Verified Solution • 2025–26 Session moderate

QJEEM21Mac0cb: JEE Main 2021 Mathematics - Class 12 JEEM21Mac0cb Solution: The function f(x), that satisfies the condition , is :

The function f(x), that satisfies the condition
f(x)=x+∫0π/2sin⁡x.cos⁡y f(y) dyf(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos y\,f(y)\,dy} , is :

Multiple Choice Options

  • A x+23(π−2)sin⁡xx + {2 \over 3}(\pi - 2)\sin x
  • B x+(π+2)sin⁡xx + (\pi + 2)\sin x
  • C x+π2sin⁡xx + {\pi \over 2}\sin x
  • D x+(π−2)sin⁡xx + (\pi - 2)\sin x

Step-by-Step Verified Solution

f(x)=x+∫0π/2sin⁡xcos⁡y f(y) dyf(x) = x + \int\limits_0^{\pi /2} {\sin x\cos y\,f(y)\,dy}

f(x)=x+sinx∫0π/2cos⁡y f(y) dy⏟Kf(x) = x + sinx\underbrace {\int_0^{\pi /2} {\cos y\,f(y)\,dy} }_K

⇒f(x)=x+Ksin⁡x\Rightarrow f(x) = x + K\sin x

⇒f(y)=y+Ksin⁡y\Rightarrow f(y) = y + K\sin y

Now, K=∫0π/2ycos⁡y dyApply IBP+K∫0π/2cos⁡ysin⁡y dyPut sin⁡y=tK = \int_0^{\pi /2} {\mathop {y\cos y\,dy}\limits_{Apply\,IBP} } + K\int_0^{\pi /2} {\mathop {\cos y\sin y\,dy}\limits_{Put\,\sin y = t} }

K=(ysin⁡y)0π/2−∫0π/2sin⁡ydy+K∫01t dtK = \left( {y\sin y} \right)_0^{\pi /2} - \int_0^{\pi /2} {\sin y dy + K\int_0^1 {t\,dt} }

⇒K=π2−1+K(12)\Rightarrow K = {\pi \over 2} - 1 + K\left( {{1 \over 2}} \right)

⇒K=π−2\Rightarrow K = \pi - 2

So, f(x)=x+(π−2)sin⁡xf(x) = x + (\pi - 2)\sin x

Option (d)

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