Question

Let $f\left(x\right)$ be a non-negative continuous and differentiable function such that the area bounded by the curve $y=f\left(x\right)$ x axis and the ordinate $x=\frac{\pi }{4}$ and $x=\beta >\frac{\pi }{4}$ is $(\beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta )$ then $f^1\left(0\right)$ is

• A. $2-\frac{\pi }{4}$
• B. $2+\frac{\pi }{4}$
• C. $1+\frac{\pi }{4}$
• D. $0$

Answer 1

Answer: (A)

$2-\frac{\pi }{4}$

Given that

${\int }_{\pi /4}^{\beta }f\left(x\right)dx=\beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta$

Differentiating both sides w.r.t $\beta$ we get,

$\Rightarrow f\left(\beta \right)=\beta \cos \beta +\sin \beta -\frac{\pi }{4}\sin \beta +\sqrt{2}$

$\Rightarrow f^{\prime }\left(\beta \right)=-\beta \sin \beta +\cos \beta +\cos \beta -\frac{\pi }{4}\cos \beta$

$\Rightarrow f^{\prime }\left(0\right)=\cos \left(0\right)+\cos \left(0\right)-\frac{\pi }{4}\cos \left(0\right)$

$\Rightarrow f^{\prime }\left(0\right)=1+1-\frac{\pi }{4}$

$\Rightarrow f^{\prime }\left(0\right)=2-\frac{\pi }{4}$