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JEE Main 2022 25th July Shift 2 Mathematics Question Paper and Solutions
SECTION โ 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. For
\(\begin{array}{l}z \in \mathbb{C}\ \text{if the minimum value of}\ \left(|z-3\sqrt{2}| + |z-p\sqrt{2}i|\right)\end{array} \)
is 5โ2, then a value of p is _________.
(A) 3
\(\begin{array}{l}(\text{B})\ \frac{7}{2}\end{array} \)
(C) 4
\(\begin{array}{l}(\text{D})\ \frac{9}{2}\end{array} \)
Answer (C)
Sol.
It is sum of distance of z from (3โ2, 0) and (0, pโ2).
For minimising, z should lie on AB and
\(\begin{array}{l}AB=5\sqrt{2}\end{array} \)
\(\begin{array}{l}(AB)^2=18+2p^2\end{array} \)
\(\begin{array}{l}p=\pm4\end{array} \)
2. The number of real values of ฮป, such that the system of linear equations
2x โ 3y + 5z = 9
x + 3y โ z = โ18
3x โ y + (ฮป2 โ | ฮป |)z = 16
has no solutions, is
(A) 0
(B) 1
(C) 2
(D) 4
Answer (C)
Sol.
\(\begin{array}{l}\Delta = \begin{vmatrix}2 & -3 & 5 \\1 & 3 & -1 \\3 & -1 & \lambda^2-|\lambda| \\\end{vmatrix} =2(3\lambda^2-3|\lambda|-1)+3(\lambda^2-|\lambda|+3)+5(-1-9)\end{array} \)
= 9ฮป2 โ 9 | ฮป | โ 43
= 9 | ฮป |2 โ 9 | ฮป | โ 43
ฮ = 0 for 2 values of | ฮป | out of which one is โve and other is +ve
So, 2 values of ฮป satisfy the system of equations to obtain no solution.
3. The number of bijective functions f : {1, 3, 5, 7, โฆ, 99} โ {2, 4, 6, 8, โฆ.., 100} such that
\(\begin{array}{l}f(3)\ge f(9)\ge f(15)\ge f(21)\ge โฆ \ge f(99)\end{array} \)
is_____.
\(\begin{array}{l}(\text{A})\ ^{50}P_{17}\end{array} \)
\(\begin{array}{l}(\text{B})\ ^{50}P_{33}\end{array} \)
\(\begin{array}{l}(\text{C})\ 33!\times 17!\end{array} \)
\(\begin{array}{l}(\text{D})\ \frac{50!}{2}\end{array} \)
Answer (B)
Sol. As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction
f(3) > f(9) > f(15) โฆ.. > f(99)
So number of ways
\(\begin{array}{l}=^{50}C_{17}\cdot 1\cdot 33! = ^{50}P_{33}\end{array} \)
4. The remainder when (11)1011 + (1011)11 is divided by 9 is
(A) 1
(B) 4
(C) 6
(D) 8
Answer (D)
Sol.
\(\begin{array}{l}Re\left( \frac{(11)^{1011}+(1011)^{11}}{9} \right) = Re\left(\frac{2^{1011}+3^{11}}{9}\right)\end{array} \)
\(\begin{array}{l}\text{For } Re \left(\frac{2^{1011}}{9}\right)\end{array} \)
\(\begin{array}{l}2^{1011}=(9-1)^{337}=^{337}C_09^{337}(-1)^0 +^ {337}C_1 9^{336}(-1)^1+^{337}C_2 9^{335}(-1)^2 + โฆ + ^{337}C_{337} 9^{0}(-1)^{337}\end{array} \)
so, remainder is 8.
and
\(\begin{array}{l}Re\left(\frac{3^{11}}{9}\right)=0\end{array} \)
So, remainder is 8.
5. The sum
\(\begin{array}{l}\sum_{n=1}^{21}\frac{3}{(4n-1)(4n+3)}\end{array} \)
is equal to
\(\begin{array}{l}(\text{A})\ \frac{7}{87}\end{array} \)
\(\begin{array}{l}(\text{B})\ \frac{7}{29}\end{array} \)
\(\begin{array}{l}(\text{C})\ \frac{14}{87}\end{array} \)
\(\begin{array}{l}(\text{D})\ \frac{21}{29}\end{array} \)
Answer (B)
Sol.
\(\begin{array}{l}\sum_{n=1}^{21}\frac{3}{(4n-1)(4n+3)}=\frac{3}{4}\sum_{n=1}^{21}\frac{1}{4n-1}-\frac{1}{4n+3}\end{array} \)
\(\begin{array}{l}=\frac{3}{4}\left[\left(\frac{1}{3}-\frac{1}{7}\right) + \left(\frac{1}{7}-\frac{1}{11}\right)+โฆ+ \left(\frac{1}{83}-\frac{1}{87}\right)\right]\end{array} \)
\(\begin{array}{l}=\frac{3}{4}\left[\frac{1}{3}-\frac{1}{87}\right]=\frac{3}{4}~\frac{84}{3.87}= \frac{7}{29}\end{array} \)
6.
\(\begin{array}{l}\displaystyle \lim_{ x\to \frac{x}{4}}\frac{8\sqrt{2}-(\cos x + \sin x)^7}{\sqrt{2}-\sqrt{2}\sin 2x}\end{array} \)
is equal to
(A) 14
(B) 7
\(\begin{array}{l}(\text{C})\ 14\sqrt{2}\end{array} \)
\(\begin{array}{l}(\text{D})\ 7\sqrt{2}\end{array} \)
Answer (A)
Sol.
