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# JEE Main 2022 June 24 – Shift 1 Maths Question Paper with Solutions

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JEE Main 2022 June 24 – Shift 1 Maths Question Paper with 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.

1. Let A = {zC : 1 ≤ |z – (1 + i)| ≤ 2} and B = {zA : |z – (1 – i)| = 1}. Then, B :

(A) Is an empty set

(B) Contains exactly two elements

(C) Contains exactly three elements

(D) Is an infinite set

Sol. Set A represents region 1 i.e.R1 and clearly set B has infinite points in it.

2. The remainder when 32022 is divided by 5 is :

(A) 1

(B) 2

(C) 3

(D) 4

Sol. 32022 = (10 – 1)1011

= 1011C0(10)1011 (–1)0 + 1011C1(10)1010 (–1)1 + ….. + 1011C1010(10)1 (–1)1010 + 1011C1011(10)0 (–1)1011

= 5k –1, where kI

So when divided by 5, it leaves remainder 4.

3. The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is :

(A) 9

(B) 10

(C) 11

(D) 12

Sol. S = 4πr2

$$\begin{array}{l}\frac{dS}{dt}=8\pi r\frac{dr}{dt}\end{array}$$
$$\begin{array}{l}\frac{dS}{dt}=\text{ constant so} \Rightarrow r\frac{dr}{dt}=k \left ( \text{Let} \right )\end{array}$$
$$\begin{array}{l}r dr=k dt \Rightarrow \frac{r^2}{2}=kt+C\end{array}$$

at t = 0, r = 3

$$\begin{array}{l}\frac{9}{2}=C \end{array}$$

at t = 5,

$$\begin{array}{l}\frac{49}{2}=k\cdot5+\frac{9}{2}\Rightarrow k =4\end{array}$$

At t = 9,

$$\begin{array}{l}\frac{r^2}{2}=\frac{81}{2}\end{array}$$

So, r = 9

4. Bag A contains 2 white, 1 black and 3 red balls and bas B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is

$$\begin{array}{l}\frac{6}{11}\end{array}$$
then n is equal to _______.

(A) 13

(B) 6

(C) 4

(D) 3

Sol. $$\begin{array}{l}P\left ( 1R \text{ and } 1B \right )=P\left ( A \right )\cdot P\left ( \frac{1R 1B}{A} \right )+P\left ( B \right )\cdot P\left ( \frac{1R 1B}{B} \right )\end{array}$$
$$\begin{array}{l}=\frac{1}{2}\cdot\frac{^3C_1\cdot^1C_1}{^6C_2}+\frac{1}{2}\cdot\frac{^2C_1\cdot^3C_1}{^{n+5}C_1}\end{array}$$
$$\begin{array}{l}P\left ( \frac{1R 1B}{A} \right )=\frac{\frac{1}{2}\cdot\frac{3}{15}}{\frac{1}{2}\cdot\frac{3}{15}+\frac{1}{2}\cdot\frac{6\cdot2}{\left ( n+5 \right )\left ( n+4 \right )}}=\frac{6}{11}\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{\frac{1}{10}}{\frac{1}{10}+\frac{6}{(n+5)(n+4)}} =\frac{6}{11}\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{11}{10}=\frac{6}{10}+\frac{36}{(n+5)(n+4)}\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{5}{10 \times 36} =\frac{1}{(n+5)(n+4)}\end{array}$$

n2 + 9n – 52 = 0

n = 4 is only possible value

5. Let

$$\begin{array}{l}x^2+y^2+Ax+By+C=0\end{array}$$
be a circle passing through (0, 6) and touching the parabola y = x2 at (2, 4). Then A + C is equal to ________.

(A) 16

(B)

$$\begin{array}{l}\frac{88}{5}\end{array}$$

(C) 72

(D) –8

Sol. For tangent to parabola y = x2 at (2, 4)

$$\begin{array}{l}\left. \frac{dy}{dx} \right|_{(2,4)}=4\end{array}$$

Equation of tangent is

y – 4 = 4(x – 2)

⇒ 4xy – 4 = 0

Family of circle can be given by

(x – 2)2 + (y – 4)2 + λ(4xy – 4) = 0

As it passes through (0, 6)

22 + 22 + λ(–10) = 0

$$\begin{array}{l}\Rightarrow \lambda = \frac{4}{5}\end{array}$$

Equation of circle is

$$\begin{array}{l}\left (x-2 \right )^2+\left ( y-4 \right )^2+\frac{4}{5}\left ( 4x-y-4 \right )=0\end{array}$$
$$\begin{array}{l}\Rightarrow \left ( x^2+y^2-4x-8y+20 \right )+\left ( \frac{16}{5}x -\frac{4}{5}y-\frac{16}{5} \right )=0\end{array}$$
$$\begin{array}{l}A=-4+\frac{16}{5},C=20-\frac{16}{5}\end{array}$$

So, A + C = 16

6. The number of values of α for which the system of equations :

x + y + z = α

αx + 2αy + 3z = –1

x + 3αy + 5z = 4

is inconsistent, is

(A) 0

(B) 1

(C) 2

(D) 3

Sol.

