A) 6
B) 4
C) 2
D) 1
Correct Answer: A
Solution :
\[f(x)=a\cos (\pi x)\,+b\] \[\therefore \,\,f'(x)\,=-a\pi \sin (\pi x)\] So, \[f'\left( \frac{1}{2} \right)\,=\pi \Rightarrow \,-a\pi =\pi \] \[\Rightarrow \,a=-1\] Also, \[\int\limits_{\frac{1}{2}}^{\frac{3}{2}}{\,f(x)dx=\frac{2}{\pi }+1\Rightarrow \,\int\limits_{\frac{1}{2}}^{\frac{3}{2}}{(a\cos (\pi x)\,+b)\,dx}}\] \[\Rightarrow \,\left( \frac{a}{\pi }\sin \,(\pi x)\,+bx \right)_{\frac{1}{2}}^{\frac{3}{2}}\,=\frac{2}{\pi }+1\] \[\Rightarrow \left( \frac{-a}{\pi }+\frac{3b}{2} \right)\,-\left( \frac{a}{\pi }+\frac{b}{2} \right)\,=\frac{2}{\pi }+1\] \[\Rightarrow \,\frac{-2a}{\pi }+b=\frac{2}{\pi }+1\] As \[a=-1\Rightarrow \,b=1\] So, \[\frac{-12}{\pi }({{\sin }^{-1}}a+{{\cos }^{-1}}b)=\frac{-12}{\pi }\,({{\sin }^{-1}}(-1)\,+{{\cos }^{-1}}(1))\]\[=\frac{-12}{\pi }\left( \frac{-\pi }{2}+0 \right)=6\].
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