\(\begin{array}{l}\displaystyle \lim_{ x\to\frac{x}{4} }\frac{8\sqrt{2}-(\cos x + sin x)^7}{\sqrt{2}-\sqrt{2}\sin 2x}~~~~~~~\left(\frac{0}{0} \text{ form}\right)\end{array} \)
\(\begin{array}{l}=\displaystyle \lim_{ x\to\frac{x}{4} }\frac{-7(\cos x + \sin x)^6(-\sin x + \cos x)}{-2\sqrt{2}\cos 2x}\end{array} \)
using LโH Rule
\(\begin{array}{l}=\displaystyle \lim_{ x\to\frac{x}{4} }\frac{56(\cos x โ \sin x)}{2\sqrt{2} \cos 2x} ~~~~~\left(\frac{0}{0}\right)\end{array} \)
\(\begin{array}{l}=\displaystyle \lim_{ x\to\frac{x}{4} }\frac{-56(\sin x + \cos x)}{-4\sqrt{2} \sin 2x}\end{array} \)
using LโH Rule
\(\begin{array}{l}=7\sqrt{2}\cdot \sqrt{2}=14\end{array} \)
7.
\(\begin{array}{l}\displaystyle \lim_{n \to \infty}\frac{1}{2^n}\left(\frac{1}{\sqrt{1-\frac{1}{2^n}}} + \frac{1}{\sqrt{1-\frac{2}{2^n}}} + \frac{1}{\sqrt{1-\frac{3}{2^n}}} + โฆ + \frac{1}{\sqrt{1-\frac{2^n -1}{2^n}}}\right)\end{array} \)
is equal to
(A)
(B) 1
(C) 2
(D) โ2
Answer (C)
Sol.
\(\begin{array}{l}I = \displaystyle \lim_{n \to \infty}\frac{1}{2^n}\left(\frac{1}{\sqrt{1-\frac{1}{2^n}}} + \frac{1}{\sqrt{1-\frac{2}{2^n}}} + \frac{1}{\sqrt{1-\frac{3}{2^n}}} + โฆ + \frac{1}{\sqrt{1-\frac{2^n -1}{2^n}}}\right)\end{array} \)
Let 2n = t and if n โ โ then t โ โ
\(\begin{array}{l}I=\displaystyle \lim_{n \to \infty} \frac{1}{t}\left( \sum_{r=1}^{t=1}\frac{1}{\sqrt{1-\frac{r}{t}}}\right)\end{array} \)
\(\begin{array}{l}l=\int_{0}^{1}\frac{dx}{\sqrt{1-x}}=\int_{0}^{1}\frac{dx}{\sqrt{x}}~~~~~~~~ \int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx\end{array} \)
\(\begin{array}{l}=\left[2x^{\frac{1}{2}}\right]_0^1=2\end{array} \)
8. If A and B are two events such that
\(\begin{array}{l}P(A)=\frac{1}{3}, P(B)=\frac{1}{5}\ \text{and}\ P(A\cup B)=\frac{1}{2},\ \text{then}\end{array} \)
\(\begin{array}{l}P(A|Bโ) + P(B|Aโ|)\end{array} \)
is equal to
\(\begin{array}{l} (\text{A})\ \frac{3}{4}\end{array} \)
\(\begin{array}{l} (\text{B})\ \frac{5}{8}\end{array} \)
\(\begin{array}{l} (\text{C})\ \frac{5}{4}\end{array} \)
\(\begin{array}{l}\text{D})\ \frac{7}{8}\end{array} \)
Answer (B)
Sol.
\(\begin{array}{l}P(A)=\frac{1}{3}, P(B)=\frac{1}{5} \text{ and } P(A \cup B)=\frac{1}{2}\end{array} \)
\(\begin{array}{l}\therefore P(A \cap B) = \frac{1}{3}+\frac{1}{5}-\frac{1}{2}=\frac{1}{30}\end{array} \)
\(\begin{array}{l}\text{Now, } P(A|Bโ) +P(B|Aโ)=\frac{P(A \cap Bโ)}{P(Bโ)}+\frac{P(B \cap Aโ)}{P(Aโ)}\end{array} \)
\(\begin{array}{l}=\frac{\frac{9}{30}}{\frac{4}{5}}+\frac{\frac{5}{30}}{\frac{2}{3}}=\frac{5}{8}\end{array} \)
9. Let [t] denote the greatest integer less than or equal to t. Then the value of the integral
\(\begin{array}{l}\int_{-3}^{101}\left([\sin (\pi x)]+e^{[\cos(2\pi x)]}\right)dx\end{array} \)
is equal to
\(\begin{array}{l}(\text{A})\ \frac{52(1-e)}{e}\end{array} \)
\(\begin{array}{l}(\text{B})\ \frac{52}{e}\end{array} \)
\(\begin{array}{l}(\text{C})\ \frac{52(2+e))}{e}\end{array} \)
\(\begin{array}{l}(\text{D})\ \frac{104}{e}\end{array} \)
Answer (B)
Sol.
\(\begin{array}{l}I = \int_{-3}^{101}\left(\left[sin(\pi x)\right]+e^{\left[cos(2\pi x)\right]}\right)dx\end{array} \)
\(\begin{array}{l}\left[\sin \pi x\right]\ \text{is periodic with period 2}\end{array} \)
and \(\begin{array}{l}e^{\left[\cos 2\pi x\right]}\ \text{is periodic with period 1.}\end{array} \)
So,
\(\begin{array}{l}I = 52\int_{0}^{2}\left(\left[\sin(\pi x)\right]+e^{\left[\cos 2\pi x\right]}\right)dx\end{array} \)
\(\begin{array}{l}=52\left\{\int_{1}^{2}-1dx + \int_{1/4}^{3/4}e^{-1}dx + \int_{5/4}^{7/4}e^{-1}dx + \int_{0}^{1/4}e^0 dx + \int_{3/4}^{5/4}e^0 dx + \int_{7/4}^{2}e^0 dx\right\} \end{array} \)
\(\begin{array}{l}=\frac{52}{e} \end{array} \)
10. Let the point P(ฮฑ, ฮฒ) be at a unit distance from each of the two lines L1 : 3x โ 4y + 12 = 0 and L2 : 8x + 6y + 11 = 0. If P lies below L1 and above L2, then 100(ฮฑ + ฮฒ) is equal to
(A) โ14
(B) 42
(C) โ22
(D) 14
Answer (D)
Sol.
