Differential Equations jee mains questions | JEE Main Maths Differential Equations Previous Year Question

Q 11 :

If the solution curve, of the differential equation $\frac{d y}{d x} = \frac{x + y - 2}{x - y}$ passing through the point (2, 1) is $\tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{β} \log_{e} (α + {(\frac{y - 1}{x - 1})}^{2}) = \log_{e}$ $| x - 1 |$ , then $5 β + α$ is equal to __________. [2024]

(11)

We have, $\frac{d y}{d x} = \frac{x + y - 2}{x - y}$

Substitute x = X + h and y = Y + k

$\Rightarrow \frac{d Y}{d X} = \frac{X + Y + h + k - 2}{X - Y + h - k}$

Put h + k – 2 = 0 and h – k = 0

$\Rightarrow$ h = 1 and k = 1

$∴ \frac{d Y}{d X} = \frac{X + Y}{X - Y}$

Now, put Y = vX $\Rightarrow \frac{d Y}{d X} = v + X \frac{d v}{d X}$

$∴ v + X \frac{d v}{d X} = \frac{X + v X}{X - v X} = \frac{1 + v}{1 - v}$

$\Rightarrow X \frac{d v}{d X} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v + v^{2}}{1 - v} = \frac{1 + v^{2}}{1 - v}$

$\Rightarrow \int \frac{1 - v}{1 + v^{2}} d v = \int \frac{d X}{X} \Rightarrow \int \frac{1}{1 + v^{2}} d v - \int \frac{v}{1 + v^{2}} d v = \int \frac{d X}{X}$

$\Rightarrow \tan^{- 1} v - \frac{1}{2} \log_{e} | 1 + v^{2} | = \log | X | + c$

$\Rightarrow \tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{2} \log_{e}$ $| 1 + {(\frac{y - 1}{x - 1})}^{2} |$ $= \log$ $| x - 1 |$ $+ c$ ... (i)

Equation (i) passing through (2, 1), then

$\tan^{- 1} (\frac{1 - 1}{2 - 1}) - \frac{1}{2} \log_{e}$ $| 1 + {(\frac{1 - 1}{2 - 1})}^{2} |$ $= \log_{e} (2 - 1) + c$

$\Rightarrow \log_{e} 1 + c = 0 \Rightarrow c = 0$

$∴ \tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{2} \log_{e} | 1 + {(\frac{y - 1}{x - 1})}^{2} | = \log_{e} | x - 1 |$

By comparing with

$\tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{β} \log_{e}$ $| α + {(\frac{y - 1}{x - 1})}^{2} |$ $= \log_{e}$ $| x - 1 |$

$β = 2, α = 1$

$∴ 5 β + α = 5 \times 2 + 1 = 11$ .

Q 12 :

If the solution curve y = y(x) of the differential equation $(1 + y^{2}) (1 + \log_{e} x) d x + x d y = 0$ , x > 0 passes through the point (1, 1) and $y (e) = \frac{α - \tan (\frac{3}{2})}{β + \tan (\frac{3}{2})}$ then $α + 2 β$ is __________. [2024]

(3)

We have, $(1 + y^{2}) (1 + \log_{e} x) d x + x d y = 0$

$\Rightarrow \frac{d y}{1 + y^{2}} = - \frac{(1 + \log_{e} x) d x}{x}$

Integrating both sides, we get

$\tan^{- 1} y = - \frac{1}{2} {(1 + \log_{e} x)}^{2} + C$ ...(i)

At (1, 1), $\frac{π}{4} = \frac{- 1}{2} + C \Rightarrow C = \frac{π}{4} + \frac{1}{2}$

$∴ y = \tan {\frac{π}{4} + \frac{1}{2} [1 - {(1 + \log_{e} x)}^{2}]}$

$\Rightarrow y (e) = \tan (\frac{π}{4} + \frac{1}{2} - 2)$

$= \tan [\frac{π}{4} - \frac{3}{2}] = \frac{\tan \frac{π}{4} - \tan \frac{3}{2}}{1 + \tan \frac{π}{4} \tan \frac{3}{2}} = \frac{1 - \tan \frac{3}{2}}{1 + \tan \frac{3}{2}}$

So, $α$ = 1 and $β$ = 1

$∴ α + 2 β = 3$ .

Q 13 :

Let $f (x) = \sqrt{\lim_{r \to x} {\frac{2 r^{2} [(f {(r))}^{2} - f (x) f (r)]}{r^{2} - x^{2}} - r^{3} e^{\frac{f (r)}{r}}}}$ be differentiable in $(- \infty, 0) \cup (0, \infty)$ and f(1) = 1. Then the value of ea, such that f(a) = 0, is equal to __________. [2024]

(2)

$f (x) = \sqrt{\lim_{r \to x} {\frac{2 r^{2} ((f {(r))}^{2} - f (x) f (r))}{r^{2} - x^{2}} - r^{3} e^{f (r) / r}}}$

Using L'Hospital rule, we have

$f (x) = \sqrt{\lim_{r \to x} {\frac{2 r^{2} [2 f (r) \cdot f' (r) - f (x) \cdot f' (r)] + [f {(r)}^{2} - f (x) \cdot f (r)] \cdot 4 r}{2 r} - r^{3} e^{\frac{f (r)}{r}}}}$

$= \sqrt{\frac{2 x^{2} [2 f (x) \cdot f' (x) - f (x) \cdot f' (x)] + 0}{2 x} - x^{3} e^{\frac{f (x)}{x}}}$

$∴ f (x) = \sqrt{x f (x) \cdot f' (x) - x^{3} \cdot e^{\frac{f (x)}{x}}}$

Let y = f(x)

$y^{2} = x y \cdot \frac{d y}{d x} - x^{3} e^{\frac{y}{x}} \Rightarrow \frac{d y}{d x} = \frac{y^{2} + x^{3} e^{y / x}}{x y}$

Put $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

$\Rightarrow \frac{v^{2} x^{2} + x^{3} e^{\frac{v x}{x}}}{x \cdot v x} = v + x \frac{d v}{d x} \Rightarrow \frac{v^{2} + x e^{v}}{v} - v = x \frac{d v}{d x}$

$\Rightarrow \frac{e^{v}}{v} = \frac{d v}{d x} \Rightarrow d x = v e^{- v} d v$

$\Rightarrow \int d x = \int v \cdot e^{- v} d v \Rightarrow x + c = - e^{- v} (v + 1)$

$\Rightarrow x + c = - e^{- y / x} (\frac{y}{x} + 1)$ ... (i)

$∵ y (1) = 1$

$\Rightarrow 1 + c = - e^{- 1} (2)$

$\Rightarrow c = - 1 - \frac{2}{e}$ ... (ii)

Now, f(a) = 0

$∴ a + c = - e^{- 0 / a} (\frac{0}{a} + 1)$ [Using (i)]

a + c = –1

$a = - 1 - c = - 1 + 1 + \frac{2}{e} = \frac{2}{e}$ [Using (ii)]

$∴$ ea = 2.

Q 14 :

Let y = y(x) be the solution of the differential equation $(1 - x^{2}) d y = [x y + (x^{3} + 2) \sqrt{3 (1 - x^{2})}] d x$ , –1 < x < 1, y(0) = 0. If $y (\frac{1}{2}) = \frac{m}{n}$ , m and n are co-prime numbers, then m + n is equal to __________. [2024]

(97)

We have, $(1 - x^{2}) d y = [x y + (x^{3} + 2) \sqrt{3 (1 - x^{2})}] d x$

$\Rightarrow \frac{d y}{d x} = \frac{x}{(1 - x^{2})} y + \frac{(x^{3} + 2) \sqrt{3 - 3 x^{2}}}{(1 - x^{2})}$

$\Rightarrow \frac{d y}{d x} - \frac{x}{1 - x^{2}} y = \frac{(x^{3} + 2) \sqrt{3 - 3 x^{2}}}{1 - x^{2}}$

$I . F . = e^{\int \frac{- x}{1 - x^{2}} d x}$

Put $1 - x^{2} = t \Rightarrow - x d x = \frac{d t}{2}$

$\Rightarrow I . F . = e^{\int \frac{d t}{2 t}} = e^{\frac{1}{2} \ln t} = e^{\frac{1}{2} \ln (1 - x^{2})}$

$= e^{\ln {(1 - x^{2})}^{\frac{1}{2}}} = {(1 - x^{2})}^{\frac{1}{2}} = \sqrt{1 - x^{2}}$

$∴ y \sqrt{(1 - x^{2})} = \int \frac{(x^{3} + 2) \sqrt{3 (1 - x^{2})}}{(1 - x^{2})} \sqrt{1 - x^{2}} d x$

$\Rightarrow y \sqrt{1 - x^{2}} = \sqrt{3} \int (x^{3} + 2) d x$

$\Rightarrow y \sqrt{1 - x^{2}} = \sqrt{3} [\frac{x^{4}}{4} + 2 x] + c$

Since, y(0) = 0, so $0 = c \Rightarrow y = \sqrt{\frac{3}{1 - x^{2}}} (\frac{x^{4}}{4} + 2 x)$

Now, $y (\frac{1}{2}) = \sqrt{\frac{3}{1 - \frac{1}{4}}} (\frac{{(\frac{1}{2})}^{4}}{4} + 2 \times \frac{1}{2}) = 2 (\frac{1}{64} + 1)$

$= 2 (\frac{65}{64}) = \frac{65}{32} = \frac{m}{n}$

$∴$ m = 65, n = 32 $\Rightarrow$ m + n = 65 +32 = 97.

