Question

If $f(x)=a\cos (\pi x)+b,f^{\prime }\left(\frac{1}{2}\right)=\pi$ and $\underset{\frac{1}{2}}{\overset{\frac{3}{2}}{\int }}f(x)dx=\frac{2}{\pi }+1$, then the value of $\frac{-12}{\pi }\left({\sin }^{-1}a+{\cos }^{-1}b\right)$ is equal to

• A. 6
• B. 4
• C. 2
• D. 1

Answer 1

Answer: (A)

$f(x)=a\cos (\pi x)+b$
$\therefore f^{\prime }(x)=-a\pi \sin (\pi x)$
So, $f^{\prime }\left(\frac{1}{2}\right)=\pi \Rightarrow -a\pi =\pi \Rightarrow a=-1$
Also, $\underset{\frac{1}{2}}{\overset{\frac{3}{2}}{\int }}f(x)dx=\frac{2}{\pi }+1\Rightarrow \underset{\frac{1}{2}}{\overset{\frac{3}{2}}{\int }}(a\cos (\pi x)+b)dx=\frac{2}{\pi }+1\Rightarrow {\left(\frac{a}{\pi }\sin (\pi x)+bx\right)}_{\frac{1}{2}}^{\frac{3}{2}}=\frac{2}{\pi }+1$
$\Rightarrow \left(\frac{-a}{\pi }+\frac{3b}{2}\right)-\left(\frac{a}{\pi }+\frac{b}{2}\right)=\frac{2}{\pi }+1\Rightarrow \frac{-2a}{\pi }+b=\frac{2}{\pi }+1$
As $a=-1\Rightarrow b=1$.
So, $\frac{-12}{\pi }\left({\sin }^{-1}a+{\cos }^{-1}b\right)=\frac{-12}{\pi }\left({\sin }^{-1}(-1)+{\cos }^{-1}(1)\right)=\frac{-12}{\pi }\left(\frac{-\pi }{2}+0\right)=6$.