L1 : 3x โ 4y + 12 = 0
L2 : 8x + 6y + 11 = 0
Equation of angle bisector of L1 and L2 of angle containing origin
2(3x โ 4y + 12) = 8x + 6y + 11
2x + 14y โ 13 = 0 โฆ(i)
\(\begin{array}{l}\frac{3\alpha โ 4\beta + 12}{5}=1 \end{array} \)
\(\begin{array}{l}\Rightarrow 3\alpha -4 \beta + 7 =0~~~~~โฆ(ii)\end{array} \)
Solution of 2x + 14y โ 13 = 0 and 3x โ 4y + 7 = 0 gives the required point
\(\begin{array}{l}P\left(\alpha,\beta\right),\alpha = \frac{-23}{25},\beta = \frac{53}{50}\end{array} \)
\(\begin{array}{l}100(\alpha +\beta) = 14\end{array} \)
11. Let a smooth curve y = f(x) be such that the slope of the tangent at any point (x, y) on it is directly proportional to (-y/x). If the curve passes through the points (1, 2) and (8, 1), then
\(\begin{array}{l}\left|y\left(\frac{1}{8}\right)\right|\ \text{ is equal to}\end{array} \)
(A) 2 loge2
(B) 4
(C) 1
(D) 4 loge2
Answer (B)
Sol.
\(\begin{array}{l}\frac{dy}{dx}\propto \frac{-y}{x}\end{array} \)
\(\begin{array}{l}\frac{dy}{dx}= \frac{-ky}{x} \Rightarrow \int \frac {dy}{y}=-K\int\frac{dx}{x}\end{array} \)
ln | y | = -Kln | x | + C
If the above equation satisfies (1, 2) and (8, 1)
\(\begin{array}{l}ln2 = -K \times 0 + C \Rightarrow C = ln2\end{array} \)
\(\begin{array}{l} \ln1 = -K \ln8+ \ln2 \Rightarrow K= \frac{1}{3}\end{array} \)
\(\begin{array}{l}\text{So, at } x= \frac{1}{8}\end{array} \)
\(\begin{array}{l}\ln \left|y\right| = -\frac{1}{3}\ln\left(\frac{1}{8}\right)+\ln2 = 2\ln2\end{array} \)
|y| = 4
12. If the ellipse
\(\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\ \text{meets the line}\ \frac{x}{7}+\frac{y}{2\sqrt{6}} =1\end{array} \)
on the x-axis and the line \(\begin{array}{l}\frac{x}{7}-\frac{y}{2\sqrt{6}} =1\end{array} \)
on the y-axis, then the eccentricity of the ellipse is
\(\begin{array}{l}(\text{A})\ \frac{5}{7}\end{array} \)
\(\begin{array}{l}(\text{B})\ \frac{2\sqrt{6}}{7}\end{array} \)
\(\begin{array}{l}(\text{C})\ \frac{3}{7}\end{array} \)
\(\begin{array}{l}(\text{D})\ \frac{2\sqrt{5}}{7}\end{array} \)
Answer (A)
Sol.
\(\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\ \text{meets the line}\ \frac{x}{7} + \frac{7}{2\sqrt{6}}=1\ \text{on the x-axis.}\end{array} \)
So, a = 7
and
\(\begin{array}{l}\frac{x^2}{a^2} + \frac{y^2}{b^2}=1\ \text{meets the line}\ \frac{x}{7}-\frac{y}{2\sqrt{6}}=1\ \text{on the y-axis.}\end{array} \)
\(\begin{array}{l}\text{So, }b=2\sqrt{6}\end{array} \)
\(\begin{array}{l}\therefore e^2 = 1- \frac{b^2}{a^2}=1-\frac{24}{49}\end{array} \)
\(\begin{array}{l}e=\frac{5}{7}\end{array} \)
13. The tangents at the points A(1, 3) and B(1, โ1) on the parabola y2 โ 2x โ 2y = 1 meet at the point P. Then the area (in unit2) of the triangle PAB is :
(A) 4
(B) 6
(C) 7
(D) 8
Answer (D)
Sol. Given curve : y2 โ 2x โ 2y = 1.
Can be written as
\(\begin{array}{l}(y-1)^2 = 2(x + 1)\end{array} \)
And, the given information can be plotted as shown in figure.
Tangent at A : 2y โ x โ 5 = 0 {using T = 0}
Intersection with y = 1 is x = โ3
Hence, point P is (โ3, 1)
Taking advantage of symmetry
\(\begin{array}{l}\text{Area of } \Delta PAB = 2 \times \frac{1}{2}\times(1-(-3))\times(3-1)= 8 ~\text{sq. units}\end{array} \)
14. Let the foci of the ellipse
\(\begin{array}{l}\frac{x^2}{16}+\frac{y^2}{7}=1\ \text{and the hyperbola}\ \frac{x^2}{144}-\frac{y^2}{\alpha}=\frac{1}{25}\end{array} \)
coincide. Then the length of the latus rectum of the hyperbola is :
\(\begin{array}{l}(\text{A})\ \frac{32}{9}\end{array} \)
\(\begin{array}{l}(\text{B})\ \frac{18}{5}\end{array} \)
\(\begin{array}{l}(\text{C})\ \frac{27}{4}\end{array} \)
\(\begin{array}{l}(\text{D})\ \frac{27}{10}\end{array} \)
Answer (D)
Sol.
