Question

Let $f\left(x\right)={\int }_0^x\frac{3^t}{1+t^2}dt,$ where $x>0,$ then

• A. for $0<\alpha <\beta ,f\left(\alpha \right)<f\left(\beta \right)$
• B. for $0<\alpha <\beta ,f\left(\alpha \right)>f\left(\beta \right)$
• C. $f\left(x\right)+\frac{\pi }{4}<ta{n}^{-1}x,\forall x\ge 1$
• D. $f\left(x\right)+\frac{\pi }{4}>ta{n}^{-1}x,\forall x\ge 1$

Answer 1

Answer: (A), (D)

The correct options are
A for $0<\alpha <\beta ,f\left(\alpha \right)<f\left(\beta \right)$
D $f\left(x\right)+\frac{\pi }{4}>ta{n}^{-1}x,\forall x\ge 1$

$f^{\prime }\left(x\right)=\frac{3^x}{1+x^2}>0\forall x>0\Rightarrow f^{\prime }\left(x\right)=\frac{3^x}{1+x^2}>\frac{1}{1+x^2}\forall x\ge 1$
$\Rightarrow {\int }_0^x{f}^{\prime }\left(x\right)dx>{\int }_1^x\frac{1}{1+x^2}dx$
$\Rightarrow f\left(x\right)>ta{n}^{-1}x-ta{n}^{-1}1\Rightarrow f\left(x\right)+\frac{\pi }{4}>ta{n}^{-1}x$