Q 15 :

Let Y = Y(X) be a curve lying in the first quadrant such that the area enclosed by the line Y – y = Y'(x)(X – x) and the co-ordinate axes, where (x, y) is any point on the curve, is always $\frac{- y^{2}}{2 Y' (x)} + 1$ , Y'(X) $\neq$ 0. If Y(1) = 1, then 12Y(2) equals __________. [2024]

(20)

We have, curve : Y = y(x)

Line : Y – y = Y'(x)(X – x)

Area = $\frac{1}{2} (x - \frac{y}{Y' (x)}) (y - x Y' (x))$

$\Rightarrow 1 - \frac{y^{2}}{2 Y' (x)} = \frac{(x Y' (x) - y) (y - x Y' (x))}{2 Y' (x)}$

$\Rightarrow \frac{2 Y' (x) - y^{2}}{2 Y' (x)} = \frac{(x Y' (x) - y) (y - x Y' (x))}{2 Y' (x)}$

$\Rightarrow 2 Y' (x) - y^{2} = x y Y' (x) - x^{2} {[Y' (x)]}^{2} - y^{2} + x y Y' (x)$

$\Rightarrow x^{2} {[Y' (x)]}^{2} + Y' (x) [2 - x y - x y] = 0$

$\Rightarrow Y' (x) [x^{2} Y' (x) + 2 - 2 x y] = 0$

$\Rightarrow x^{2} Y' (x) + 2 - 2 x y = 0 or Y' (x) = \frac{2 x y - 2}{x^{2}}$

$\Rightarrow \frac{d y}{d x} = (\frac{2}{x}) y - \frac{2}{x^{2}} \Rightarrow \frac{d y}{d x} - (\frac{2}{x}) y = \frac{- 2}{x^{2}}$

$I . F . = e^{- \int \frac{2}{x} d x} = e^{- 2 \log x} = \frac{1}{x^{2}}$

Solution is

$y \cdot \frac{1}{x^{2}} = \int \frac{- 2}{x^{2}} \times \frac{1}{x^{2}} d x = - 2 \int \frac{1}{x^{4}} d x or \frac{y}{x^{2}} = \frac{2}{3 x^{3}} + c$

$y = \frac{2}{3 x} + c x^{2}$ ... (i)

$1 = \frac{2}{3} + c \Rightarrow c = \frac{1}{3}$

At x = 2,

$∴ y = \frac{2}{3 \times 2} + \frac{1}{3} \times 4 = \frac{1}{3} + \frac{4}{3} = \frac{5}{3} \Rightarrow 12 y = 20$ .

Q 16 :

Let y = y(x) be the solution of the differential equation $\sec^{2} x d x + (e^{2 y} \tan^{2} x + \tan x) d y = 0$ , $0 < x < \frac{π}{2}$ , $y (\frac{π}{4}) = 0$ . If $y (\frac{π}{6}) = α$ , then $e^{8 α}$ is equal to __________. [2024]

(9)

We have, $\sec^{2} x d x + (e^{2 y} \tan^{2} x + \tan x) d y = 0$

Put $\tan x = t$

$\Rightarrow \sec^{2} x \frac{d x}{d y} = \frac{d t}{d y}$

Now, $\sec^{2} x \frac{d x}{d y} + (e^{2 y} \tan^{2} x + \tan x) = 0$

$\Rightarrow \frac{d t}{d y} + e^{2 y} t^{2} + t = 0 \Rightarrow \frac{d t}{d y} + t = - t^{2} e^{2 y}$

$\frac{1}{t^{2}} \frac{d t}{d y} + \frac{1}{t} = - e^{2 y}$

Again put $\frac{1}{t} = u \Rightarrow \frac{- 1}{t^{2}} \frac{d t}{d y} = \frac{d u}{d y}$

$∴ - \frac{d u}{d y} + u = - e^{2 y} \Rightarrow \frac{d u}{d y} - u = e^{2 y}$

$∴ I . F . = e^{- \int d y} = e^{- y}$

$∴ u e^{- y} = \int e^{- y} e^{2 y} d y$

$\Rightarrow u e^{- y} = \int e^{y} d y \Rightarrow \frac{1}{t} e^{- y} = e^{y} + c$

$\Rightarrow \frac{1}{\tan x} e^{- y} = e^{y} + c$ ... (i)

When, $x = \frac{π}{4}$ , y = 0

$∴ \frac{1}{\tan (\frac{π}{4})} e^{- 0} = e^{0} + c \Rightarrow 1 = 1 + c \Rightarrow c = 0$

Also, when $x = \frac{π}{6}, y = α$

$∴$ From (i), we have

$\frac{1}{\tan \frac{π}{6}} e^{- α} = e^{α} + 0 \Rightarrow \sqrt{3} = e^{2 α} \Rightarrow {(e^{2 α})}^{4} = {(\sqrt{3})}^{4}$

$\Rightarrow e^{8 α} = 9$ .

Q 17 :

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function. If $10 \int_{1}^{x} f (t) d t = 5 x f (x) - x^{5} - 9$ for all $x \geq 1$ , then the value of f(3) is : [2025]

32
18
22
26

1

2

3

4

(1)

We have, $10 \int_{1}^{x} f (t) d t = 5 x f (x) - x^{5} - 9$

On differentiating both sides, we get

$10 f (x) = 5 f (x) + 5 x f' (x) - 5 x^{4}$

$\Rightarrow 5 f (x) = 5 x f' (x) - 5 x^{4}$

$\Rightarrow f (x) = x f' (x) - x^{4}$

$\Rightarrow f' (x) - \frac{1}{x} f (x) = x^{3}$

$\Rightarrow \frac{d y}{d x} - \frac{y}{x} = x^{3}$ [Taking y = f(x)]

This is a linear differential equation

$∴ I . F . = e \int_{}^{\frac{- 1}{x} d x} = e^{- \ln x} = \frac{1}{x}$

Solution is $\frac{y}{x} = \int \frac{x^{3}}{x} d x + c$

$\Rightarrow \frac{y}{x} = \frac{x^{3}}{3} + c$

$\Rightarrow y = \frac{x^{4}}{3} + c x$

$\Rightarrow f (x) = \frac{x^{4}}{3} + c x$ ... (i)

Now, using original equation, we have

$10 \int_{1}^{x} f (t) d t = 5 x f (x) - x^{5} - 9$

Put x = 1, we get 0 = 5f(1) – 10 $\Rightarrow$ f(1) = 2

Substituting this value in equation (i), we get

$\Rightarrow 2 = \frac{1}{3} + c \Rightarrow c = \frac{5}{3}$ $∴ f (x) = \frac{x^{4}}{3} + \frac{5}{3} x$

$\Rightarrow f (3) = 27 + 5 = 32$

Q 18 :

Let g be differentiable function such that $\int_{0}^{x} g (t) d t = x - \int_{0}^{x} t g (t) d t, x \geq 0$ and let y = y(x) satisfy the differential equation $\frac{d y}{d x} - y \tan x = 2 (x + 1) \sec x g (x), x \in [0, \frac{π}{2})$ . If y(0) = 0, then $y (\frac{π}{3})$ is equal to [2025]

$\frac{2 π}{3 \sqrt{3}}$
$\frac{2 π}{3}$
$\frac{4 π}{3 \sqrt{3}}$
$\frac{4 π}{3}$

1

2

3

4

(4)

we have, $\int_{0}^{x} g (t) d t = x - \int_{0}^{x} t g (t) d t$ ... (i)

On differentiating equation (i), we get g(x) = 1 – xg(x)

$\Rightarrow g (x) = \frac{1}{1 + x}$

Now, $\frac{d y}{d x} - y \tan x = 2 (x + 1) \sec x g (x)$

$\Rightarrow \frac{d y}{d x} - y \tan x = 2 \sec x$

$∴ I . F . = e^{- \int \tan x d x} = e^{\log_{e} \cos x} = \cos x$

Solution of D.E. is given by

$y \cos x = \int 2 \cos x \cdot \sec x d x + C$

$\Rightarrow y \cos x = 2 x + C \Rightarrow y (0) = 0 \Rightarrow C = 0$

$∴ y = \frac{2 x}{\cos x}$

$\Rightarrow y = 2 x \sec x \Rightarrow y (\frac{π}{3}) = 2 \cdot \frac{π}{3} \cdot 2 = \frac{4 π}{3}$

Q 19 :

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} + 3 (\tan^{2} x) y + 3 y = \sec^{2} x, y (0) = \frac{1}{3} + e^{3}$ . Then $y (\frac{π}{4})$ is equal to [2025]

$\frac{4}{3} + e^{3}$
$\frac{4}{3}$
$\frac{2}{3}$
$\frac{2}{3} + e^{3}$

1

2

3

4

(2)

Given, $\frac{d y}{d x} + 3 (\tan^{2} x) y + 3 y = \sec^{2} x$

$\Rightarrow \frac{d y}{d x} + 3 (\sec^{2} x) y = \sec^{2} x$

$I . F . = e^{3 \int \sec^{2} x d x} = e^{3 \tan x}$

$∴$ Solution is given by $e^{3 \tan x} y = \int e^{3 \tan x} \sec^{2} x d x$

$\Rightarrow e^{3 \tan x} y = \frac{e^{3 \tan x}}{3} + C$ ... (i)

$∵ y (0) = \frac{1}{3} + e^{3} \Rightarrow C = e^{3}$

Using equation (i), we get

$e^{3 \tan x} y = \frac{e^{3 \tan x}}{3} + e^{3}$

$∴ y (\frac{π}{4}) = \frac{\frac{e^{3}}{3} + e^{3}}{e^{3}} = \frac{4}{3}$

Q 20 :

If a curve y = y(x) passes through the point $(1, \frac{π}{2})$ and satisfies the differential equation $(7 x^{4} \cot y - e^{x} cosec y) \frac{d x}{d y} = x^{5}, x \geq 1$ , then at x = 2, the value of cos y is: [2025]

$\frac{2 e^{2} + e}{128}$
$\frac{2 e^{2} - e}{128}$
$\frac{2 e^{2} - e}{64}$
$\frac{2 e^{2} + e}{64}$

1

2

3

4

(2)

We have, $(7 x^{4} \cot y - e^{x} cosec y) \frac{d x}{d y} = x^{5}$

$\Rightarrow \frac{d y}{d x} = \frac{7 x^{4} \cot y}{x^{5}} - \frac{e^{x} cosec y}{x^{5}}$

$\Rightarrow \frac{d y}{d x} - \frac{7}{x} \cot y = \frac{- e^{x}}{x^{5}} cosec y$

$\Rightarrow \sin y \cdot \frac{d y}{d x} - \cos y (\frac{7}{x}) = - \frac{e^{x}}{x^{5}}$

