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JEE Main 2022 July 26th Shift 2 Mathematics Question Paper and Solutions
SECTION – A
Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.
Choose the correct answer :
1. The minimum value of the sum of the squares of the roots of x2 + (3 – a)x + 1 = 2a is
(A) 4
(B) 5
(C) 6
(D) 8
Answer (C)
Sol.
α + β = a – 3, αβ = 1 – 2a
⇒ α2 + β2 = (a – 3)2 – 2(1– 2a)
= a2 – 6a + 9 – 2 + 4a
= a2 – 2a + 7
= (a – 1)2 + 6
So,
\(\begin{array}{l} \alpha^2+\beta^2\geq6 \end{array} \)
2. If z = x + iy satisfies | z | – 2 = 0 and |z – i| – | z + 5i| = 0, then
(A) x + 2y – 4 = 0
(B) x2 + y – 4 = 0
(C) x + 2y + 4 = 0
(D) x2 – y + 3 = 0
Answer (C)
Sol.
\(\begin{array}{l} \left|z-i\right|=\left|z+5i\right|\end{array} \)
So, z lies on ⊥r bisector of (0, 1) and (0, –5)
i.e., liney = –2
as |z| = 2
⇒ z = –2i
x = 0 and y = –2
so, x + 2y + 4 = 0
3. Let \(\begin{array}{l} A=\begin{bmatrix}1 \\ 1\\1\end{bmatrix}\text{ and }B=\begin{bmatrix}9^2 & -10^2 & 11^2 \\12^2 & 13^2 & -14^2 \\-15^2 & 16^2 & 17^2 \\\end{bmatrix},\end{array} \)
then the value of A′BA is
(A) 1224
(B) 1042
(C) 540
(D) 539
Answer (D)
Sol.
\(\begin{array}{l} A’BA=\begin{bmatrix}1 & 1 & 1 \\\end{bmatrix}\begin{bmatrix}9^2 & -10^2 & 11^2 \\12^2 & 13^2 & -14^2 \\-15^2 & 16^2 & 17^2 \\\end{bmatrix}A\end{array} \)
\(\begin{array}{l} =\begin{bmatrix}9^2+12^2-15^2 & -10^2+13^2+16^2 & 11^2-14^2+17^2 \\\end{bmatrix} \begin{bmatrix}1 \\1 \\1\end{bmatrix}\end{array} \)
\(\begin{array}{l} =\left[9^2+12^2-15^2-10^2+13^2+16^2+11^2-14^2+17^2\right]\end{array} \)
= [(92 – 102) + (112 + 122) + (132 – 142) + (162 – 152) + 172]
= [–19 + 265 + (–27) + 31 + 289]
= [585 – 46] = [539]
4. \(\begin{array}{l} \sum_{\underset{i\neq j}{i, j=0}}^{n}{^n}C_i\ ^nC_j\end{array} \)
is equal to
\(\begin{array}{l}\left(A\right) 2^{2n} – ^{2n}C_n \end{array} \)
\(\begin{array}{l}\left(B\right) 2^{2n-1} – ^{2n-1}C_{n-1} \end{array} \)
\(\begin{array}{l} (\text{C})\ 2^{2n}-\frac{1}{2}\ ^{2n}C_n\end{array} \)
\(\begin{array}{l} (\text{D})\ 2^{n-1}+2^{2n-1}C_n\end{array} \)
Answer (A*)
Sol.
\(\begin{array}{l} \sum_{\underset{i\neq j}{i, j=0}}^{n} {^n}C_i\ ^nC_j=\sum_{i, j=0}^{n}{^n}C_i\ ^nC_j-\sum_{i=j}^{n}{^n}C_i\ ^nC_j\end{array} \)
\(\begin{array}{l} =\sum_{j=0}^{n}{^n}C_i\sum_{j=0}^{n}{^n}C_j-\sum_{i=0}^{n}{^n}C_i\ C_i\end{array} \)
\(\begin{array}{l} =2^n\cdot2^n-\ ^{2n}C_n\end{array} \)
\(\begin{array}{l} =2^{2n}-^{2n}C_n\end{array} \)
5. Let P and Q be any points on the curves (x – 1)2 + (y + 1)2 = 1 and y = x2, respectively. The distance between P and Q is minimum for some value of the abscissa of P in the interval
\(\begin{array}{l} (\text{A})\ \left(0,\frac{1}{4}\right) \end{array} \)
\(\begin{array}{l} (\text{B})\ \left(\frac{1}{2},\frac{3}{4}\right) \end{array} \)
\(\begin{array}{l} (\text{C})\ \left(\frac{1}{4},\frac{1}{2}\right)\end{array} \)
\(\begin{array}{l} (\text{D})\ \left(\frac{3}{4},1\right)\end{array} \)
Answer (C)
Sol.
\(\begin{array}{l} y = mx+2a+\frac{1}{m^2}\text{(Equation of normal to}~ x^2 = 4ay ~\text{in slope form) through} (1, -1).\end{array} \)
\(\begin{array}{l}4m^3 + 6m^2 + 1 = 0\end{array} \)
\(\begin{array}{l} \Rightarrow m\simeq -16 \end{array} \)
\(\begin{array}{l} \text{Slope of normal} \simeq \frac{-8}{5}=\tan\theta\end{array} \)
\(\begin{array}{l} \Rightarrow\ \cos\theta\simeq\frac{-5}{\sqrt{89}},\sin\theta\simeq\frac{8}{\sqrt{89}} \end{array} \)
\(\begin{array}{l} x_p=1+\cos\theta\simeq 1-\frac{5}{\sqrt{89}}\in\left(\frac{1}{4},\frac{1}{2}\right)\end{array} \)
6. If the maximum value of a, for which the function \(\begin{array}{l} f_a\left(x\right)=\tan^{-1}2x-3ax+7\end{array} \)
\(\begin{array}{l}\text{is non-decreasing in}\ \left(-\frac{\pi}{6},\frac{\pi}{6}\right),\ \text{is}\ \overline{a},\ \text{then}\ f_{\overline{a}}\left(\frac{\pi}{8}\right)\end{array} \)
is equal to
\(\begin{array}{l} (\text{A})\ 8-\frac{9\pi}{4\left(9+\pi^2\right)} \end{array} \)
\(\begin{array}{l} (\text{B})\ 8-\frac{4\pi}{9\left(4+\pi^2\right)} \end{array} \)
\(\begin{array}{l} (\text{C})\ 8\left(\frac{1+\pi^2}{9+\pi^2}\right)\end{array} \)
\(\begin{array}{l} (\text{D})\ 8-\frac{\pi}{4}\end{array} \)
Answer (*)
Sol.
