Question

Let $f(x)=\underset{0}{\overset{x}{\int }}\left(5\ln \left(1+y^2\right)-10y{\tan }^{-1}y+16\sin y\right)dy$, then which of the following hold(s) good?

• A. $f^{\prime }(x)$ is increasing for all $x\in (0,1)$.
• B. $f(x)$ is positive for all $x\in (0,1)$.
• C. $f(x)$ is negative for all $x\in (0,1)$.
• D. $\underset{0}{\overset{x}{\int }}f(y)dy$ is increasing for all $x\in (0,1)$.

Answer 1

Answer: (A), (C), (D)

$f^{\prime }(x)=5\ln \left(1+x^2\right)-10x{\tan }^{-1}x+16\sin x$
$f^{\prime \prime }(x)=2\left(8\cos x-5{\tan }^{-1}x\right)$
$f^{\prime \prime \prime }(x)=-2\left(8\sin x+\frac{5}{1+x^2}\right)<0\forall x\in (0,1)$
So, $f^{\prime \prime }(x)$ is decreasing function $\forall x\in (0,1)$
$\therefore f^{\prime \prime }(x)>f^{\prime \prime }(1)\Rightarrow f^{\prime \prime }(x)>0$
$\Rightarrow f^{\prime }(x)$ is increasing function for $x>0$
$f^{\prime }(x)>f^{\prime }(0)$
$f^{\prime }(x)>0.$
$\Rightarrow f(x)$ is increasing function for $x>0$
Hence, $f(x)>f(0)$
$\Rightarrow f(x)>0$
$\therefore \underset{0}{\overset{x}{\int }}f(y)dy\text{ is increasing for all }x\in (0,1)$