If $- \cos y = t \Rightarrow \sin y \cdot \frac{d y}{d x} = \frac{d t}{d x}$

$\Rightarrow \frac{d t}{d x} + t (\frac{7}{x}) = - \frac{e^{x}}{x^{5}}$

Here, $I . F . = x^{7}$

$\Rightarrow t \cdot x^{7} = - \int x^{2} \cdot e^{x} d x \Rightarrow \cos y \cdot x^{7} = x^{2} e^{x} - 2 \int x e^{x} d x$

$\Rightarrow \cos y \cdot x^{7} = x^{2} e^{x} - 2 x e^{x} + 2 e^{x} + c$

at point $(1, \frac{π}{2})$ , we get c = –e

Thus, the value of cos y at x = 2 is

$\cos y = \frac{2 e^{2} - e}{128}$

Q 21 :

Let y = y(x) be the solution curve of the differential equation $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0, x > 0$ , passing through the point (1, 0). Then y(2) is equal to [2025]

$\frac{2}{2 - e^{2}}$
$\frac{4}{4 + e^{2}}$
$\frac{4}{4 - e^{2}}$
$\frac{2}{2 + e^{2}}$

1

2

3

4

(2)

We have, $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0$

$\Rightarrow x (x^{2} + e^{x}) \frac{d y}{d x} + e^{x} (x - 2) y = x^{3}$

$\Rightarrow \frac{d y}{d x} + \frac{e^{x} (x - 2)}{x (x^{2} + e^{x})} y = \frac{x^{3}}{(x^{2} + e^{x}) x}$

Where, $P = \frac{e^{x} (x - 2)}{x (x^{2} + e^{x})}$ and $Q = \frac{x^{2}}{(x^{2} + e^{x})}$

$I . F . = e^{\int \frac{e^{x} (x - 2)}{x (x^{2} + e^{x})} d x} = e^{\int \frac{e^{x} + 2 x}{e^{x} + x^{2}} d x - \int \frac{2}{x} d x} = e^{\ln | e^{x} + x^{2} | - 2 \ln x}$

$I . F . = 1 + \frac{e^{x}}{x^{2}}$

Now, $y \cdot (1 + \frac{e^{x}}{x^{2}}) = \int \frac{x^{2}}{x^{2} + e^{x}} \cdot \frac{x^{2} + e^{x}}{x^{2}} d x + C$

$\Rightarrow y (1 + \frac{e^{x}}{x^{2}}) = x + C$

It is passing through the point (1, 0), then

C = –1

So, $y (1 + \frac{e^{x}}{x^{2}}) = x - 1$

Then $y (2) = \frac{4}{4 + e^{2}}$

Q 22 :

Let y = y(x) be the solution of the differential equation $(x^{2} + 1) y' - 2 x y = (x^{4} + 2 x^{2} + 1) \cos x$ , y(0) = 1. Then $\int_{- 3}^{3} y (x) d x$ is : [2025]

18
36
30
24

1

2

3

4

(4)

We have, $(x^{2} + 1) \frac{d y}{d x} - 2 x y = (x^{4} + 2 x^{2} + 1) \cos x$

$\Rightarrow \frac{d y}{d x} - (\frac{2 x}{x^{2} + 1}) y = \frac{{(x^{2} + 1)}^{2} \cos x}{(x^{2} + 1)} = (x^{2} + 1) \cos x$ (Linear D.E.)

Here, $P = \frac{- 2 x}{(x^{2} + 1)}, Q = (x^{2} + 1) \cos x$

$I . F . = e^{\int P d x} = e^{\int \frac{- 2 x}{(x^{2} + 1)} d x} = \frac{1}{(x^{2} + 1)}$

Thus, the solution of linear D.E. is given by

$\Rightarrow y \cdot \frac{1}{(x^{2} + 1)} = \int (x^{2} + 1) \cos x \cdot \frac{1}{(x^{2} + 1)} d x + c$

$\Rightarrow \frac{y}{x^{2} + 1} = \sin x + c \Rightarrow \frac{1}{1} = 0 + c \Rightarrow c = 1$ [ $∵$ y(0) = 1]

$∴ y = (x^{2} + 1) (\sin x + 1)$

$\int_{- 3}^{3} y (x) d x = \int_{- 3}^{3} (x^{2} + 1) (\sin x + 1) d x$

$= \int_{- 3}^{3} (x^{2} \sin x + x^{2} + \sin x + 1) d x$

$\int_{- 3}^{3} x^{2} \sin x d x + \int_{- 3}^{3} x^{2} d x + \int_{- 3}^{3} \sin x d x + \int_{- 3}^{3} 1 d x$

= 0 + 18 + 0 + 6 = 24

Q 23 :

Let f(x) = x – 1 and $g (x) = e^{x}$ for $x \in R$ . If $\frac{d y}{d x} = (e^{- 2 \sqrt{x}} g (f (f (x))) - \frac{y}{\sqrt{x}})$ , y(0) = 0, then y(1) is [2025]

$\frac{e - 1}{e^{4}}$
$\frac{2 e - 1}{e^{3}}$
$\frac{1 - e^{3}}{e^{4}}$
$\frac{1 - e^{2}}{e^{4}}$

1

2

3

4

(1)

f(x) = x – 1 and $g (x) = e^{x}$

f(f(x)) = f(x – 1) = (x – 1) – 1 = x – 2

$g (f (f (x))) = g (x - 2) = e^{x - 2}$

Now, $\frac{d y}{d x} + \frac{y}{\sqrt{x}} = e^{- 2 \sqrt{x}} e^{x - 2} \Rightarrow \frac{d y}{d x} + \frac{y}{\sqrt{x}} = e^{x - 2 - 2 \sqrt{x}}$

Which is a linear differential equation

$∴ I . F . = e^{\int \frac{1}{\sqrt{x}} d x} = e^{2 \sqrt{x}}$

$∴$ Solution of differential equation is given by

$y e^{2 \sqrt{x}} = \int e^{x - 2 - 2 \sqrt{x}} e^{2 \sqrt{x}} d x$

$\Rightarrow y e^{2 \sqrt{x}} = \int e^{x - 2} d x \Rightarrow y e^{2 \sqrt{x}} = e^{x - 2} + C$

Now, y(0) = 0

$\Rightarrow 0 = e^{- 2} + C \Rightarrow C = - e^{- 2}$

$∴ y (x) = e^{x - 2 - 2 \sqrt{x}} - e^{- 2 - 2 \sqrt{x}}$

Now, $y (1) = e^{1 - 2 - 2} - e^{- 2 - 2} = e^{- 3} - e^{- 4} = \frac{e - 1}{e^{4}}$

Q 24 :

Let x = x(y) be the solution of the differential equation $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If x(1) = 1, then $x (\frac{1}{2})$ is : [2025]

3 + e
$\frac{3}{2} + e$
$\frac{1}{2} + e$
3 – e

1

2

3

4

(4)

We have, $y^{2} d x + (x - \frac{1}{y}) d y = 0$

$\Rightarrow y^{2} d x = (\frac{1}{y} - x) d y \Rightarrow \frac{d x}{d y} + \frac{x}{y^{2}} = \frac{1}{y^{3}}$ ,

which is a linear D.E. with $P = \frac{1}{y^{2}}$ and $Q = \frac{1}{y^{3}}$

$I . F . = e^{\int P d y} = e^{\int \frac{1}{y^{2}} d y} = e^{\frac{- 1}{y}}$

$∴$ Solution is given by $x \cdot e^{- 1 / y} = \int \frac{1}{y^{3}} \cdot e^{- 1 / y} d y + c$

Putting $\frac{1}{y} = t \Rightarrow \frac{- 1}{y^{2}} d y = d t$

$\Rightarrow x \cdot e^{\frac{- 1}{y}} = - \int t \cdot e^{- t} d y + c \Rightarrow x \cdot e^{\frac{- 1}{y}} = - e^{- t} (- t - 1) + c$

$\Rightarrow x \cdot e^{\frac{- 1}{y}} = e^{\frac{- 1}{y}} (1 + \frac{1}{y}) + c \Rightarrow x = 1 + \frac{1}{y} + c e^{1 / y}$

Now, at x(1) = 1, i.e., when x = 1, y = 1

$\Rightarrow 1 = 1 + 1 + c e \Rightarrow c = \frac{- 1}{e}$

$∴$ Solution is, $x = 1 + \frac{1}{y} - \frac{1}{e} (e^{1 / y})$

$∴ x (\frac{1}{2}) = 1 + 2 - \frac{1}{e} \times e^{2} = 3 - e$

Q 25 :

If x = f(y) is the solution of the differential equation $(1 + y^{2}) + (x - 2 e^{\tan^{— 1} y}) \frac{d y}{d x} = 0, y \in (- \frac{π}{2}, \frac{π}{2})$ with f(0) = 1, then $f (\frac{1}{\sqrt{3}})$ is equal to : [2025]

$e^{π / 6}$
$e^{π / 12}$
$e^{π / 3}$
$e^{π / 4}$

1

2

3

4

(1)

$\frac{d x}{d y} + \frac{x}{1 + y^{2}} = \frac{2 e^{\tan^{- 1} y}}{1 + y^{2}}$ , which is linear differential equation of first order.

$I . F . = e^{\int \frac{d y}{1 + y^{2}}} = e^{\tan^{— 1} y}$

Thus, the required solution is given by

$x e^{\tan^{— 1} y} = \int \frac{2 {(e^{\tan^{— 1} y})}^{2}}{1 + y^{2}} d y$

Put $\tan^{— 1} y = t, \frac{d y}{1 + y^{2}} = d t$ $∴ x e^{\tan^{- 1} y} = e^{2 \tan^{- 1} y} + c$

Given, f(0) = 1, we get 1 = 1 + c $\Rightarrow$ c = 0

Therefore, $x = e^{\tan^{- 1} y}$

$\Rightarrow f (\frac{1}{\sqrt{3}}) = e^{\frac{π}{6}}$

Q 26 :

Let a curve y = f(x) pass through the points (0, 5) and $(\log_{e} 2, k)$ . If the curve satisfies the differential equation $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$ , then k is equal to [2025]

16
32
8
4

1

2

3

4

(3)

Given, $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$

$\Rightarrow \frac{d y}{d x} = \frac{2 (3 + y) e^{2 x}}{(7 + e^{2 x})}$

$\Rightarrow \frac{d y}{d x} - \frac{2 e^{2 x}}{(7 + e^{2 x})} \times y = \frac{6 e^{2 x}}{(7 + e^{2 x})}$

$∴ I . F . = e^{- \int (\frac{2 e^{2 x}}{7 + e^{2 x}}) d x} = \frac{1}{7 + e^{2 x}}$

$∴ y \times \frac{1}{7 + e^{2 x}} = \int \frac{6 e^{2 x}}{{(7 + e^{2 x})}^{2}} d x \Rightarrow \frac{y}{7 + e^{2 x}} = \frac{- 3}{(7 + e^{2 x})} + c$

Since, f(x) pass through the points (0, 5) and $(\log_{e} 2, k)$ .

