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JEE Main 2022 29th July Shift 1 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. Let R be a relation from the set {1, 2, 3, ….., 60} to itself such that R = {(a, b) : b = pq, where p, q≥ 3 are prime numbers}. Then, the number of elements in R is :
(A) 600
(B) 660
(C) 540
(D) 720
Answer (B)
Sol. b can take its values as 9, 15, 21, 33, 39, 51, 57, 25, 35, 55, 49
b can take these 11 values
and a can take any of 60 values
So, number of elements in R = 60 × 11
= 660
2. If z = 2 + 3i, then \(\begin{array}{l}z^5+(\bar{z})^5\end{array} \)
is equal to :
(A) 244
(B) 224
(C) 245
(D) 265
Answer (A)
Sol. z = (2 + 3i)
\(\begin{array}{l} \Rightarrow\ z^5=\left(2+3i\right)\left(\left(2+3i\right)^2\right)^2\end{array} \)
= (2 + 3i) (-5 + 12i)2
= (2 + 3i)(-119 – 120i)
= – 238 – 240i – 357i + 360
= 122 – 597i
\(\begin{array}{l} \overline{z}^5=122+597i\end{array} \)
\(\begin{array}{l} z^5+\overline{z}^5=244\end{array} \)
3. Let A and B be two 3 × 3 non-zero real matrices such that AB is a zero matrix. Then
(A) the system of linear equations AX = 0 has a unique solution
(B) the system of linear equations AX = 0 has infinitely many solutions
(C) B is an invertible matrix
(D) adj(A) is an invertible matrix
Answer (B)
Sol. AB is zero matrix
⇒ |A| = |B| = 0
So neither A nor B is invertible
If |A| = 0
⇒ |adj A| = 0 so adj A is not invertible
AX = 0 is homogeneous system and |A| = 0
So, it is having infinitely many solutions.
4. If \(\begin{array}{l} \frac{1}{\left(20-a\right)\left(40-a\right)}+\frac{1}{\left(40-a\right)\left(60-a\right)}+\cdots + \frac{1}{\left(180-a\right)\left(200-a\right)}=\frac{1}{256},\end{array} \)
then the maximum value of a is :
(A) 198
(B) 202
(C) 212
(D) 218
Answer (C)
Sol.
\(\begin{array}{l} \frac{1}{20}\left(\frac{1}{20-a}-\frac{1}{40-a}+\frac{1}{40-a}-\frac{1}{60-a}+\cdots + \frac{1}{180-a}-\frac{1}{200-a}\right)=\frac{1}{256}\end{array} \)
\(\begin{array}{l} \Rightarrow\ \frac{1}{20}\left(\frac{1}{20-a}-\frac{1}{200-a}\right)=\frac{1}{256} \end{array} \)
\(\begin{array}{l} \Rightarrow\ \frac{1}{20}\left(\frac{180}{\left(20-a\right)\left(200-a\right)}\right)=\frac{1}{256}\end{array} \)
⇒ (20 – a)(200 – a) = 9.256
OR a2 – 220a + 1696 = 0
⇒ a = 212, 8
5. If \(\begin{array}{l} \displaystyle \lim_{x \to 0}\frac{\alpha e^x+\beta e^{-x}+\gamma\sin x}{x\sin^2 x}=\frac{2}{3},\end{array} \)
where α, β, γ∈R, then which of the following is NOT correct?
(A) α2 + β2 + γ2 = 6
(B) αβ + βγ + γα + 1 = 0
(C) αβ2 + βγ2 + γα2 + 3 = 0
(D) α2 – β2 + γ2 = 4
Answer (C)
Sol.
\(\begin{array}{l} \displaystyle \lim_{x \to 0}\frac{\alpha e^x+\beta e^{-x}+\gamma\sin x}{x\sin^2 x}=\frac{2}{3}\end{array} \)
⇒ α + β = 0 (to make indeterminant form) …(i)
Now,
\(\begin{array}{l} \displaystyle \lim_{x \to 0}\frac{\alpha e^x-\beta e^{-x}+\gamma\cos x}{3x^2}=\frac{2}{3}\end{array} \)
(Using L-H Rule)
⇒ α – β + γ = 0 (to make indeterminant form) …(ii)
Now,
\(\begin{array}{l} \displaystyle \lim_{x \to 0}\frac{\alpha e^x+\beta e^{-x}-\gamma\sin x}{6x}=\frac{2}{3} \end{array} \)
(Using L-H Rule)
\(\begin{array}{l} \Rightarrow \frac{\alpha-\beta-\gamma}{6}=\frac{2}{3}\end{array} \)
⇒ α – β – γ = 4 …(iii)
⇒ γ = –2
and (i) + (ii)
2α = –γ
⇒ α = 1 and β = –1
and αβ2 + βγ2 + γα2 + 3 = 1 – 4 – 2 + 3 = –2
6. The integral \(\begin{array}{l} \displaystyle\int\limits_0^\frac{\pi}{2}\frac{1}{3+2\sin x+\cos x}dx \end{array} \)
is equal to
(A) tan–1(2)
\(\begin{array}{l} (\text{B})\ \tan^{-1}\left(2\right)-\frac{\pi}{4}\end{array} \)
\(\begin{array}{l} (\text{C})\ \frac{1}{2}\tan^{-1}\left(2\right)-\frac{\pi}{8}\end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{1}{2}\end{array} \)
Answer (B)
Sol.