$$\begin{array}{l}\Delta =\begin{vmatrix}1 & 1 & 1 \\\alpha & 2\alpha & 3 \\1 & 3\alpha & 5 \\\end{vmatrix}\end{array}$$

= 1(10α – 9α) – 1 (5α – 3) + 1 (3α2 – 2α)

= α – 5α + 3 + 3α2 – 2α

= 3α2 – 6α + 3

For inconsistency Δ = 0 i.e. α = 1

Now check for α = 1

x + y + z = 1 …(i)

x + 2y + 3z = –1 …(ii)

x + 3y + 5z = 4 …(iii)

By (ii) × 2 – (i) × 1

x + 3y + 5z = –3

so equations are

inconsistent for α = 1

7. If the sum of the squares of the reciprocals of the roots α and β of the equation 3x2 + λx – 1 = 0 is 15, then 6(α3 + β3)2 is equal to :

(A) 18

(B) 24

(C) 36

(D) 96

Sol.

$$\begin{array}{l}\frac{1}{\alpha^2}+\frac{1}{\beta^2}=15 \Rightarrow \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha^2\beta^2}=15\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{\frac{\lambda^2}{9}+\frac{2}{3}}{\frac{1}{9}}=15\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{\lambda^2}{9}=1 \Rightarrow \lambda^2=9\end{array}$$

α3 + β3 = (α + β) (α2 + β2 – αβ)

$$\begin{array}{l}=\left ( \frac{-\lambda}{3} \right )\left ( \frac{\lambda^2}{9}-3 \left ( \frac{-1}{3} \right ) \right )=\left(\frac{-\lambda}{3}\right) \left( \frac{\lambda^2}{9}+1 \right )=\frac{-2\lambda}{3}\end{array}$$
$$\begin{array}{l}6\left ( \alpha^3+\beta^3 \right)^2=6\cdot\frac{4\lambda^2}{9}=24\end{array}$$

8. The set of all values of k for which

$$\begin{array}{l}\left ( \tan^{-1}x \right )^3+\left ( \cot^{-1}x \right )^3=k\pi ^3, x \in R,\end{array}$$
is the interval:

(A)

$$\begin{array}{l}\left [ \frac{1}{32},\frac{7}{8} \right )\end{array}$$

(B)

$$\begin{array}{l}\left ( \frac{1}{24},\frac{13}{16} \right )\end{array}$$

(C)

$$\begin{array}{l}\left [ \frac{1}{48},\frac{13}{16} \right ]\end{array}$$

(D)

$$\begin{array}{l}\left [ \frac{1}{32},\frac{9}{8} \right )\end{array}$$

Sol. Let

$$\begin{array}{l}\tan^{-1}x=t\in \left ( \frac{-\pi}{2},\frac{\pi}{2} \right )\end{array}$$
$$\begin{array}{l}\cot^{-1}x=\frac{\pi}{2}-t\end{array}$$
$$\begin{array}{l}f\left ( t \right )=t^3+\left ( \frac{\pi}{2}-t \right )^3 \Rightarrow f’\left ( t \right )=3t^2-3\left ( \frac{\pi}{2}-t\right )^2\end{array}$$
$$\begin{array}{l}f’\left ( t \right )=0 \text{ at } t=\frac{\pi}{4}\end{array}$$
$$\begin{array}{l}f\left ( t\right )\left| \frac{}{}\right._{\min}=\frac{\pi^{3}}{64} + \frac{\pi^{3}}{64}=\frac{\pi^{3}}{32}\end{array}$$

Max will occur around

$$\begin{array}{l}t=-\frac{\pi}{2}\end{array}$$

Range of

$$\begin{array}{l}f(t) =\left[\frac{\pi^3}{32},\frac{7\pi^3}{8} \right )\end{array}$$
$$\begin{array}{l}k \in \left[\frac{1}{32},\frac{7}{8} \right )\end{array}$$

9. Let

$$\begin{array}{l}S=\left\{ \sqrt{n}:1\le n \le 50 \text{ and } n \text{ is odd} \right\}\end{array}$$
.

Let

$$\begin{array}{l}a\in S \text{ and } A=\begin{bmatrix}1 & 0 & a \\-1 & 1 & 0 \\-a & 0 & 1 \\\end{bmatrix}\end{array}$$

If

$$\begin{array}{l}\sum_{a~\in~S} \det \left ( adj A \right ) = 100 \lambda\end{array}$$
, then λ is equal to :

(A) 218

(B) 221

(C) 663

(D) 1717

Sol. |A| = a2 + 1

|adj A| = (a2 + 1)2

$$\begin{array}{l}S=\left\{ 1, \sqrt{3},\sqrt{5},\sqrt{7}, \dots, \sqrt{49} \right\} \end{array}$$
$$\begin{array}{l}\sum_{a~\in~S} \det \left ( adj A \right )=\left ( 1^2 + 1 \right )^2 + \left ( 3 + 1 \right )^2 +\left ( 5 + 1 \right )^2 + \dots + \left ( 49 + 1 \right )^2\end{array}$$

= 22 (12 + 22 + 32 + … + 252)

$$\begin{array}{l}=4\cdot\frac{25\cdot26\cdot51}{6}=100\cdot221\end{array}$$

λ = 221

10. For the function

f(x) = 4 loge(x – 1) – 2x2 + 4x + 5, x > 1, which one of the following is NOT correct?

