Download JEE Main 2023 Question Paper (12 Apr - Shift 1) 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
- The number of five digit numbers, greater than 40000 and divisible by 5, which can be formed using the digits 0, 1, 3, 5, 7 and 9 without repetition, is equal to
- 120
- 132
- 72
- 96
- Let $\alpha$, $\beta$ be the roots of the quadratic equation $x^2$+$\sqrt{6}x$+3=0. Then $\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}$ is equal to
- 729
- 72
- 81
- 9
- Let <$a_n$> be a sequence such that $a_1$+$a_2$+........+$a_n$=$\frac{n^2+3n}{(n+1)(b+2)}$. If $28\sum \limits_{k=1}^{10}\frac{1}{a_k}$=$p_1$$p_2$$p_3$........$p_m$, where $p_1$, $p_2$,.......,$p_m$ are the first $m$ prime numbers, then $m$ is equal to.........
- 7
- 6
- 5
- 8
- Let the lines $l_1$:$\frac{x+5}{3}$=$\frac{y+4}{1}$=$\frac{z-\alpha}{-2}$ and $l_2$:$3x$+$2y$+$z-2$=0=$x-3$+$2z-13$ be coplanar If the point $P(a, b, c)$ on $l_1$ is nearest to the point $Q(-4, -3, 2)$, then $|a| + |b|+|c|$ is equal to
- 12
- 14
- 10
- 8
- Let $P\left(\frac{2\sqrt{3}}{\sqrt{7}},\frac{6}{\sqrt{7}}\right)$, $Q$, $R$ and $S$ be four points on the ellipse $9x^2$+$4y^2$+36. Let $PQ$ and $RS$ be mutually perpendicular and pass through the origin. If $\frac{1}{(PQ)^2}$+$\frac{1}{(RS)^2}$=$\frac{p}{q}$, where $p$ and $q$ are coprime, then $p + q$ is equal to
- 143
- 137
- 157
- 147
- Let $a$, $b$, $c$ be three distinct real numbers, none equal to one. If the vectors $a\hat{i}$+$\hat{j}$+$\hat{k}$, $\hat{i}$+$b\hat{j}$+$\hat{k}$ and $\hat{i}$+$\hat{j}$+$c\hat{k}$ are coplanar, then $\frac{1}{1-a}$+$\frac{1}{1-b}$+$\frac{1}{1-c}$ is equal to
- 1
- $-1$
- $-2$
- 2
- If the local maximum value of the function $f(x)$=$\left(\frac{\sqrt{3e}}{2\sin x}\right)^{\sin^2x}$, $x \in \left(0, \frac{\pi}{2}\right)$, is $\frac{k}{e}$, then $\left(\frac{k}{e}\right)^{8}$+$\frac{k^8}{e^5}$+$k^8$ is equal to
- $e^5$+$e^6$+$e^{11}$
- $e^3$+$e^5$+$e^{11}$
- $e^3$+$e^6$+$e^{11}$
- $e^3$+$e^6$+$e^{10}$
- Let $D$ be the domain of the function $f(x)$=$\sin^{-1}\left(\log_{3x}\left(\frac{6+2\log_3x}{-5x}\right)\right)$. If the range of the function $g:D\to R$ defined by $g(x)$=$x-[x]$, ($[x]$ is the greatest integer function), is $(\alpha, \beta)$, then $\alpha^2+\frac{5}{\beta}$ is equal to
- 46
- 135
- 136
- 45
- Let $y = y(x)$, $y$ > 0, be a solution curve of the differential equation (1 + $x^2$) $dy$ = $y (x – y) dx$. If $y (0)$ = 1 and $y(2 \sqrt{2})$, = $\beta$,then
- $e^{3\beta^{-1}}$=$e(3+2\sqrt{2})$
- $e^{\beta^{-1}}$=$e^{-2}(5+\sqrt{2})$
- $e^{\beta^{-1}}$=$e^{-2}(3+2\sqrt{2})$
- $e^{3\beta^{-1}}$=$e(5+\sqrt{2})$
- Among the two statements
(S1) : $(p=>q) \bigwedge (q \bigwedge(~q))$ is a contradiction and
(S2) : $(p \bigwedge q) \bigvee ((\text{~}p) \bigwedge q) \bigvee (p \bigwedge(\text{~}q)) \bigvee ((\text{~}p) \bigwedge (\text{~}q))$ is a tautology
- only (S2) is true
- only (S1) is true
- both are false
- both are true
- If $\lambda \in Z$, $\vec{a}$=$\lambda \hat{i}$+$\hat{j}$$-\hat{k}$ and $\vec{b}$=$3\hat{i}$$-\hat{j}$+2$\hat{k}$. Let $\vec{c}$ be a vector such that $(\vec{a}+\vec{b}+\vec{c})$×$\vec{c}$=0, $\vec{a}$•$\vec{c}$=$-17$ and $\vec{b}$•$\vec{c}$=$-20$. Then $|\vec{c}×(\lambda \hat{i}+\hat{j}+\hat{k})|^2$ is equal to
- 62
- 46
- 53
- 49
- The sum, of the coefficients of the first 50 terms in the binomial expansion of $(1 – x)^{100}$, is equal to
- $-{}^{101} C_{50}$
- ${}^{99} C_{49}$
- $-{}^{99} C_{49}$
- ${}^{101} C_{50}$
- The area of the region enclosed by the curve $y = x^3$and its tangent at the point (-1, -1) is
- $\frac{27}{4}$
- $\frac{19}{4}$
- $\frac{23}{4}$
- $\frac{31}{4}$
- Let $A$ = $\begin{equation*} \begin{bmatrix} 1 & \frac{1}{51} \\ 0 & 1 \end{bmatrix} \end{equation*}$. If $B$=$\begin{equation*} \begin{bmatrix} 1 & 2 \\ -1 & -1\end{bmatrix} \begin{bmatrix} -1 & -2 \\ 1 & 1\end{bmatrix} \end{equation*}$, then the sum of all the elements of the matrix $\sum \limits_{n=1}^{50} B^n$ is equal to
