Question

If $f(x)=a \cos (\pi x)+b, f^{\prime}\left(\frac{1}{2}\right)=\pi$ and $\int\limits_{\frac{1}{2}}^{\frac{3}{2}} f(x) d x=\frac{2}{\pi}+1$, 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, $\int\limits_{\frac{1}{2}}^{\frac{3}{2}} f(x) d x=\frac{2}{\pi}+1 \Rightarrow \int\limits_{\frac{1}{2}}^{\frac{3}{2}}(a \cos (\pi x)+b) d x=\frac{2}{\pi}+1 \Rightarrow\left(\frac{a}{\pi} \sin (\pi x)+b x\right)_{\frac{1}{2}}^{\frac{3}{2}}=\frac{2}{\pi}+1$
$\Rightarrow \left(\frac{- a }{\pi}+\frac{3 b }{2}\right)-\left(\frac{ a }{\pi}+\frac{ b }{2}\right)=\frac{2}{\pi}+1 \Rightarrow \frac{-2 a }{\pi}+ b =\frac{2}{\pi}+1$
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$.