\(\begin{array}{l} f_a\left(x\right)=\tan^{-1}2x-3ax+7 \end{array} \)
\(\begin{array}{l}\because f_a(x) ~\text{is non}\mbox{-}\text{decreasing in} \left(-\frac{\pi}{6},\frac{\pi}{6}\right)\end{array} \)
\(\begin{array}{l} \therefore\ f_a’\left(x\right)\geq 0 \Rightarrow \frac{2}{1+4x^2}-3a\geq0\end{array} \)
\(\begin{array}{l} \Rightarrow\ 3a\leq\frac{2}{1+4x^2} \end{array} \)
\(\begin{array}{l}\text{So,}~ a_\text{max}=\frac{2}{3}\left(\frac{1}{1+4\times\frac{\pi^2}{36}}\right)=\frac{6}{9+\pi^2}=\overline{a} \end{array} \)
\(\begin{array}{l} \therefore\ f_{\overline{a}}\left(\frac{\pi}{8}\right)=\tan^{-1}\frac{\pi}{4}-3\frac{\pi}{8}\cdot\frac{6}{9+\pi^2}+7\end{array} \)
7. Let \(\begin{array}{l} \beta=\displaystyle \lim_{x \to 0}\frac{\alpha x-\left(e^{3x}-1\right)}{\alpha x\left(e^{3x}-1\right)}\ \text{for some}\ \alpha\ \in\ \mathbb{R}.\end{array} \)
Then the value of α + β is
\(\begin{array}{l} (\text{A})\ \frac{14}{5} \end{array} \)
\(\begin{array}{l} (\text{B})\ \frac{3}{2} \end{array} \)
\(\begin{array}{l} (\text{C})\ \frac{5}{2} \end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{7}{2} \end{array} \)
Answer (C)
Sol.
\(\begin{array}{l} \beta=\displaystyle \lim_{x \to 0}\frac{\alpha x-\left(e^{3x}-1\right)}{\alpha x\left(e^{3x}-1\right)}, \alpha\ \in\ \mathbb{R}\end{array} \)
\(\begin{array}{l} =\displaystyle \lim_{x \to 0} \frac{\frac{\alpha}{3}-\left(\frac{e^{3x}-1}{3x}\right)}{\alpha x\left(\frac{e^{3x-1}}{3x}\right)}\end{array} \)
So, α = 3 (to make indeterminant form)
\(\begin{array}{l} \beta=\displaystyle \lim_{x \to 0}\frac{1-\left(\frac{e^{3x}-1}{3x}\right)}{3x}=\frac{1-\frac{\left(3x+\frac{9x^2}{2}+\cdots\right)}{3x}}{3x}\end{array} \)
\(\begin{array}{l} =\frac{-\left(\frac{9}{2}x^2+\frac{\left(3x\right)^3}{31}+\cdots\right)}{9x^2} =\frac{-1}{2}\end{array} \)
\(\begin{array}{l} \therefore\ \alpha +\beta=3-\frac{1}{2}=\frac{5}{2}\end{array} \)
8. The value of \(\begin{array}{l} \text{log}_e2\frac{d}{dx}\left(\text{log}_{\cos x}\text{cosec x}\right) \textup{ at }x=\frac{\pi}{4}\end{array} \)
is
\(\begin{array}{l} (\text{A})\ -2\sqrt{2}\end{array} \)
\(\begin{array}{l} (\text{B})\ 2\sqrt{2}\end{array} \)
\(\begin{array}{l} (\text{C})\ -4\end{array} \)
\(\begin{array}{l} (\text{D})\ 4\end{array} \)
Answer (D)
Sol. Let
\(\begin{array}{l} f\left(x\right)=\text{log}_{\cos x} \text{cosec}x\end{array} \)
\(\begin{array}{l} =\frac{\text{log }\text{cosec}x}{\text{log cos}x}\end{array} \)
\(\begin{array}{l} f’\left(x\right)=\frac{\text{log cos}x\cdot\sin x\cdot \left(-\text{cosec}x\cot x-\text{log cosec}x\cdot\frac{1}{\cos x}\cdot-\sin x\right)}{\left(\text{log cos}x\right)^2}\end{array} \)
at x = π/4
\(\begin{array}{l}f’\left(\frac{\pi}{4}\right)= \frac{-log (\frac{1}{\sqrt2})+log\sqrt2}{(log \frac{1}{\sqrt2})^2}\end{array} \)
\(\begin{array}{l}=\frac{2}{log \sqrt2}\end{array} \)
\(\begin{array}{l} \therefore\ \text{log}_e\ 2f’\left(x\right)\text{ at }x=\frac{\pi}{4}=4\end{array} \)
9. \(\begin{array}{l} \displaystyle\int\limits_0^{20\pi}\left(\left|\sin x\right|+\left|\cos x\right|\right)^2 dx\end{array} \)
is equal to
\(\begin{array}{l} (\text{A})\ 10\left(\pi+4\right)\end{array} \)
\(\begin{array}{l} (\text{B})\ 10\left(\pi+2\right)\end{array} \)
\(\begin{array}{l} (\text{C})\ 20\left(\pi-2\right)\end{array} \)
\(\begin{array}{l} (\text{D})\ 20\left(\pi+2\right)\end{array} \)
Answer (D)
Sol.