(A) f is increasing in (1, 2) and decreasing in (2, )

(B) f(x) = –1 has exactly two solutions

(C) f′(e) – f′′(2) < 0

(D) f(x) = 0 has a root in the interval (e, e + 1)

Sol.

$$\begin{array}{l}f'(x)=\frac{4}{x-1}-4x+4=\frac{4\left ( 2x-x^2 \right )}{x-1}\end{array}$$ So maxima occurs at x = 2

f(2) = 4⋅0 – 2⋅22 + 4⋅2 + 5 = 5

so clearly f(x) = –1 has

exactly 2 solutions

$$\begin{array}{l}f”(x)=\frac{4(2-2x)(x-1)}{(x-1)^2}-\left( 2x-x^2 \right )\end{array}$$

so f′(e) – f ′′(2) > 0

so option c is not correct

11. If the tangent at the point (x1, y1) on the curve y = x3 + 3x2 + 5 passes through the origin, then (x1, y1) does NOT lie on the curve :

(A)

$$\begin{array}{l}x^2+\frac{y^2}{81}=2\end{array}$$

(B)

$$\begin{array}{l}\frac{y^2}{9}-x^2=8 \end{array}$$

(C) y = 4x2 + 5

(D)

$$\begin{array}{l}\frac{x}{3}-y^2=2\end{array}$$

Sol. mopmTangent

$$\begin{array}{l}\frac{y_1}{x_1}=3x_1^2+6x_1\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{x_1^3+3x_1^2+5}{x_1}=3x_1^2+6x_1\end{array}$$
$$\begin{array}{l}\Rightarrow x_1^3+3x_1^2+5=3x_1^3+6x_1^2\end{array}$$
$$\begin{array}{l}\Rightarrow 2x_1^3+3x_1^2-5=0\end{array}$$
$$\begin{array}{l}\Rightarrow \left ( x_1-1 \right )\left ( 2x_1^2+5x_1+5 \right )=0\end{array}$$

So, (x1, y1) = (1, 9)

12. The sum of absolute maximum and absolute minimum values of the function f(x) = |2x2 + 3x – 2| + sinx cosx in the interval [0, 1] is :

(A)

$$\begin{array}{l}3+\frac{\sin(1)\cos^2\left ( \frac{1}{2} \right )}{2} \end{array}$$

(B)

$$\begin{array}{l}3+\frac{1}{2}\left( 1+2\cos(1)\right )\sin(1)\end{array}$$

(C)

$$\begin{array}{l}5+\frac{1}{2}\left ( \sin(1)+sin(2) \right )\end{array}$$

(D)

$$\begin{array}{l}2+\sin\left ( \frac{1}{2} \right )\cos\left ( \frac{1}{2} \right )\end{array}$$

Sol.

$$\begin{array}{l}f(x)=\left| \left ( 2x-1 \right )\left ( x+2 \right )\right|+\frac{\sin2x}{2}\end{array}$$
$$\begin{array}{l}\underline{0\le x \le \frac{1}{2}}~~~~f(x)=(1-2x)(x+2)+\frac{\sin2x}{2}\end{array}$$

f‘(x) = –4x – 3 + cos2x <0

For

$$\begin{array}{l}x\ge\frac{1}{2}~:~~f'(x)=4x+3+\cos2x>0\end{array}$$

So, minima occurs at

$$\begin{array}{l}x=\frac{1}{2}\end{array}$$
$$\begin{array}{l}f(x)|_{min} = \left| 2\left ( \frac{1}{2} \right )^2 + \frac{3}{2}-2\right|+\sin\left ( \frac{1}{2} \right )\cdot\cos\left ( \frac{1}{2} \right )\end{array}$$
$$\begin{array}{l}=\frac{1}{2}\sin1\end{array}$$

So, maxima is possible at x = 0 or x = 1

Now checking for x = 0 and x = 1, we can see it attains its maximum value at x = 1

$$\begin{array}{l}f(x)|_{max} =\left| 2+3-2\right|+\frac{\sin2}{2}\end{array}$$
$$\begin{array}{l}=3+\frac{1}{2}sin2\end{array}$$

Sum of absolute maximum and minimum value

$$\begin{array}{l}=3+\frac{1}{2}\left(\sin1 + \sin2 \right)\end{array}$$

13. If

$$\begin{array}{l}\left\{ a_i\right\}_{i=1}^n\end{array}$$
, where n is an even integer, is an arithmetic progression with common difference 1, and
$$\begin{array}{l}\sum_{i=1}^{n}a_i = 192,\sum_{i=1}^{n/2}a_{2i}= 120\end{array}$$
, then n is equal to :

