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

Q 31 :
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]

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(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 32 :
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]

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(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 33 :
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]

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(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 34 :
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]

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(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 35 :
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]

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(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 36 :
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]

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(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 37 :
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
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26

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(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 38 :
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]

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

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(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 39 :
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]

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

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(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 40 :
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]

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

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(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

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

Topic Question Set

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Q 32 :
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]

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Q 33 :
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]

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Q 34 :
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]

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Q 35 :
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]

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Q 36 :
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]

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Q 37 :
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 39 :
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 40 :
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]

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Topic Question Set

Q 31 : 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]

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Q 32 : 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]

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Q 33 : 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]

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Q 34 : 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]

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Q 35 : 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]

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Q 36 : 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]

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Q 37 : 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 38 : 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 39 : 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 40 : 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]

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Q 32 :
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 33 :
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 34 :
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 35 :
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 36 :
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 37 :
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 39 :
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 40 :
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]