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# JEE Main 2022 June 25 - Shift 1 Maths Question Paper with Solutions

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JEE Main 2022 June 25 β 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 circle C touch the lines L1 : 4x – 3y +K1 = 0 and L2 : 4x – 3y + K2 = 0, K1, K2R. If a line passing through the centre of the circle C intersects L1 at (–1, 2) and L2 at (3, –6), then the equation of the circle Cis :

(A) (x – 1)2 + (y – 2)2 = 4

(B) (x + 1)2 + (y – 2)2 = 4

(C) (x – 1)2 + (y + 2)2 = 16

(D) (x – 1)2 + (y – 2)2 = 16

Sol.

Co-ordinate of centre

$$\begin{array}{l}C=\left ( \frac{3+\left ( -1 \right )}{2},\frac{-6+2}{2} \right )\equiv \left ( 1,-2 \right )\end{array}$$

L1 is passing through A

⇒ –4 – 6 + K1 = 0

K1 = 10

L2 is passing through B

⇒ 12 + 18 + K2 = 0

K2 = –30

Equation of L1 : 4x – 3y + 10 = 0

Equation of L1 : 4x – 3y – 30 = 0

Diameter of circle

$$\begin{array}{l}=\left|\frac{10+30}{\sqrt{4^2+\left ( -3 \right )^2}} \right|=8\end{array}$$

Equation of circle (x – 1)2 + (y + 2)2 = 16

2. The value of

$$\begin{array}{l}\displaystyle\int\limits_0^\pi\frac{e^{\cos x}\sin x}{\left ( 1+\cos^2x \right )\left ( e^{\cos x}+e^{-\cos x} \right )}dx\end{array}$$
is equal to :

(A)

$$\begin{array}{l}\frac{\pi^2}{4}\end{array}$$

(B)

$$\begin{array}{l}\frac{\pi^2}{2}\end{array}$$

(C)

$$\begin{array}{l}\frac{\pi}{4}\end{array}$$

(D)

$$\begin{array}{l}\frac{\pi}{2}\end{array}$$

Sol.

$$\begin{array}{l}\displaystyle\int\limits_0^\pi\frac{e^{\cos x}\sin x}{\left ( 1+\cos^2x \right )\left ( e^{\cos x}+e^{-\cos x} \right )}dx\end{array}$$

Let cosx = t

sinxdx = dt

$$\begin{array}{l}=\displaystyle\int \limits_1^{-1}\frac{-e^tdt}{\left ( 1+t^2 \right )\left ( e^t+e^{-t} \right )}\end{array}$$
$$\begin{array}{l}I=\displaystyle\int \limits_{-1}^1\frac{e^{-t}}{\left ( 1+t^2 \right )\left ( e^{t}+e^{-t} \right )}dt\dots\left ( i \right )\end{array}$$
$$\begin{array}{l}I=\displaystyle\int \limits_{-1}^1\frac{e^{-t}}{\left ( 1+t^2 \right )\left ( e^{-t}+e^t \right )}dt\dots\left ( ii \right )\end{array}$$

$$\begin{array}{l}2I=\displaystyle\int \limits_{-1}^1\frac{dt}{1+t^2}\end{array}$$
$$\begin{array}{l}\left.\begin{matrix}2I = \tan^{-t}\end{matrix}\right]_{-1}^{1}\end{array}$$
$$\begin{array}{l}2I=\frac{\pi}{4}-\left ( -\frac{\pi}{4} \right )\end{array}$$
$$\begin{array}{l}2I=\frac{\pi}{2}\end{array}$$
$$\begin{array}{l}I=\frac{\pi}{4}\end{array}$$

3. Let a, b and c be the length of sides of a triangle ABC such that

$$\begin{array}{l}\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}\end{array}$$
. If r and R are the radius of incircle and radius of circumcircle of the triangle ABC, respectively, then the value of
$$\begin{array}{l}\frac{R}{r}\end{array}$$
is equal to :

(A)

$$\begin{array}{l}\frac{5}{2}\end{array}$$

(B) 2

(C)

$$\begin{array}{l}\frac{3}{2}\end{array}$$

(D) 1

Sol.

$$\begin{array}{l}\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=\lambda\end{array}$$
$$\begin{array}{l}\begin{array}{c} a + b = 7 \lambda\\ b + c = 8 \lambda\\c + a = 9 \lambda\\\hline a + b + c 12\lambda \\ \end{array}\end{array}$$

a = 4λ, b = 3λ, c = 5λ

$$\begin{array}{l}s=\frac{4\lambda+3\lambda+5\lambda}{2} =6\lambda\end{array}$$
$$\begin{array}{l}\Delta=\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}=\sqrt{\left ( 6\lambda \right )\left ( 2\lambda \right )\left ( 3\lambda \right )\left ( \lambda \right )}\end{array}$$

= 6λ2

$$\begin{array}{l}R=\frac{abc}{4\Delta}=\frac{\left ( 4\lambda \right )\left ( 3\lambda \right )\left ( 5\lambda \right )}{4\left ( 6\lambda^2 \right )}= \frac{5}{2}\lambda\end{array}$$
$$\begin{array}{l}r=\frac{\Delta}{s}=\frac{6\lambda^2}{6\lambda}=\lambda\end{array}$$
$$\begin{array}{l}\frac{R}{r}=\frac{\frac{5}{2}\lambda}{\lambda}=\frac{5}{2}\end{array}$$

4. Let ƒ : NR be a function such that ƒ(x + y) = 2ƒ(x) ƒ(y) for natural numbers x and y. If ƒ(1) = 2, then the value of α for which

$$\begin{array}{l}\sum_{k=1}^{10}f\left ( \alpha+k \right )=\frac{512}{3}\left ( 2^{20}-1 \right )|\end{array}$$

holds, is :

(A) 2

(B) 3

(C) 4

(D) 6

Sol. ƒ(x + y) = 2ƒ(x)ƒ(y) & ƒ(1) = 2

x = y = 1

$$\begin{array}{l}\left.\begin{matrix}\Rightarrow f\left ( 2 \right )=2^3\\~~~~~~x=2,y=1\\\Rightarrow f\left ( 3)=2^5 \right )\end{matrix}\right\}\! f\left ( x \right )=2^{\left ( 2x-1 \right )}\end{array}$$

Now,

$$\begin{array}{l}\displaystyle\sum\limits_{k=1}^{10}f(\alpha+k)=\frac{512}{3}\left ( 2^{20}-1 \right ) \end{array}$$
$$\begin{array}{l}2\displaystyle\sum\limits_{k=1}^{10}f(\alpha)f\left ( k \right )=\frac{512}{3}\left ( 2^{20}-1 \right ) \end{array}$$
$$\begin{array}{l}2f\left ( \alpha \right )\left [ f\left ( 1 \right )+f\left ( 2 \right )+\cdots+f\left ( 10 \right ) \right ]=\frac{512}{3}(2^{20}-1)\end{array}$$
$$\begin{array}{l}2f\left ( \alpha \right )\left [ 2+2^3+2^5+\cdots \text{upto~10~terms} \right ]=\frac{512}{3}(2^{20}-1)\end{array}$$
$$\begin{array}{l}2f\left ( \alpha \right )\cdot2\left ( \frac{2^{20}-1}{4-1}\right )=\frac{512}{3}(2^{20}-1)\end{array}$$

