Question

If $f\left(x\right)=\underset{1}{\overset{x}{\int }}\frac{3^t}{1+t^2}dt$, where $x>1$, then which of the following is/are correct

• A. For $1<\alpha <\beta ,f\left(\alpha \right)<f\left(\beta \right)$
• B. For $1<\alpha <\beta ,f\left(\alpha \right)>f\left(\beta \right)$
• C. $f\left(x\right)+\frac{\pi }{4}<{\tan }^{-1}x,\forall x\ge 1$
• D. $f\left(x\right)+\frac{\pi }{4}>{\tan }^{-1}x,\forall x>1$

Answer 1

Answer: (A), (D)

$f\left(x\right)+\frac{\pi }{4}>{\tan }^{-1}x,\forall x>1$

Given: $f\left(x\right)=\underset{1}{\overset{x}{\int }}\frac{3^t}{1+t^2}dt$
$\Rightarrow f^{\prime }\left(x\right)=\frac{3^x}{1+x^2}>0\forall x>1$
$\therefore f\left(x\right)$ is increasing.
Hence for $1<\alpha <\beta ,f\left(\alpha \right)<f\left(\beta \right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{3^x}{1+x^2}>\frac{1}{1+x^2}\forall x>1$
$\Rightarrow \underset{1}{\overset{x}{\int }}{f}^{\prime }\left(x\right)dx>\underset{1}{\overset{x}{\int }}\frac{1}{1+x^2}dx$
$\Rightarrow f\left(x\right)>{\tan }^{-1}x-{\tan }^{-1}1[\because f(1)=0]\Rightarrow f\left(x\right)+\frac{\pi }{4}>{\tan }^{-1}x$