Q106 / 155 · WBJEE 2023 · Mathematics
Let \(f\) be a non-negative function defined on \(\left[0, \frac{\pi}{2}\right]\). If \(\int_0^x\left(f^{\prime}(t)-\sin 2 t\right) d t=\int_x^0 f(t) \tan t d t, f(0)=1\), then \(\int_0^{\frac{\pi}{2}} f(x) d x\) is
- A3
- Correct answerB\(3-\frac{\pi}{2}\)
- C\(3+\frac{\pi}{2}\)
- D\(\frac{\pi}{2}\)
Solution
\(\begin{aligned}
\text {Hint } & f^{\prime}(x)-\sin 2 x=-f(x) \tan x \\
& \Rightarrow f^{\prime}(x)+\tan x f(x)=\sin 2 x \\
& \Rightarrow \frac{d f(x)}{d x}+\tan x . f(x)=\sin 2 x
\end{aligned}\)
\(\text { If }=e^{f \tan x d x}=e^{\ln |\sec x|}=|\sec x|=\sec x \quad \text { As, } x \in\left[0, \frac{\pi}{2}\right]\)
Solution is,
\(\begin{aligned}
& f(x) \times \sec x=\int \sin 2 x \sec x d x=\int 2 \sin x \cos x \sec x d x=2 \int \sin x d x \\
& f(x) \sec x=-2 \cos x+C \\
& \text { Put } x=0 ; f(0)=1 \\
& f(0) \sec 0=-2 \cos 0+c \\
& \Rightarrow 1=-2+C \quad \therefore C=3
\end{aligned}\)
\(\therefore\) Required solution is;
\(\begin{aligned}
& f(x) \sec x=-2 \cos x+3 \\
& f(x)=-2 \cos ^2 x+3 \cos x
\end{aligned}\)
\(\begin{aligned}
& =-2\left(\frac{1+\cos 2 x}{2}\right)+3 \cos x=-1-\cos 2 x+3 \cos x \\
& f(x)=-1-\cos 2 x+3 \cos x
\end{aligned}\)
Now \(\int_0^{\frac{\pi}{2}} f(x) d x\)
\(\int_0^{\frac{\pi}{2}}(-1-\cos 2 x+3 \cos x) d x=\left[-x-\frac{\sin 2 x}{2}+3 \sin x\right]_0 ^{-\frac{\pi}{2}}=\left(-\frac{\pi}{2}+3\right)-0=3-\frac{\pi}{2}\)
\(\left[-x-\frac{\sin 2 x}{2}+\left.3 \sin x\right|_0 ^{\frac{\pi}{2}}\right.\)
Hence, the correct answer is option: 'B'.