\(\begin{array}{l} I=\displaystyle\int\limits_0^{20\pi}\left(\left|\sin x\right|+\left|\cos x\right|\right)^2 dx\end{array} \)
\(\begin{array}{l} =20\displaystyle\int\limits_0^\pi\left(1+\left|\sin 2x\right|\right)dx\end{array} \)
\(\begin{array}{l} =40\displaystyle\int\limits_0^\frac{\pi}{2}\left(1+\sin 2x\right)dx\end{array} \)
\(\begin{array}{l} \left. =40\left(x-\frac{\cos 2x}{2}\right)\right|_0^{\frac{\pi}{2}}\end{array} \)
\(\begin{array}{l} =40\left(\frac{\pi}{2}+\frac{1}{2}+\frac{1}{2}\right)=20\left(\pi+2\right)\end{array} \)
10. Let the solution curve y = f(x) of the differential equation \(\begin{array}{l} \frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt{1-x^2}},x \in \left(-1, 1\right)\end{array} \)
pass through the origin. Then \(\begin{array}{l} \displaystyle\int\limits_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}f\left(x\right)dx\end{array} \)
is
\(\begin{array}{l} (\text{A})\ \frac{\pi}{3}-\frac{1}{4}\end{array} \)
\(\begin{array}{l} (\text{B})\ \frac{\pi}{3}-\frac{\sqrt{3}}{4}\end{array} \)
\(\begin{array}{l} (\text{C})\ \frac{\pi}{6}-\frac{\sqrt{3}}{4}\end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{\pi}{6}-\frac{\sqrt{3}}{2}\end{array} \)
Answer (B)
Sol.
\(\begin{array}{l} \frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt{1-x^2}}\end{array} \)
which is first order linear differential equation.
\(\begin{array}{l}\text{Integrating factor (I.F.) = } e^{\int\frac{x}{x^2-1}dx} \end{array} \)
\(\begin{array}{l} =e^{\frac{1}{2}\text{ln}\left|x^2-1\right|}=\sqrt{\left|x^2-1\right|}\end{array} \)
\(\begin{array}{l} =\sqrt{1-x^2}\ \because x\in \left(-1, 1\right)\end{array} \)
Solution of differential equation
\(\begin{array}{l} y\sqrt{1-x^2}=\int\left(x^4+2x\right)dx=\frac{x^5}{5}+x^2+c\end{array} \)
Curve is passing through origin, c = 0
\(\begin{array}{l} y=\frac{x^5+5x^2}{5\sqrt{1-x^2}} \end{array} \)
\(\begin{array}{l} \displaystyle\int\limits_\frac{-\sqrt{3}}{2}^{\frac{\sqrt{3}}{2}}\frac{x^5+5x^2}{5\sqrt{1-x^2}}dx=0+2\displaystyle\int\limits_0^{\frac{\sqrt{3}}{2}}\frac{x^2}{\sqrt{1-x^2}}dx\end{array} \)
\(\begin{array}{l} \text{put}\ x = sin\theta \\dx = cos\theta~ d\theta\end{array} \)
\(\begin{array}{l} I=2\displaystyle\int\limits_0^\frac{\pi}{3}\frac{\sin^2\theta\cdot\cos\theta d\theta}{\cos\theta}\end{array} \)
\(\begin{array}{l} =\displaystyle\int\limits_0^\frac{\pi}{3}\left(1-\cos 2\theta \right)d\theta\end{array} \)
\(\begin{array}{l} \left.=\left(\theta-\frac{\sin 2\theta}{2}\right)\right| _0^\frac{\pi}{3}\end{array} \)
\(\begin{array}{l} =\frac{\pi}{3}-\frac{\sqrt{3}}{4} \end{array} \)
11. The acute angle between the pair of tangents drawn to the ellipse 2x2 + 3y2 = 5 from the point (1, 3) is
\(\begin{array}{l} (\text{A})\ \tan^{-1}\left(\frac{16}{7\sqrt{5}}\right)\end{array} \)
\(\begin{array}{l} (\text{B})\ \tan^{-1}\left(\frac{24}{7\sqrt{5}}\right)\end{array} \)
\(\begin{array}{l} (\text{C})\ \tan^{-1}\left(\frac{32}{7\sqrt{5}}\right)\end{array} \)
\(\begin{array}{l} (\text{D})\ \tan^{-1}\left(\frac{3+8\sqrt{5}}{35}\right)\end{array} \)
Answer (B)
Sol. 2x2 + 3y2 = 5
Equation of tangent having slope m.