\(\begin{array}{l} I=\displaystyle\int\limits_0^{\pi/2}\frac{1}{3+2\sin x+\cos x}dx\end{array} \)
\(\begin{array}{l} =\displaystyle\int\limits_0^{\pi/2}\frac{\left(1+\tan^2x/2\right)dx}{3\left(1+\tan^2 x/2\right)+2\left(2\tan x/2\right)+\left(1-\tan^2 x/2\right)}\end{array} \)
Let tan x/2 = t ⇒ sec2x/2 dx = 2 dt
\(\begin{array}{l} I=\displaystyle\int\limits_0^1\frac{2dt}{4+2t^2+4t}\end{array} \)
\(\begin{array}{l} =\displaystyle\int\limits_0^1\frac{dt}{t^2+2t+2}=\displaystyle\int\limits_0^1\frac{dt}{\left(t+1\right)^2+1} \end{array} \)
\(\begin{array}{l} \left.=\tan^{-1}\left(t+1\right)\right|_0^1=\tan^{-1}2-\frac{\pi}{4}\end{array} \)
7. Let the solution curve y = y(x) of the differential equation \(\begin{array}{l} \left(1+e^{2x}\right)\left(\frac{dy}{dx}+y\right)=1\end{array} \)
pass through the point (0, π/2). Then, \(\begin{array}{l} \displaystyle \lim_{x \to \infty}e^xy\left(x\right)\end{array} \)
is equal to
\(\begin{array}{l} (\text{A})\ \frac{\pi}{4} \end{array} \)
\(\begin{array}{l} (\text{B})\ \frac{3\pi}{4} \end{array} \)
\(\begin{array}{l} (\text{C})\ \frac{\pi}{2} \end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{3\pi}{2} \end{array} \)
Answer (B)
Sol.
\(\begin{array}{l} \text{D.E.}\left(1+e^{2x}\right)\left(\frac{dy}{dx}+y\right)=1\end{array} \)
\(\begin{array}{l} \Rightarrow\ \frac{dy}{dx}+y=\frac{1}{1+e^{2x}}\end{array} \)
\(\begin{array}{l} \text{I.F.}=e^{\int 1.dx}=e^x \end{array} \)
∴ Solution
\(\begin{array}{l} e^xy\left(x\right)=\int\frac{e^x}{1+e^{2x}}dx\end{array} \)
\(\begin{array}{l} \Rightarrow\ e^xy\left(x\right)=\tan^{-1}\left(e^x\right)+C \end{array} \)
∵ It passes through
\(\begin{array}{l} \left(0,\frac{\pi}{2}\right), C=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}\end{array} \)
\(\begin{array}{l} \therefore\ \displaystyle \lim_{x \to \infty}e^xy\left(x\right)=\displaystyle \lim_{x \to \infty}\tan^{-1}\left(e^x\right)+\frac{\pi}{4}\end{array} \)
\(\begin{array}{l} =\frac{3\pi}{4}\end{array} \)
8. Let a line L pass through the point intersection of the lines bx + 10y – 8 = 0 and \(\begin{array}{l}2x – 3y = 0, b \in R – \left\{ \frac{4}{3}\right\}.\end{array} \)
If the line L also passes through the point (1, 1) and touches the circle 17(x2 + y2) = 16, then the eccentricity of the ellipse (x2/5) + (y2/5) = 1 is
\(\begin{array}{l} (\text{A})\ \frac{2}{\sqrt{5}} \end{array} \)
\(\begin{array}{l} (\text{B})\ \sqrt{\frac{3}{5}} \end{array} \)
\(\begin{array}{l} (\text{C})\ \frac{1}{\sqrt{5}}\end{array} \)
\(\begin{array}{l} (\text{D})\ \sqrt{\frac{2}{5}}\end{array} \)
Answer (B)
Sol. L1 :bx + 10y – 8 = 0, L2 : 2x – 3y = 0
then L : (bx + 10y – 8) + λ(2x – 3y) = 0
∵ It passes through (1, 1)
∴ b + 2 – λ = 0 ⇒ λ = b + 2
and touches the circle x2 + y2 = 16/17
\(\begin{array}{l} \left|\frac{8^2}{\left(2\lambda+b\right)^2+\left(10-3\lambda\right)^2}\right|=\frac{16}{17}\end{array} \)
\(\begin{array}{l} \Rightarrow\ 4\lambda^2+b^2+4b\lambda+100+9\lambda^2-60\lambda=68\end{array} \)
\(\begin{array}{l} \Rightarrow\ 13\left(b+2\right)^2+b^2+4b\left(b+2\right)-60\left(b+2\right)+32=0 \end{array} \)
\(\begin{array}{l} \Rightarrow\ 18b^2=36\ \therefore b^2=2\end{array} \)
\(\begin{array}{l}\therefore \text{Eccentricity of ellipse}\ \frac{x^2}{5}+\frac{y^2}{b^2}=1\ \text{is}\end{array} \)
\(\begin{array}{l}e=\sqrt{1-\frac{2}{5}}=\sqrt{\frac{3}{5}}\end{array} \)
9. If the foot of the perpendicular from the point A(–1, 4, 3) on the plane P : 2x + my + nz = 4, is (-2, 7/2, 3/2), then the distance of the point A from the plane P, measured parallel to a line with direction ratios 3, –1, –4, is equal to
(A) 1
\(\begin{array}{l} (\text{B})\ \sqrt{26}\end{array} \)
\(\begin{array}{l} (\text{C})\ 2\sqrt{2}\end{array} \)
\(\begin{array}{l} (\text{D})\ \sqrt{14}\end{array} \)
Answer (B)
Sol. (-2, 7/2, 3/2) satisfies the plane P : 2x + my + nz = 4