\(\begin{array}{l}\text{Ellipse} \colon \frac{x^2}{16}+\frac{y^2}{7}=1\end{array} \)
\(\begin{array}{l}\text{Eccentricity} =\sqrt{1-\frac{7}{16}}=\frac{3}{4}\end{array} \)
\(\begin{array}{l}\text{Foci }\equiv(\pm ae, 0)\equiv(\pm 3, 0)\end{array} \)
\(\begin{array}{l}\text{Hyperbola }\colon \frac{x^2}{\left(\frac{144}{25}\right)}-\frac{y^2}{\left(\frac{\alpha}{25}\right)}=1\end{array} \)
\(\begin{array}{l}\text{Eccentricity }=\sqrt{1+\frac{\alpha}{144}}=\frac{1}{12}\sqrt{144+ \alpha}\end{array} \)
\(\begin{array}{l}\text{Foci } \equiv (\pm ae, 0)\equiv\left(\pm \frac{12}{5}\cdot \frac{1}{12}\sqrt{144 + \alpha},0\right)\end{array} \)
If foci coincide then
\(\begin{array}{l}3=\frac{1}{5}\sqrt{144+\alpha}\Rightarrow \alpha = 81\end{array} \)
Hence, hyperbola is \(\begin{array}{l}\frac{x^2}{\left(\frac{12}{5}\right)^2}-\frac{y^2}{\left(\frac{9}{5}\right)^2}=1\end{array} \)
Length of latus rectum
\(\begin{array}{l}=2\cdot \frac{\frac{81}{25}}{\frac{12}{5}}=\frac{27}{10}\end{array} \)
15. A plane E is perpendicular to the two planes 2x โ 2y + z = 0 and x โ y + 2z = 4, and passes through the point P(1, โ1, 1). If the distance of the plane E from the point Q(a, a, 2) is 3โ2, then (PQ)2 is equal to
(A) 9
(B) 12
(C) 21
(D) 33
Answer (C)
Sol. First plane, P1 = 2x โ 2y + z = 0,
\(\begin{array}{l}\text{normal vector} \equiv \overline{n}_1=(2,-2,1)\end{array} \)
Second plane, P2 โก x โ y + 2z = 4,
\(\begin{array}{l}\text{normal vector}\equiv \overline{n}_2=(1,-1,2)\end{array} \)
\(\begin{array}{l}\text{Plane perpendicular to}\ P_1\ \text{and}\ P_2\ \text{will have normal vector}\ \overline{n}_3\end{array} \)
Where
\(\begin{array}{l}\overline{n}_3=(\overline{n}_1 \times \overline{n}_2)\end{array} \)
Hence,
\(\begin{array}{l}\overline{n}_3=(-3,-3,0)\end{array} \)
\(\begin{array}{l}\text{Equation of plane E through}\ P(1,-1,1)\ \text{and}\ \overline{n}_3\ \text{as normal vector}\end{array} \)
\(\begin{array}{l}(x-1, y+1, z-1)\cdot(-3,-3,0)=0\end{array} \)
\(\begin{array}{l}\Rightarrow x+y = 0 \equiv E\end{array} \)
Distance of PQ(a, a, 2) from E = |2a/โ2|
as given,
\(\begin{array}{l}\left|\frac{2a}{\sqrt{2}}\right|=3\sqrt{2}\Rightarrow a = \pm 3\end{array} \)
\(\begin{array}{l}\text{Hence, }Q\equiv (\pm 3, \pm 3, 2)\end{array} \)
Distance PQ
\(\begin{array}{l}=\sqrt{21}\Rightarrow (PQ)^2 = 21\end{array} \)
16. The shortest distance between the lines
\(\begin{array}{l}\frac{x+7}{-6}=\frac{y-6}{7}=z\ \text{and}\ \frac{7-x}{2}=y-2=z-6\end{array} \)
is
\(\begin{array}{l}(\text{A})\ 2\sqrt{29}\end{array} \)
(B) 1
\(\begin{array}{l}(\text{C})\ \sqrt{\frac{37}{29}}\end{array} \)
\(\begin{array}{l}(\text{D})\ \sqrt{\frac{29}{2}}\end{array} \)
Answer (A)
Sol.
\(\begin{array}{l}L_1:\frac{x+7}{-6}=\frac{y-6}{7}=\frac{z-0}{1}\end{array} \)
Any point on it
\(\begin{array}{l}\vec{a}_1(-7,6,0)\end{array} \)
and L1 is parallel to
\(\begin{array}{l}\vec{b}_1(-6,7,1)\end{array} \)
\(\begin{array}{l}L_2:\frac{x-7}{-2}=\frac{y-2}{1}=\frac{z-6}{1}\end{array} \)
Any point on it,
\(\begin{array}{l}\vec{a}_2(7,2,6)\end{array} \)
and L2 is parallel to
\(\begin{array}{l}\vec{b}_2(-2,1,1)\end{array} \)
Shortest distance between L1 and L2
\(\begin{array}{l}=\left|\frac{(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|=\left|\frac{(-14,4,-6)\cdot(3,2,4)}{\sqrt{9+4+16}}\right|\end{array} \)
\(\begin{array}{l}=2\sqrt{29}\end{array} \)
17. Let
\(\begin{array}{l}\vec{a}=\hat{i}-\hat{j}+2\hat{k}\end{array} \)
\(\begin{array}{l}\text{and let}\ \vec{b}\ \text{be a vector such that}\end{array} \)
\(\begin{array}{l}\vec{a}\times \vec{b} =2\hat{i}-\hat{k}\ \text{and}\ \vec{a}\cdot\vec{b}=3\end{array} \)
Then the projection of \(\begin{array}{l}\vec{b}\ \text{on the vector}\ \vec{a}- \vec{b}\ \text{is}:\end{array} \)
\(\begin{array}{l}(\text{A})\ \frac{2}{\sqrt{21}}\end{array} \)
\(\begin{array}{l}(\text{B})\ 2\sqrt{\frac{3}{7}}\end{array} \)
\(\begin{array}{l}(\text{C})\ \frac{2}{3}\sqrt{\frac{7}{3}}\end{array} \)
\(\begin{array}{l}(\text{D})\ \frac{2}{3}\end{array} \)
Answer (A)
Sol.
