Question

If the area bounded by $f\left(x\right)={\left(tan\right)}^{3}x+tanx$ from $x=0$ to $x=\frac{\pi }{4}$ is $k$ square units, then the maximum value of $g\left(x\right)=ksinx$ is $\left(\forall x\in \left[0,\frac{\pi }{4}\right]\right)$

• A. $2$
• B. $\frac{1}{2\sqrt{2}}$
• C. $4$
• D. $8$

Answer 1

Answer: (B)

The required area is
$A={\int }_0^{\frac{\pi }{4}}\left({\left(tan\right)}^{3}x+tanx\right)dx$
$={\int }_0^{\frac{\pi }{4}}tanx\left({\left(tan\right)}^{2}x+1\right)dx$
$={\int }_0^{\frac{\pi }{4}}tanx\cdot se{c}^{2}xdx$
Let $tanx=t$
$\Rightarrow se{c}^{2}xdx=dt$
$\Rightarrow I={\int }_0^{1}tdt={\left[\frac{t^2}{2}\right]}_0^1$
$=\frac{1}{2}$
$\therefore g\left(x\right)=\frac{1}{2}sinx\le \frac{1}{2\cdot \sqrt{2}}$