Then, $\frac{5}{8} = \frac{- 3}{8} + c \Rightarrow c = 1$

$∴ y = - 3 + (7 + e^{2 x}) \Rightarrow y = e^{2 x} + 4$

$∴ k = e^{2 \log_{e} 2} + 4$

$\Rightarrow k = e^{\log_{e} 2^{2}} + 4 \Rightarrow k = 4 + 4 = 8$

Q 27 :

Let x = x(y) be the solution of the differential equation $y = (x - y \frac{d x}{d y}) \sin (\frac{x}{y})$ , y > 0 and $x (1) = \frac{π}{2}$ . Then cos (x(2)) is equal to : [2025]

$2 {(\log_{e} 2)}^{2} - 1$
$1 - 2 (\log_{e} 2)$
$1 - 2 {(\log_{e} 2)}^{2}$
$2 (\log_{e} 2) - 1$

1

2

3

4

(1)

We have, $y d y = (x d y - y d x) \sin (\frac{x}{y})$

$\Rightarrow \frac{d y}{y} = (\frac{x d y - y d x}{y^{2}}) \sin (\frac{x}{y})$

$\Rightarrow \frac{d y}{y} = \sin (\frac{x}{y}) d (\frac{x}{y})$

$\Rightarrow \log y = \cos \frac{x}{y} + C$

$\Rightarrow 0 = \cos \frac{π}{2} + C$ $[∵ x = \frac{π}{2} and y = 1]$

$\Rightarrow C = 0$

$∴ \log y = \cos \frac{x}{y}$

Put y = 2 in equation (i), we get

$\Rightarrow \cos \frac{x}{2} = \log 2$

Now, $\cos x = 2 \cos^{2} \frac{x}{2} - 1 = 2 {(\log_{e} 2)}^{2} - 1$ .

Q 28 :

Let y = y(x) be the solution of the differential equation $(x y - 5 x^{2} \sqrt{1 + x^{2}}) d x + (1 + x^{2}) d y = 0$ , y(0) = 0. Then $y (\sqrt{3})$ is equal to [2025]

$\sqrt{\frac{15}{2}}$
$\sqrt{\frac{14}{3}}$
$\frac{5 \sqrt{3}}{2}$
$2 \sqrt{2}$

1

2

3

4

(3)

We have, $(x y - 5 x^{2} \sqrt{1 + x^{2}}) d x + (1 + x^{2}) d y = 0$

$\Rightarrow \frac{d y}{d x} = \frac{5 x^{2} \sqrt{1 + x^{2}} - x y}{(1 + x^{2})}$

$\Rightarrow \frac{d y}{d x} + \frac{x y}{(1 + x^{2})} = \frac{5 x^{2}}{\sqrt{1 + x^{2}}}$

$I . F . = \int e^{\frac{x}{1 + x^{2}}} d x = \sqrt{1 + x^{2}}$

$y \times \sqrt{1 + x^{2}} = \int \frac{5 x^{2}}{\sqrt{1 + x^{2}}} \times \sqrt{1 + x^{2}} d x + C$

$y \sqrt{1 + x^{2}} = \frac{5 x^{3}}{3} + C$

$∵ y (0) = 0$

$∴ C = 0$

$\Rightarrow y = \frac{5 x^{3}}{3 \sqrt{1 + x^{2}}}$

$∴ y (\sqrt{3}) = \frac{5 \times 3 \times \sqrt{3}}{3 \sqrt{1 + 3}} = \frac{5 \sqrt{3}}{2}$

Q 29 :

Let for some function y = f(x), $\int_{0}^{x} t f (t) d t = x^{2} f (x), x > 0$ and f(2) = 3. Then f(6) is equal to [2025]

3
6
1
2

1

2

3

4

(3)

We have, $\int_{0}^{x} t f (x) d t = x^{2} f (x), x > 0$

Differentiate the given equation, we get

$x f (x) = 2 x f (x) + x^{2} f' (x)$

$\Rightarrow x f (x) = - x^{2} f' (x) \Rightarrow f (x) = - x f' (x)$

$\Rightarrow \int \frac{f' (x)}{f (x)} d x = - \int \frac{1}{x} d x \Rightarrow \log f (x) = - \log x + \log C$

$\Rightarrow \log f (x) = - \log x + \log C \Rightarrow f (x) = \frac{C}{x}$

We have, $f (2) = 3; f (2) = 3 = \frac{C}{2} \Rightarrow C = 6$

So, $f (x) = \frac{6}{x} \Rightarrow f (6) = \frac{6}{6} = 1$

Q 30 :

Let $f : R \to R$ be a twice differentiable function such that f(2) = 1. If F(x) = xf(x) for all $x \in R$ . $\int_{0}^{2} x F' (x) d x = 6$ and $\int_{0}^{2} x^{2} F'' (x) d x = 40$ , then $F' (2) + \int_{0}^{2} F (x) d x$ is equal to : [2025]

13
9
11
15

1

2

3

4

(3)

We have, $\int_{0}^{2} x F' (x) d x = 6$ ... (i)

$\Rightarrow {[x F (x)]}_{0}^{2} - \int_{0}^{2} F (x) d x = 6$

$\Rightarrow 2 F (2) - \int_{0}^{2} F (x) d x = 6$

$\Rightarrow 4 - \int_{0}^{2} F (x) d x = 6$ $[∵ F (2) = 2 f (2) = 2]$

$\Rightarrow \int_{0}^{2} F (x) d x = - 2$

Also, $\Rightarrow \int_{0}^{2} x^{2} F'' (x) d x = 40$

$\Rightarrow {[x^{2} F' (x)]}_{0}^{2} - 2 \int_{0}^{2} x F' (x) d x = 40$

$\Rightarrow 4 F' (2) - 2 (6) = 40 \Rightarrow F' (2) = 13$ [Using (i)]

Thus, $F' (2) + \int_{0}^{2} F (x) d x = 13 - 2 = 11$

Q 31 :

Let y = y(x) be the solution of the differential equation $\cos x (\log_{e} {(\cos x))}^{2} d y + (\sin x - 3 y \sin x \log_{e} (\cos x)) d x = 0$ , $x \in (0, \frac{π}{2})$ . If $y (\frac{π}{4}) = \frac{- 1}{\log_{e} 2}$ , then $y (\frac{π}{6})$ is equal to: [2025]

$- \frac{1}{\log_{e} (4)}$
$\frac{1}{\log_{e} (3) - \log_{e} (4)}$
$\frac{2}{\log_{e} (3) - \log_{e} (4)}$
$\frac{1}{\log_{e} (4) - \log_{e} (3)}$

1

2

3

4

(2)

We have, $\cos x (\ln {(\cos x))}^{2} d y + (\sin x - 3 y (\sin x) \ln (\cos x)) d x = 0$

$\Rightarrow \cos x (\ln {(\cos x))}^{2} \frac{d y}{d x} - 3 \sin x \cdot \ln (\cos x) y = - \sin x$

$\Rightarrow \frac{d y}{d x} - \frac{3 \tan x}{\ln (\cos x)} y = \frac{- \tan x}{(\ln {(\cos x))}^{2}}$

$\Rightarrow \frac{d y}{d x} + \frac{3 \tan x}{\ln (\sec x)} y = \frac{- \tan x}{(\ln {(\sec x))}^{2}}$

$I . F . = e^{\int \frac{3 \tan x}{\ln (\sec x)} d x} = \ln {(\sec x))}^{3}$

$∴$ Solution is $y \times (\ln {(\sec x))}^{3} = - \int \frac{\tan x}{(\ln {(\sec x))}^{2}} (\ln {(\sec x))}^{3} d x + C$

$\Rightarrow y \times (\ln {(\sec x))}^{3} = - \frac{1}{2} (\ln {(\sec x))}^{2} + C$

At $x = \frac{π}{4}, y = - \frac{1}{\ln 2}$

$∴ \frac{- 1}{\ln 2} \times {(\ln \sqrt{2})}^{3} = - \frac{1}{2} {(\ln \sqrt{2})}^{2} + C$

$\Rightarrow \frac{- 1}{8 \ln 2} \times {(\ln 2)}^{3} = \frac{- 1}{2} \times \frac{1}{4} {(\ln 2)}^{2} + C$

$\Rightarrow - \frac{1}{8} {(\ln 2)}^{2} = \frac{- 1}{8} {(\ln 2)}^{2} + C \Rightarrow C = 0$

$∴ y (\ln {(\sec x))}^{3} = \frac{- 1}{2} (\ln {(\sec x))}^{2} + 0$

$\Rightarrow y = \frac{- 1}{2 \ln (\sec x)} \Rightarrow y = \frac{1}{2 \ln (\cos x)}$

$∴ y (\frac{π}{6}) = \frac{1}{2 \ln (\cos \frac{π}{6})} = \frac{1}{2 \ln (\frac{\sqrt{3}}{2})}$

$= \frac{1}{2 (\frac{1}{2} \ln 3 - \ln 2)} = \frac{1}{\ln 3 - \ln 4}$

Q 32 :

If for the solution curve y = f(x) of the differential equation $\frac{d y}{d x} + (\tan x) y = \frac{2 + \sec x}{{(1 + 2 \sec x)}^{2}}$ , $x \in (\frac{- π}{2}, \frac{π}{2})$ , $f (\frac{π}{3}) = \frac{\sqrt{3}}{10}$ , then $f (\frac{π}{4})$ is equal to: [2025]