- 100
- 50
- 75
- 125
- Let the plane $P$ : $4x$$ – y$ + $z$ = 10 be rotated by an angle $\frac{\pi}{2}$ about its line of intersection with the plane $x$ + $y $$– z$ = 4. If $\alpha$ is the distance of the point
(2, 3, − 4) from the new position of the plane $P$, then 35$\alpha$ is
- 90
- 85
- 105
- 126
- If $\frac{1}{n+1}{}^n C_n$+$\frac{1}{n}{}^n C_{n-1}$+....+$\frac{1}{2}{}^n C_1$+${}^n C_0$=$\frac{1023}{10}$, then $n$ is equal to
- 6
- 9
- 8
- 7
- Let $C$ be the circle in the complex plane with centre $z_0$=$\frac{1}{2}(1+3i)$and radius $r$ = 1. Let $z_1$ = $1+ i$ and the complex number $z_2$ be outside the circle $C$ such that $|z_1 – z_0| $$|z_2 – z_0|$ = 1. If $z_0$, $z_1$ and $z_2$ are collinear, then the smaller value of $|z_2|^2$is equal to
- $\frac{13}{2}$
- $\frac{5}{2}$
- $\frac{3}{2}$
- $\frac{7}{2}$
- If the point $\left(\alpha, \frac{7\sqrt{3}}{3}\right)$ lies on the curve traced by the mid-points of the line segments of the lines $x \cos\theta $+$y \sin \theta $=7, $\theta \in \left(0, \frac{\pi}{2}\right)$ between the co-
ordinates axes, then $\alpha$ is equal to
- 7
- $-7$
- $-7\sqrt{3}$
- $7\sqrt{3}$
- Two dice $A$ and $B$ are rolled, Let the numbers obtained on $A$ and $B$ be $\alpha$ and $\beta$ respectively. If the variance of $\alpha – \beta$ is $\frac{p}{q}$ where $p$ and $q$ are co-prime, then the sum of the positive divisors of $p$ is equal to
- 36
- 48
- 31
- 72
- In a triangle $ABC$, if $cos A$ + 2 $cos B$ + $cos C$ = 2 and the lengths of the sides opposite to the angles $A$ and $C$ are 3 and 7 respectively, then $cos A – cos C$ is equal to
- $\frac{3}{7}$
- $\frac{9}{7}$
- $\frac{10}{7}$
- $\frac{5}{7}$
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.
- A fair $n (n > 1)$ faces die is rolled repeatedly until a number less than $n$ appears. If the mean of the number of tosses required is $\frac{n}{9}$, then $n$ is equal to __________.
- Let the digits $a$, $b$, $c$ be in $A.P.$ Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in $A.P.$ at least once. How many such numbers can be formed?
- Let $[x]$ be the greatest integer $\leq x$. Then the number of points in the interval (–2,1), where the function $f(x)$ = $| [x] |$ + $\sqrt{x – [x]}$ is discontinuous is________.
- Let the plane $x$ + $3y – 2z$ + 6 = 0 meet the co-ordinate axes at the points $A$,$B$,$C$. If the orthocentre of the triangle $ABC$ is $\left(\alpha, \beta, \frac{6}{7}\right)$, then $98 (\alpha + \beta)^2$is equal to______.
- Let $I(x)$=$\int \sqrt{\frac{x+7}{x}}dx$ and $I(9)$=12+$7\log_e7$. If $I(1)$=$\alpha$+7$\log_e(1+2\sqrt{2})$, then $\alpha^4$ is equal to.........
- Let $D_k$=$\begin{equation*} \begin{vmatrix} 1 & 2k & 2k-1 \\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2 \end{vmatrix} \end{equation*}$. If $\sum \limits_{k=1}^{n} D_k$=96, then $n$ is equal to........
- Let the positive numbers $a_1$, $a_2$, $a_3$, $a_4$ and $a_5$ be in a $G.P.$ Let their mean and variance be $\frac{31}{10}$ and $m$and $n$ respectively, where $m$ and $n$ are co-prime. If the mean of their reciprocals is $\frac{31}{40}$and $a_3$ + $a_4$ + $a_5$ = 14, then $m + n$ is equal to________.
- The number of relations, on the set {1,2,3} containing (1,2) and (2,3), which are reflexive and transitive but not symmetric, is_____
- If $\int \limits_{-0.15}^{0.15}|100x^2-1|dx$=$\frac{k}{3000}$, then $k$ is equal to........
- Two circles in the first quadrant of radii $r_1$ and $r_2$touch the coordinate axes. Each of them cuts off an intercept of 2 units with the line $x$ + $y$ = 2. Then $r_1^2$+$r_2^2$–$r_1r_2$ is equal to________.
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