\(\begin{array}{l} -4+\frac{7m}{2}+\frac{3n}{2}=4\Rightarrow 7m+3n=16\cdots \left(i\right)\end{array} \)
Line joining A(–1, 4, 3) and (-2, 7/2, 3/2) is perpendicular to P : 2x + my + nz = 4
\(\begin{array}{l} \frac{1}{2}=\frac{\frac{1}{2}}{m}=\frac{\frac{3}{2}}{n}\Rightarrow m=1~\&~n=3\end{array} \)
Plane P : 2x + y + 3z = 4
Distance of P from A(–1, 4, 3) parallel to the line
\(\begin{array}{l} \frac{x+1}{3}=\frac{y-4}{-1}=\frac{z-3}{-4}:L\end{array} \)
for point of intersection of P&L
2(3r – 1) + (–r + 4) + 3(–4r + 3) = 4 ⇒r = 1
Point of intersection : (2, 3, –1)
Required distance
\(\begin{array}{l} =\sqrt{3^2+1^2+4^2} \end{array} \)
\(\begin{array}{l} =\sqrt{26}\end{array} \)
10. Let \(\begin{array}{l} \vec{a}=3\hat{i}+\hat{j}\ \text{and}\ \vec{b}=\hat{i}+2\hat{j}+\hat{k}.\end{array} \)
\(\begin{array}{l}\text{Let}\ \vec{c}\ \text{be a vector satisfying}\ \vec{a}\times\left(\vec{b}\times\vec{c}\right)=\vec{b}+\lambda\vec{c}.\end{array} \)
\(\begin{array}{l}\text{If}\ \vec{b}\ \text{and}\ \vec{c}\ \text{are non-parallel},\end{array} \)
then the value of λ is
(A) –5
(B) 5
(C) 1
(D) –1
Answer (A)
Sol.
\(\begin{array}{l} \vec{a}=3\hat{i}+\hat{j}~\& \vec{b}=\hat{i}+2\hat{j}+\hat{k}\end{array} \)
\(\begin{array}{l} \vec{a}\times\left(\vec{b}\times\vec{c}\right) =\left(\vec{a}\cdot\vec{c}\right)\vec{b}-\left(\vec{a}\cdot\vec{b}\right)\vec{c}=\vec{b}+\lambda\vec{c}\end{array} \)
\(\begin{array}{l}\text{If}\ \vec{b}~\& \vec{c}\ \text{are non-parallel},\ \text{then}\end{array} \)
\(\begin{array}{l} \vec{a}\cdot\vec{c}=1~\&~\vec{a}\cdot \vec{b}=-\lambda\end{array} \)
but
\(\begin{array}{l} \vec{a}\cdot\vec{b}=5\Rightarrow \lambda =-5\end{array} \)
11. The angle of elevation of the top of a tower from a point A due north of it is α and from a point B at a distance of 9 units due west of A is \(\begin{array}{l}\cos^{-1}\left(\frac{3}{\sqrt{13}}\right).\end{array} \)
If the distance of the point B from the tower is 15 units, then cot α is equal to :
\(\begin{array}{l} (\text{A})\ \frac{6}{5}\end{array} \)
\(\begin{array}{l} (\text{B})\ \frac{9}{5}\end{array} \)
\(\begin{array}{l} (\text{C})\ \frac{4}{3}\end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{7}{3}\end{array} \)
Answer (A)
Sol.
\(\begin{array}{l} NA=\sqrt{15^2-9^2}=12\end{array} \)
\(\begin{array}{l} \frac{h}{15}=\tan\theta=\frac{2}{3}\end{array} \)
h = 10 units
\(\begin{array}{l} \cot\alpha=\frac{12}{10}=\frac{6}{5}\end{array} \)
12. The statement (p∧q) ⇒ (p∧r) is equivalent to :
(A) q⇒ (p∧r)
(B) p⇒ (p∧r)
(C) (p∧r) ⇒ (p∧q)
(D) (p∧q) ⇒ r
Answer (D)
Sol.
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A |
B |
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| p |
q |
r |
p ∧ q |
p ∧ r |
A → B |
q → B |
p → B |
B → A |
A → r |
| T |
T |
T |
T |
T |
T |
T |
T |
T |
T |
| T |
F |
T |
F |
T |
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T |
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F |
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| F |
T |
T |
F |
F |
T |
F |
T |
T |
T |
| F |
F |
T |
F |
F |
T |
T |
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T |
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| T |
T |
F |
T |
F |
F |
F |
F |
T |
F |
| T |
F |
F |
F |
F |
T |
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F |
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| F |
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F |
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| F |
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T |
(p∧ q) ⇒ (p∧ r) is equivalent to (p∧ q) ⇒ r
13. Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be P(1, 1). If the line AP intersects the line BC at the point Q(k1, k2), then k1 + k2 is equal to :
(A) 2
\(\begin{array}{l} (\text{B})\ \frac{4}{7} \end{array} \)
\(\begin{array}{l} (\text{B})\ \frac{2}{7} \end{array} \)
(D) 4
Answer (B)
Sol.