\(\begin{array}{l} y=mx\pm\sqrt{\frac{5}{2}m^2+\frac{5}{3}}\end{array} \)
which passes through (1, 3)
\(\begin{array}{l} 3=m\pm\sqrt{\frac{5}{2}m^2+\frac{5}{3}}\end{array} \)
\(\begin{array}{l} \frac{5}{2}m^2+\frac{5}{3}=9+m^2-6m \end{array} \)
\(\begin{array}{l} \frac{3}{2}m^2+6m-\frac{22}{3}=0\end{array} \)
\(\begin{array}{l} 9m^2+36m-44=0\end{array} \)
\(\begin{array}{l} m_1+m_2=-4, m_1m_2=-\frac{44}{9} \end{array} \)
\(\begin{array}{l} \left(m_1-m_2\right)^2=16+4\times\frac{44}{9}=\frac{320}{9} \end{array} \)
Acute angle between the tangents is given by
\(\begin{array}{l} \alpha = \tan^{-1}\left|\frac{m_1-m_2}{1+m_1m_2}\right| \end{array} \)
\(\begin{array}{l} =\tan^{-1}\left|\frac{\frac{8\sqrt{5}}{3}}{1-\frac{44}{9}}\right| \end{array} \)
\(\begin{array}{l} =\tan^{-1}\left(\frac{24\sqrt{5}}{35}\right)\end{array} \)
\(\begin{array}{l} \alpha =\tan^{-1}\left(\frac{24}{7\sqrt{5}}\right)\end{array} \)
12. The equation of a common tangent to the parabolas y = x2 and y = –(x – 2)2 is
(A) y = 4(x – 2)
(B) y = 4(x – 1)
(C) y = 4(x + 1)
(D) y = 4(x + 2)
Answer (B)
Sol. Equation of tangent of slope m to y = x2
\(\begin{array}{l} y=mx-\frac{1}{4}m^2 \end{array} \)
Equation of tangent of slope m to y = –(x – 2)2
\(\begin{array}{l} y=m\left(x-2\right)+\frac{1}{4}m^2\end{array} \)
If both equation represent the same line
\(\begin{array}{l} \frac{1}{4}m^2-2m=-\frac{1}{4}m^2\end{array} \)
m = 0, 4
So, equation of tangent
\(\begin{array}{l} y=4x-4\end{array} \)
13. Let the abscissae of the two points P and Q on a circle be the roots of x2 – 4x – 6 = 0 and the ordinates of P and Q be the roots of y2 + 2y – 7 = 0. If PQ is a diameter of the circle x2 + y2 + 2ax + 2by + c = 0, then the value of (a + b – c) is
(A) 12
(B) 13
(C) 14
(D) 16
Answer (A)
Sol. Abscissae of PQ are roots of x2 – 4x – 6 = 0
Ordinates of PQ are roots of y2 + 2y – 7 = 0
and PQ is diameter
\(\begin{array}{l} \Rightarrow \text{Equation of circle is}~x^2+y^2-4x+2y-13=0\end{array} \)
But, given
\(\begin{array}{l} x^2+y^2+2ax+2by+c=0\end{array} \)
By comparison a = –2, b = 1, c = –13
⇒ a + b – c = –2 + 1 + 13 = 12
14. If the line x – 1 = 0 is a directrix of the hyperbola kx2 – y2 = 6, then the hyperbola passes through the point
\(\begin{array}{l} (\text{A})\ \left(-2\sqrt{5},6\right)\end{array} \)
\(\begin{array}{l} (\text{B})\ \left(-\sqrt{5},3\right)\end{array} \)
\(\begin{array}{l} (\text{C})\ \left(\sqrt{5},-2\right)\end{array} \)
\(\begin{array}{l} (\text{D})\ \left(2\sqrt{5},3\sqrt{6}\right)\end{array} \)
Answer (C)
Sol.
\(\begin{array}{l}\text{Given hyperbola :} \frac{x^2}{6/k}-\frac{y^2}{6}=1 \end{array} \)
\(\begin{array}{l}\text{Eccentricity }= e = \sqrt{1+\frac{6}{6/k}}=\sqrt{1+k}\end{array} \)
\(\begin{array}{l}\text{Directrices }: x=\pm\frac{a}{e}\Rightarrow x=\pm \frac{\sqrt{6}}{\sqrt{k}\sqrt{k+1}}\end{array} \)
\(\begin{array}{l}\text{As given} : \frac{\sqrt{6}}{\sqrt{k}\sqrt{k+1}}=1\end{array} \)
\(\begin{array}{l} \Rightarrow k = 2 \end{array} \)
\(\begin{array}{l}\text{Here hyperbola is } \frac{x^2}{3}-\frac{y^2}{6}=1 \end{array} \)
Checking the option gives
\(\begin{array}{l} \left(\sqrt{5},-2\right)\text{satisfies it.}\end{array} \)
15. A vector \(\begin{array}{l}\vec{a}\end{array} \)
is parallel to the line of intersection of the plane determined by the vectors \(\begin{array}{l} \hat{i},\hat{i}+\hat{j}\end{array} \)
and the plane determined by the vectors \(\begin{array}{l} \hat{i}-\hat{j},\hat{i}+\hat{k}. \end{array} \)
The obtuse angle between\(\begin{array}{l}\vec{a}\ \text{and the vector}\ \vec{b}=\hat{i}-2\hat{j}+2\hat{k}\end{array} \)
is
\(\begin{array}{l} (\text{A})\ \frac{3\pi}{4} \end{array} \)
\(\begin{array}{l} (\text{B})\ \frac{2\pi}{3} \end{array} \)
\(\begin{array}{l} (\text{C})\ \frac{4\pi}{5} \end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{5\pi}{6}\end{array} \)
Answer (A)
Sol.
\(\begin{array}{l}\text{If}\ \vec{n}_1\ \text{is a vector normal to the plane determined by}\ \hat{i}\text{ and }\hat{i}+\hat{j}\end{array} \)
then
\(\begin{array}{l} \vec{n}_1=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\1 & 0 & 0 \\1 & 1 & 0 \\\end{vmatrix}=\hat{k}\end{array} \)
\(\begin{array}{l}\text{If}\ \vec{n}_2\ \text{is a vector normal to the plane determined by}\ \hat{i}-\hat{j}\ \text{and}\ \hat{i}+\hat{k}\end{array} \)
then
\(\begin{array}{l} \vec{n}_2=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\1 & -1 & 0 \\1 & 0 & 1 \\\end{vmatrix}=-\hat{i}-\hat{j}+\hat{k}\end{array} \)
\(\begin{array}{l}\text{Vector}\ \vec{a}\ \text{is parallel to}\ \vec{n}_1\times \vec{n}_2\end{array} \)
i.e.
