Question

Let $f\left(x\right)={\int }_1^x\frac{3^t}{1+t^2}dt,$ where x > 0. Then

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

Answer 1

Answer: (A), (D)

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

$f\left(x\right)={\int }_1^x\frac{3^t}{1+t^2}dt$
Let $I=\frac{3^x}{1+x^2}>\forall x>0\Rightarrow I=\frac{3^x}{1+x^2}>\frac{1}{1+x^2}\forall x>0$
Gives
$f\left(x\right)={\int }_1^{x}Idx>{\int }_1^x\frac{1}{1+x^2}dx={\left[{tan}^{-1}x\right]}_1^x\Rightarrow f\left(x\right)>{tan}^{-1}x-{tan}^{-1}1\Rightarrow f\left(x\right)+\frac{\pi }{4}>{tan}^{-1}x$
Hence, options 'A' and 'D' are correct.