Let D be mid-point of AC, then
\(\begin{array}{l} \frac{b+3}{2}=1\Rightarrow b=-1\end{array} \)
Let E be mid-point of BC,
\(\begin{array}{l} \frac{5-b}{b-a}\cdot\frac{\frac{\left(3+b\right)}{2}}{\frac{a+b}{2}-1}=-1 \end{array} \)
On Putting b = –1, we get a = 5 or –3
But a = 5 is rejected as ab> 0
A(–3, 3), B(–1, 5), C(–3, –1), P(1, 1)
Line BC ⇒ y = 3x + 8
\(\begin{array}{l}Line\ AP\Rightarrow y=\frac{3-x}{2}\end{array} \)
Point of intersection (-13/7, 17/7)
14. Let \(\begin{array}{l} \hat{a}\ \text{and}\ \hat{b}\end{array} \)
be two unit vectors such that the angle between them is π/4. If θ is the angle between the vectors \(\begin{array}{l} \left(\hat{a}+\hat{b}\right)\ \text{and}\ \left(\hat{a}+2\hat{b}+2\left(\hat{a}\times\hat{b}\right)\right),\end{array} \)
then the value of 164 cos2θ is equal to :
\(\begin{array}{l} (\text{A})\ 90+27\sqrt{2}\end{array} \)
\(\begin{array}{l} (\text{B})\ 45+18\sqrt{2}\end{array} \)
\(\begin{array}{l} (\text{C})\ 90+3\sqrt{2}\end{array} \)
\(\begin{array}{l} (\text{D})\ 54+90\sqrt{2}\end{array} \)
Answer (A)
Sol.
\(\begin{array}{l} \hat{a}\cdot\hat{b}=\frac{1}{\sqrt{2}}\text{ and }=\left|\overrightarrow{a}\times\overrightarrow{b}\right|=\frac{1}{\sqrt{2}}\end{array} \)
\(\begin{array}{l} \frac{\left(\hat{a}+\hat{b}\right)\cdot\left(\hat{a}+2\hat{b}+2\left(\hat{a}\times\hat{b}\right)\right)}{\left|\hat{a}+\hat{b}\right|\left|\hat{a}+2\hat{b}+2\left(\hat{a}\times\hat{b}\right)\right|}=\cos\theta\end{array} \)
\(\begin{array}{l} \Rightarrow\ \cos\theta=\frac{1+3\hat{a}\hat{b}+2}{\left|\hat{a}+\hat{b}\right|\left|\hat{a}+2\hat{b}+2\left(\hat{a}\times\hat{b}\right)\right|} \end{array} \)
\(\begin{array}{l} \left|\hat{a}+\hat{b}\right|^2=2+\sqrt{2}\end{array} \)
\(\begin{array}{l} \left|\hat{a}+2\hat{b}+2\left(\hat{a}\times\hat{b}\right)\right|^2=1+4+4\left|\hat{a}\times\hat{b}\right|^2+4\hat{a}\hat{b}\end{array} \)
\(\begin{array}{l} =5+4\cdot\frac{1}{2}+\frac{4}{\sqrt{2}}=7+2\sqrt{2}\end{array} \)
So,
\(\begin{array}{l} \cos^2\theta=\frac{\left(3+\frac{3}{\sqrt{2}}\right)^2}{\left(2+\sqrt{2}\right)\left(7+2\sqrt{2}\right)}=\frac{9\sqrt{2}\left(5\sqrt{2}+3\right)}{164} \end{array} \)
\(\begin{array}{l} \Rightarrow 164\cos^2\theta=90+27\sqrt{2}\end{array} \)
15. If \(\begin{array}{l} f\left(\alpha\right)=\displaystyle\int\limits_1^\alpha\frac{\text{log}_{10}t}{1+t}dt, \alpha > 0,\end{array} \)
then f(e3) + f(e–3) is equal to :
(A) 9
\(\begin{array}{l} (\text{B})\ \frac{9}{2}\end{array} \)
\(\begin{array}{l} (\text{C})\ \frac{9}{\text{log}_e\left(10\right)}\end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{9}{2\text{log}_e\left(10\right)}\end{array} \)
Answer (D)
Sol.
\(\begin{array}{l} f\left(\alpha\right)=\int_1^\alpha\frac{\text{log}_{10}t}{1+t}dt\cdots\left(i\right)\end{array} \)
\(\begin{array}{l} f\left(\frac{1}{\alpha}\right)=\int_1^\frac{1}{\alpha}\frac{\text{log}_{10}t}{1+t}dt\end{array} \)
Substituting t = 1/p
\(\begin{array}{l} f\left(\frac{1}{\alpha}\right)=\int_1^\alpha\frac{\text{log}_{10}\left(\frac{1}{p}\right)}{1+\frac{1}{p}} \left(\frac{-1}{p^2}\right)dp\end{array} \)
\(\begin{array}{l} =\int_1^\alpha\frac{\text{log}_{10}\ p}{p\left(p+1\right)}dp=\int_1^\alpha\left(\frac{\text{log}_{10}t}{t}-\frac{\text{log}_{10}t}{t+1}\right)dt\cdots\left(ii\right)\end{array} \)
By (i) + (ii)
\(\begin{array}{l} f\left(\alpha\right)+f\left(\frac{1}{\alpha}\right)=\int_1^\alpha\frac{\text{log}_{10}t}{t}dt=\int_1^\alpha\frac{\text{ln}t}{t}\cdot\text{log}_{10}\ e\ dt\end{array} \)
\(\begin{array}{l} =\frac{\left(\text{ln}\alpha\right)^2}{2\text{log}_e10} \end{array} \)
\(\begin{array}{l} \alpha=e^3\Rightarrow f\left(e^3\right)+f\left(e^{-3}\right)=\frac{9}{2\text{log}_e10}\end{array} \)
16. The area of the region \(\begin{array}{l} \left\{\left(x,y\right);\left|x-1\right|\leq y \leq \sqrt{5-x^2} \right\}\end{array} \)
is equal to
\(\begin{array}{l} (\text{A})\ \frac{5}{2}\sin^{-1}\left(\frac{3}{5}\right)-\frac{1}{2}\end{array} \)
\(\begin{array}{l} (\text{B})\ \frac{5\pi}{4}-\frac{3}{2}\end{array} \)
\(\begin{array}{l} (\text{C})\ \frac{3\pi}{4}+\frac{3}{2}\end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{5\pi}{4}-\frac{1}{2}\end{array} \)
Answer (D)
Sol.