\(\begin{array}{l}\vec{a}\ \text{is parallel to}\ \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\0 & 0 & 1 \\-1 & -1 & 1 \\\end{vmatrix} =\hat{i}-\hat{j}\end{array} \)
Given
\(\begin{array}{l} \vec{b}=\hat{i}-2\hat{j}+2\hat{k}\end{array} \)
Cosine of acute angle between
\(\begin{array}{l} \vec{a}\text{ and }\vec{b}=\left|\frac{\vec{a}\cdot\vec{b}}{\left|\vec{a}\right|\cdot\left|\vec{b}\right|}\right|=\frac{1}{\sqrt{2}}\end{array} \)
Obtuse angle between
\(\begin{array}{l} \vec{a}\text{ and }\vec{b}=\frac{3\pi}{4}\end{array} \)
16. If \(\begin{array}{l} 0<x<\frac{1}{\sqrt{2}}\ \text{and}\ \frac{\sin^{-1}x}{\alpha}=\frac{\cos^{-1}x}{\beta}\end{array} \)
then a value of \(\begin{array}{l} \sin\left(\frac{2\pi\alpha}{\alpha+\beta}\right)\end{array} \)
is
\(\begin{array}{l} (\text{A})\ 4\sqrt{\left(1-x^2\right)}\left(1-2x^2\right)\end{array} \)
\(\begin{array}{l} (\text{B})\ 4x\sqrt{\left(1-x^2\right)}\left(1-2x^2\right)\end{array} \)
\(\begin{array}{l} (\text{C})\ 2x\sqrt{\left(1-x^2\right)}\left(1-4x^2\right)\end{array} \)
\(\begin{array}{l} (\text{D})\ 4\sqrt{\left(1-x^2\right)}\left(1-4x^2\right)\end{array} \)
Answer (B)
Sol. Let
\(\begin{array}{l} \frac{\sin^{-1}x}{\alpha}=\frac{\cos^{-1}x}{\beta}=k\Rightarrow \sin^{-1}x+\cos^{-1}x=k\left(\alpha+\beta\right)\end{array} \)
\(\begin{array}{l} \Rightarrow \alpha+\beta=\frac{\pi}{2k}.\end{array} \)
Now,
\(\begin{array}{l} \frac{2\pi\alpha}{\alpha+\beta}=\frac{2\pi\alpha}{\frac{\pi}{2k}}=4k\alpha=4\sin^{-1}x \end{array} \)
Here
\(\begin{array}{l} \sin\left(\frac{2\pi\alpha}{\alpha+\beta}\right)=\sin\left(4\sin^{-1}x\right)\end{array} \)
Let
\(\begin{array}{l} \sin^{-1}x = \theta \because x \in \left(0,\frac{1}{\sqrt{2}}\right)\Rightarrow \theta \in \left(0, \frac{\pi}{4}\right)\end{array} \)
\(\begin{array}{l} \Rightarrow x = sin \theta\end{array} \)
\(\begin{array}{l} \Rightarrow \cos\theta=\sqrt{1-x^2} \end{array} \)
\(\begin{array}{l} \Rightarrow \sin2\theta=2x\cdot\sqrt{1-x^2} \end{array} \)
\(\begin{array}{l} \Rightarrow \cos2\theta=\sqrt{1-4x^2\left(1-x^2\right)}=\sqrt{\left(2x^2-1\right)^2}=1-2x^2\end{array} \)
\(\begin{array}{l} \left(\because \cos2\theta > 0 \text{ as } 2\theta \in \left(0,\frac{\pi}{2}\right)\right) \end{array} \)
\(\begin{array}{l} \Rightarrow \sin4\theta=2\cdot 2x\sqrt{1-x^2}\left(1-2x^2\right)\end{array} \)
\(\begin{array}{l} =4x\sqrt{1-x^2}\left(1-2x^2\right) \end{array} \)
17. Negation of the Boolean expression p⇔ (q⇒p) is
(A) (~ p) ∧q (B) p∧ (~ q)
(C) (~ p) ∨ (~ q) (D) (~ p) ∧ (~ q)
Answer (D)
Sol. p⇔ (q⇒p)
~ (p⇔ (q⇔p))
≡p⇔ ~ (q⇒p)
≡p⇔ (q∧ ~ p)
≡ (p⇒ (q∧ ~ p)) ∧ ((q∧ ~ p) ⇒p))
≡ (~ p∨ (q∧ ~ p)) ∧ ((~ q∨p) ∨p))
≡ ((~ p∨q) ∧ ~ p) ∧ (~ q∨p)
≡ ~ p∧ (~ q∨p)
≡ (~ p∧ ~ q) ∨ (~ p∧p)
≡ (~ p∧ ~ q) ∨c
≡ (~ p∧ ~ q)
18. Let X be a binomially distributed random variable with mean 4 and variance 4/3. Then, 54 P(X≤ 2) is equal to
\(\begin{array}{l} (\text{A})\ \frac{73}{27}\end{array} \)
\(\begin{array}{l} (\text{B})\ \frac{146}{27}\end{array} \)
\(\begin{array}{l} (\text{C})\ \frac{146}{81}\end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{126}{81}\end{array} \)
Answer (B)
Sol. Mean = 4 = μ = np
\(\begin{array}{l}Variance = \sigma^2=np\left(1-P\right)=\frac{4}{3}\end{array} \)
\(\begin{array}{l} 4\left(1-P\right)=\frac{4}{3}\end{array} \)
\(\begin{array}{l} P=\frac{2}{3} \end{array} \)
\(\begin{array}{l} n\times\frac{2}{3}=4 \\n = 6\end{array} \)
\(\begin{array}{l} P\left(X=k\right)=\ ^nC_k\ P^k\left(1-P\right)^{n-k} \end{array} \)
\(\begin{array}{l} P\left(X\leq 2\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)\end{array} \)
