\pi/2}^{\pi/2}\left(f(x)+\frac{x^2 \cos x}{1+e^x}\right)dx$=$\left(\frac{\pi}{\alpha}\right)^2$$-\alpha$, then $\alpha$ is equal to...............
- Consider a circle $(x - \alpha)^2$+$(y-\beta)^2$=50, where $\alpha, \beta > 0$. If the circle touches the line $y + x$ = 0 at the point $P$, whose distance from the origin is $4 \sqrt{2}$, then $(\alpha+\beta)^2$ is equal to…………….
- The lines $\frac{x-2}{2}$=$\frac{y}{-2}$=$\frac{z-7}{16}$ and $\frac{x+3}{4}$=$\frac{y+2}{3}$=$\frac{z+2}{1}$ intersect at the point $P$. If the distance of $P$ from the line $\frac{x+1}{2}$=$\frac{y-1}{3}$=$\frac{z-1}{1}$ is $l$, then $14l^2$ is equal to……………..
- Let the complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on the circles $|z-z_0|^2$=4 and $|z-z_0|^2$=16 respectively, where $z_0=1+i$. Then, the value of $100|\alpha|2$ is……………
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