ƒ(α) = 128 = 2 – 1

= 2α – 1 = 7

⇒ α = 4

5. Let A be a 3 × 3 real matrix such that

$$\begin{array}{l}A\begin{pmatrix}1 \\1 \\0\end{pmatrix}= \begin{pmatrix}1 \\1 \\0\end{pmatrix};~~A \begin{pmatrix}1 \\0 \\1\end{pmatrix}= \begin{pmatrix}-1 \\0 \\1\end{pmatrix}\end{array}$$
and
$$\begin{array}{l}A\begin{pmatrix}0 \\0 \\1\end{pmatrix}= \begin{pmatrix}1 \\1 \\2\end{pmatrix}\end{array}$$

If X = (x1, x2, x3)T and I is an identity matrix of order 3, then the system

$$\begin{array}{l}\left ( A-2I \right )X=\begin{pmatrix} 4\\ 1\\1\end{pmatrix}\end{array}$$
has :

(A) No solution

(B) Infinitely many solutions

(C) Unique solution

(D) Exactly two solutions

Sol. Let

$$\begin{array}{l}A=\begin{bmatrix}a & b & c \\d & e & f \\g & h & i \\\end{bmatrix}\end{array}$$
$$\begin{array}{l}A\begin{bmatrix}1 \\1 \\0\end{bmatrix}=\begin{bmatrix}1 \\1 \\0\end{bmatrix}\Rightarrow\begin{bmatrix} a& b & c \\d & 3 & f \\g & h & i \\\end{bmatrix}=\begin{bmatrix}1 \\1 \\0\end{bmatrix}\Rightarrow\begin{matrix}a + b =1 \\d + e =1 \\g +h =0 \\\end{matrix} \end{array}$$
$$\begin{array}{l}A\begin{bmatrix} 1\\ 0\\ 1\end{bmatrix}=\begin{bmatrix}-1 \\ 0\\1\end{bmatrix}\Rightarrow\begin{bmatrix} a& b& c \\ d& e& f \\ g& h& i \\\end{bmatrix}\begin{bmatrix} 1\\ 0\\1\end{bmatrix}=\begin{bmatrix} -1\\ 0\\1\end{bmatrix}\Rightarrow\begin{matrix}a+c=-1\\d+f=0\\g+i=1\end{matrix}\end{array}$$
$$\begin{array}{l}A\begin{bmatrix} 0\\ 0\\ 1\end{bmatrix}=\begin{bmatrix}1 \\1 \\2\end{bmatrix}\Rightarrow\begin{bmatrix} a& b& c \\ d& e& f \\ g& h& i \\\end{bmatrix}\begin{bmatrix} 0\\ 0\\1\end{bmatrix}=\begin{bmatrix} 1\\ 1\\2\end{bmatrix}\Rightarrow\begin{matrix}c=1\\f=1\\i=2\end{matrix}\end{array}$$

Solving will get

a = –2, b = 3, c = 1, d = –1, e = 2, f = 1, g = –1,
h = 1, i = 2

$$\begin{array}{l}A=\begin{bmatrix}-2 & 3 & 1 \\-1 & 2 & 1 \\-1 & 1 & 2 \\\end{bmatrix}\Rightarrow A=2I=\begin{bmatrix}-4 & 3 & 1 \\-1 & 0 & 1 \\-1 & 1 & 0 \\\end{bmatrix} \end{array}$$
$$\begin{array}{l}\left ( A-2I \right )x=\begin{bmatrix}4 \\1 \\1\end{bmatrix}\end{array}$$

⇒ –4x1 + 3x2 + x3 = 4 …(i)

x1 + x3 = 1 …(ii)

x1 + x2 = 1 …(iii)

So 3(iii) + (ii) = (i)

∴Infinite solution

6. Let ƒ : RR be defined as

ƒ(x) = x3 + x – 5

If g(x) is a function such that ƒ(g(x)) =

$$\begin{array}{l}x,\forall^{‘}x^{‘}\epsilon \textbf{R}\end{array}$$
then g′(63) is equal to_____.

(A)

$$\begin{array}{l}\frac{1}{49}\end{array}$$

(B)

$$\begin{array}{l}\frac{3}{49}\end{array}$$

(C)

$$\begin{array}{l}\frac{43}{49}\end{array}$$

(D)

$$\begin{array}{l}\frac{91}{49}\end{array}$$

Sol.ƒ(x) = 3x2 + 1

ƒ′(x) is bijective function

and ƒ(g(x)) = xg(x) is inverse of ƒ(x)

g(ƒ(x)) = x

g′(f(x)).f′(x) = 1

$$\begin{array}{l}g’\left ( f\left ( x \right ) \right )=\frac{1}{3x^2+1}\end{array}$$

Put x = 4 we get

$$\begin{array}{l}g’\left ( 63 \right )=\frac{1}{49}\end{array}$$

7. Consider the following two propositions :

P1 : ~ (p → ~ q)

P2: (p ~q) ((-~p) ∨ q)

If the proposition p → ((~p) q) is evaluated as FALSE, then :

(A) P1 is TRUE and P2 is FALSE

(B) P1 is FALSE and P2 is TRUE

(C) Both P1 and P2 are FALSE

(D) Both P1 and P2 are TRUE

Sol. Given p → (~ pq) is false

⇒ ~ p q is false and p is true

Now p = True.

~ T q = F

F q = Fq is false

P1 : ~ (T → ~ F) ≡ ~ (TT) ≡ False.

