Verified Solution • 2025–26 Session moderate

QJEEM22Mab9b2: JEE Main 2022 Mathematics - Class 12 JEEM22Mab9b2 Solution: If and , then the maximum value of is :

If {{dy} \over {dx}} + 2y\tan x = \sin x,\,0 < x < {\pi \over 2} and y(π3)=0y\left( {{\pi \over 3}} \right) = 0 , then the maximum value of y(x)y(x) is :

Multiple Choice Options

  • A 18{1 \over 8}
  • B 34{3 \over 4}
  • C 14{1 \over 4}
  • D 38{3 \over 8}

Step-by-Step Verified Solution

dydx+2ytan⁡x=sin⁡x{{dy} \over {dx}} + 2y\tan x = \sin x

which is a first order linear differential equation.

Integrating factor (I. F.) =e∫2tan⁡x dx= {e^{\int {2\tan x\,dx} }}

=e2ln⁡∣sec⁡x∣=sec⁡2x= {e^{2\ln |\sec x|}} = {\sec ^2}x

Solution of differential equation can be written as

y . sec⁡2x=∫sin⁡x . sec⁡2x dx=∫sec⁡ x . tan⁡x dxy\,.\,{\sec ^2}x = \int {\sin x\,.\,{{\sec }^2}x\,dx = \int {\sec \,x\,.\,\tan x\,dx} }

y sec⁡2x=sec⁡x+Cy~{\sec ^2}x = \sec x + C

y(π3)=0,0=sec⁡π3+C⇒    C=−2y\left( {{\pi \over 3}} \right) = 0,0 = \sec {\pi \over 3} + C \Rightarrow \,\,\,\,C = - 2

y=sec⁡x−2sec⁡2x=cos⁡x−2cos⁡2xy = {{\sec x - 2} \over {{{\sec }^2}x}} = \cos x - 2{\cos ^2}x

=18−2(cos⁡x−14)2= {1 \over 8} - 2{\left( {\cos x - {1 \over 4}} \right)^2}

ymax⁡=18{y_{\max }} = {1 \over 8}

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