Question

Let $F\left(x\right)=\underset{x}{\overset{x^2+\frac{\pi }{6}}{\int }}2{\cos }^{2}tdt$ for all $x\in R$ and $f:[0,\frac{1}{2}]\to [0,\infty )$ be a continuous function. For $a\in [0,\frac{1}{2}],$ 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

$F\left(x\right)=\underset{x}{\overset{x^2+\frac{\pi }{6}}{\int }}2{\cos }^{2}tdt\text{Using Newton-Leibnitz's theorem,}{F}^{\prime }\left(x\right)=2{\cos }^2(x^2+\frac{\pi }{6})\times 2x-2{\cos }^{2}x\times 1=4x{\cos }^2(x^2+\frac{\pi }{6})-2{\cos }^{2}x\text{Given,}\underset{0}{\overset{a}{\int }}f\left(x\right)dx=F^{\prime }\left(a\right)+2\Rightarrow \underset{0}{\overset{a}{\int }}f\left(x\right)dx=4a{\cos }^2(a^2+\frac{\pi }{6})-2{\cos }^{2}a+2$
Differentiating w.r.t. '$a$', we get
$f\left(a\right)=4a\times 2\cos (a^2+\frac{\pi }{6})[-\sin (a^2+\frac{\pi }{6})]\times 2a$
$+4{\cos }^2(a^2+\frac{\pi }{6})-4\cos a\times (-\sin a)\Rightarrow f\left(0\right)=0+4{\cos }^2\left(\frac{\pi }{6}\right)-0\Rightarrow f\left(0\right)=4\times {\left(\frac{\sqrt{3}}{2}\right)}^2=3$