P2 : (T~ F) (~ T F) ≡ (TT) (FF)

TF ≡ False

8. If

$$\begin{array}{l}\frac{1}{2\cdot3^{10}}+\frac{1}{2^2\cdot3^9}+\dots+\frac{1}{2^{10}\cdot3}=\frac{K}{2^{10}\cdot3^{10}}\end{array}$$
then the remainder when K is divided by 6 is :

(A) 1

(B) 2

(C) 3

(D) 5

Sol.

$$\begin{array}{l}\frac{1}{2\cdot3^{10}}+\frac{1}{2^2\cdot3^9}+\dots+\frac{1}{2^{10}\cdot3}=\frac{K}{2^{10}\cdot3^{10}}\end{array}$$
$$\begin{array}{l}\frac{1}{2\cdot3^{10}}\left [ \frac{\left ( \frac{3}{2}^{10} \right )-1}{\frac{3}{2}-1} \right ]=\frac{K}{2^{10}\cdot3^{10}}\end{array}$$
$$\begin{array}{l}=\frac{3^{10}-2^{10}}{2^{10}\cdot3^{10}}=\frac{K}{2^{10}\cdot3^{10}}\Rightarrow K=3^{10}-2^{10}\end{array}$$

Now K = (1 + 2)10 – 210

= 10C0 + 10C1 2 + 10C2 23 + … + 10C10 210 – 210

= 10C0 + 10C1 2 + 6λ + 10C9⋅ 29

= 1 + 20 + 5120 + 6λ

= 5136 + 6λ + 5

= 6μ + 5

λ, μ ∈ N

∴ remainder = 5

9. Let ƒ(x) be a polynomial function such that ƒ(x) + ƒ′(x) + ƒ′′(x) = x5 + 64. Then, the value of

$$\begin{array}{l}\displaystyle \lim_{ x\to 1}\frac{f\left ( x \right )}{x-1}\end{array}$$
is equal to :

(A) –15 (B) –60

(C) 60 (D) 15

Sol.

$$\begin{array}{l}\displaystyle \lim_{ x\to 1}\frac{f\left ( x \right )}{x-1}\end{array}$$

ƒ(x) + ƒ′(x) + ƒ′′(x) = x5 + 64

Let ƒ(x) = x5 + ax4 + bx3 + cx2 + dx + e

ƒ′(x) = 5x4 + 4ax3 + 3bx2 + 2cx + d

ƒ′′(x) = 20x3 + 12ax2 + 6bx + 2c

x5 + (a + 5)x4 + (b + 4a + 20) x3 + (c + 3b + 12a) x2 + (d + 2c + 6b) x + e + d + 2c = x5 + 64

a + 5 = 0

b + 4a + 20 = 0

c + 3b + 12a = 0

d + 2c + 6b = 0

e + d + 2c = 64

a = – 5, b = 0, c = 60, d = –120,e = 64

ƒ(x) = x5 – 5x4 + 60x2 – 120x + 64

Now,

$$\begin{array}{l}\displaystyle \lim_{ x\to 1}\frac{x^5-5x^4+60x^2-120x+64}{x-1}~ \text{is}\left ( \frac{0}{0}~\textup{from} \right )\end{array}$$

By L′ Hopital rule

$$\begin{array}{l}\displaystyle \lim_{ x\to 1}\frac{5x^4-20x^3+120x-120}{1}\end{array}$$

= –15

10. Let E1 and E2 be two events such that the conditional probabilities

$$\begin{array}{l}P\left ( E_1|E_2 \right )=\frac{1}{2},\end{array}$$
$$\begin{array}{l}P\left ( E_2|E_1 \right )=\frac{3}{4}\end{array}$$
and
$$\begin{array}{l}P\left ( E_1\cap E_2 \right )=\frac{1}{8}\cdot\end{array}$$
Then:

(A)

$$\begin{array}{l}P\left ( E_1\cap E_2 \right )=P\left ( E_1 \right )\cdot P\left ( E_2 \right )\end{array}$$

(B)

$$\begin{array}{l}P\left ( E_1^{‘}\cap E_2^{‘} \right )=P\left ( E_1^{‘} \right )\cdot P\left ( E_2 \right )\end{array}$$

(C)

$$\begin{array}{l}P\left ( E_1\cap E_2^{‘} \right )=P\left ( E_1 \right )\cdot P\left ( E_2 \right )\end{array}$$

(D)

$$\begin{array}{l}P\left ( E_1^{‘}\cap E_2 \right )=P\left ( E_1 \right )\cdot P\left ( E_2 \right )\end{array}$$

Sol.

$$\begin{array}{l}P\left ( \frac{E_1}{E_2} \right )=\frac{1}{2}\Rightarrow\frac{P\left ( E_1\cap E_2 \right )}{P\left ( E_2 \right )}=\frac{1}{2}\end{array}$$
$$\begin{array}{l}P\left ( \frac{E_2}{E_1} \right )=\frac{3}{4}\Rightarrow\frac{P\left ( E_2\cap E_1 \right )}{P\left ( E_1 \right )}=\frac{3}{4}\end{array}$$
$$\begin{array}{l}P\left ( E_1\cap E_2 \right )=\frac{1}{8}\end{array}$$
$$\begin{array}{l}P\left ( E_2 \right )=\frac{1}{4},P\left ( E_1 \right )=\frac{1}{6}\end{array}$$

(A)

$$\begin{array}{l}P\left ( E_1\cap E_2 \right )=\frac{1}{8}~\text{and}~P\left ( E_1 \right )\cdot P\left ( E_2 \right )=\frac{1}{24}\end{array}$$

$$\begin{array}{l}P\left ( E_1\cap E_2 \right )\neq P\left (E_1 \right ).P\left ( E_2 \right )\end{array}$$

(B)

$$\begin{array}{l}P\left ( E_1^{‘}\cap E_2^{‘} \right )=1-P\left ( E_1\cup E_2 \right )\end{array}$$
$$\begin{array}{l}=1-\left [ \frac{1}{4}+\frac{1}{6}-\frac{1}{8} \right ]=\frac{17}{24}\end{array}$$
$$\begin{array}{l}P\left ( E_1^{‘} \right )=\frac{3}{4}\Rightarrow P\left ( E_1^{‘} \right )P\left ( E_2 \right )=\frac{3}{24}\end{array}$$

$$\begin{array}{l}P\left ( E_1^{‘}\cap E_2^{‘} \right )\neq P\left ( E_1^{‘} \right )\cdot P\left ( E_2 \right )\end{array}$$

(C)

$$\begin{array}{l}P\left ( E_1\cap E_2^{‘} \right )= P\left ( E_1 \right )- P\left (E_1\cap E_2 \right )\end{array}$$
$$\begin{array}{l}=\frac{1}{6}-\frac{1}{8}=\frac{1}{24}\end{array}$$
$$\begin{array}{l}P\left ( E_1 \right )\cdot P\left ( E_2 \right )=\frac{1}{24}\end{array}$$

$$\begin{array}{l}P\left ( E_1\cap E_2^{‘} \right )=P\left ( E_1 \right )\cdot P\left ( E_2 \right )\end{array}$$

(D)

$$\begin{array}{l}P\left ( E_1^{‘}\cap E_2 \right )=P\left ( E_2 \right )-P\left ( E_1\cap E_2 \right )\end{array}$$
$$\begin{array}{l}=\frac{1}{4}-\frac{1}{8}=\frac{1}{8}\end{array}$$
$$\begin{array}{l}P\left ( E_1 \right )P\left ( E_2 \right )=\frac{1}{24}\end{array}$$

$$\begin{array}{l}P\left ( E_1^{‘}\cap E_2 \right )\neq P\left ( E_1 \right ).P\left ( E_2 \right )\end{array}$$