\(\begin{array}{l} A=\displaystyle\int\limits_{-1}^1\left(\sqrt{5-x^2}-\left(1-x\right)\right)dx+\displaystyle\int\limits_{1}^2\left(\sqrt{5-x^2}-\left(x-1\right)\right)dx\end{array} \)
\(\begin{array}{l} \left.A=2\left(\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}\sin^{-1}\frac{x}{\sqrt{5}}\right)-2x\right|_0^1 \left.+\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}\sin^{-1}\frac{x}{\sqrt{5}}-\frac{x^2}{2}+x\right|_1^2\end{array} \)
\(\begin{array}{l} =\left(\frac{5\pi}{4}-\frac{1}{2}\right)\text{ sq. units} \end{array} \)
17. Let the focal chord of the parabola P :y2 = 4x along the line L : y = mx + c, m> 0 meet the parabola at the points M and N. Let the line L be a tangent to the hyperbola H :x2 – y2 = 4. If O is the vertex of P and F is the focus of H on the positive x-axis, then the area of the quadrilateral OMFN is
\(\begin{array}{l} (\text{A})\ 2\sqrt{6} \end{array} \)
\(\begin{array}{l} (\text{B})\ 2\sqrt{14} \end{array} \)
\(\begin{array}{l} (\text{C})\ 4\sqrt{6} \end{array} \)
\(\begin{array}{l} (\text{D})\ 4\sqrt{14} \end{array} \)
Answer (B)
Sol.
\(\begin{array}{l} H:\frac{x^2}{4}-\frac{y^2}{4}=1\end{array} \)
Focus (ae, 0)
\(\begin{array}{l} F\left(2\sqrt{2},0\right) \end{array} \)
y = mx + c passes through (1, 0)
0 = m + c …(i)
L is tangent to hyperbola
\(\begin{array}{l} c=\pm\sqrt{4m^2-4}\end{array} \)
\(\begin{array}{l} -m=\pm\sqrt{4m^2-4}\end{array} \)
m2 = 4m2 – 4
\(\begin{array}{l} m=\frac{2}{\sqrt{3}}\end{array} \)
\(\begin{array}{l} c=\frac{-2}{\sqrt{3}} \end{array} \)
\(\begin{array}{l} T:y=\frac{2}{\sqrt{3}}x-\frac{2}{\sqrt{3}} \end{array} \)
P :y2 = 4x
\(\begin{array}{l} y^2=4\left(\frac{\sqrt{3}y+2}{2}\right)\end{array} \)
\(\begin{array}{l} y^2-2\sqrt{3}y-4 =0\end{array} \)
\(\begin{array}{l}\text{Area}= \frac{1}{2}\begin{vmatrix}0 & 0 \\x_1 & y_1 \\2\sqrt{2} & 0 \\x_2 & y_2 \\0 & 0 \\\end{vmatrix}\end{array} \)
\(\begin{array}{l} =\left|\frac{1}{2}\left(-2\sqrt{2}y_1+2\sqrt{2}y_2\right)\right|\end{array} \)
\(\begin{array}{l} =\sqrt{2}\left|y_2-y_1\right|=\sqrt{2}\sqrt{\left(y_1+y_2\right)^2-4y_1y_2}\end{array} \)
\(\begin{array}{l} =\sqrt{56}\end{array} \)
\(\begin{array}{l} =2\sqrt{14}\end{array} \)
18. The number of points, where the function f: ℝ → ℝ, f(x) = |x – 1|cos|x – 2|sin|x – 1| + (x – 3)|x2 – 5x + 4|, is NOT differentiable, is
(A) 1
(B) 2
(C) 3
(D) 4
Answer (B)
Sol. f :R→R.
f(x) = |x – 1| cos |x – 2| sin |x – 1| + (x – 3) |x2 – 5x + 4|
= |x – 1| cos |x – 2| sin |x – 1| + (x – 3) |x – 1||x – 4|
= |x – 1| [cos |x – 2| sin |x – 1| + (x – 3) |x – 4|]
Sharp edges at x = 1 and x = 4
∴ Non-differentiable at x = 1 and x = 4.
19. Let S = {1, 2, 3, …, 2022}. Then the probability that a randomly chosen number n from the set S such that HCF (n, 2022) = 1, is
\(\begin{array}{l} (\text{A})\ \frac{128}{1011} \end{array} \)
\(\begin{array}{l} (\text{B})\ \frac{166}{1011} \end{array} \)
\(\begin{array}{l} (\text{C})\ \frac{127}{337} \end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{112}{337} \end{array} \)
Answer (D)
Sol. S = {1, 2, 3, …… 2022}
HCF (n, 2022) = 1
⇒n and 2022 have no common factor
Total elements = 2022
2022 = 2 × 3 × 337
M : numbers divisible by 2.