\(\begin{array}{l}\vec{a}=\hat{i}-\hat{j}+2\hat{k}\end{array} \)
\(\begin{array}{l}\vec{a}\times \vec{b}=2\hat{i}-\hat{k}\end{array} \)
\(\begin{array}{l}\vec{a}\cdot \vec{b}=3\end{array} \)
\(\begin{array}{l}\left|\vec{a}\times \vec{b}\right|^2+\left|\vec{a}\cdot \vec{b}\right|^2=\left|\vec{a}\right|^2\cdot \left|\vec{b}\right|^2\end{array} \)
\(\begin{array}{l}\Rightarrow 5 + 9 = 6 \left|\vec{b}\right|^2\end{array} \)
\(\begin{array}{l}\Rightarrow \left|\vec{b}\right|^2 = \frac{7}{3}\end{array} \)
\(\begin{array}{l}\left|\vec{a} โ \vec{b}\right|=\sqrt{\left|\vec{a}\right|^2 + \left|\vec{b}\right|^2-2\vec{a}\cdot\vec{b}} = \sqrt{\frac{7}{3}} \end{array} \)
\(\begin{array}{l}\text{projection of}|\vec{b}| \text{ on } \vec{a}-\vec{b}=\frac{\vec{b}\cdot(\vec{a}-\vec{b})}{|\vec{a}-\vec{b}|}\end{array} \)
\(\begin{array}{l}=\frac{\vec{b}\cdot \vec{a}-|\vec{b}|^2}{|\vec{a}-\vec{b}|}=\frac{3-\frac{7}{3}}{\sqrt{\frac{7}{3}}}\end{array} \)
\(\begin{array}{l}=\frac{2}{\sqrt{21}}\end{array} \)
18. If the mean deviation about median for the number 3, 5, 7, 2k, 12, 16, 21, 24 arranged in the ascending order, is 6 then the median is
(A) 11.5
(B) 10.5
(C) 12
(D) 11
Answer (D)
Sol.
\(\begin{array}{l}\text{Median } = \frac{2k+12}{2}=k+6\end{array} \)
\(\begin{array}{l}\text{Mean deviation}=\sum\frac{|x_i-M|}{n}=6\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{(k+3)+(k+1)+(k-1)+(6-k)+(6-k)+(10-k)+(15-k)+(18-k)}{8}\end{array} \)
\(\begin{array}{l}\therefore \frac{58-2k}{8}= 6\\k = 5\end{array} \)
\(\begin{array}{l}\text{Median } = \frac{2\times 5 + 12 }{2}=11\end{array} \)
19.
\(\begin{array}{l}2\sin\left(\frac{\pi}{22}\right)\sin\left(\frac{3\pi}{22}\right)\sin\left(\frac{5\pi}{22}\right)\sin\left(\frac{7\pi}{22}\right)\sin\left(\frac{9\pi}{22}\right)\end{array} \)
is equal to :
\(\begin{array}{l}\left(A\right)\frac{3}{16}\\\left(B\right)\frac{1}{16}\\\left(C\right)\frac{1}{32}\\\left(D\right)\frac{9}{32}\end{array} \)
Answer (B)
Sol.
\(\begin{array}{l}2\sin\left(\frac{\pi}{22}\right)\sin\left(\frac{3\pi}{22}\right)\sin\left(\frac{5\pi}{22}\right)\sin\left(\frac{7\pi}{22}\right)\sin\left(\frac{9\pi}{22}\right)\end{array} \)
\(\begin{array}{l}=2\sin\left(\frac{11\pi-10\pi}{22}\right)\sin\left(\frac{11\pi-8\pi}{22}\right)\sin\left(\frac{11\pi-6\pi}{22}\right)\sin\left(\frac{11\pi-4\pi}{22}\right)\sin\left(\frac{11\pi-2\pi}{22}\right)\end{array} \)
\(\begin{array}{l}=2\cos\frac{\pi}{11}\cos\frac{2\pi}{11}\cos\frac{3\pi}{11}\cos\frac{4\pi}{11}\cos\frac{5\pi}{11}\end{array} \)
\(\begin{array}{l}=\frac{2\sin\frac{32\pi}{11}}{2^5\sin\frac{\pi}{11}}\end{array} \)
\(\begin{array}{l}=\frac{1}{16}\end{array} \)
20. Consider the following statements :
P : Ramu is intelligent.
Q : Ramu is rich.
R : Ramu is not honest.