11. Let

$$\begin{array}{l}A=\begin{bmatrix}0 & -2 \\2 & 0 \\\end{bmatrix}\end{array}$$
If M and N are two matrices given by
$$\begin{array}{l}M=\displaystyle\sum\limits_{k=1}^{10} A^{2k}\end{array}$$
and
$$\begin{array}{l}N=\displaystyle\sum\limits_{k=1}^{10}A^{2k-1}\end{array}$$
then MN2is :

(A) a non-identity symmetric matrix

(B) a skew-symmetric matrix

(C) neither symmetric nor skew-symmetric matrix

(D) an identity matrix

Sol.

$$\begin{array}{l}A=\begin{bmatrix}0 & -2 \\2 & 0 \\\end{bmatrix}\end{array}$$
$$\begin{array}{l}A^2=\begin{bmatrix}0 & -2 \\2 & 0 \\\end{bmatrix}\begin{bmatrix}0 & -2 \\2 & 0 \\\end{bmatrix}=\begin{bmatrix}-4 & 0 \\0 & -4 \\\end{bmatrix}=-4I \end{array}$$

M = A2 + A4 + A6 + … + A20

= –4I + 16l – 64I + … upto 10 terms

= –I [4 – 16 + 64 … + upto 10 terms]

$$\begin{array}{l}=-I\cdot 4\left [ \frac{\left ( -4 \right )^{10}-1}{-4-1} \right ]=\frac{4}{5}\left ( 2^{20}-1 \right )I\end{array}$$

= A – 4A + 16A + … upto 10 terms

$$\begin{array}{l}=A\left ( \frac{\left ( -4 \right )^{10}-1}{-4-1} \right )=- \left ( \frac{2^{20}-1}{5} \right )A\end{array}$$
$$\begin{array}{l}N^2=\frac{(2^{20}-1)^2}{2^5}A^2=\frac{-4}{25}\left ( 2^{20}-1 \right )^2t\end{array}$$
$$\begin{array}{l}MN^2=\frac{-16}{125}\left ( 2^{20}-1 \right )^3I=KI~~~~\left ( K\neq\pm 1 \right )\end{array}$$

(MN2)T = (KI)T = KI

∴ A is correct

12. Let g : (0, ∞) →R be a differentiable function such that

$$\begin{array}{l}\int\left ( \frac{x\left ( \cos x-\sin x \right )}{e^x+1}+\frac{g\left ( x \right )\left ( e^x+1-xe^x \right )}{\left ( e^x+1 \right )^2} \right ) dx=\frac{xg\left ( x \right )}{e^x+1}+c,\end{array}$$
for all x> 0, where c is an arbitrary constant. Then :

(A) g is decreasingin

$$\begin{array}{l}\left ( 0, \frac{\pi}{4} \right )\end{array}$$

(B) g′ is increasing in

$$\begin{array}{l}\left ( 0, \frac{\pi}{4} \right )\end{array}$$

(C) g + g′ is increasing in

$$\begin{array}{l}\left ( 0, \frac{\pi}{2} \right )\end{array}$$

(D) gg′ is increasing in

$$\begin{array}{l}\left ( 0, \frac{\pi}{2} \right )\end{array}$$

Sol.

$$\begin{array}{l}\int\left ( \frac{x\left ( \cos x-\sin x \right )}{e^x+1}+\frac{g\left ( x \right )\left ( e^x+1-xe^x \right )}{\left ( e^x+1 \right )^2} \right ) dx=\frac{xg\left ( x \right )}{e^x+1}+c\end{array}$$

Differentiating on both sides

$$\begin{array}{l}\frac{x\left ( \cos x-\sin x \right )}{e^x+1}+\frac{g\left ( x \right )\left ( e^x+1-xe^x \right )}{\left ( e^x-1 \right )^2}\end{array}$$
$$\begin{array}{l}=\frac{\left ( e^x+1 \right )\left ( g\left ( x \right )+xg^\frac{1}{x} \right )-xg\left ( x \right )e^x}{\left ( e^x+1 \right )^2}\end{array}$$
$$\begin{array}{l}=\frac{g\left ( x \right )\left [ e^x+1-xe^x \right ]}{\left ( e^x+1 \right )^2}+\frac{xg^{‘}\left ( x \right )\left ( e^x+1 \right )}{\left ( e^x+1 \right )^2}\end{array}$$
$$\begin{array}{l}=\frac{x\left ( \cos x-\sin x \right )}{e^x+1}=\frac{xg^{‘}\left ( x \right )}{e^x+1}\end{array}$$

$$\begin{array}{l}g^{‘}\left ( x \right )=\cos x-\sin x > 0~\textup{in}\left ( 0,\frac{\pi}{4} \right )\end{array}$$

⇒ g(x) is increasing in

$$\begin{array}{l}\left ( 0,\frac{\pi}{4} \right )\end{array}$$
⇒A is wrong

Now,

$$\begin{array}{l}g^{”}\left ( x \right )=-\sin x -\cos x<0~\textup{in}~\left ( 0,\frac{\pi}{4} \right )\end{array}$$

g(x) is increasing in

$$\begin{array}{l}\left ( 0,\frac{\pi}{4} \right )\end{array}$$
B is wrong

Let h(x) = g(x) + g′(x)

$$\begin{array}{l}h^{‘}\left ( x \right )=g^{‘}\left ( x \right )+g^{”}\left ( x \right )=-2\sin x < 0~\textup{in}~x\epsilon\left ( 0,\frac{\pi}{2} \right )\end{array}$$

g + g′is decreasing in

$$\begin{array}{l}\left ( 0,\frac{\pi}{2} \right )\end{array}$$

c is wrong

Let J(x) = g(x) – g′(x)

$$\begin{array}{l}J^{‘}\left ( x \right )=g^{‘}\left ( x \right )-g^{”}\left ( x \right )=2\cos x>0~\text{in}~\left ( 0,\frac{\pi}{2} \right )\end{array}$$

gg′ is increasing in

$$\begin{array}{l}\left ( 0,\frac{\pi}{2} \right )\Rightarrow~\textup{correct}\end{array}$$

13. Let f :RR and g : RR be two functions defined by f(x) = loge(x2 + 1) – ex + 1 and

$$\begin{array}{l}g\left ( x \right )=\frac{1-2e^{2x}}{e^x}\end{array}$$
. Then, for which of the following range of α, the inequality
$$\begin{array}{l}f\left ( g\left ( \frac{\left ( \alpha-1 \right )^2}{3} \right ) \right )>f\left ( g\left ( \alpha-\frac{5}{3} \right ) \right )\end{array}$$
holds?

