Question

Let $F\left(x\right)={\int }_{sinx}^{cosx}{e}^{(1+arcsint{)}^2}dt$ on $[0,\frac{\pi }{2}]$ then

• A. $F^{\prime \prime }\left(c\right)=0$ for all $cϵ(0,\frac{\pi }{4})$
• B. $F^{\prime \prime }\left(c\right)=0$ for some $cϵ(0,\frac{\pi }{2})$
• C. $F^{\prime }\left(c\right)\ne 0$ for some $cϵ(0,\frac{\pi }{2})$
• D. $F\left(c\right)\ne 0$ for all $cϵ(0,\frac{\pi }{2})$

Answer 1

The correct option is B $F^{\prime \prime }\left(c\right)=0$ for some $cϵ(0,\frac{\pi }{2})$

$F^{\prime }\left(x\right)$
$=\frac{d}{dx}\left[cos\right(x)e^{(1-x+\frac{\pi }{2}{)}^2}-e^{(1+x{)}^2}.sin(x\left)\right]$
$=-2(1-x+\frac{\pi }{2})e^{(1-x+\frac{\pi }{2}{)}^2}.cos\left(x\right)-sin\left(x\right).(1-x+\frac{\pi }{2}{)}^2-2(1+x)e^{(1+x{)}^2}.sin\left(x\right)-e^{(1+x{)}^2}.cos\left(x\right)$
Hence
$F^{\prime }\left(0\right)=F^{\prime }\left(\frac{\pi }{2}\right)=-(1+2e^{1+\frac{\pi }{2}})$.
Therefore by Rolle's Theorem,
$F^{\prime \prime }\left(c\right)=0$ for $0<c<\frac{\pi }{2}$.