QJEEM21Ma211b: JEE Main 2021 Mathematics - Class 12 JEEM21Ma211b Solution: Let y = y(x) satisfies the equation , for all x > 0, where A =…

Question

Verified Solution • 2025–26 Session moderate

Let y = y(x) satisfies the equation

{{dy} \over {dx}} - |A| = 0

, for all x > 0, where

A = [\begin{matrix} y & \sin x & 1 \\ 0 & - 1 & 1 \\ 2 & 0 & \frac{1}{x} \end{matrix}]

. If

y(\pi ) = \pi + 2

, then the value of

y\left( {{\pi \over 2}} \right)

is :

Accepted Answer

|A| = - {y \over x} + 2\sin x + 2

{{dy} \over {dx}} = |A|

{{dy} \over {dx}} = - {y \over x} + 2\sin x + 2

{{dy} \over {dx}} + {y \over x} = 2\sin x + 2

I.F. = {e^{\int {{1 \over x}dx} }} = x

\Rightarrow yx = \int {x(2\sin x + 2)dx}

xy = {x^2} - 2x\cos x + 2\sin x + c

..... (i)

Now, x =

\pi

, y =

\pi

+ 2

Use in (i)

c = 0

Now, (i) becomes

xy = {x^2} - 2x\cos x + 2\sin x

put

x = \pi /2

{\pi \over 2}y = {\left( {{\pi \over 2}} \right)^2} - 2.{\pi \over 2}\cos {\pi \over 2} + 2\sin {\pi \over 2}

y({\pi \over 2}) = {{{\pi ^2}} \over 4} + 2

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