(A) (2, 3)

(B) (–2, –1)

(C) (1, 2)

(D) (–1, 1)

Sol.

$$\begin{array}{l}f\left ( x \right )=log_e\left ( x^2+1 \right )-e^{-x}+1\end{array}$$
$$\begin{array}{l}f^{‘}\left ( x \right )=\frac{2x}{x^2+1}+e^{-x}\end{array}$$
$$\begin{array}{l}=\frac{2}{x+\frac{1}{x}}+e^{-x}>0~~\forall x\epsilon R\end{array}$$
$$\begin{array}{l}g\left ( x \right )=e^{-x}-2e^x\end{array}$$
$$\begin{array}{l}g^{‘}\left ( x \right )=-e^{-x}-2e^x<0~~~\forall x\epsilon R\end{array}$$

f(x) is increasing and g(x) is decreasing function.

$$\begin{array}{l}f\left ( g\left ( \frac{\left ( \alpha-1 \right )^2}{3} \right ) \right )>f\left ( g\left ( \alpha-\frac{5}{3} \right ) \right )\end{array}$$
$$\begin{array}{l}\Rightarrow\frac{\left ( \alpha-1 \right )^2}{3}<\alpha-\frac{5}{3}\end{array}$$

= α2 – 5α + 6 < 0

= (α – 2)(α – 3) < 0

= α∈(2, 3)

14. Let

$$\begin{array}{l}\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}a_i>0, i=1,~2,~3\end{array}$$
be a vector which makes equal angles with the coordinate axes OX, OY and OZ. Also, let the projection of
$$\begin{array}{l}\vec{a}\end{array}$$
on the vector
$$\begin{array}{l}3\hat{i}+4\hat{j}\end{array}$$
be 7. Let
$$\begin{array}{l}\vec{b}\end{array}$$
be a vector obtained by rotating
$$\begin{array}{l}\vec{a}\end{array}$$
with 90°. If
$$\begin{array}{l}\vec{a},\vec{b}\end{array}$$
and x-axis are coplanar, then projection of a vector
$$\begin{array}{l}\vec{b}\end{array}$$
on
$$\begin{array}{l}3\hat{i}+4\hat{j}\end{array}$$
is equal to :

(A)

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

(B)

$$\begin{array}{l}\sqrt{2}\end{array}$$

(C) 2

(D) 7

Sol.

$$\begin{array}{l}\cos^2~\alpha+cos^2~\beta+\cos^2~\gamma=1~\Rightarrow~\cos^2~\alpha=\frac{1}{3}\end{array}$$
$$\begin{array}{l}\Rightarrow\cos\alpha=\frac{1}{\sqrt{3}}\end{array}$$
$$\begin{array}{l}\vec{a}=\frac{\lambda}{3}\left ( \hat{i}+\hat{j}+\hat{k} \right ),\lambda>0\end{array}$$
$$\begin{array}{l}\frac{\lambda}{\sqrt{3}}\frac{\left ( \hat{i}+\hat{j}+\hat{k} \right )\cdot\left ( 3\hat{i}+4\hat{j} \right )}{\sqrt{3^2+4^2}}=7\end{array}$$
$$\begin{array}{l}\Rightarrow\frac{\lambda}{\sqrt{3}}\left ( 3+4 \right )=7\times5\end{array}$$

$$\begin{array}{l}\lambda=5\sqrt{3}\end{array}$$
$$\begin{array}{l}\vec{a}=5\left ( \hat{i}+\hat{j}+\hat{k} \right )\end{array}$$

Let

$$\begin{array}{l}\vec{b}=p\hat{i}+q\hat{j}+r\hat{k}\end{array}$$
$$\begin{array}{l}\vec{a}\cdot\vec{b}=0~\text{and}~\left [ \vec{a}~\vec{b}~\hat{i} \right ]=0\end{array}$$

p + q + r = 0 …(i)

&

$$\begin{array}{l}\begin{vmatrix}p & q & r \\1 & 1 & 1 \\1 & 0 & 0 \\\end{vmatrix}=0\end{array}$$
$$\begin{array}{l}\Rightarrow\vec{b}=-2r\hat{i}+r\hat{j}+r\hat{k}[\end{array}$$
$$\begin{array}{l}\vec{b}=r\left ( -2\hat{i}+\hat{j}+\hat{k} \right )\end{array}$$

Now

$$\begin{array}{l}\left|\vec{a} \right|=\left|\vec{b} \right|\end{array}$$
$$\begin{array}{l}5\sqrt{3}=\left|r \right|\sqrt{b}\Rightarrow~\left|r \right|=\frac{5}{\sqrt{2}}\end{array}$$

⇒ Projection of

$$\begin{array}{l}\vec{b}\end{array}$$
on
$$\begin{array}{l}3\hat{i}+4\hat{j}=\left|\frac{\vec{b}\cdot\left ( 3\hat{i}+4\hat{j} \right )}{\sqrt{3^2+4^2}} \right|\end{array}$$
$$\begin{array}{l}=\left|r \right|\frac{\left ( -6+4 \right )}{5}=\left|\frac{-2}{5} \right|\end{array}$$

Projection

$$\begin{array}{l}=\frac{2}{5}\times\frac{5}{\sqrt{2}}=\sqrt{2}\end{array}$$

∴ B is correct

15. Let y = y(x) be the solution of the differential equation (x + 1)y′ – y = e3x(x + 1)2, with

$$\begin{array}{l}y\left ( 0 \right )=\frac{1}{3}\end{array}$$
Then, the point
$$\begin{array}{l}x=-\frac{4}{3}\end{array}$$
for the curve y = y(x)is :

(A) not a critical point

(B) a point of local minima

(C) a point of local maxima

(D) a point of inflection

Sol.

$$\begin{array}{l}\left ( x+1 \right )\frac{dy}{dx}-y=e^{3x}\left ( x+1 \right )^2\end{array}$$
$$\begin{array}{l}\frac{dy}{dx}-\frac{y}{x+1}=e^{3x}\left ( x+1 \right )\end{array}$$

If

$$\begin{array}{l}e^{-\int\frac{1}{x+1}x }=e^{-log\left ( x+1 \right )}=\frac{1}{x+1}\end{array}$$

$$\begin{array}{l}y\left ( \frac{1}{x+1} \right )=\int\frac{e^{3x}\left ( x+1 \right )}{x+1}dx\end{array}$$
$$\begin{array}{l}\frac{y}{x+1}=\int e^{3x}dx\end{array}$$
$$\begin{array}{l}\frac{y}{x+1}=\frac{e^{3x}}{3}+c \end{array}$$
$$\begin{array}{l}\because y\left ( 0 \right )=\frac{1}{3}\end{array}$$
$$\begin{array}{l}\frac{1}{3}=\frac{1}{3}+c\end{array}$$
• c = 0