{2, 4, 6, ….., 2022} n(M) = 1011
N : numbers divisible by 3.
{3, 6, 9, ….., 2022} n(N) = 674
L : numbers divisible by 6.
{6, 12, 18, ….., 2022} n(L) = 337
n(M∪N) = n(M) + n(N) –n(L)
= 1011 + 674 – 337
= 1348
0 = Number divisible by 337 but not in M∪N
{337, 1685}
Number divisible by 2, 3 or 337
= 1348 + 2 = 1350
Required probability
\(\begin{array}{l} =\frac{2022-1350}{2022}\end{array} \)
\(\begin{array}{l} =\frac{672}{2022} \end{array} \)
\(\begin{array}{l} =\frac{112}{337} \end{array} \)
20. Let \(\begin{array}{l} f\left(x\right)=3^{\left(x^2-2\right)^3+4},x\in \mathbb{R}.\end{array} \)
Then which of the following statements are true?
P :x = 0 is a point of local minima of f
Q: x = √2 is a point of inflection of f
R :f ′ is increasing for x > √2
(A) Only P and Q
(B) Only P and R
(C) Only Q and R
(D) All P, Q and R
Answer (D)
Sol.
\(\begin{array}{l} f\left(x\right)=3^{\left(x^2-2\right)^3+4}, x\in R \end{array} \)
\(\begin{array}{l} f\left(x\right)=81.3^{\left(x^2-2\right)^3} \end{array} \)
\(\begin{array}{l} f’\left(x\right)=81.3^{\left(x^2-2\right)^3}\text{ln}2.3\left(x^2-2\right)2x\end{array} \)
\(\begin{array}{l} =\left(486\ \text{ln}2\right)\left(3^{\left(x^2-2\right)^3}\left(x^2-2\right)x\right)\end{array} \)
⇒ x = 0 is the local minima.
f′′(x) = (486 ln2)
\(\begin{array}{l} \begin{pmatrix}3^{\left(x^2-2\right)^3}\cdot\left(x^2-2\right) \\\left(5x^2-2+6x^2\text{ln}3\left(x^2-2\right)\right)\end{pmatrix}\end{array} \)
\(\begin{array}{l} f”\left(x\right)=0~~~~~x=\sqrt{2}\end{array} \)
\(\begin{array}{l} f”\left(\sqrt{2}^+\right)>0\end{array} \)
\(\begin{array}{l} f”\left(\sqrt{2}^-\right)<0\end{array} \)
\(\begin{array}{l} \Rightarrow x=\sqrt{2}\end{array} \)
is point of inflection
\(\begin{array}{l} f”\left(x\right)>0~\forall~x>\sqrt{2}\end{array} \)
⇒ f′(x) is increasing for x > √2
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 S = {θ ∈ (0, 2π) : 7 cos2θ – 3 sin2θ – 2 cos22θ = 2}. Then, the sum of roots of all the equations x2 – 2 (tan2θ + cot2θ) x + 6 sin2θ = 0, θ ∈ S, is _______.
Answer (16)
Sol. 7 cos2θ – 3 sin2θ – 2 cos22θ = 2
\(\begin{array}{l} \Rightarrow 4\left(\frac{1+\cos2\theta}{2}\right)+3\cos2\theta-2\cos^22\theta=2 \end{array} \)
⇒ 2 + 5 cos2θ – 2 cos2 2θ = 2
⇒ cos 2θ = 0 or 5/2 (rejected)
\(\begin{array}{l} \Rightarrow \cos2\theta=0=\frac{1-\tan^2\theta}{1+\tan^2\theta}\Rightarrow \tan^2\theta =1\end{array} \)
∴ Sum of roots = 2 (tan2θ + cot2θ) = 2 × 2 = 4
But as tanθ = ±1 for π/4, 3π/4, 5π/4, 7π/4 in the interval (0, 2π)
∴ Four equations will be formed
Hence, sum of roots of all the equations
= 4 × 4 = 16.
2. Let the mean and the variance of 20 observations x1, x2,…., x20 be 15 and 9, respectively. For a ∈ R, if the mean of (x1 + α)2, (x2 + α)2,….,(x20 + α)2 is 178, then the square of the maximum value of α is equal to ___________.
Answer (4)
Sol. Given
\(\begin{array}{l} \displaystyle\sum\limits_{\frac{i=1}{20}}^{20}x_i=15\Rightarrow\ \displaystyle\sum\limits_{i=1}^{20}x_i=300\cdots\left(1\right)\end{array} \)
and
\(\begin{array}{l} \displaystyle\sum\limits_{\frac{i=1}{20}}^{20}x_i^2-\left(\overline{x}\right)^2=9\Rightarrow \displaystyle\sum\limits_{i=1}^{20}x_i^2=4680\cdots\left(2\right)\end{array} \)
\(\begin{array}{l} \text{Mean}=\frac{\left(x_1+\alpha\right)^2+\left(x_2+\alpha\right)^2+\cdots+\left(x_{20}+\alpha\right)^2}{20} \end{array} \)
= 178
\(\begin{array}{l} \Rightarrow\ \frac{\displaystyle\sum\limits_{i=1}^{20}x_i^2+2\alpha\displaystyle\sum\limits_{i=1}^{20}x_i+20\alpha^2}{20}=178\end{array} \)
⇒ 4680 + 600α + 20α2 = 3560
⇒ α2 + 30α + 56 = 0
⇒ α2 + 28α + 2α + 56 = 0
⇒ (α + 28)(α + 2) = 0
αmax = – 2
\(\begin{array}{l}\Rightarrow \alpha_{\text{max}}^2=4\end{array} \)
3. Let a line with direction ratios a, – 4a, –7 be perpendicular to the lines with direction ratios 3, – 1, 2b and b, a, – 2. If the point of intersection of the line \(\begin{array}{l} \frac{x+1}{a^2+b^2}=\frac{y-2}{a^2-b^2}=\frac{z}{1} \end{array} \)
and the plane x – y + z = 0 is (α, β, γ), then α + β + γ is equal to ______.