\(\begin{array}{l} =\ ^6C_0P^0\left(1-p\right)^6+\ ^6C_1P^1\left(1-P\right)^5+\ ^6C_2P^2\left(1-P\right)^4\end{array} \)
\(\begin{array}{l} =\ ^6C_0\left(\frac{1}{3}\right)^6+\ ^6C_1\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)^5+\ ^6C_2\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^4\end{array} \)
\(\begin{array}{l} =\left(\frac{1}{3}\right)^6\left[1+12+60\right]=\frac{73}{3^6}\end{array} \)
\(\begin{array}{l}54P(X\le 2)=\frac{73}{3^6}\times 54 =\frac{146}{27} \end{array} \)
19. The integral \(\begin{array}{l} \int\frac{\left(1-\frac{1}{\sqrt{3}}\right)\left(\cos x-\sin x\right)}{\left(1+\frac{2}{\sqrt{3}}\sin 2x\right)}dx\end{array} \)
is equal to
\(\begin{array}{l} (\text{A})\ \frac{1}{2}\text{log}_e\left|\frac{\tan\left(\frac{x}{2}+\frac{\pi}{12}\right)}{\tan\left(\frac{x}{2}+\frac{\pi}{6}\right)}\right|+C\end{array} \)
\(\begin{array}{l} (\text{B})\ \frac{1}{2}\text{log}_e\left|\frac{\tan\left(\frac{x}{2}+\frac{x}{6}\right)}{\tan\left(\frac{x}{2}+\frac{\pi}{3}\right)}\right|+C\end{array} \)
\(\begin{array}{l} (\text{C})\ \text{log}_e\left|\frac{\tan\left(\frac{x}{2}+\frac{\pi}{6}\right)}{\tan\left(\frac{x}{2}+\frac{\pi}{12}\right)}\right|+C\end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{1}{2}\text{log}_e\left|\frac{\tan\left(\frac{x}{2}-\frac{\pi}{12}\right)}{\tan\left(\frac{x}{2}-\frac{\pi}{6}\right)}\right|+C\end{array} \)
Answer (A)
Sol.
\(\begin{array}{l} =\int\frac{\left(1-\frac{1}{\sqrt{3}}\right)\left(\cos x-\sin x\right)}{\left(1+\frac{2}{\sqrt{3}}\sin 2x\right)}dx\end{array} \)
\(\begin{array}{l} =\int\frac{\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)\sqrt{2}\sin\left(\frac{\pi}{4}-x\right)}{\left(\frac{2}{\sqrt{3}}\right)\left(\sin\frac{\pi}{3}+\sin 2x\right)}dx\end{array} \)
\(\begin{array}{l} =\int\frac{\frac{\left(\sqrt{3}-1\right)}{\sqrt{2}}\sin \left(\frac{\pi}{4}-x\right)}{\left(\sin\frac{\pi}{3}+\sin2x\right)}dx\end{array} \)
\(\begin{array}{l} =\int\frac{\frac{\sqrt{3}-1}{2\sqrt{2}}\sin\left(\frac{\pi}{4}-x\right)}{\sin\left(\frac{\pi}{6}+x\right)\cos\left(\frac{\pi}{6}-x\right)}dx\end{array} \)
\(\begin{array}{l} =\frac{1}{2}\int\frac{2\sin \frac{\pi}{12}\sin\left(\frac{\pi}{4}-x\right)}{\sin\left(\frac{\pi}{6}+x\right)\cos \left(\frac{\pi}{6}-x\right)}dx\end{array} \)
\(\begin{array}{l} =\frac{1}{2}\int\frac{\cos\left(\frac{\pi}{6}-x\right)-\cos\left(\frac{\pi}{3}-x\right)}{\sin\left(\frac{\pi}{6}+x\right)\cos\left(\frac{\pi}{6}-x\right)}dx\end{array} \)
\(\begin{array}{l} =\frac{1}{2}\left[\int\text{cosec}\left(\frac{\pi}{6}+x\right)dx-\int\sec\left(\frac{\pi}{6}-x\right)dx\right]\end{array} \)
\(\begin{array}{l} =\frac{1}{2}\left[\text{In}\left|\tan\left(\frac{\pi}{12}+\frac{x}{2}\right)\right|-\int\text{cosec}\left(\frac{\pi}{3}-x\right)dx\right]\end{array} \)
\(\begin{array}{l} =\frac{1}{2}\left[\text{In}\left|\tan\left(\frac{\pi}{12}+\frac{x}{2}\right)\right|-\text{In}\left|\frac{\pi}{6}+\frac{x}{2}\right|\right]+C\end{array} \)
\(\begin{array}{l} =\frac{1}{2}\text{In}\left|\frac{\tan\left(\frac{\pi}{12}+\frac{x}{2}\right)}{\tan\left(\frac{\pi}{6}+\frac{x}{2}\right)}\right|+C\end{array} \)
20. The area bounded by the curves y = |x2 – 1| and y = 1 is
\(\begin{array}{l} (\text{A})\ \frac{2}{3}\left(\sqrt{2}+1\right)\end{array} \)
\(\begin{array}{l} (\text{B})\ \frac{4}{3}\left(\sqrt{2}-1\right)\end{array} \)
\(\begin{array}{l} (\text{C})\ 2\left(\sqrt{2}-1\right)\end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{8}{3}\left(\sqrt{2}-1\right)\end{array} \)
Answer (D)
Sol.