So,

$$\begin{array}{l}y=\frac{e^{3x}}{3}(x+1) \end{array}$$

$$\begin{array}{l}y^{‘}=e^{3x}\left ( x+1 \right )+\frac{e^{3x}}{3}=e^{3x}\left ( x+\frac{4}{3} \right )\end{array}$$
$$\begin{array}{l}y^{”}=3e^{3x}\left ( x+\frac{4}{3} \right )+e^{3x}=e^{3x}\left ( 3x+5 \right )\end{array}$$
$$\begin{array}{l}y^{‘}=0~\textup{at}~x=\frac{-4}{3}~\&~y^{”}=e^{-4}\left ( 1 \right )>0~\textup{at}~x=\frac{-4}{3}\end{array}$$

$$\begin{array}{l}x=\frac{-4}{3}\end{array}$$
is point of local minima

16. If y = m1x + c1 and y = m2x + c2, m1m2 are two common tangents of circle x2 + y2 = 2 and parabola y2 = x, then the value of 8|m1m2| is equal to :

(A)

$$\begin{array}{l}3+4\sqrt{2}\end{array}$$

(B)

$$\begin{array}{l}-5+6\sqrt{2}\end{array}$$

(C)

$$\begin{array}{l}-4+3\sqrt{2}\end{array}$$

(D)

$$\begin{array}{l}7+6\sqrt{2}\end{array}$$

Sol. Let tangent to y2 = x be

$$\begin{array}{l}y=mx+\frac{1}{4m}\end{array}$$

For it being tangent to circle.

$$\begin{array}{l}\left|\frac{\frac{1}{4}m}{\sqrt{1+m^2}} \right|=\sqrt{2}\end{array}$$

⇒ 32m4 + 32m2 – 1 = 0

$$\begin{array}{l}m^2=\frac{-32\pm\sqrt{\left ( 32 \right )^2+4\left ( 32 \right )}}{64}\end{array}$$

$$\begin{array}{l}8m_1m_2=-4+3\sqrt{2}\end{array}$$

17. Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S: x + y + z = 5. If a line L passing through (1, –1, –1), parallel to the line PQ meets the plane S at R, then QR2 is equal to :

(A) 2 (B) 5

(C) 7 (D) 11

Sol. As L is parallel to PQ d.r.s of S is <1, 1, 1>

$$\begin{array}{l}L\equiv \frac{x-1}{1}=\frac{y+1}{1}=\frac{z+1}{1}\end{array}$$

Point of intersection of L and S be λ

⇒ (λ + 1) + (λ – 1) + (λ – 1) = S

⇒ λ = 2

R≡ (3, 1, 1)

Let Q(α, β, γ)

$$\begin{array}{l}\frac{\alpha-1}{1}=\frac{\beta}{1}=\frac{\gamma-1}{1}=\frac{-2\left ( 3 \right )}{3}\end{array}$$

⇒ α = 3, β = 2, γ = 3

Q≡ (3, 2, 3)

(QR)2 = 02 + (1)2 + (2)2 = 5

18. If the solution curve y = y(x) of the differential equation y2dx + (x2xy + y2)dy = 0, which passes through the point (1,1) and intersects the line

$$\begin{array}{l}y=\sqrt{3}x\end{array}$$
at the point
$$\begin{array}{l}\left ( \alpha,\sqrt{3}\alpha \right )\end{array}$$
, then value of
$$\begin{array}{l}log_e\left ( \sqrt{3}\alpha \right )\end{array}$$
is equal to :

(A)

$$\begin{array}{l}\frac{\pi}{3}\end{array}$$

(B)

$$\begin{array}{l}\frac{\pi}{2}\end{array}$$

(C)

$$\begin{array}{l}\frac{\pi}{12}\end{array}$$

(D)

$$\begin{array}{l}\frac{\pi}{6}\end{array}$$

Sol.

$$\begin{array}{l}\frac{dy}{dx}=\frac{y^2}{xy-x^2-y^2}\end{array}$$

Put y = vx we get

$$\begin{array}{l}v+x\frac{dv}{dx}=\frac{v^2}{v-1-v^2}\end{array}$$

$$\begin{array}{l}x\frac{dv}{dx}=\frac{v^2-v^2+v+v^3}{v-1-v^2}\end{array}$$

$$\begin{array}{l}\int\frac{v-1-v^2}{v\left ( 1+v^2 \right )}dv=\int\frac{dx}{x}\end{array}$$
$$\begin{array}{l}\tan^{-1}\left ( \frac{y}{x} \right )-\textup{In}\left ( \frac{y}{x} \right )=\textup{In}~x+c\end{array}$$

As it passes through (1, 1)

$$\begin{array}{l}c=\frac{\pi}{4}\end{array}$$

$$\begin{array}{l}\tan^{-1}\left ( \frac{y}{x} \right )-\textup{In}\left ( \frac{y}{x} \right )=\textup{In}~x+\frac{\pi}{4}\end{array}$$

Put

$$\begin{array}{l}y=\sqrt{3}x\end{array}$$
we get

$$\begin{array}{l}\frac{\pi}{3}-\textup{In}\sqrt{3}=\textup{In}~x+\frac{\pi}{4}\end{array}$$

$$\begin{array}{l}\textup{In}~x=\frac{\pi}{12}-\textup{In}\sqrt{3}=\textup{In}~\alpha\end{array}$$

$$\begin{array}{l}\textup{In}\left ( \sqrt{3}\alpha \right )=\textup{In}\sqrt{3}+\textup{In}~\alpha\end{array}$$
$$\begin{array}{l}=\textup{In}\sqrt{3}+\frac{\pi}{12}-\textup{In}\sqrt{3}=\frac{\pi}{12}\end{array}$$

19. Let

$$\begin{array}{l}x=2t,y=\frac{t^2}{3}\end{array}$$
be a conic. Let S be the focus and B be the point on the axis of the conic such that SABA, where A is any point on the conic. If k is the ordinate of the centroid of the ΔSAB, then
$$\begin{array}{l}\displaystyle \lim_{ t\to 1}k\end{array}$$
is equal to

(A)

$$\begin{array}{l}\frac{17}{18}\end{array}$$

(B)

$$\begin{array}{l}\frac{19}{18}\end{array}$$

(C)

$$\begin{array}{l}\frac{11}{18}\end{array}$$

(D)

$$\begin{array}{l}\frac{13}{18}\end{array}$$

Sol.

$$\begin{array}{l}x=2t,y=\frac{2}{3}\end{array}$$
$$\begin{array}{l}t\rightarrow1~~A\equiv\left ( 2,\frac{1}{3} \right )\end{array}$$

Given conic is x2 = 12y S ≡ (0, 3)

Let B ≡ (0, β)

GivenSABA

$$\begin{array}{l}\left ( \frac{\frac{1}{3}}{2-3} \right ) \left ( \frac{\beta-\frac{1}{3}}{-2} \right )=-1\end{array}$$

$$\begin{array}{l}\left ( \beta-\frac{1}{3} \right )\frac{1}{3}=-2\end{array}$$

$$\begin{array}{l}\beta=\frac{1}{3}\left ( \frac{-17}{3} \right )\end{array}$$
$$\begin{array}{l}\underset{\left ( \textup{as}~t\rightarrow1 \right )}{\textup{Ordinate~of~centroid}}=k=\frac{\beta+\frac{1}{3}+3}{3}\end{array}$$
$$\begin{array}{l}=\frac{\frac{-17}{9}+\frac{10}{3}}{3}=\frac{13}{18}\end{array}$$

