Verified Solution • 2025–26 Session moderate

QJEEM22Maa168: JEE Main 2022 Mathematics - Class 12 JEEM22Maa168 Solution: Let . Then the value of is _____________.

Let f(θ)=sin⁡θ+∫−π/2π/2(sin⁡θ+tcos⁡θ)f(t)dtf(\theta ) = \sin \theta + \int\limits_{ - \pi /2}^{\pi /2} {(\sin \theta + t\cos \theta )f(t)dt} . Then the value of ∣∫0π/2f(θ)dθ∣\left| {\int_0^{\pi /2} {f(\theta )d\theta } } \right| is _____________.

Step-by-Step Verified Solution

f(θ)=sin⁡θ(1+∫−π/2π/2f(t)dt)+cos⁡θ(∫−π/2π/2tf(t)dt)f(\theta)=\sin \theta\left(1+\int_{-\pi / 2}^{\pi / 2} f(t) d t\right)+\cos \theta\left(\int_{-\pi / 2}^{\pi / 2} t f(t) d t\right)

Clearly f(θ)=asin⁡θ+bcos⁡θf(\theta)=a \sin \theta+b \cos \theta

Where a=1+∫−π/2π/2(asin⁡t+bcos⁡t)dt⇒a=1+2b...(i)a=1+\int_{-\pi / 2}^{\pi / 2}(a \sin t+b \cos t) d t \Rightarrow a=1+2 b\quad\quad...(i)

and b=∫−π/2π/2(atsin⁡t+btcos⁡t)dt⇒b=2a...(ii)b=\int_{-\pi / 2}^{\pi / 2}(a t \sin t+b t \cos t) d t \Rightarrow b=2 a\quad\quad...(ii)

from (i) and (ii) we get

a=−13 and b=−23a=-\frac{1}{3} \text { and } b=-\frac{2}{3}

So f(θ)=−13(sin⁡θ+2cos⁡θ)f(\theta)=-\frac{1}{3}(\sin \theta+2 \cos \theta)

⇒∣∫0π/2f(θ)dθ∣=13(1+2×1)=1\Rightarrow\left|\int_{0}^{\pi / 2} f(\theta) d \theta\right|=\frac{1}{3}(1+2 \times 1)=1

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