\(\begin{array}{l} \text{ Area } =2\displaystyle\int\limits_0^{\sqrt{2}}\left(1-\left|x^2-1\right|\right)dx\end{array} \)
\(\begin{array}{l} =2\left[\displaystyle\int\limits_0^1\left(1-\left(1-x^2\right)\right)dx+\displaystyle\int\limits_1^{\sqrt{2}}\left(2-x^2\right)dx\right]\end{array} \)
\(\begin{array}{l} =2\left.\left[\left[\frac{x^3}{3}\right]_0^1+\left[2x-\frac{x^3}{3}\right]_1^{\sqrt{2}}\right]\right]\end{array} \)
\(\begin{array}{l} =2\left(\frac{4\sqrt{2}-4}{3}\right)=\frac{8}{3}\left(\sqrt{2}-1\right) \end{array} \)
SECTION – B
Numerical Value Type Questions: This section contains 10 questions. In Section B, attempt any five questions out of 10. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.
1. Let A = {1, 2, 3, 4, 5, 6, 7} and B = {3, 6, 7, 9}. Then the number of elements in the set {C ⊆ A :Cߏ∩B ≠ φ} is _____________.
Answer (112)
Sol. As Cߏ∩B≠φ, c must be not be formed by {1, 2, 4, 5}
∴ Number of subsets of A = 27 = 128
and number of subsets formed by {1, 2, 4, 5} = 16
∴ Required no. of subsets = 27 – 24 = 128 – 16
= 112
2. The largest value of a, for which the perpendicular distance of the plane containing the lines
\(\begin{array}{l} \vec{r}=\left(\hat{i}+\hat{j}\right)+\lambda\left(\hat{i}+a\hat{j}-\hat{k}\right)\text{ and }\vec{r}=\left(\hat{i}+\hat{j}\right)+\mu \left(-\hat{i}+\hat{j}-a\hat{k}\right)\end{array} \)
from the point (2, 1, 4) is √3, is _____________.
Answer (2*)
Sol. Normal to plane
\(\begin{array}{l} =\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\1 & a & -1 \\-1 & 1 & -a \\\end{vmatrix} \end{array} \)
\(\begin{array}{l} =\hat{i}\left(1-a^2\right)-\hat{j}\left(-a-1\right)+\hat{k}\left(1+a\right)\end{array} \)
\(\begin{array}{l} =\left(1-a\right)\hat{i}+\hat{j}+\hat{k}\end{array} \)
∴ Plane (1 – a) (x – 1) + (y – 1) + z = 0
\(\begin{array}{l}\text{Distance from} (2, 1, 4) ~\text{is} \sqrt{3}\end{array} \)
i.e.
\(\begin{array}{l} \Rightarrow \left|\frac{\left(1-a\right)+0+4}{\sqrt{\left(1-a\right)^2+1+1}}\right|=\sqrt{3}\end{array} \)
\(\begin{array}{l} \Rightarrow 25 + a^2 – 10a = 3a^2 – 6a + 9\\ \Rightarrow 2a^2 + 4a – 16 = 0\\ \Rightarrow a^2 + 2a – 8 = 0\\ a = 2~ or -4\\ \therefore a_{max} = 2\end{array} \)
3. Numbers are to be formed between 1000 and 3000, which are divisible by 4, using the digits 1, 2, 3, 4, 5 and 6 without repetition of digits. Then the total number of such numbers is ______________.
Answer (30)
Sol. Number must start by 1 or 2 and for divisibility by
4 last two digits shall be divisible by 4
⇒ Total 30 numbers
4. If \(\begin{array}{l} \sum_{k=1}^{10}\frac{k}{k^4+k^2+1}=\frac{m}{n}\end{array} \)
, where m and n are co-prime, then m + n is equal to
Answer (166)
Sol.
\(\begin{array}{l} \sum_{k=1}^{10}\frac{k}{k^4+k^2+1}\end{array} \)
\(\begin{array}{l} =\frac{1}{2}\left[\sum_{k=1}^{10}\right.\left(\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right)\end{array} \)
\(\begin{array}{l} =\frac{1}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\cdots+\frac{1}{91}-\frac{1}{111}\right]\end{array} \)
\(\begin{array}{l} =\frac{1}{2}\left[1-\frac{1}{111}\right]=\frac{110}{2.111}=\frac{55}{111}=\frac{m}{n}\end{array} \)
∴ m + n = 55 + 111 = 166
5. If the sum of solutions of the system of equations 2sin2θ – cos2θ = 0 and 2cos2θ + 3sinθ = 0 in the interval [0, 2π] is kπ, then k is equal to _______.
Answer (3)
Sol. Equation (1)
\(\begin{array}{l} 2sin^2\theta = 1 – 2sin^2\theta \end{array} \)
\(\begin{array}{l}\Rightarrow sin^2\theta = \frac{1}{4}\end{array} \)
\(\begin{array}{l} \Rightarrow \sin\theta=\pm\frac{1}{2}\end{array} \)
\(\begin{array}{l} \Rightarrow \theta=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}\end{array} \)
\(\begin{array}{l} \text{Equation}(2)~2cos^2\theta + 3sin\theta = 0\\ \Rightarrow 2sin^2\theta – 3sin\theta – 2 = 0\\ \Rightarrow 2sin^2\theta – 4sin\theta + sin\theta – 2 = 0\\ \Rightarrow (sin\theta – 2) (2sin\theta + 1) = 0 \end{array} \)
\(\begin{array}{l} \Rightarrow \sin\theta=\frac{-1}{2}\end{array} \)
\(\begin{array}{l} \Rightarrow\ \theta=\frac{7\pi}{6},\frac{11\pi}{6}\end{array} \)
\(\begin{array}{l} \therefore \text{Common solutions = } \frac{7\pi}{6};\frac{11\pi}{6}\end{array} \)
\(\begin{array}{l} \text{Sum of solutions =}\frac{7\pi+11\pi}{6}=\frac{18\pi}{6}=3\pi\\ \therefore k = 3\end{array} \)
6. The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If σ is the standard deviation of the data after omitting the two wrong observations from the data, then 38σ2 is equal to ______.