(A) 48

(B) 96

(C) 92

(D) 104

Sol.

$$\begin{array}{l}a_1+a_2+\ldots+a_n = 192 \Rightarrow\frac{n}{2}\left ( a_1+a_n \right )=192….(1)\end{array}$$

a2 + a4 + a6 + … + an = 120

$$\begin{array}{l}\Rightarrow \frac{n}{4}\left ( a_1+1+a_n \right )=120….(2)\end{array}$$
From (2) & (1)

$$\begin{array}{l}\frac{480}{n}-\frac{384}{n}=1\end{array}$$

n = 96

14. If x = x(y) is the solution of the differential equation

$$\begin{array}{l}y\frac{dx}{dy}=2x+y^3(y+1)e^y, x(1)=0;\end{array}$$
then x(e) is equal to :

(A) e3(ee – 1)

(B) ee(e3 – 1)

(C) e2(ee + 1)

(D) ee(e2 – 1)

Sol.

$$\begin{array}{l}\frac{dx}{dy}-\frac{2x}{y}=y^2(y+1)e^y\end{array}$$

If

$$\begin{array}{l}=e^{\int-\frac{2}{y}dy}=e^{-2\ln y}=\frac{1}{y^2}\end{array}$$

Solution is given by

$$\begin{array}{l}x\cdot\frac{1}{y^2}=\int y^2\left ( y+1 \right )e^y\cdot\frac{1}{y^2}dy\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{x}{y^2}=\int(y+1)e^ydy\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{x}{y^2}=ye^y + c\end{array}$$

x = y2 (yey + c)

at, y = 1, x = 0

⇒ 0 = 1(1.e1 + c) ⇒ c = –e

at y = e,

x = e2(e.eee)

15. Let λx – 2y = μ be a tangent to the hyperbola a2x2y2 = b2. Then

$$\begin{array}{l}\left ( \frac{\lambda}{a} \right )^2-\left ( \frac{\mu}{b} \right )^2\end{array}$$
is equal to :

(A) –2

(B) –4

(C) 2

(D) 4

Sol.

$$\begin{array}{l}\frac{x^2}{\left ( \frac{b^2}{a^2} \right )}-\frac{y^2}{b^2}=1\end{array}$$

Tangent in slope form ⇒

$$\begin{array}{l}y=mx\pm\sqrt{\frac{b^2}{a^2}m^2-b^2}\end{array}$$

i.e., same as

$$\begin{array}{l}y=\frac{\lambda x}{2}-\frac{\mu}{2}\end{array}$$

Comparing coefficients,

$$\begin{array}{l}m=\frac{\lambda}{2}, \frac{b^2}{a^2}m^2-b^2=\frac{\mu^2}{4}\end{array}$$

Eliminating

$$\begin{array}{l}m,\frac{b^2}{a^2}\cdot\frac{\lambda^2}{4}-b^2=\frac{\mu^2}{4}\end{array}$$
$$\begin{array}{l} \Rightarrow\frac{\lambda^2}{a^2}-\frac{\mu^2}{b^2}=4\end{array}$$

16. Let

$$\begin{array}{l}\hat{a}, \hat{b}\end{array}$$
be unit vectors. If
$$\begin{array}{l}\vec{c}\end{array}$$
be a vector such that the angle between
$$\begin{array}{l}\hat{a}\end{array}$$
and
$$\begin{array}{l}\vec{c}\end{array}$$
is
$$\begin{array}{l}\frac{\pi}{12}\end{array}$$
and
$$\begin{array}{l}\hat{b}=\vec{c}+2\left ( \vec {c}\times\hat{a} \right )\end{array}$$
then
$$\begin{array}{l}\left|6 \vec{c}\right|^2\end{array}$$
is equal to:

(A)

$$\begin{array}{l}6\left ( 3-\sqrt{3} \right )\end{array}$$

(B)

$$\begin{array}{l}3+\sqrt{3}\end{array}$$

(C)

$$\begin{array}{l}6\left ( 3+\sqrt{3} \right )\end{array}$$

(D)

$$\begin{array}{l}6\left ( \sqrt{3} + 1 \right )\end{array}$$

Sol.

$$\begin{array}{l}\because \hat{b}=\vec{c}+2\left( \vec{c}\times \hat{a} \right )\end{array}$$
$$\begin{array}{l}\Rightarrow \hat{b}\cdot\vec{c}=\left|\vec{c}\right|^2\end{array}$$
…(i)

$$\begin{array}{l}\therefore \hat{b}-\vec{c}=2\left ( \vec{c}\times \vec{a}\right )\end{array}$$
$$\begin{array}{l}\Rightarrow\left|\hat{b}\right|^2 +\left| \vec{c}\right|^2-2\hat{b}\cdot\vec{c}=4\left|\vec{c}\right|^2\left|\vec{a}\right|^2\sin^2\frac{\pi}{12}\end{array}$$
$$\begin{array}{l} \Rightarrow 1+\left| \vec{c}\right|^2-2\left| \vec{c}\right|^2 =4\left| \vec{c}\right|^2\left ( \frac{\sqrt{3}-1}{2\sqrt{2}} \right )^2\end{array}$$
$$\begin{array}{l}\Rightarrow 1=\left| \vec{c}\right|^2\left ( 3-\sqrt{3} \right )\end{array}$$
$$\begin{array}{l}\Rightarrow 36\left| \vec{c}\right|^2=\frac{36}{3-\sqrt{3}}=6\left ( 3+\sqrt{3} \right )\end{array}$$