$\frac{9 \sqrt{3} + 3}{10 (4 + \sqrt{3})}$
$\frac{\sqrt{3} + 1}{10 (4 + \sqrt{3})}$
$\frac{4 - \sqrt{2}}{14}$
$\frac{5 - \sqrt{3}}{2 \sqrt{2}}$

1

2

3

4

(3)

We have, $\frac{d y}{d x} + (\tan x) y = \frac{2 + \sec x}{{(1 + 2 \sec x)}^{2}}$

Here, P = tan x and $Q = \frac{2 + \sec x}{{(1 + 2 \sec x)}^{2}}$

$I . F . = e^{\int P d x} = e^{\int \tan x d x} = e^{\log | \sec x |} = \sec x$

$∴$ Solution is given by,

$y \cdot \sec x = \int \frac{(2 + \sec x)}{{(1 + 2 \sec x)}^{2}} \cdot \sec x d x + c$

$\Rightarrow y \cdot \sec x = \int \frac{2 \cos x + 1}{{(\cos x + 2)}^{2}} d x + c$

Put $\cos x = \frac{1 - t^{2}}{1 + t^{2}} \Rightarrow - \sin x d x = \frac{- 4 t}{{(1 + t^{2})}^{2}} d t$

$∴ y \cdot \sec x = \int \frac{2 (\frac{1 - t^{2}}{1 + t^{2}}) + 1}{{(\frac{1 - t^{2}}{1 + t^{2}} + 2)}^{2}} \cdot \frac{2 d t}{(1 + t^{2})} + c$

$\Rightarrow y \cdot \sec x = \int \frac{2 - 2 t^{2} + 1 + t^{2}}{{(1 - t^{2} + 2 + 2 t^{2})}^{2}} \cdot 2 d t + c$

$\Rightarrow y \cdot \sec x = 2 \int \frac{3 - t^{2}}{{(t^{2} + 3)}^{2}} d t + c$

Put $t + \frac{3}{t} = u \Rightarrow (1 - \frac{3}{t^{2}}) d t = d u$

$∴ y \cdot \sec x = - 2 \int \frac{d u}{u^{2}} + c$

$\Rightarrow y \cdot \sec x = \frac{2}{u} + c \Rightarrow y \cdot \sec x = \frac{2}{t + 3 / t} + c$

Now, at $x = \frac{π}{3}, t = \frac{1}{\sqrt{3}}, y = \frac{\sqrt{3}}{10}$

$\Rightarrow \frac{2 \sqrt{3}}{10} = \frac{2}{1 / \sqrt{3} + 3 \sqrt{3}} + c \Rightarrow c = 0$

Hence, $y \cdot \sec x = \frac{2}{t + 3 / t} \Rightarrow y = \frac{2}{\sec x \cdot (t + 3 / t)}$

At $x = \frac{π}{4}, t = \sqrt{2} - 1$

$∴ f (\frac{π}{4}) = \frac{1}{\sqrt{2}} \cdot \frac{2 (\sqrt{2} - 1)}{{(\sqrt{2} - 1)}^{2} + 3}$

$= \frac{\sqrt{2} (\sqrt{2} - 1)}{6 - 2 \sqrt{2}} = \frac{2 - \sqrt{2}}{6 - 2 \sqrt{2}} \times \frac{6 + 2 \sqrt{2}}{6 + 2 \sqrt{2}}$

$= \frac{12 - 2 \sqrt{2} - 4}{36 - 8} = \frac{8 - 2 \sqrt{2}}{28} = \frac{4 - \sqrt{2}}{14}$

Q 33 :

Let $f : R \to R$ be a thrice differentiable odd function satisfying $f' (x) \geq 0, f'' (x) = f (x), f (0) = 0, f' (0) = 3$ . Then $9 f (\log_{e} 3)$ is equal to _________. [2025]

(36)

Since, f(x) is a thrice differentiable odd function,

$f'' (x) = f (x) \Rightarrow f' (x) \cdot f'' (x) = f' (x) \cdot f (x)$

$\Rightarrow \frac{(f' {(x))}^{2}}{2} = \frac{(f {(x))}^{2}}{2} + C \Rightarrow (f' {(x))}^{2} = (f {(x))}^{2} + C'$

$∵ (f' {(x))}^{2} = (f {(x))}^{2} + 9$ [ $∵ f' (0) = 3, f (0) = 0$ ]

$y = f (x) \Rightarrow \frac{d y}{d x} = f' (x) = \sqrt{(f {(x))}^{2} + 9}$

$= \int d x \Rightarrow \log | y + \sqrt{y^{2} + 9} | = x + C$

$\Rightarrow f (0) = 0 \Rightarrow C = \log 3 \Rightarrow y + \sqrt{y^{2} + 9} = 3 e^{x}$

At x = log 3; y = 4

$∴ 9 f (\log_{e} 3) = 36$

Q 34 :

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} + 2 y \sec^{2} x = 2 \sec^{2} x + 3 \tan x \cdot \sec^{2} x$ such that $y (0) = \frac{5}{4}$ . Then $12 (y (\frac{π}{4}) - e^{- 2})$ is equal to __________. [2025]

(21)

We have, $\frac{d y}{d x} + (2 \sec^{2} x) y = 2 \sec^{2} x + 3 \tan x \sec^{2} x$

$I . F . = e^{\int 2 \sec^{2} x d x} = e^{2 \tan x}$

$∴$ Solution of differential equation is given by

$y \cdot e^{2 \tan x} = \int 2 \sec^{2} x \cdot e^{2 \tan x} d x + \int 3 \tan x \sec^{2} x \cdot e^{2 \tan x} d x$

Put $\tan x = t \Rightarrow \sec^{2} x d x = d t$

$∴ y e^{2 \tan x} = \int 2 e^{2 t} d t + 3 \int t e^{2 t} d t$

$= \frac{2 e^{2 t}}{2} + 3 [\frac{t e^{2 t}}{2} - \int \frac{e^{2 t}}{2} d t]$

$= e^{2 t} + \frac{3}{2} t e^{2 t} - \frac{3}{4} e^{2 t} + c$

$\Rightarrow y e^{2 \tan x} = e^{2 \tan x} + \frac{3}{2} \tan x e^{2 \tan x} - \frac{3}{4} e^{2 \tan x} + c$

$\Rightarrow y = \frac{1}{4} + \frac{3}{2} \tan x + c e^{- 2 \tan x}$

Now, $y (0) = \frac{5}{4} \Rightarrow \frac{5}{4} = \frac{1}{4} + c \Rightarrow c = 1$

$∴ y = \frac{1}{4} + \frac{3}{2} \tan x + e^{- 2 \tan x}$

$\Rightarrow y (\frac{π}{4}) = \frac{1}{4} + \frac{3}{2} + e^{- 2} = \frac{7}{4} + e^{- 2}$

$∴ 12 (y (\frac{π}{4}) - e^{- 2}) = 12 (\frac{7}{4} + e^{- 2} - e^{- 2}) = 21$

Q 35 :

Let y = f(x) be the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{x^{6} + 4 x}{\sqrt{1 - x^{2}}}$ , – 1 < x < 1 such that f(0) = 0. If $6 \int_{- 1 / 2}^{1 / 2} f (x) d x = 2 π - α$ , then $α^{2}$ is equal to __________. [2025]

(27)

From given differential equation, we have

$I . F . = e^{- \frac{1}{2} \int \frac{2 x}{1 - x^{2}} d x} = e^{\frac{1}{2} \ln (1 - x^{2})} = \sqrt{1 - x^{2}}$

Hence, solution of given D.E. is

$y \times \sqrt{1 - x^{2}} = \int (x^{6} + 4 x) d x = \frac{x^{7}}{7} + 2 x^{2} + c$

Given, y(0) = 0, then c = 0

$∴ y = \frac{\frac{x^{7}}{7} + 2 x^{2}}{\sqrt{1 - x^{2}}}$

Let $I = 6 \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{\frac{x^{7}}{7} + 2 x^{2}}{\sqrt{1 - x^{2}}} d x$ ... (i)

$\Rightarrow I = 6 \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{\frac{- x^{7}}{7} + 2 x^{2}}{\sqrt{1 - x^{2}}} d x$ ... (ii)

Adding (i) & (ii), we get

$2 I = 24 \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{x^{2}}{\sqrt{1 - x^{2}}} d x = 2 \times 24 \int_{0}^{\frac{1}{2}} \frac{x^{2}}{\sqrt{1 - x^{2}}} d x$

Put $x = \sin θ \Rightarrow d x = \cos θ d θ$

$∴ I = 24 \int_{0}^{\frac{π}{6}} \frac{\sin^{2} θ}{\cos θ} \cos θ d θ$

$= 24 \int_{0}^{\frac{π}{6}} (\frac{1 - \cos 2 θ}{2}) d θ = 12 {[θ - \frac{\sin 2 θ}{2}]}_{0}^{\frac{π}{6}}$

$= 12 (\frac{π}{6} - \frac{\sqrt{3}}{4}) = 2 π - 3 \sqrt{3}$

On comparing, we get $α^{2} = {(3 \sqrt{3})}^{2} = 27$ .