Answer (238)
Sol.
\(\begin{array}{l} \mu=\frac{\sum x_i}{40}=30\ \Rightarrow \sum x_i=1200\end{array} \)
\(\begin{array}{l} \sigma^2=\frac{\sum x_i^2}{40}-\left(30\right)^2=25\ \Rightarrow\ \sum x_i^2 = 37000\end{array} \)
After omitting two wrong observations
\(\begin{array}{l} \sum y_i=1200-12-10=1178\end{array} \)
\(\begin{array}{l} \sum y_i^2=37000-144-100=36756 \end{array} \)
Now
\(\begin{array}{l} \sigma^2=\frac{\sum y_i^2}{38}-\left(\frac{\sum y_i}{38}\right)^2\end{array} \)
\(\begin{array}{l} =\frac{36756}{38}-\left(\frac{1178}{38}\right)^2=-31^2\end{array} \)
\(\begin{array}{l} 38\sigma ^2 = 36756 – 36518 = 238\end{array} \)
7. The plane passing through the line L :lx – y + 3(1 – l) z = 1, x + 2y – z = 2 and perpendicular to the plane 3x + 2y + z = 6 is 3x – 8y + 7z = 4. If θ is the acute angle between the line L and the y-axis, then 415 cos2θ is equal to ________.
Answer (125)
Sol. L :lx – y + 3 (1 – l)z = 1, x + 2y – z = 2
and plane containing the line p : 3x – 8y + 7z = 4
Let
\(\begin{array}{l}\vec{n}\end{array} \)
be the vector parallel to L.
then
\(\begin{array}{l}\vec{n}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ l& -1 & 3\left(1-l\right) \\1 & 2 & -1 \\\end{vmatrix}\end{array} \)
\(\begin{array}{l} =\left(6l-5\right)\hat{i}+\left(3-2l\right)\hat{j}+\left(2l+1\right)\hat{k}\end{array} \)
∵ R containing L
3(6l – 5) – 8(3 – 2l) + 7 (2l + 1) = 0
18l + 16l + 14l – 15 – 24 + 7 = 0
\(\begin{array}{l} \therefore l=\frac{32}{48}=\frac{2}{3} \end{array} \)
Let θ be the acute angle between L and y-axis
\(\begin{array}{l} \therefore\ \cos\theta=\frac{\frac{5}{3}}{\sqrt{1+\frac{25}{9}+\frac{49}{9}}}=\frac{5}{\sqrt{83}}\end{array} \)
\(\begin{array}{l} \therefore\ 415\cos^2\theta=125\end{array} \)
8. Suppose y = y(x) be the solution curve to the differential equation \(\begin{array}{l} \frac{dy}{dx}-y=2-e^{-x}\ \text{such that}\ \displaystyle \lim_{x \to \infty}y\left(x\right)\end{array} \)
is finite. If a and bare respectively the x – and y – intercepts of the tangent to the curve at x = 0, then the value of a – 4b is equal to _____.
Answer (3)
Sol.
\(\begin{array}{l}\text{If} = e^{-x}\\y\cdot e^{-x}=-2e^{-x}+\frac{e^{-2x}}{2}+C \end{array} \)
\(\begin{array}{l}\Rightarrow y = -2 + e^{-x}+Ce^x\\ \displaystyle \lim_{x \to \infty} y\left(x\right)\text{is finite so}~C = 0\end{array} \)
y = –2 + e–x
\(\begin{array}{l} \Rightarrow \underset{~~~~~~~~~~~~~~~~~~~~~~~x=0}{\left.\frac{dy}{dx}=-e^{-x}\Rightarrow \frac{dy}{dx}\right|} =-1\end{array} \)
Equation of tangent
y + 1 = –1 (x – 0)
or y + x = –1
Soa = –1, b = –1
⇒ a–4b = 3
9. Different A.P.’s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.’s having at least 3 terms and at most 33 terms is ______.
Answer (53)
Sol.
\(\begin{array}{l} d_1=\frac{199-100}{2}\notin I\end{array} \)
\(\begin{array}{l} d_2=\frac{199-100}{3}=33\end{array} \)
\(\begin{array}{l} d_3=\frac{199-100}{4}\notin I\end{array} \)
\(\begin{array}{l} d_n=\frac{199-100}{i+1}\in I \end{array} \)
di= 33 + 11, 9
Sum of CD’s = 33 + 11 + 9
= 53
10. The number of matrices \(\begin{array}{l} A=\begin{pmatrix}a & b \\c & d \\\end{pmatrix},\ \text{where}\ a, b, c, d\in{-1, 0, 1, 2, 3, …..,10},\end{array} \)
such that A = A–1, is ______.
Answer (50)
Sol.
\(\begin{array}{l} \because\ A=\begin{bmatrix}a & b \\c & d \\\end{bmatrix}\text{ then }A^2=\begin{bmatrix}a^2+bc & b\left(a+d\right) \\c\left(a+d\right) & bc+d^2 \\\end{bmatrix}\end{array} \)
For A–1 must exist ad – bc≠ 0 …(i)
and A = A–1⇒A2 = I
∴ a2 + bc = d2 + bc = 1 …(ii)
and b(a + d) = c(a + d) = 0 …(iii)
Case I : When a = d = 0, then possible values of
(b, c) are (1, 1), (–1, 1) and (1, –1) and (–1, 1).
Total four matrices are possible.
Case II : When a = –d then (a, d) be (1, –1) or
(–1, 1).
Then total possible values of (b, c) are
(12 + 11) × 2 = 46.
∴ Total possible matrices = 46 + 4 = 50.
JEE Main 2022 July 26th Shift 2 Paper Analysis with Solutions
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