17. If a random variable X follows the Binomial distribution B(33, p) such that 3P(X = 0) = P(X = 1), then the value of

$$\begin{array}{l}\frac{P\left ( X=15 \right )}{P\left ( X=18 \right )}-\frac{P\left ( X=16 \right )}{P\left ( X=17 \right )}\end{array}$$
is equal to:

(A) 1320

(B) 1088

(C)

$$\begin{array}{l}\frac{120}{1331}\end{array}$$

(D)

$$\begin{array}{l}\frac{1088}{1089}\end{array}$$

Sol. 3P(X = 0) = P(X = 1)

$$\begin{array}{l}3\cdot^nC_0P^0\left ( 1-P \right )^n=~^nC_1P^1\left ( 1-P \right )^{n-1}\end{array}$$
$$\begin{array}{l}\frac{3}{n}=\frac{P}{1-P}\Rightarrow \frac{1}{11}=\frac{P}{1-P}\end{array}$$

⇒ 1 – P = 11P

$$\begin{array}{l}\Rightarrow P=\frac{1}{12}\end{array}$$
$$\begin{array}{l}\frac{P(X=15)}{P(X=18)}-\frac{P(X=16)}{P(X=17)}\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{^{33}C_{15}P^{15}(1-P)^{18}}{^{33}C_{18}P^{18}(1-P)^{15}}-\frac{^{33}C_{16}P^{16}(1-P)^{17}}{^{33}C_{17}P^{17}(1-P)^{16}}\end{array}$$
$$\begin{array}{l}\Rightarrow \left ( \frac{1-P}{P} \right )^3-\left ( \frac{1-P}{P} \right )\end{array}$$

⇒ 113 – 11 = 1320

18. The domain of the function

$$\begin{array}{l}f(x)=\frac{\cos^{-1}\left ( \frac{x^2-5x+6}{x^2-9} \right )}{\log_e\left ( x^2-3x+2 \right )}\end{array}$$
is:

(A)

$$\begin{array}{l}\left ( -\infty,1 \right )\cup\left ( 2, \infty \right )\end{array}$$

(B) (2, ∞)

(C)

$$\begin{array}{l}\left [ -\frac{1}{2},1 \right )\cup\left ( 2,\infty \right )\end{array}$$

(D)

$$\begin{array}{l}\left [ -\frac{1}{2},1 \right )\cup\left ( 2,\infty \right )-\left\{ \frac{3+\sqrt{5}}{2},\frac{3-\sqrt{5}}{2}\right\}\end{array}$$

Sol.

$$\begin{array}{l}-1\le \frac{x^2-5x+6}{x^2-9}\le 1\end{array}$$
and x2 – 3x + 2 > 0, ≠ 1

$$\begin{array}{l}\frac{\left ( x-3 \right )\left ( 2x+1 \right )}{x^2-9}\ge 0 \left|\frac{5(x-3)}{x^2-9} \right.\ge 0\end{array}$$

Solution to this inequality is

$$\begin{array}{l}x\in \left [ \frac{-1}{2},\infty \right )-\{3\}\end{array}$$

for x2 – 3x + 2 > 0 and ≠ 1

$$\begin{array}{l}x\in\left ( -\infty,1 \right )\cup \left ( 2,\infty \right )-\left\{ \frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2}\right\}\end{array}$$

Combining the two solution sets (taking intersection)

$$\begin{array}{l}x\in \left [ \frac{-1}{2},1 \right )\cup\left ( 2,\infty \right )-\left\{ \frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2}\right\}\end{array}$$

19. Let

$$\begin{array}{l}S=\left\{ \theta \in \left [ -\pi, \pi \right ]-\left\{ \pm\frac{\pi}{2}\right\}: \sin\theta\tan\theta + \tan\theta=\sin2\theta\right\}\end{array}$$
. If
$$\begin{array}{l}T=\sum_{\theta = S}\cos2\theta\end{array}$$
then T + n(S) is equal to:

(A)

$$\begin{array}{l}7+\sqrt{3}\end{array}$$

(B) 9

(C)

$$\begin{array}{l}8+\sqrt{3}\end{array}$$

(D) 10

Sol. tanθ (sinθ + 1) – sin2θ = 0

tanθ (sinθ + 1 – 2cos2θ) = 0

⇒ tanθ = 0 or 2sin2θ + sinθ – 1 = 0

⇒ (2sinθ + 1)(sinθ – 1) = 0

$$\begin{array}{l}\sin\theta=\frac{-1}{2}\end{array}$$
or 1

But, sinθ = 1 not possible

$$\begin{array}{l}\theta = 0, \pi, – \pi, -\frac{\pi}{6},\frac{-5\pi}{6}\end{array}$$

n(S) = 5

$$\begin{array}{l}T=\sum\cos2\theta=cos0^\circ + \cos2\pi + \cos(-2\pi)+\cos\left ( -\frac{5\pi}{3} \right )+\cos\left ( -\frac{\pi}{3} \right )\end{array}$$

= 4

20. The number of choices for

$$\begin{array}{l}\Delta \in \left\{ \land, \lor, \Rightarrow, \Leftrightarrow \right\}\end{array}$$
such that (p Δ q) ⇒ ((p Δ ~ q) ∨ ((~p) Δ q)) is a tautology, is

(A) 1

(B) 2

(C) 3

(D) 4

Sol. Let x : (pΔq) ⇒ (pΔ ~ q) ∨ (~pΔq)

Case-I

When Δ is same as ∨

Then (pΔ~q) ∨ (~pΔq) becomes

(p ∨ ~q) ∨ ( ~p q) which is always true, so x becomes a tautology.

Case-II

When Δ is same as ∧

Then (pq) ⇒ (p ∧ ~q) ∨ (~pq)

If pq is T, then (p ∧ ~ q) ∨ (~ pq) is F

so x cannot be a tautology.

Case-III

When Δ is same as ⇒

Then (p ⇒ ~ q) ∨ (~ pq) is same at (~ p ∨ ~ q) ∨ (pq), which is always true, so x becomes a tautology.

Case-IV

When Δ is same as ⇔

Then (p ⇔ q) ⇒ (p ⇔ ~ q) ∨ (~ pq)

pq is true when p and q have same truth values, then p ⇔ ~ q and ~pq both are false. Hence x cannot be a tautology.

So finally x can be ∨ or ⇒.

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 andw the on-screen virtual numeric keypad in the place designated to enter the answer.

1. The number of one-one functions ƒ : {a, b, c, d} → {0, 1, 2, …, 10} such that 2ƒ(a) – ƒ(b) + 3ƒ(c) + ƒ(d) = 0 is ________.

Sol. 3ƒ(c) + 2ƒ(a) + ƒ(d) = ƒ(b)

Value of ƒ(c)

Value of ƒ(a)

Number of functions

0

1

7

2

5

3

3

4

2

1

0

6

2

2

3

1

2

0

3

1

1

3

0

1

Total Number of functions =

31

2. In an examination, there are 5 multiple choice questions with 3 choices, out of which exactly one is correct. There are 3 marks for each correct answer, –2 marks for each wrong answer and 0 mark if the question is not attempted. Then, the number of ways a student appearing in the examination gets 5 marks is _____.

Sol. Let student marks x correct answers and y incorrect. So

3x – 2y = 5 and x + y 5 where x, y W

Only possible solution is (x, y) = (3, 2)

Student can mark correct answer by only one choice but for incorrect answer, there are two choices. So total number of ways of scoring 5 marks = 5C3(1)3. (2)2 = 40

3. Let

$$\begin{array}{l}A\left ( \frac{3}{\sqrt{a}},\sqrt{a} \right ), a > 0\end{array}$$
, be a fixed point in the xy-plane, The image of A in y-axis be B and the image of B in x-axis be C. If D(3cosθ, asinθ) is a point in the fourth quadrant such that the maximum area of ΔACD is 12 square units, then a is equal to ______.

Sol. Clearly B is

$$\begin{array}{l}\left ( -\frac{3}{\sqrt{a}},+\sqrt{a} \right )\end{array}$$
and C is
$$\begin{array}{l}\left ( -\frac{3}{\sqrt{a}},-\sqrt{a} \right )\end{array}$$

Area of

$$\begin{array}{l}\Delta ACD =\frac{1}{2}\begin{vmatrix}\frac{3}{\sqrt{a}} & \sqrt{a} & 1 \\-\frac{3}{\sqrt{a}} & -\sqrt{a} & 1 \\3\cos\theta & a\sin\theta & 1 \\\end{vmatrix}\end{array}$$

$$\begin{array}{l}\Delta =\begin{vmatrix}0 & 0 & 1 \\-\frac{3}{\sqrt{a}} & -\sqrt{a} & 1 \\3\cos\theta & a\sin\theta & 1 \\\end{vmatrix}\end{array}$$
$$\begin{array}{l}\Rightarrow \Delta = \left| 3\sqrt{a}\sin\theta +3\sqrt{a}\cos\theta\right|=3\sqrt{a}\left|\sin\theta + \cos\theta \right|\end{array}$$
$$\begin{array}{l}\Rightarrow \Delta_{\max} = 3\sqrt{a}\cdot\sqrt{2}=12 \Rightarrow a = \left ( 2\sqrt{2} \right )^2=8\end{array}$$

4. Let a line having direction ratios 1, –4, 2 intersect the lines

$$\begin{array}{l}\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}\end{array}$$
and
$$\begin{array}{l}\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}\end{array}$$
at the points A and B. Then (AB)2 is equal to _________.