20. Let a circle C in complex plane pass through the points z1 = 3 + 4i, z2 = 4 + 3i and z3 = 5i. If z(≠z1) is a point on C such that the line through z and z1 is perpendicular to the line through z2 and z3, then arg(z) is equal to:

(A)

$$\begin{array}{l}\tan^{-1}\left (\frac{2}{\sqrt{5}} \right )-\pi\end{array}$$

(B)

$$\begin{array}{l}\tan^{-1}\left ( \frac{24}{7} \right )-\pi\end{array}$$

(C) tan–1 (3) – π

(D)

$$\begin{array}{l}\tan^{-1}\left (\frac{3}{4} \right )-\pi\end{array}$$

Sol. z1 = 3 + 4i, z2 = 4 + 3i and z3 = 5i

Clearly C x2 + y2 = 25

Let z(x, y)

$$\begin{array}{l}\left ( \frac{y-4}{x-3}\right )\left ( \frac{2}{-4} \right )=-1\end{array}$$

y = 2x – 2 ≡ L

z is intersection of C&L

$$\begin{array}{l}z\equiv\left ( \frac{-7}{5},\frac{-24}{5} \right )\end{array}$$

$$\begin{array}{l}\textup{Arg}\left ( z \right )=-\pi+\tan^{-1}\left ( \frac{24}{7} \right )\end{array}$$

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 Cr denote the binomial coefficient of xr in the expansion of (1 + x)10. If for α, β∈R, C1 + 3⋅2 C2 + 5⋅3 C3 + … upto 10 terms

$$\begin{array}{l}=\frac{\alpha\times2^{11}}{2^\beta-1}\left ( C_0+\frac{C_1}{2}+\frac{C_2}{3}+\dots\textup{upto~10~terms} \right )\end{array}$$
then the value of α + β is equal to _______.

Sol. Given that C1 + 2⋅3C2 + 5⋅3C3 + … 10 terms

$$\begin{array}{l}=\frac{\alpha\cdot2^{11}}{2^\beta-1}\left ( C_1+\frac{C_2}{2}+\dots \right )\end{array}$$

$$\begin{array}{l}\displaystyle\sum\limits_{r=1}^{10}r\left ( 2r-1 \right )C_r=\frac{\alpha\cdot 2^{11}}{2^\beta-1}\left ( \displaystyle\sum\limits_{r=0}^{10}\frac{C_r}{r} \right )\end{array}$$

Using C1 + 2C2 + …. + nCn = n.2n – 1,

12C1 + 22C2 + … + n2Cn = n.2n – 1 + n(n – 1)2n – 2

and

$$\begin{array}{l}C_0+\frac{C_1}{2}+\dots\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\end{array}$$
we get

⇒ 2(10.29 + 10.9.28) – 10.29

$$\begin{array}{l}=\frac{\alpha2^{11}}{2^\beta-1} \frac{\left ( 2^{11}-1 \right )}{11}\end{array}$$

Comparing both side we get

$$\begin{array}{l}2^{11}.25=\frac{\alpha.2^{11}}{2^\beta-1}\frac{\left ( 2^{11}-1 \right )}{11}\end{array}$$

⇒ α = 25 × 11 = 275 &β = 11

⇒ α + β = 286

(*RHS shall have II-terms)

2. The number of 3-digit odd numbers, whose sum of digits is a multiple of 7, is ________.

Sol. For odd number unit place shall be 1, 3, 5, 7 or 9.

xy1, xy3, xy 5, xy 7, xy 9 are the type of numbers.

If xy 1 then

x + y = 6, 13, 20 … Cases are required

i.e., 6 + 6 + 0 + … = 12 ways

If xy 3 then

x + y = 4, 11, 18, …. Cases are required

i.e., 4 + 8 + 1 + 0 … = 13 ways

Similarly for xy 5, we have

x + y = 2, 9, 16, …

i.e., 2 + 9 + 3 = 14 ways

for xy 7 we have

x + y = 0, 7, 14, ….

i.e., 0 + 7 + 5 = 12 ways

And for xy 9 we have

x + y = 5, 12, 19 …

i.e., 5 + 7 + 0 … = 12 ways

∴ Total 63 ways

3. Let θ be the angle between the vectors

$$\begin{array}{l}\vec{a}\end{array}$$
and
$$\begin{array}{l}\vec{b}\end{array}$$
, where
$$\begin{array}{l}\left|\vec{a} \right|=4,\left|\vec{b} \right|=3\end{array}$$
and
$$\begin{array}{l}\theta\epsilon\left ( \frac{\pi}{4},\frac{\pi}{3} \right )\end{array}$$
. Then
$$\begin{array}{l}\left|\left ( \vec{a}-\vec{b} \right )\times\left ( \vec{a}+\vec{b} \right ) \right|^2+4\left ( \vec{a}.\vec{b} \right )^2\end{array}$$
is equal to ______.

Sol.

$$\begin{array}{l}\left|\left ( \vec{a}-\vec{b} \right )\times\left ( \vec{a}+\vec{b} \right ) \right|^2+4\left ( \vec{a}\cdot\vec{b} \right )^2\end{array}$$

$$\begin{array}{l}\left|\vec{a}\times\vec{a}+\vec{a}\times\vec{b}-\vec{b}\times\vec{a}-\vec{b}\times\vec{b} \right|^2+4\left ( \vec{a}\cdot\vec{b} \right )^2\end{array}$$

$$\begin{array}{l}\left|2\left ( \vec{a}\times\vec{b} \right ) \right|^2+4\left ( \vec{a}\cdot\vec{b} \right )^2\end{array}$$

$$\begin{array}{l}4\left ( \vec{a}\times\vec{b} \right )^2+\left ( \vec{a}\cdot\vec{b} \right )^2\end{array}$$

$$\begin{array}{l}4\left|\vec{a} \right|^2\left|\vec{b} \right|^2=4.16.9=576\end{array}$$

4. Let the abscissae of the two points P and Q be the roots of 2x2rx + p = 0 and the ordinates of P and Q be the roots of x2sxq = 0. If the equation of the circle described on PQ as diameter is 2(x2 + y2) – 11x – 14y – 22 = 0, then 2r + s – 2q + p is equal to _________.