The negation of the statement โRamu is intelligent and honest if and only if Ramu is not richโ can be expressed as :
(A) ((P โง (~ R)) โง Q) โง ((~ Q) โง ((~ P) โจ R))
(B) ((P โง R) โง Q) โจ ((~ Q) โง ((~ P) โจ (~ R)))
(C) ((P โง R) โง Q) โง ((~ Q) โง (( ~ P) โจ (~ R)))
(D) ((P โง (~ R)) โง Q) โจ ((~ Q) โง ((~ P) โง R))
Answer (D)
Sol. P : Ramu is intelligent
Q : Ramu is rich
R : Ramu is not honest
Given statement, โRamu is intelligent and honest if and only if Ramu is not richโ
= (P โง ~ R) โ ~ Q
So, negation of the statement is
~ [(P โง ~ R) โ ~ Q]
= ~ [{~ (P โง ~ R) โจ ~ Q} โง {Q โจ (P โง ~ R)}]
= ((P โง ~ R) โง Q) โจ (~ Q โง (~ P โจ R))
SECTION โ 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, โ00.33, โ00.30, 30.27, โ27.30) 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}. Define
\(\begin{array}{l}B=\{T \subseteq A :\text{ either } 1 \notin T\ or\ 2 \in T\}\end{array} \)
and \(\begin{array}{l}C=\{T \subseteq A : T\text{ the sum of all the elements of } T \text{ is a prime number} \}\end{array} \)
. Then the number of elements in the set B โช C is ______.
Answer (107)
Sol. โต (B โช C)โฒ = Bโฒ โฉ Cโฒ
Bโฒ is a set containing sub sets of A containing element 1 and not containing 2.
And Cโฒ is a set containing subsets of A whose sum of elements is not prime.
So, we need to calculate number of subsets of
{3, 4, 5, 6, 7} whose sum of elements plus 1 is composite.
Number of such 5 elements subset = 1
Number of such 4 elements subset = 3 (except selecting 3 or 7)
Number of such 3 elements subset = 6 (except selecting {3, 4, 5}, {3, 6, 7}, {4, 5, 7} or {5, 6, 7})
Number of such 2 elements subset = 7 (except selecting {3, 7}, {4, 6}, {5, 7})
Number of such 1 elements subset = 3 (except selecting {4} or {6})
Number of such 0 elements subset = 1
n(BโฒโฉCโฒ) = 21 โ n(BโชC) = 27 โ 21 = 107
2. Let f(x) be a quadratic polynomial with leading coefficient 1 such that f(0) = p, p โ 0, and f(1) = 1/3 . If the equations f(x) = 0 and fo fo fo f(x) = 0 have a common real root, then f(โ3) is equal to ______.
Answer (25)
Sol. Let f(x) = (x โ ฮฑ) (x โ ฮฒ)
It is given that f(0) = p โ ฮฑฮฒ = p
and
\(\begin{array}{l}f(1)=\frac{1}{3}\Rightarrow (1-\alpha)(1-\beta)=\frac{1}{3}\end{array} \)
Now, let us assume that ฮฑ is the common root of f(x) = 0 and fofofof(x) = 0
fofofof(x) = 0
โ fofof(0) = 0
โ fof(p) = 0
So, f(p) is either ฮฑ or ฮฒ.
(p โ ฮฑ) (p โ ฮฒ) = ฮฑ
(ฮฑฮฒ โ ฮฑ) (ฮฑฮฒ โ ฮฒ) = ฮฑ โ (ฮฒ โ 1) (ฮฑ โ 1) ฮฒ = 1
So, ฮฒ = 3 (โต ฮฑ โ 0)
\(\begin{array}{l}(1-\alpha)(1-3)=\frac{1}{3}\end{array} \)
\(\begin{array}{l}\alpha=\frac{7}{6}\end{array} \)
\(\begin{array}{l}f(x)=\left(x-\frac{7}{6}\right)(x-3)\end{array} \)
\(\begin{array}{l}f(-3)=\left(-3-\frac{7}{6}\right)(3-3)=25\end{array} \)
3. Let
\(\begin{array}{l}A = \begin{bmatrix}1&a&a\\0&1&b\\0&0&1\\\end{bmatrix},a,b \in \mathbb{R}\end{array} \)
. If for some \(\begin{array}{l}n\in N, A^n = \begin{bmatrix}1 & 48 & 2160 \\0 & 1 & 96 \\0 & 0 & 1 \\\end{bmatrix}\end{array} \)
then n + a + b is equal to __________.
Answer (24)
Sol.
\(\begin{array}{l}A=\begin{bmatrix} 1& 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \\\end{bmatrix} + \begin{bmatrix}0 & a & a \\0 & 0 & b \\0 & 0 & 0 \\\end{bmatrix} = I + B\end{array} \)
\(\begin{array}{l}B^2=\begin{bmatrix} 0& a & a \\0 & 0 & b \\0 & 0 & 0 \\\end{bmatrix}+\begin{bmatrix}0 & a & a \\0 & 0 & b \\0 & 0 & 0 \\\end{bmatrix}=\begin{bmatrix} 0& 0 & ab \\0 & 0 & 0 \\0 & 0 & 0 \\\end{bmatrix}\end{array} \)
B3 = 0
โด An = (1 + B)n = nC0I + nC1 B + nC2 B2 + nC3 B3 + โฆ.
\(\begin{array}{l}=\begin{bmatrix} 1& 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \\\end{bmatrix}+\begin{bmatrix}0 & na & na \\0 & 0 & nb \\0 & 0 & 0 \\\end{bmatrix}+\begin{bmatrix} 0& 0 & \frac{n(n-1)ab}{2} \\0 & 0 & 0 \\0 & 0 & 0 \\\end{bmatrix}\end{array} \)
\(\begin{array}{l}=\begin{bmatrix} 1& na & na+\frac{n(n-1)}{2}ab \\0 & 1 & nb \\0 & 0 & 1 \\\end{bmatrix}=\begin{bmatrix}1 & 48 & 2160 \\0 & 1 & 48 \\0 & 0 & 1 \\\end{bmatrix}\end{array} \)
On comparing we get na = 48, nb = 96 and
\(\begin{array}{l}na+\frac{n(n-1)}{2}ab=2160\end{array} \)
โ a = 4, n = 12 and b = 8
n + a + b = 24
4. The sum of the maximum and minimum values of the function f(x) = |5x โ 7| + [x2 + 2x] in the interval [5/4, 2], where [t] is the greatest integer โค t, is ____.