Sol. Let A(3λ + 7, –λ + 1, λ – 2) and B (2μ, 3μ + 7, μ)

So, DR’s of AB ∝ 3λ – 2μ + 7, – (λ + 3μ + 6), λ – μ – 2

Clearly

$$\begin{array}{l}\frac{3\lambda-2\mu+7}{1}=\frac{\lambda+3\mu+6}{4}=\frac{\lambda-\mu-2}{2}\end{array}$$

⇒ 5λ – 3μ = –16 …(i)

And λ – 5μ = 10 …(ii)

From (i) and (ii) we get λ = –5, μ = –3

So, A is (–8, 6, –7) and B is (–6, –2, –3)

$$\begin{array}{l}AB=\sqrt{4+64+16}\Rightarrow \left ( AB \right )^2=84\end{array}$$

5. The number of points where the function

$$\begin{array}{l}f(x)=\begin{cases}\left| 2x^2-3x-7\right| & \text{if } x\le-1 \\\left [ 4x^2-1 \right ] &\text{if }-1<x<1 \\\left| x+1\right| + \left| x-2\right| & \text{if } x \ge 1 \\\end{cases} \end{array}$$
[t] denotes the greatest integer ≤ t, is discontinuous is _____________.

Sol. f (–1) = 2 and f (1) = 3

For x ∈ (–1, 1), (4x2 – 1) ∈ [–1, 3)

hence f(x) will be discontinuous at x = 1 and also

whenever 4x2 – 1 = 0, 1 or 2

$$\begin{array}{l}\Rightarrow x=\pm\frac{1}{2}, \pm\frac{1}{\sqrt{2}} \text{ and }\pm\frac{\sqrt{3}}{2}\end{array}$$

So there are total 7 points of discontinuity.

6. Let

$$\begin{array}{l}f(\theta)=sin\theta + \int_{-\pi/2}^{\pi/2}\left ( \sin\theta + t\cos\theta \right )f(t)dt\end{array}$$
. Then the value of
$$\begin{array}{l}\left|\int_{0}^{\pi/2}f(\theta)d\theta \right|\end{array}$$
is _________.

Sol.

$$\begin{array}{l}f(\theta) = \sin\theta\left ( 1+ \int_{-\pi/2}^{\pi/2}f(t)dt\right )+cos\theta\left (\int_{-\pi/2}^{\pi/2}tf(t)dt\right)\end{array}$$

Clearly f(θ) = asinθ + bcosθ

Where

$$\begin{array}{l}a=1+\int_{-\pi/2}^{\pi/2}\left ( a\sin t + b\cos t \right )dt \Rightarrow a= 1+ 2b ~~~~~\dots (1)\end{array}$$

and

$$\begin{array}{l}b=\int_{-\pi/2}^{\pi/2}\left ( at\sin t + bt\cos t \right )dt \Rightarrow b= 2a ~~~~~\dots (2)[/latex from (1) and (2) we get \(\begin{array}{l}a=-\frac{1}{3} \text{ and } b = -\frac{2}{3}\end{array}$$

So

$$\begin{array}{l}f(\theta)=-\frac{1}{3}\left ( \sin\theta + 2 \cos\theta \right )\end{array}$$
$$\begin{array}{l}\Rightarrow \left| \int_{0}^{\pi/2}f(\theta)d\theta\right| = \frac{1}{3}\left ( 1+2 \times 1 \right )=1\end{array}$$

7. Let

$$\begin{array}{l}\underset{0 \le x \le 2}{\text{Max}}\left\{ \frac{9-x^2}{5-x}\right\}=\alpha \end{array}$$
and
$$\begin{array}{l}\underset{0 \le x \le 2}{\text{Min}}\left\{ \frac{9-x^2}{5-x}\right\}=\beta\end{array}$$
.

If

$$\begin{array}{l}\int_{\beta-\frac{8}{3}}^{2\alpha -1} \text{Max}\left\{ \frac{9-x^2}{5-x},x\right\}dx =\alpha_1 + \alpha_2 \log_e\left ( \frac{8}{15} \right )\end{array}$$
then
$$\begin{array}{l}\alpha_1 + \alpha_2\end{array}$$
is equal to __________.

