Question

Let $f(x)=\frac{4x^2-3x+1}{\underset{0}{\overset{2\pi x}{\int }}{\sin }^{4}tdt}$. If $f^{\prime }\left(\frac{1}{2}\right)=\frac{p}{q\pi },p,q\in N$, then $(p-q)$ equal

• A. 6
• B. 5
• C. 4
• D. 3

Answer 1

Answer: (B)

$f(x)=\frac{4x^2-4x+1+x}{\underset{0}{\overset{2\pi x}{\int }}{\sin }^{2}tdt}=\frac{(2x-1{)}^2}{\underset{0}{\overset{2\pi x}{\int }}{\sin }^{4}tdt}+\frac{x}{\underset{0}{\overset{2\pi x}{\int }}{\sin }^{4}tdt}$
$f^{\prime }(x)=g^{\prime }(x)+\frac{\underset{0}{\overset{2\pi x}{\int }}{\sin }^{4}tdt\cdot 1-x{\sin }^4(2\pi x)\cdot 2\pi }{{\left(\underset{0}{\overset{2\pi x}{\int }}{\sin }^{4}tdt\right)}^2}$
$f^{\prime }\left(\frac{1}{2}\right)=g\cdot \left(\frac{1}{2}\right)+\frac{1}{\underset{0}{\overset{\pi }{\int }}{\sin }^{4}tdt}=0+\frac{1}{2\cdot \frac{3\pi }{16}}=\frac{8}{3\pi }\Rightarrow p-q=5$