Q 36 :

Let f be a differentiable function such that $2 {(x + 2)}^{2} f (x) - 3 {(x + 2)}^{2} = 10 \int_{0}^{x} (t + 2) f (t) d t, x \geq 0$ . Then f(2) is equal to _________. [2025]

(19)

We have,

$2 {(x + 2)}^{2} f (x) - 3 {(x + 2)}^{2} = 10 \int_{0}^{x} (t + 2) f (t) d t, x \geq 0$ ... (i)

On differentiating, we get

$4 (x + 2) f (x) + 2 {(x + 2)}^{2} f' (x) - 6 (x + 2) = 10 (x + 2) f (x)$

$\Rightarrow 2 {(x + 2)}^{2} f' (x) - 6 (x + 2) f (x) = 6 (x + 2)$

$\Rightarrow (x + 2) f' (x) - 3 f (x) = 3$

$\Rightarrow (x + 2) \frac{d y}{d x} - 3 y = 3$ $[∵ f' (x) = \frac{d y}{d x} and f (x) = y]$

$\Rightarrow \frac{d y}{d x} = \frac{3 (1 + y)}{(x + 2)} \Rightarrow \frac{d y}{3 (1 + y)} = \frac{d x}{(x + 2)}$

Integrating on both sides, we get

$\int \frac{d y}{(1 + y)} = 3 \int \frac{d x}{(x + 2)} \Rightarrow \ln (1 + y) = 3 \ln (x + 2) + \ln C$

$\Rightarrow (1 + y) = C {(x + 2)}^{3}$ ... (ii)

Put x = 0 in equation (i), we get

$2 {(2)}^{2} f (0) - 3 {(2)}^{2} = 0 \Rightarrow f (0) = \frac{3}{2}$

From (ii), $1 + \frac{3}{2} = C {(2)}^{3} \Rightarrow C = \frac{5}{16}$

$∴ y = \frac{5}{16} {(x + 2)}^{3} - 1 = f (x)$

$∴ f (2) = \frac{5}{16} \times 64 - 1 = 19$

Q 37 :

Let y = y(x) be the solution of the differential equation $2 \cos x \frac{d y}{d x} = \sin 2 x - 4 y \sin x$ , $x \in (0, \frac{π}{2})$ . If $y (\frac{π}{3}) = 0$ , then $y' (\frac{π}{4}) + y (\frac{π}{4})$ is equal to _________. [2025]

(1)

Given $2 \cos x \frac{d y}{d x} = \sin 2 x - 4 y \sin x$ , $x \in (0, \frac{π}{2})$

The simplified is given by $\frac{d y}{d x} = \frac{2 \sin x \cdot \cos x}{2 \cos x} - \frac{4 \sin x}{2 \cos x} y$

$\Rightarrow \frac{d y}{d x} + 2 y \tan x = \sin x$

The integrating factor is given by

$I . F . = e^{\int 2 \tan x d x} = 2 \sec^{2} x$

$∴$ The solution is given by

$\Rightarrow 2 y \sec^{2} x = 2 \int \frac{\sin x}{\cos^{2} x} d x$

$\Rightarrow y \sec^{2} x = \int \tan x \cdot \sec x d x$

$\Rightarrow y \sec^{2} x = \sec x + c$

Since, $y (\frac{π}{3}) = 0 \Rightarrow 0 = 2 + c \Rightarrow c = - 2$

$∴ y = \cos x - 2 \cos^{2} x & y' = - \sin x + 4 \sin x \cos x$

Here, $y (\frac{π}{4}) = \frac{1}{\sqrt{2}} - 1 & y' (\frac{π}{4}) = \frac{- 1}{\sqrt{2}} + 2$

Hence, $y' (\frac{π}{4}) + y (\frac{π}{4}) = \frac{1}{\sqrt{2}} - 1 - \frac{1}{\sqrt{2}} + 2 = 1$ .

Q 38 :

If y = y(x) is the solution of the differential equation $\sqrt{4 - x^{2}} \frac{d y}{d x} = ((\sin^{- 1} {(\frac{x}{2}))}^{2} - y) \sin^{- 1} (\frac{x}{2})$ , $- 2 \leq x \leq 2, y (2) = \frac{π^{2} - 8}{4}$ , then $y^{2} (0)$ is equal to _________. [2025]

(4)

From the given D.E.

$\frac{d y}{d x} + \frac{(\sin^{- 1} \frac{x}{2})}{\sqrt{4 - x^{2}}} y = \frac{{(\sin^{- 1} \frac{x}{2})}^{3}}{\sqrt{4 - x^{2}}}$

$I . F . = e^{\int \frac{\sin^{- 1} \frac{x}{2}}{\sqrt{4 - x^{2}}} d x} = e^{\frac{{(\sin^{- 1} \frac{x}{2})}^{2}}{2}}$

$∴ y e^{\frac{{(\sin^{- 1} \frac{x}{2})}^{2}}{2}} = \int \frac{{(\sin^{- 1} \frac{x}{2})}^{3}}{\sqrt{4 - x^{2}}} e^{\frac{{(\sin^{- 1} \frac{x}{2})}^{2}}{2}} d x$

$\Rightarrow y = {(\sin^{- 1} \frac{x}{2})}^{2} - 2 + c e^{- \frac{{(\sin^{- 1} \frac{x}{2})}^{2}}{2}}$

Now, $y (2) = \frac{π^{2}}{4} - 2 + c e^{- \frac{π^{2}}{8}}$

On comparing with $y (2) = \frac{π^{2} - 8}{4}$ , we get c = 0

$∴ y = {(\sin^{- 1} \frac{x}{2})}^{2} - 2$

Hence, $y (0) = - 2 \Rightarrow y^{2} (0) = 4$

Q 39 :

Let $f : (0, \infty) \to R$ be a twice differentiable function. If for some $a \neq 0, \int_{0}^{1} f (λ x) d λ = a f (x)$ , f(1) = 1 and $f (16) = \frac{1}{8}$ , then $16 - f' (\frac{1}{16})$ is equal to __________. [2025]

(112)

We have $\int_{0}^{1} f (λ x) d λ = a f (x)$

$\Rightarrow \frac{1}{x} \int_{0}^{x} f (t) d t = a f (x) [Put λ x = t \Rightarrow d λ = \frac{1}{x} d t]$

$\Rightarrow \int_{0}^{x} f (t) d t = a x f (x)$

$\Rightarrow f (x) = a (x f' (x) + f (x)) \Rightarrow (1 - a) f (x) = a x f' (x)$

$\Rightarrow \frac{f' (x)}{f (x)} = \frac{(1 - a)}{a} \frac{1}{x}$ $\Rightarrow \ln f (x) = \frac{1 - a}{a} \ln x + c$ (On integrating)

Now, x = 1, f(1) = 1 $\Rightarrow$ c = 0

Again x = 16, $f (16) = \frac{1}{8} \Rightarrow \frac{1}{8} = {(16)}^{\frac{1 - a}{a}} \Rightarrow - 3 = \frac{4 - 4 a}{a} \Rightarrow a = 4$

Therefore f(x) = $x^{- 3 / 4}$

$\Rightarrow f' (x) = - \frac{3}{4} x^{- \frac{7}{4}}$

Hence, $16 - f' (\frac{1}{16}) = 16 - (- \frac{3}{4} {(2^{- 4})}^{- 7 / 4})$

= 16 + 96 = 112.

Q 40 :

If the solution curve $f (x, y) = 0$ of the differential equation $(1 + \log_{e} x) \frac{d x}{d y} - x \log_{e} x = e^{y}, x > 0,$ passes through the points (1, 0) and $(α, 2)$ , then $α^{α}$ is equal to [2023]

$e^{e^{2}}$
$e^{\sqrt{2} e^{2}}$
$e^{2 e^{\sqrt{2}}}$
$e^{2 e^{2}}$

1

2

3

4

(4)

$The given differential equation is,$

$(1 + \log x) \frac{d x}{d y} - x \log x = e^{y}$

Put $x \log x = t \Rightarrow (x \frac{1}{x} + \log x) d x = d t$

$\Rightarrow (1 + \log x) d x = d t$

$∴$ $Given equation becomes: \frac{d t}{d y} - t = e^{y},$ which is a linear differential equation.

$∴$ $I . F . = e^{\int - d y} = e^{- y}$

So, required solution is given by

$e^{- y} \cdot t = \int e^{y} \cdot e^{- y} d y = y + C$

$\Rightarrow t = (y + C) e^{y}$

$\Rightarrow x \log x = (y + C) e^{y} \dots (i)$

Now, (i) passes through (1, 0) and $(α, 2)$

$∴ C = 0 and α \log α = 2 e^{2}$

$\Rightarrow \log α^{α} = 2 e^{2} \Rightarrow α^{α} = e^{2 e^{2}}$

Topic Question Set

Q 12 :

If the solution curve y = y(x) of the differential equation $(1 + y^{2}) (1 + \log_{e} x) d x + x d y = 0$ , x > 0 passes through the point (1, 1) and $y (e) = \frac{α - \tan (\frac{3}{2})}{β + \tan (\frac{3}{2})}$ then $α + 2 β$ is __________. [2024]

Q 13 :

Let $f (x) = \sqrt{\lim_{r \to x} {\frac{2 r^{2} [(f {(r))}^{2} - f (x) f (r)]}{r^{2} - x^{2}} - r^{3} e^{\frac{f (r)}{r}}}}$ be differentiable in $(- \infty, 0) \cup (0, \infty)$ and f(1) = 1. Then the value of ea, such that f(a) = 0, is equal to __________. [2024]

Q 14 :

Let y = y(x) be the solution of the differential equation $(1 - x^{2}) d y = [x y + (x^{3} + 2) \sqrt{3 (1 - x^{2})}] d x$ , –1 < x < 1, y(0) = 0. If $y (\frac{1}{2}) = \frac{m}{n}$ , m and n are co-prime numbers, then m + n is equal to __________. [2024]

Q 15 :

Let Y = Y(X) be a curve lying in the first quadrant such that the area enclosed by the line Y – y = Y'(x)(X – x) and the co-ordinate axes, where (x, y) is any point on the curve, is always $\frac{- y^{2}}{2 Y' (x)} + 1$ , Y'(X) $\neq$ 0. If Y(1) = 1, then 12Y(2) equals __________. [2024]

Q 16 :

Let y = y(x) be the solution of the differential equation $\sec^{2} x d x + (e^{2 y} \tan^{2} x + \tan x) d y = 0$ , $0 < x < \frac{π}{2}$ , $y (\frac{π}{4}) = 0$ . If $y (\frac{π}{6}) = α$ , then $e^{8 α}$ is equal to __________. [2024]

Q 17 :

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function. If $10 \int_{1}^{x} f (t) d t = 5 x f (x) - x^{5} - 9$ for all $x \geq 1$ , then the value of f(3) is : [2025]

Q 19 :

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} + 3 (\tan^{2} x) y + 3 y = \sec^{2} x, y (0) = \frac{1}{3} + e^{3}$ . Then $y (\frac{π}{4})$ is equal to [2025]

Q 20 :

If a curve y = y(x) passes through the point $(1, \frac{π}{2})$ and satisfies the differential equation $(7 x^{4} \cot y - e^{x} cosec y) \frac{d x}{d y} = x^{5}, x \geq 1$ , then at x = 2, the value of cos y is: [2025]

Q 21 :

