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

Q 51 :
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]

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

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(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 52 :
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]

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

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(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 53 :
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]

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(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 54 :
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]

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(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 55 :
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]

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(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 56 :
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]

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(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 57 :
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]

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(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 58 :
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]

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(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 59 :
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]

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(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.

Topic Question Set

Q 51 :
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 52 :
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 53 :
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]

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Q 54 :
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]

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Q 55 :
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]

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Q 56 :
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]

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Q 57 :
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]

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Q 58 :
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]

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Q 59 :
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]

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Topic Question Set

Q 51 : 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 52 : 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 53 : 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]

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Q 54 : 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]

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Q 55 : 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]

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Q 56 : 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]

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Q 57 : 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]

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Q 58 : 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]

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Q 59 : 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]

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Q 51 :
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 52 :
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 53 :
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 54 :
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 55 :
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 56 :
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 57 :
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 58 :
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 59 :
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]