Question

Let $F\left(x\right)={\int }_x^{x^2+\frac{\pi }{6}}2co{s}^{2}tdt$ for all $xϵR$ and $f:[0,\frac{1}{2}]\to [0,\infty )$ be a continuous function. For $\alpha ϵ[0,\frac{1}{2}]$, if $F^{\prime }\left(a\right)+2$ is the area of the region bounded by $x=0,y=0,y=f\left(x\right)$ and $x=a$, then $f\left(0\right)$ is

Answer 1

$F\left(x\right)={\int }_x^{x^2+\frac{\pi }{6}}2{\cos }^{2}tdt$
Differentiating both side w.r.t $x$
$F^{\prime }\left(x\right)=2{\cos }^2(x^2+\frac{\pi }{6})\cdot 2x-2{\cos }^{2}x$
$\therefore F^{\prime }\left(a\right)=4a{\cos }^2(a^2+\frac{\pi }{6})-2{\cos }^{2}a$
Now given, $F^{\prime }\left(a\right)+2={\int }_0^{a}f\left(x\right)dx$
$4a{\cos }^2(a^2+\frac{\pi }{6})-2{\cos }^{2}a+2={\int }_0^{a}f\left(x\right)dx$
Differentiating both side w.r.t $a$
$4{\cos }^2(a^2+\frac{\pi }{6})-8a\cos (a^2+\frac{\pi }{6})\sin (a^2+\frac{\pi }{6})+4\cos a\sin a=f\left(a\right)$
$\therefore f\left(0\right)=4{\cos }^2\left(\frac{\pi }{6}\right)=4\times \frac{3}{4}=3$