Answer (10)
Sol. Given a.3 + (– 4a)(–1) + (–7) 2b = 0 …(1)
and ab –4a2 + 14 = 0 …(2)
⇒ a2 = 4 and b2 = 1
\(\begin{array}{l} \therefore\ \text{Line }L\equiv\frac{x+1}{5}=\frac{y-2}{3}=\frac{z}{1}=\lambda(\text{ say})\end{array} \)
⇒ General point on line is (5λ – 1, 3λ + 2, λ)
for finding point of intersection with x – y + z = 0
we get (5λ – 1) – (3λ + 2) + (λ) = 0
⇒ 3λ – 3 = 0 ⇒λ = 1
∴ Point at intersection (4, 5, 1)
∴ α + β + γ = 4 + 5 + 1 = 10
4. Let a1, a2, a3,…. be an A.P. If \(\begin{array}{l} \displaystyle\sum\limits_{r=1}^\infty\frac{a_r}{2^r}=4,\end{array} \)
then 4a2 is equal to ________.
Answer (16)
Sol. Given
\(\begin{array}{l} S=\frac{a_1}{2}+\frac{a_2}{2^2}+\frac{a_3}{2^3}+\frac{a_4}{2^4}+\cdots\infty\end{array} \)
\(\begin{array}{l} \frac{\frac{1}{2}S=~~\frac{a_1}{2^2}+\frac{a_2}{2^3}+\cdots\infty}{\frac{S}{2}~~=\frac{a_1}{2}+\frac{\left(a_2+a_1\right)}{2^2}+\frac{\left(a_3+a_2\right)}{2^3}}+\cdots\infty\end{array} \)
\(\begin{array}{l} \Rightarrow\ \frac{S}{2}=\frac{a_1}{2}+\frac{d}{2}\end{array} \)
⇒ a1 + d = a2 = 4 ⇒ 4a2 = 16
5. Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of \(\begin{array}{l} \left(\sqrt[4]2{}+\frac{1}{\sqrt[4]{3}}\right)^n,\ \text{in the increasing powers of}\ \frac{1}{\sqrt[4]{3}}\ \text{be}\ \sqrt[4]{6}:1.\end{array} \)
\(\begin{array}{l}\text{If the sixth term from the beginning is}\ \frac{\alpha}{\sqrt[4]{3}},\end{array} \)
then α is equal to ___________.
Answer (84)
Sol. Fifth term from beginning
\(\begin{array}{l} =^nC_4\left(2^\frac{1}{4}\right)^{n-4}\left(3^\frac{-1}{4}\right)^4\end{array} \)
Fifth term from end = (n – 5 + 1)th term from begin
\(\begin{array}{l} =^nC_{n-4}\left(2^\frac{1}{4}\right)^3\left(3^\frac{-1}{4}\right)^{n-4} \end{array} \)
Given
\(\begin{array}{l} \frac{^nC_42^{\frac{n-4}{4}}\cdot3^{-1}}{^nC_{n-3}2^\frac{4}{4}\cdot3^{-\left(\frac{n-4}{4}\right)}}=6^\frac{1}{4} \end{array} \)
\(\begin{array}{l} \Rightarrow 6^\frac{n-8}{4}=6^\frac{1}{4}\end{array} \)
\(\begin{array}{l} \Rightarrow \frac{n-8}{4}=\frac{1}{4}~~\Rightarrow n=9\end{array} \)
\(\begin{array}{l} T_6=T_{5+1}={^9C_5}\left(2^\frac{1}{4}\right)^4\left(3^\frac{-1}{4}\right)^5\end{array} \)
\(\begin{array}{l} =\frac{^9C_5\cdot2}{3^\frac{1}{4}\cdot 3}=\frac{84}{3^\frac{1}{4}}=\frac{\alpha}{3^\frac{1}{4}} \end{array} \)
⇒ α = 84.
6. The number of matrices of order 3 × 3, whose entries are either 0 or 1 and the sum of all the entries is a prime number, is _________.
Answer (282)
Sol. In a 3 × 3 order matrix there are 9 entries.
These nine entries are zero or one.
The sum of positive prime entries are 2, 3, 5 or 7.
Total possible matrices
\(\begin{array}{l} =\frac{9!}{2!\cdot7!}+\frac{9!}{3!\cdot6!}+\frac{9!}{5!\cdot4!}+\frac{9!}{7!\cdot2!}\end{array} \)
= 36 + 84 + 126 + 36
= 282
7. Let p and p + 2 be prime numbers and let
\(\begin{array}{l} \Delta=\begin{vmatrix}p! & \left(p+1\right)! & \left(p+2\right)! \\\left(p+1\right)! & \left(p+2\right)! & \left(p+3\right)! \\\left(p+2\right)! & \left(p+3\right)! & \left(p+4\right)! \\\end{vmatrix} \end{array} \)
Then the sum of the maximum values of α and β, such that pα and (p + 2)β divide Δ, is _______.