Let y = y(x) be the solution curve of the differential equation $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0, x > 0$ , passing through the point (1, 0). Then y(2) is equal to [2025]

Q 22 :

Let y = y(x) be the solution of the differential equation $(x^{2} + 1) y' - 2 x y = (x^{4} + 2 x^{2} + 1) \cos x$ , y(0) = 1. Then $\int_{- 3}^{3} y (x) d x$ is : [2025]

Q 23 :

Let f(x) = x – 1 and $g (x) = e^{x}$ for $x \in R$ . If $\frac{d y}{d x} = (e^{- 2 \sqrt{x}} g (f (f (x))) - \frac{y}{\sqrt{x}})$ , y(0) = 0, then y(1) is [2025]

Q 24 :

Let x = x(y) be the solution of the differential equation $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If x(1) = 1, then $x (\frac{1}{2})$ is : [2025]

Q 25 :

If x = f(y) is the solution of the differential equation $(1 + y^{2}) + (x - 2 e^{\tan^{— 1} y}) \frac{d y}{d x} = 0, y \in (- \frac{π}{2}, \frac{π}{2})$ with f(0) = 1, then $f (\frac{1}{\sqrt{3}})$ is equal to : [2025]

Q 26 :

Let a curve y = f(x) pass through the points (0, 5) and $(\log_{e} 2, k)$ . If the curve satisfies the differential equation $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$ , then k is equal to [2025]

Q 27 :

Let x = x(y) be the solution of the differential equation $y = (x - y \frac{d x}{d y}) \sin (\frac{x}{y})$ , y > 0 and $x (1) = \frac{π}{2}$ . Then cos (x(2)) is equal to : [2025]

Q 28 :

Let y = y(x) be the solution of the differential equation $(x y - 5 x^{2} \sqrt{1 + x^{2}}) d x + (1 + x^{2}) d y = 0$ , y(0) = 0. Then $y (\sqrt{3})$ is equal to [2025]

Q 29 :

Let for some function y = f(x), $\int_{0}^{x} t f (t) d t = x^{2} f (x), x > 0$ and f(2) = 3. Then f(6) is equal to [2025]

Q 30 :

Let $f : R \to R$ be a twice differentiable function such that f(2) = 1. If F(x) = xf(x) for all $x \in R$ . $\int_{0}^{2} x F' (x) d x = 6$ and $\int_{0}^{2} x^{2} F'' (x) d x = 40$ , then $F' (2) + \int_{0}^{2} F (x) d x$ is equal to : [2025]

Q 31 :

Let y = y(x) be the solution of the differential equation $\cos x (\log_{e} {(\cos x))}^{2} d y + (\sin x - 3 y \sin x \log_{e} (\cos x)) d x = 0$ , $x \in (0, \frac{π}{2})$ . If $y (\frac{π}{4}) = \frac{- 1}{\log_{e} 2}$ , then $y (\frac{π}{6})$ is equal to: [2025]

Q 32 :

If for the solution curve y = f(x) of the differential equation $\frac{d y}{d x} + (\tan x) y = \frac{2 + \sec x}{{(1 + 2 \sec x)}^{2}}$ , $x \in (\frac{- π}{2}, \frac{π}{2})$ , $f (\frac{π}{3}) = \frac{\sqrt{3}}{10}$ , then $f (\frac{π}{4})$ is equal to: [2025]

Q 33 :

Let $f : R \to R$ be a thrice differentiable odd function satisfying $f' (x) \geq 0, f'' (x) = f (x), f (0) = 0, f' (0) = 3$ . Then $9 f (\log_{e} 3)$ is equal to _________. [2025]

Q 34 :

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} + 2 y \sec^{2} x = 2 \sec^{2} x + 3 \tan x \cdot \sec^{2} x$ such that $y (0) = \frac{5}{4}$ . Then $12 (y (\frac{π}{4}) - e^{- 2})$ is equal to __________. [2025]

Q 35 :

Let y = f(x) be the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{x^{6} + 4 x}{\sqrt{1 - x^{2}}}$ , – 1 < x < 1 such that f(0) = 0. If $6 \int_{- 1 / 2}^{1 / 2} f (x) d x = 2 π - α$ , then $α^{2}$ is equal to __________. [2025]

Q 36 :

Let f be a differentiable function such that $2 {(x + 2)}^{2} f (x) - 3 {(x + 2)}^{2} = 10 \int_{0}^{x} (t + 2) f (t) d t, x \geq 0$ . Then f(2) is equal to _________. [2025]

Q 37 :

Let y = y(x) be the solution of the differential equation $2 \cos x \frac{d y}{d x} = \sin 2 x - 4 y \sin x$ , $x \in (0, \frac{π}{2})$ . If $y (\frac{π}{3}) = 0$ , then $y' (\frac{π}{4}) + y (\frac{π}{4})$ is equal to _________. [2025]

Q 38 :

If y = y(x) is the solution of the differential equation $\sqrt{4 - x^{2}} \frac{d y}{d x} = ((\sin^{- 1} {(\frac{x}{2}))}^{2} - y) \sin^{- 1} (\frac{x}{2})$ , $- 2 \leq x \leq 2, y (2) = \frac{π^{2} - 8}{4}$ , then $y^{2} (0)$ is equal to _________. [2025]

Q 39 :

Let $f : (0, \infty) \to R$ be a twice differentiable function. If for some $a \neq 0, \int_{0}^{1} f (λ x) d λ = a f (x)$ , f(1) = 1 and $f (16) = \frac{1}{8}$ , then $16 - f' (\frac{1}{16})$ is equal to __________. [2025]

Q 40 :

If the solution curve $f (x, y) = 0$ of the differential equation $(1 + \log_{e} x) \frac{d x}{d y} - x \log_{e} x = e^{y}, x > 0,$ passes through the points (1, 0) and $(α, 2)$ , then $α^{α}$ is equal to [2023]

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 11 : If the solution curve, of the differential equation dydx=x+y–2x–y passing through the point (2, 1) is tan–1(y–1x–1)–1β loge(α+(y–1x–1)2)=loge|x–1|, then 5β+α is equal to __________. [2024]

Q 12 : If the solution curve y = y(x) of the differential equation (1+y2)(1+loge x)dx+xdy=0, x > 0 passes through the point (1, 1) and y(e)=α–tan(32)β+tan(32) then α+2β is __________. [2024]

Q 13 : Let f(x)=limr→x{2r2[(f(r))2–f(x)f(r)]r2–x2–r3ef(r)r} be differentiable in (–∞,0)∪(0,∞) and f(1) = 1. Then the value of ea, such that f(a) = 0, is equal to __________. [2024]

Q 14 : Let y = y(x) be the solution of the differential equation (1–x2)dy=[xy+(x3+2)3(1–x2)]dx, –1 < x < 1, y(0) = 0. If y(12)=mn, m and n are co-prime numbers, then m + n is equal to __________. [2024]

Q 15 : Let Y = Y(X) be a curve lying in the first quadrant such that the area enclosed by the line Y – y = Y'(x)(X – x) and the co-ordinate axes, where (x, y) is any point on the curve, is always –y22Y'(x)+1, Y'(X) ≠ 0. If Y(1) = 1, then 12Y(2) equals __________. [2024]

Q 16 : Let y = y(x) be the solution of the differential equation sec2xdx+(e2ytan2x+tan x)dy=0, 0<x<π2, y(π4)=0. If y(π6)=α, then e8α is equal to __________. [2024]

Q 17 : Let f:[1,∞)→[2,∞) be a differentiable function. If 10∫1xf(t)dt=5xf(x)–x5–9 for all x≥1, then the value of f(3) is : [2025]

Q 18 : Let g be differentiable function such that ∫0xg(t)dt=x–∫0xtg(t)dt, x≥0 and let y = y(x) satisfy the differential equation dydx–y tan x=2(x+1) sec xg(x), x∈[0,π2). If y(0) = 0, then y(π3) is equal to [2025]

Q 19 : Let y = y(x) be the solution of the differential equation dydx+3(tan2x)y+3y=sec2x, y(0)=13+e3. Then y(π4) is equal to [2025]

Q 20 : If a curve y = y(x) passes through the point (1,π2) and satisfies the differential equation (7x4cot y–excosec y)dxdy=x5, x≥1, then at x = 2, the value of cos y is: [2025]

Q 21 : Let y = y(x) be the solution curve of the differential equation x(x2+ex)dy+(ex(x–2)y–x3)dx=0, x>0, passing through the point (1, 0). Then y(2) is equal to [2025]

Q 22 : Let y = y(x) be the solution of the differential equation (x2+1)y'–2xy=(x4+2x2+1) cos x, y(0) = 1. Then ∫–33y(x)dx is : [2025]

Q 23 : Let f(x) = x – 1 and g(x)=ex for x∈R. If dydx=(e–2xg(f(f(x)))–yx), y(0) = 0, then y(1) is [2025]

Q 24 : Let x = x(y) be the solution of the differential equation y2dx+(x–1y)dy=0. If x(1) = 1, then x(12) is : [2025]

Q 25 : If x = f(y) is the solution of the differential equation (1+y2)+(x–2etan—1y)dydx=0, y∈(–π2,π2) with f(0) = 1, then f(13) is equal to : [2025]

Q 26 : Let a curve y = f(x) pass through the points (0, 5) and (loge2,k). If the curve satisfies the differential equation 2(3+y)e2xdx–(7+e2x)dy=0, then k is equal to [2025]

Q 27 : Let x = x(y) be the solution of the differential equation y=(x–ydxdy)sin(xy), y > 0 and x(1)=π2. Then cos (x(2)) is equal to : [2025]

Q 28 : Let y = y(x) be the solution of the differential equation (xy–5x21+x2)dx+(1+x2)dy=0, y(0) = 0. Then y(3) is equal to [2025]

Q 29 : Let for some function y = f(x), ∫0xtf(t)dt=x2f(x), x>0 and f(2) = 3. Then f(6) is equal to [2025]

Q 30 : Let f : R→R be a twice differentiable function such that f(2) = 1. If F(x) = xf(x) for all x∈R. ∫02xF'(x)dx=6 and ∫02x2F''(x)dx=40, then F'(2)+∫02F(x)dx is equal to : [2025]