Sol. Let P(x1, y1) &Q(x2, y2)

Solve 2x2rx + p = 0 for x1 and x2

Solve x2sxq = 0 for y1 and y2

∴ Equation of circle ≡ (xx1) (xx2) + (yy1) (yy2) = 0

x2 – (x1 + x2)x + x1x2 + y2 – (y1 + y2)y + y1y2 = 0

$$\begin{array}{l}x^2-\frac{r}{2}x+\frac{p}{2}+y^2+sy-q=0\end{array}$$

⇒ 2x2 + 2y2rx + 2sy + p – 2q = 0

Compare with 2x2 + 2y2 – 11x – 14y – 22 = 0

We get r = 11, s = 7, p – 2q = –22

⇒ 2r + s + p – 2q = 22 + 7 – 22 = 7

5. The number of values of x in the interval

$$\begin{array}{l}\left ( \frac{\pi}{4},\frac{7\pi}{4} \right )\end{array}$$
for which 14cosec2x – 2sin2x = 21 – 4cos2x holds, is _____________.

Sol.

$$\begin{array}{l}\frac{14}{\sin^2x}-2\sin^2~x=21-4\left ( 1-\sin^2x \right )\end{array}$$

Let sin2x = t

⇒ 14 – 2t2 = 21t – 4t + 4t2

⇒ 6t2 + 17t – 14 = 0

⇒ 6t2 + 21t – 4t – 14 = 0

⇒ 3t(2t + 7) – 2(2t + 7) = 0

$$\begin{array}{l}\sin^2~x=\frac{2}{3}\end{array}$$
or
$$\begin{array}{l}-\frac{7}{3}\end{array}$$
(rejected)

$$\begin{array}{l}\sin~x=\pm\sqrt{\frac{2}{3}}\end{array}$$

$$\begin{array}{l}\sin~x=\pm\sqrt{\frac{2}{3}}\end{array}$$
has 4 solutions in
$$\begin{array}{l}\left ( \frac{\pi}{4},\frac{7\pi}{4} \right )\end{array}$$

6. For a natural number n, let αn = 19n – 12n. Then, the value of

$$\begin{array}{l}\frac{31\alpha_9-\alpha_{10}}{57\alpha_8}\end{array}$$
is ______________.

Sol. αn = 19n – 12n

Let equation of roots 12 & 19 i.e.

$$\begin{array}{l}x^2-31x+228=0\end{array}$$

$$\begin{array}{l}\left ( 31-x \right )=\frac{228}{x}\end{array}$$
(where x can be 19 or 12)

$$\begin{array}{l}\frac{31\alpha_9-\alpha_{10}}{57\alpha_8}=\frac{31\left ( 19^9-12^9 \right )-\left ( 19^{10}-12^{10} \right )}{57\left ( 19^8-12^8 \right )}\end{array}$$
$$\begin{array}{l}=\frac{19^9\left ( 31-19 \right )-12^9\left ( 31-12 \right )}{57\left ( 19^8-12^8 \right )}\end{array}$$
$$\begin{array}{l}=\frac{228\left ( 19^8-12^8 \right )}{57\left ( 19^8-12^8 \right )}=4\end{array}$$
.

7. Let f :RR be a function defined by

$$\begin{array}{l}f\left ( x \right )=\left ( 2\left ( 1-\frac{x^{25}}{2} \right )\left ( 2+x^{25} \right ) \right )^\frac{1}{50}\end{array}$$
. If the function g(x) = f (f (f (x))) + f (f (x)), then the greatest integer less than or equal to g(1) is ___________.

Sol.

$$\begin{array}{l}f\left ( x \right )=\left ( 2\left ( \frac{2-x^{25}}{2} \right )\left ( 2+x^{25} \right ) \right )^\frac{1}{50}\end{array}$$
$$\begin{array}{l}=\left ( 4-x^{50} \right )^\frac{1}{50}\end{array}$$
$$\begin{array}{l}f\left ( f\left ( x \right ) \right )=\left ( 4-\left ( \left ( 4-x^{50} \right )^\frac{1}{50} \right )^{50} \right )^\frac{1}{50}=x\end{array}$$

As f(f(x)) = x we have

g(x) = f(f(f(x))) + f(f(x)) = f(x) + x

g(x) = (4 – x50)1/50 + x

g(1) = 31/50 + 1

⇒ [g(1)] = 2

8. Let the lines

$$\begin{array}{l}L_1:\vec{r}=\lambda\left ( \hat{i}+2\hat{j}+3\hat{k} \right ),\lambda\epsilon R\end{array}$$
$$\begin{array}{l}L_2:\vec{r}=\left ( \hat{i}+3\hat{j}+\hat{k} \right )+\mu\left ( \hat{i}+\hat{j}+5\hat{k} \right );\mu\epsilon R, \end{array}$$

intersect at the point S. If a plane ax + byz + d = 0 passes through S and is parallel to both the lines L1 and L2, then the value of a + b + d is equal to _______.

Sol. As plane is parallel to both the lines we have d.r’s of normal to the plane as < 7, – 2, –1>

$$\begin{array}{l}\left ( \textup{from}\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\1 & 2 & 3 \\1 & 1 & 5 \\\end{vmatrix}=7\hat{i}-\hat{j}\left ( 2 \right )+\hat{k}\left ( -1 \right ) \right )\end{array}$$

Also point of intersection of lines is

$$\begin{array}{l}2\hat{i}+4\hat{j}+6\hat{k}\end{array}$$

∴ Equation of plane is

7(x – 2) –2 (y – 4) – 1 (z – 6) = 0

⇒ 7x – 2yz = 0

a + b + d = 7 – 2 + 0 = 5

9. Let A be a 3 × 3 matrix having entries from the set {–1, 0, 1}. The number of all such matrices A having sum of all the entries equal to 5, is ____________.

Sol. Let matrix

$$\begin{array}{l}\begin{bmatrix}a & b & c \\d & e & f \\g & h & i \\\end{bmatrix}\end{array}$$

We need

a + b + c + d + e + f + g + h + i = 5

Possible cases

Number of ways

5 → 1’s, 4 → zeroes

$$\begin{array}{l}\frac{9!}{5!4!}=126\end{array}$$

6 → 1’s, 2 → zeroes, 1 →–1

$$\begin{array}{l}\frac{9!}{6!2!}=252\end{array}$$

7 → 1’s, 2 →–1’s

$$\begin{array}{l}\frac{9!}{7!2!}=36\end{array}$$

Total ways = 126 + 252 + 36 = 414

10. The greatest integer less than or equal to the sum of first 100 terms of the sequence

$$\begin{array}{l}\frac{1}{3},\frac{5}{9},\frac{19}{27},\frac{65}{81},\dots\end{array}$$
is equal to _______________.

Sol.

$$\begin{array}{l}S=\frac{1}{3}+\frac{5}{9}+\frac{19}{27}+\frac{65}{81}+\dots\end{array}$$
$$\begin{array}{l}=\displaystyle\sum\limits_{r=1}^{100}\left ( \frac{3^r-2^r}{3^r} \right )\end{array}$$
$$\begin{array}{l}=100-\frac{2}{3}\frac{\left ( 1-\left ( \frac{2}{3} \right )^{100} \right )}{1/3}\end{array}$$
$$\begin{array}{l}=98+2\left ( \frac{2}{3} \right )^{100}\end{array}$$

[S] = 98

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