Answer (04)
Sol.
\(\begin{array}{l} \Delta=\begin{vmatrix}p! & \left(p+1\right)! & \left(p+2\right)! \\\left(p+1\right)! & \left(p+2\right)! & \left(p+3\right)! \\\left(p+2\right)! & \left(p+3\right)! & \left(p+4\right)! \\\end{vmatrix} \end{array} \)
\(\begin{array}{l} =p!\left(p+1\right)!\cdot\left(p+2\right)!\begin{vmatrix}1 & p+1 & \left(p+1\right)\left(p+2\right) \\1 & \left(p+2\right) & \left(p+2\right)\left(p+3\right) \\1 & \left(p+3\right) & \left(p+3\right)\left(p+4\right) \\\end{vmatrix}\end{array} \)
\(\begin{array}{l} =p!\left(p+1\right)!\cdot\left(p+2\right)!\begin{vmatrix}1 & p+1 & p^2+3p+2 \\0 & 1 & 2p+4 \\0 & 1 & 2p+6 \\\end{vmatrix}\end{array} \)
\(\begin{array}{l} =2\left(p!\right)\cdot\left(\left(p+1\right)!\right)\cdot\left(\left(p+2\right)!\right)\end{array} \)
\(\begin{array}{l} =2\left(p+1\right)\cdot\left(p!\right)^2\cdot\left(\left(p+2\right)!\right) \end{array} \)
\(\begin{array}{l} =2\left(p+1\right)^2\cdot\left(p!\right)^3\cdot\left(\left(p+2\right)!\right)\end{array} \)
∴ Maximum value of α is 3 and β is 1.
∴ α + β = 4
8. If \(\begin{array}{l}\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+\frac{1}{4\times5\times6}+\cdots,+\frac{1}{100\times101\times102}=\frac{k}{101},\end{array} \)
then 34 k is equal to ____________.
Answer (286)
Sol.
\(\begin{array}{l} S=\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+\frac{1}{4\times5\times6} +\cdots+\frac{1}{100\times101\times102}\end{array} \)
\(\begin{array}{l} =\frac{1}{\left(3-1\right)\cdot1}\left[\frac{1}{2\times3}-\frac{1}{101\times102}\right]\end{array} \)
\(\begin{array}{l} =\frac{1}{2}\left(\frac{1}{6}-\frac{1}{101\times102}\right)\end{array} \)
\(\begin{array}{l} =\frac{143}{102\times101}=\frac{k}{101}\end{array} \)
∴ 34k = 286
9. Let S = {4, 6, 9} and T = {9, 10, 11, …,1000}. If A = {a1 + a2 + … +ak :k∈N, a1, a2, a3, …, ak∈S}, then the sum of all the elements in the set T – A is equal to ____________.
Answer (11.00)
Sol. Here S = {4, 6, 9}
And T = {9, 10, 11, ……., 1000}.
We have to find all numbers in the form of
4x + 6y + 9z, where x, y, z∈ {0, 1, 2, …..}.
If a and b are coprime number then the least number from which all the number more than or equal to it can be express as ax + by where x, y∈ {0, 1, 2, ….} is (a – 1) · (b – 1).
Then for 6y + 9z = 3(2y + 3z)
All the number from (2 – 1) · (3 – 1) = 2 and above can be express as 2x + 3z (say t).
Now 4x + 6y + 9z = 4x + 3(t + 2)
= 4x + 3t + 6
again by same rule 4x + 3t, all the number from
(4 – 1) (3 – 1) = 6 and above can be express from 4x + 3t.
Then 4x + 6y + 9z express all the numbers from 12 and above.
again 9 and 10 can be express in form 4x + 6y + 9z.
Then set A = {9, 10, 12, 13, …., 1000}.
Then T – A = {11}
Only one element 11 is there.
Sum of elements of T – A = 11
10. Let the mirror image of a circle c1 :x2 + y2 – 2x – 6y + α = 0 in line y = x + 1 be c2 : 5x2 + 5y2 + 10gx + 10fy + 38 = 0. If r is the radius of circle c2, then α + 6r2 is equal to _________.
Answer (12)
Sol. c1: x2 + y2 – 2x – 6y + α = 0
Then centre = (1, 3) and
\(\begin{array}{l}\text{radius}\ (r)= \sqrt{10-\alpha}\end{array} \)
Image of (1, 3) w.r.t. line x – y + 1 = 0 is (2, 2)
c2: 5x2 + 5y2 + 10gx + 10fy + 38 = 0
or
\(\begin{array}{l} x^2+y^2+2gx+2fy+\frac{38}{5}=0 \end{array} \)
Then (–g, –f) = (2, 2)
∴ g = f = – 2 …(i)
\(\begin{array}{l}\text{Radius of}\ c_2=r=\sqrt{4+4-\frac{38}{5}}=\sqrt{10-\alpha} \end{array} \)
\(\begin{array}{l} \Rightarrow\ \frac{2}{5}=10-\alpha\end{array} \)
\(\begin{array}{l} \therefore\ \alpha = \frac{48}{5}\text{ and }r=\sqrt{\frac{2}{5}} \end{array} \)
\(\begin{array}{l} \therefore\ \alpha+6r^2=\frac{48}{5}+\frac{12}{5}\end{array} \)
= 12
JEE Main 2022 July 29th Shift 1 Paper Analysis
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