Answer (15)
Sol. f(x) = |5x โ 7| + [x2 + 2x]
= |5x โ 7| + [(x + 1)2] โ 1
\(\begin{array}{l}\text{Critical points of}\ f(x)=\frac{7}{5},\sqrt{5}-1, \sqrt{6}-1, \sqrt{7}-1, \sqrt{8}-1,2\end{array} \)
โด Maximum or minimum value of f(x) occur at critical points or boundary points
\(\begin{array}{l}\therefore f\left(\frac{5}{4}\right)=\frac{3}{4}+4=\frac{19}{4}\end{array} \)
\(\begin{array}{l}f\left(\frac{7}{5}\right)=0+4=4\end{array} \)
as both |5x โ 7| and x2 + 2x are increasing in nature after x = 7/5.
โด f(2) = 3 + 8 = 11
\(\begin{array}{l}\therefore f\left(\frac{7}{5}\right)_{\text{min}}=4 \text{ and }f(2)_{\text{max}}=11\end{array} \)
Sum is 4 + 11 = 15
5. Let y = y(x) be the solution of the differential equation
\(\begin{array}{l}\frac{dy}{dx}=\frac{4y^3+2yx^2}{3xy^2+x^3},y(1)=1.\end{array} \)
If for some n โ N, y(2) โ [n โ 1, n), then n is equal to _________.
Answer (3)
Sol.
\(\begin{array}{l}\frac{dy}{dx}=\frac{y}{x}\frac{(4y^2+2x^2)}{(3y^2+x^2)}\end{array} \)
Put y = vx
\(\begin{array}{l}\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\end{array} \)
\(\begin{array}{l}\Rightarrow v+x\frac{dv}{dx}=\frac{v(4v^2+2)}{(3v^2+1)}\end{array} \)
\(\begin{array}{l}\Rightarrow x\frac{dv}{dx}=v\left(\frac{(4v^2+2-3v^2-1)}{3v^2+1}\right)\end{array} \)
\(\begin{array}{l}\Rightarrow \int (3v^2+1)\frac{dv}{v^3+v}=\int\frac{dx}{x}\end{array} \)
โ ln|v3 +v| = lnx + c
\(\begin{array}{l}\Rightarrow \ln \left|\left(\frac{y}{x}\right)^3+\left(\frac{y}{x}\right)\right|=\ln x + C\end{array} \)
โ y(1) = 1
โ C = ln2
โด for y(2)
\(\begin{array}{l}\ln \left(\frac{y^3}{8}+\frac{y}{2}\right)=2\ln 2 \Rightarrow \frac{y^3}{8}+\frac{y}{2}=4\end{array} \)
โ [y(2)] = 2
โ n = 3
6. Let f be a twice differentiable function on R. If fโ(0) = 4 and
\(\begin{array}{l}f(x)+\int_{0}^{x}(x-1)f'(t)dt=(e^{2x}+e^{-2x})\cos 2x + \frac{2}{a}x,\end{array} \)
then (2a + 1)5 a2 is equal to ________________.
Answer (8)
Sol.
\(\begin{array}{l}\because f(x)+\int_{0}^{x}(x-t)f'(t)dt=(e^{2x}+e^{-2x})\cos 2x + \frac{2x}{a}~~~~~โฆ(i)\end{array} \)
Here f(0) = 2 โฆ(ii)
On differentiating equation (i) w.r.t. x we get :
\(\begin{array}{l}f'(x)+\int_{0}^{x}f'(t)dt+xf'(x)-xf'(x)\\=2(e^{2x}-e^{-2x}) \cos 2x โ 2(e^{2x}+e^{-2x})\sin 2x + \frac{2}{a}\end{array} \)
โ fโฒ(x) + f(x) โ f(0) = 2(e2x โ eโ2x)cos2x โ 2 (e2x + eโ2x) sin 2x + (2/a)
Replace x by 0 we get :
\(\begin{array}{l}\Rightarrow 4 = \frac{2}{a} \Rightarrow a = \frac{1}{2}\end{array} \)
\(\begin{array}{l}\therefore (2a+1)^5 \cdot a^2 = 2^5 \cdot \frac{1}{2^2}=2^3=8\end{array} \)
7. Let
\(\begin{array}{l}a_n=\int_{-1}^{n}\left(1+\frac{x}{2}+\frac{x^2}{3}+โฆ+ \frac{x^{n-1}}{n}\right)dx\end{array} \)
for every n โ N. Then the sum of all the elements of the set {n โ N : an โ (2, 30)} is ________________ .
Answer (5)
Sol.
\(\begin{array}{l}\because a_n=\int_{-1}^{n}\left(1+\frac{x}{2}+\frac{x^2}{3 }+ โฆ + \frac{x^{n-1}}{n}\right)dx\end{array} \)
\(\begin{array}{l}=\left[x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+โฆ+\frac{x^n}{n^2}\right]_{-1}^{n}\end{array} \)
\(\begin{array}{l}a_n = \frac{n+1}{1^2}+\frac{n^2-1}{2^2}+\frac{n^3+1}{3^2}+\frac{n^4-1}{4^2}+โฆ+\frac{n^n+(-1)^{n+1}}{n^2}\end{array} \)
\(\begin{array}{l}\text{Here } a_1 =2, a_2 =\frac{2+1}{1}+\frac{2^2-1}{2}=3+\frac{3}{2}=\frac{9}{2}\end{array} \)
\(\begin{array}{l}a_3=4+2+\frac{28}{9}=\frac{100}{9}\end{array} \)
\(\begin{array}{l}a_4 = 5 + \frac{15}{4}+\frac{65}{9}+\frac{255}{16}>31\end{array} \)
โด The required set is {2, 3}. โต an โ(2, 30)
โด Sum of elements = 5.