Q 31 : Let y = y(x) be the solution of the differential equation cos x(loge(cos x))2dy+(sin x–3y sinx loge(cos x))dx=0, x∈(0, π2). If y(π4)=–1loge2, then y(π6) is equal to: [2025]

Q 32 : If for the solution curve y = f(x) of the differential equation dydx+(tan x)y=2+sec x(1+2 sec x)2, x∈(–π2,π2), f(π3)=310, then f(π4) is equal to: [2025]

Q 33 : Let f:R→R be a thrice differentiable odd function satisfying f'(x)≥0, f''(x)=f(x), f(0)=0, f'(0)=3. Then 9f(loge3) is equal to _________. [2025]

Q 34 : Let y = y(x) be the solution of the differential equation dydx+2y sec2x=2 sec2x+3 tan x·sec2x such that y(0)=54. Then 12(y(π4)–e–2) is equal to __________. [2025]

Q 35 : Let y = f(x) be the solution of the differential equation dydx+xyx2–1=x6+4x1–x2, – 1 < x < 1 such that f(0) = 0. If 6∫–1/21/2f(x)dx=2π–α, then α2 is equal to __________. [2025]

Q 36 : Let f be a differentiable function such that 2(x+2)2f(x)–3(x+2)2=10∫0x(t+2)f(t)dt, x≥0. Then f(2) is equal to _________. [2025]

Q 37 : Let y = y(x) be the solution of the differential equation 2cosxdydx=sin2x–4ysinx, x∈(0,π2). If y(π3)=0, then y'(π4)+y(π4) is equal to _________. [2025]

Q 38 : If y = y(x) is the solution of the differential equation 4–x2dydx=((sin–1(x2))2–y)sin–1(x2), –2≤x≤2, y(2)=π2–84, then y2(0) is equal to _________. [2025]

Q 39 : Let f:(0,∞)→R be a twice differentiable function. If for some a≠0, ∫01f(λx)dλ=af(x), f(1) = 1 and f(16)=18, then 16–f'(116) is equal to __________. [2025]

Q 40 : If the solution curve f(x,y)=0 of the differential equation (1+logex)dxdy-xlogex=ey, x>0, passes through the points (1, 0) and (α,2), then αα is equal to [2023]

Want Us to Email you About Special Offers & Updates?

Q 12 :

If the solution curve y = y(x) of the differential equation $(1 + y^{2}) (1 + \log_{e} x) d x + x d y = 0$ , x > 0 passes through the point (1, 1) and $y (e) = \frac{α - \tan (\frac{3}{2})}{β + \tan (\frac{3}{2})}$ then $α + 2 β$ is __________. [2024]

Q 13 :

Let $f (x) = \sqrt{\lim_{r \to x} {\frac{2 r^{2} [(f {(r))}^{2} - f (x) f (r)]}{r^{2} - x^{2}} - r^{3} e^{\frac{f (r)}{r}}}}$ be differentiable in $(- \infty, 0) \cup (0, \infty)$ and f(1) = 1. Then the value of ea, such that f(a) = 0, is equal to __________. [2024]

Q 14 :

Let y = y(x) be the solution of the differential equation $(1 - x^{2}) d y = [x y + (x^{3} + 2) \sqrt{3 (1 - x^{2})}] d x$ , –1 < x < 1, y(0) = 0. If $y (\frac{1}{2}) = \frac{m}{n}$ , m and n are co-prime numbers, then m + n is equal to __________. [2024]

Q 16 :

Let y = y(x) be the solution of the differential equation $\sec^{2} x d x + (e^{2 y} \tan^{2} x + \tan x) d y = 0$ , $0 < x < \frac{π}{2}$ , $y (\frac{π}{4}) = 0$ . If $y (\frac{π}{6}) = α$ , then $e^{8 α}$ is equal to __________. [2024]

Q 17 :

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function. If $10 \int_{1}^{x} f (t) d t = 5 x f (x) - x^{5} - 9$ for all $x \geq 1$ , then the value of f(3) is : [2025]

Q 19 :

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} + 3 (\tan^{2} x) y + 3 y = \sec^{2} x, y (0) = \frac{1}{3} + e^{3}$ . Then $y (\frac{π}{4})$ is equal to [2025]

Q 20 :

If a curve y = y(x) passes through the point $(1, \frac{π}{2})$ and satisfies the differential equation $(7 x^{4} \cot y - e^{x} cosec y) \frac{d x}{d y} = x^{5}, x \geq 1$ , then at x = 2, the value of cos y is: [2025]

Q 21 :

Let y = y(x) be the solution curve of the differential equation $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0, x > 0$ , passing through the point (1, 0). Then y(2) is equal to [2025]

Q 22 :

Let y = y(x) be the solution of the differential equation $(x^{2} + 1) y' - 2 x y = (x^{4} + 2 x^{2} + 1) \cos x$ , y(0) = 1. Then $\int_{- 3}^{3} y (x) d x$ is : [2025]

Q 23 :

Let f(x) = x – 1 and $g (x) = e^{x}$ for $x \in R$ . If $\frac{d y}{d x} = (e^{- 2 \sqrt{x}} g (f (f (x))) - \frac{y}{\sqrt{x}})$ , y(0) = 0, then y(1) is [2025]

Q 24 :

Let x = x(y) be the solution of the differential equation $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If x(1) = 1, then $x (\frac{1}{2})$ is : [2025]

Q 25 :

If x = f(y) is the solution of the differential equation $(1 + y^{2}) + (x - 2 e^{\tan^{— 1} y}) \frac{d y}{d x} = 0, y \in (- \frac{π}{2}, \frac{π}{2})$ with f(0) = 1, then $f (\frac{1}{\sqrt{3}})$ is equal to : [2025]

Q 26 :

Let a curve y = f(x) pass through the points (0, 5) and $(\log_{e} 2, k)$ . If the curve satisfies the differential equation $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$ , then k is equal to [2025]

Q 27 :

Let x = x(y) be the solution of the differential equation $y = (x - y \frac{d x}{d y}) \sin (\frac{x}{y})$ , y > 0 and $x (1) = \frac{π}{2}$ . Then cos (x(2)) is equal to : [2025]

Q 28 :

Let y = y(x) be the solution of the differential equation $(x y - 5 x^{2} \sqrt{1 + x^{2}}) d x + (1 + x^{2}) d y = 0$ , y(0) = 0. Then $y (\sqrt{3})$ is equal to [2025]

Q 29 :

Let for some function y = f(x), $\int_{0}^{x} t f (t) d t = x^{2} f (x), x > 0$ and f(2) = 3. Then f(6) is equal to [2025]

Q 30 :

Let $f : R \to R$ be a twice differentiable function such that f(2) = 1. If F(x) = xf(x) for all $x \in R$ . $\int_{0}^{2} x F' (x) d x = 6$ and $\int_{0}^{2} x^{2} F'' (x) d x = 40$ , then $F' (2) + \int_{0}^{2} F (x) d x$ is equal to : [2025]

Q 31 :

Let y = y(x) be the solution of the differential equation $\cos x (\log_{e} {(\cos x))}^{2} d y + (\sin x - 3 y \sin x \log_{e} (\cos x)) d x = 0$ , $x \in (0, \frac{π}{2})$ . If $y (\frac{π}{4}) = \frac{- 1}{\log_{e} 2}$ , then $y (\frac{π}{6})$ is equal to: [2025]

Q 32 :

If for the solution curve y = f(x) of the differential equation $\frac{d y}{d x} + (\tan x) y = \frac{2 + \sec x}{{(1 + 2 \sec x)}^{2}}$ , $x \in (\frac{- π}{2}, \frac{π}{2})$ , $f (\frac{π}{3}) = \frac{\sqrt{3}}{10}$ , then $f (\frac{π}{4})$ is equal to: [2025]

Q 33 :

Let $f : R \to R$ be a thrice differentiable odd function satisfying $f' (x) \geq 0, f'' (x) = f (x), f (0) = 0, f' (0) = 3$ . Then $9 f (\log_{e} 3)$ is equal to _________. [2025]

Q 34 :

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} + 2 y \sec^{2} x = 2 \sec^{2} x + 3 \tan x \cdot \sec^{2} x$ such that $y (0) = \frac{5}{4}$ . Then $12 (y (\frac{π}{4}) - e^{- 2})$ is equal to __________. [2025]

Q 35 :

Let y = f(x) be the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{x^{6} + 4 x}{\sqrt{1 - x^{2}}}$ , – 1 < x < 1 such that f(0) = 0. If $6 \int_{- 1 / 2}^{1 / 2} f (x) d x = 2 π - α$ , then $α^{2}$ is equal to __________. [2025]

Q 36 :

Let f be a differentiable function such that $2 {(x + 2)}^{2} f (x) - 3 {(x + 2)}^{2} = 10 \int_{0}^{x} (t + 2) f (t) d t, x \geq 0$ . Then f(2) is equal to _________. [2025]

Q 37 :

Let y = y(x) be the solution of the differential equation $2 \cos x \frac{d y}{d x} = \sin 2 x - 4 y \sin x$ , $x \in (0, \frac{π}{2})$ . If $y (\frac{π}{3}) = 0$ , then $y' (\frac{π}{4}) + y (\frac{π}{4})$ is equal to _________. [2025]

Q 38 :

If y = y(x) is the solution of the differential equation $\sqrt{4 - x^{2}} \frac{d y}{d x} = ((\sin^{- 1} {(\frac{x}{2}))}^{2} - y) \sin^{- 1} (\frac{x}{2})$ , $- 2 \leq x \leq 2, y (2) = \frac{π^{2} - 8}{4}$ , then $y^{2} (0)$ is equal to _________. [2025]

Q 39 :

Let $f : (0, \infty) \to R$ be a twice differentiable function. If for some $a \neq 0, \int_{0}^{1} f (λ x) d λ = a f (x)$ , f(1) = 1 and $f (16) = \frac{1}{8}$ , then $16 - f' (\frac{1}{16})$ is equal to __________. [2025]

Q 40 :

If the solution curve $f (x, y) = 0$ of the differential equation $(1 + \log_{e} x) \frac{d x}{d y} - x \log_{e} x = e^{y}, x > 0,$ passes through the points (1, 0) and $(α, 2)$ , then $α^{α}$ is equal to [2023]