Question

If the area bounded by $f\left(x\right)=\left(tan\right)^{3} x+tan ⁡ x$ from $x=0$ to $x=\frac{\pi }{4}$ is $k$ square units, then the maximum value of $g\left(x\right)=ksin x$ 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
$A=\displaystyle \int _{0}^{\frac{\pi }{4}} \left(\left(tan\right)^{3} x + tan ⁡ x\right)dx$
$=\displaystyle \int _{0}^{\frac{\pi }{4}} tan x \left(\left(tan\right)^{2} ⁡ x + 1\right)dx$
$= \displaystyle \int _{0}^{\frac{\pi }{4}} tan x \cdot sec^{2} x d x$
Let $tan x=t$
$\Rightarrow sec^{2} xdx=dt$
$\Rightarrow I=\displaystyle \int _{0}^{1} tdt=\left[\frac{t^{2}}{2}\right]_{0}^{1}$
$=\frac{1}{2}$
$\therefore g \left(x\right) = \frac{1}{2} sin x \leq \frac{1}{2 \cdot \sqrt{2}}$