Question

The function $f\left(x\right)=\underset{-1}{\overset{x}{\int }}t(e^t-1)(t-1)(t-2{)}^3(t-3{)}^{5}dt$ has a maximum value at $x=k$. If $k=\cos \theta +\sec \theta$, then ${\cos }^6\theta +{\sec }^6\theta$ is equal to

• A. $32$
• B. $2$
• C. $4$
• D. $16$

Answer 1

The correct option is B $2$

$f^{\prime }\left(x\right)=x(e^x-1)(x-1)(x-2{)}^3(x-3{)}^5$
Using sign rule, the function will have maximum value at $x=2$, so the value of $k=2$
So,
$2=\cos \theta +\sec \theta \Rightarrow \cos \theta +\frac{1}{\cos \theta }=2\Rightarrow \cos \theta =1$
So, the value of ${\cos }^6\theta +{\sec }^6\theta =1+1=2$