Sol. Let

$$\begin{array}{l}f(x)=\frac{x^2-9}{x-5}\Rightarrow f'(x)=\frac{(x-1)(x-9)}{(x-5)^2}\end{array}$$

So, α = f(1) = 2 and

$$\begin{array}{l}\beta = \min\left ( f(0), f(2) \right )= \frac{5}{3}\end{array}$$

Now,

$$\begin{array}{l}\int_{-1}^{3}\max \left\{ \frac{x^2-9}{x-5}, x\right\}dx = \int_{-1}^{9/5}\frac{x^2-9}{x-5}dx+\int_{9/5}^{3}x~dx\end{array}$$
$$\begin{array}{l}=\int_{-1}^{9/5}\left ( x+5+\frac{16}{x-5} \right )dx + \frac{x^2}{2}\left. \frac{}{} \right|_{9/5}^3\end{array}$$
$$\begin{array}{l}=\frac{28}{25}+14+16 \ln\left ( \frac{8}{15} \right )+\frac{72}{25}=18+16\ln\left ( \frac{8}{15} \right )\end{array}$$

Clearly

$$\begin{array}{l}\alpha_1 = 18\end{array}$$
and
$$\begin{array}{l}\alpha_2 = 16\end{array}$$
, so
$$\begin{array}{l}\alpha_1 + \alpha_2 = 34\end{array}$$

8. If two tangents drawn from a point (α, β) lying on the ellipse 25x2 + 4y2 = 1 to the parabola
y2 = 4x are such that the slope of one tangent is four times the other, then the value of (10α + 5)2 + (16β2 + 50)2 equals ____________.

Sol. ∵ (α, β) lies on the given ellipse, 25α2 + 4β2 = 1

…(1)

Tangent to the parabola,

$$\begin{array}{l}y=mx+\frac{1}{m}\end{array}$$
passes through (α, β). So, αm2 – βm + 1 = 0 has roots m1 and 4m1,

$$\begin{array}{l}m_1+4m_1=\frac{\beta}{\alpha}\end{array}$$
and
$$\begin{array}{l}m_1\cdot4m_1=\frac{1}{\alpha}\end{array}$$

Gives that 4β2 = 25α …(2)

from (1) and (2)

25(α2 + α) = 1 …(3)

Now, (10α + 5)2 + (16β2 + 50)2

= 25(2α + 1)2 + 2500 (2α + 1)2

= 2525 (4α 2 + 4α + 1) from equation (3)

$$\begin{array}{l}=2525\left ( \frac{4}{25}+1 \right )\end{array}$$

= 2929

9. Let S be the region bounded by the curves y = x3 and y2 = x. The curve y = 2|x| divides S into two regions of areas R1 and R2.strong>

If max {R1, R2} = R2, then

$$\begin{array}{l}\frac{R_2}{R_1}\end{array}$$
is equal to _____.

Sol. C1 : y = x3

C2y2 = x

and C3 = y = 2|x|

C1 and C2 intersect at (1, 1)

C2 and C3 intersect at

$$\begin{array}{l}\left ( \frac{1}{4}, \frac{1}{2} \right )\end{array}$$

Clearly

$$\begin{array}{l}R_1 = \int_{0}^{1/4}\left ( \sqrt{x}-2x \right )dx = \frac{2}{3}\left ( \frac{1}{8} \right )-\frac{1}{16}=\frac{1}{48}\end{array}$$

and

$$\begin{array}{l}R_1 +R_2 = \int_{0}^{1}\left ( \sqrt{x}-x^3 \right )dx = \frac{2}{3}-\frac{1}{4}=\frac{5}{12}\end{array}$$

So,

$$\begin{array}{l}\frac{R_1+R_2}{R_1}=\frac{5/12}{1/48}\Rightarrow1 + \frac{R_2}{R_1}=20\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{R_2}{R_1}=19\end{array}$$

10. If the shortest distance between the lines

$$\begin{array}{l}\vec{r}=\left ( -\hat{i}+3\hat{k} \right )+\lambda\left ( \hat{i}-a\hat{j} \right )\end{array}$$
and
$$\begin{array}{l}\vec{r}=\left ( -\hat{j}+2\hat{k} \right )+\mu\left ( \hat{i}- \hat{j}+\hat{k} \right )\end{array}$$
is
$$\begin{array}{l}\sqrt{\frac{2}{3}}\end{array}$$
, then the integral value of a is equal to ______________.

Sol.

$$\begin{array}{l}\vec{b_1}\times\vec{b_2}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\1 & -a & 0 \\1 & -1 & 1 \\ & & \\\end{vmatrix} =-a\hat{i}-\hat{j}+(a-1)\hat{k}\end{array}$$
$$\begin{array}{l}\vec{a_1}-\vec{a_2}=-\hat{i} + \hat{j}+\hat{k}\end{array}$$

Shortest distance

$$\begin{array}{l}=\left| \frac{(\vec{a_1}-\vec{a_2})\cdot(\vec{b_1}\times\vec{b_2})}{\left| \vec{b_1}\times\vec{b_2}\right|}\right|\end{array}$$
$$\begin{array}{l}\Rightarrow \sqrt{\frac{2}{3}}=\frac{2(a-1)}{\sqrt{a^2+1+(a-1)^2}}\end{array}$$
$$\begin{array}{l}\Rightarrow 6\left ( a^2-2a+1 \right )=2a^2-2a+2\end{array}$$
$$\begin{array}{l}\Rightarrow (a-2)(2a-1)=0 \Rightarrow a = 2 \text{ because } a \in z\end{array}$$
.

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