8. If the circles x2 + y2 + 6x + 8y + 16 = 0 and
\(\begin{array}{l}x^2 + y^2 + 2(3-\sqrt{3})x + 2(4-\sqrt{6})y=k+6\sqrt{3}+8\sqrt{6},k > 0,\end{array} \)
touch internally at the point P(ฮฑ, ฮฒ), then \(\begin{array}{l}(\alpha + \sqrt{3})^2+(\beta + \sqrt{6})^2\end{array} \)
is equal to ______.
Answer (25)
Sol. The circle x2 + y2 + 6x + 8y + 16 = 0 has centre
(โ3, โ4) and radius 3 units.
The circle
\(\begin{array}{l}x^2 + y^2 + 2(3-\sqrt{3})x + 2(4-\sqrt{6})y = k + 6\sqrt{3}+8\sqrt{6}, k> 0\end{array} \)
\(\begin{array}{l}\text{has centre}\ (\sqrt{3}-3, \sqrt{6}-4)\end{array} \)
\(\begin{array}{l}\text{and radius}\ \sqrt{k+34}\end{array} \)
โต These two circles touch internally hence
\(\begin{array}{l}\sqrt{3+6}=\left|\sqrt{k+34}-3\right|\end{array} \)
Here, k = 2 is only possible (โต k > 0)
Equation of common tangent to two circles is
\(\begin{array}{l}2\sqrt{3}x + 2 \sqrt{6}y+16+6\sqrt{3}+8\sqrt{6}+k =0\end{array} \)
โต k = 2 then equation is
\(\begin{array}{l}x+\sqrt{2}y+3+4\sqrt{2}+3\sqrt{3}=0~~~~~~โฆ(i)\end{array} \)
โต (ฮฑ, ฮฒ) are foot of perpendicular from (โ3, โ4) to line (i) then
\(\begin{array}{l}\frac{\alpha + 3}{1}=\frac{\beta + 4}{\sqrt{2}}=\frac{-(-3-4\sqrt{2}+3+4\sqrt{2}+3\sqrt{3})}{1+2}\end{array} \)
\(\begin{array}{l}\therefore \alpha + 3 = \frac{\beta + 4}{\sqrt{2}}=-\sqrt{3}\end{array} \)
\(\begin{array}{l}\Rightarrow (\alpha + \sqrt{3})^2= 9 \text{ and }(\beta + \sqrt{6})^2 = 16\end{array} \)
\(\begin{array}{l}\therefore (\alpha + \sqrt{3})^2 + (\beta + \sqrt{6})^2=25\end{array} \)
9. Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve 4x3 โ 3xy2 + 6x2 โ 5xy โ 8y2 + 9x + 14 = 0 at the point (โ2, 3) be A. Then 8A is equal to _______.
Answer (170)
Sol. 4x3 โ 3xy2 + 6x2 โ 5xy โ 8y2 + 9x + 14 = 0 differentiating both sides we get
12x2 โ 3y2 โ 6xyyโ + 12x โ 5y -5xyโ โ 16yyโ + 9 = 0
\(\begin{array}{l}\downarrow(-2, 3)\end{array} \)
โ 48 โ 27 + 36yโ โ 24 โ 15 + 10yโ โ 48yโ + 9 = 0
โ 2yโ = โ 9
\(\begin{array}{l}\Rightarrow m_T = \frac{-9}{2}\ \& \ m_N = \frac{2}{9}\end{array} \)
\(\begin{array}{l}\therefore \text{ Area } = \frac{1}{2}\times \text{ Base } \times \text{ Hight}\end{array} \)
\(\begin{array}{l}A=\frac{1}{2}\times\left(\frac{-4}{3}+\frac{31}{2}\right)(3)=\frac{1}{2}\left(\frac{85}{6}\right)\cdot 3 = \frac{85}{4}\\= 8A = 170\end{array} \)
10. Let x = sin(2tanโ1 ฮฑ) and
\(\begin{array}{l}y=\sin\left(\frac{1}{2}\tan^{-1}\frac{4}{3}\right).\end{array} \)
\(\begin{array}{l}\text{If}\ S=\{\alpha \in \mathbb{R} : y^2 = 1 -x \},\ \text{then}\ \sum_{a\in S}16\alpha^3\end{array} \)
is equal to _______.
Answer (130)
Sol.
\(\begin{array}{l}\because x = \sin(2\tan^{-1}\alpha)=\frac{2\alpha}{1+\alpha^2}~~~~~โฆ(i)\end{array} \)
and
\(\begin{array}{l}y=\sin\left(\frac{1}{2}\tan^{-1}\frac{4}{3}\right)=\sin\left(\sin^{-1}\frac{1}{\sqrt{5}}\right)=\frac{1}{\sqrt{5}}\end{array} \)
Now,
\(\begin{array}{l}y^2 = 1 โ x\end{array} \)
\(\begin{array}{l}\frac{1}{5}=1-\frac{2\alpha}{1+\alpha^2}\end{array} \)
\(\begin{array}{l}\Rightarrow 1 + \alpha^2 = 5 + 5\alpha^2-10\alpha\end{array} \)
\(\begin{array}{l}\Rightarrow2\alpha^2 โ 5\alpha + 2 =0\end{array} \)
\(\begin{array}{l}\therefore \alpha = 2, \frac{1}{2}\end{array} \)
\(\begin{array}{l}\therefore \sum_{a\in S}16\alpha^3 = 16 \times2^3 + 16 \times \frac{1}{2^3}\end{array} \)
= 130
JEE Main 2022 July 25th Shift 2 Paper Analysis
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