Question

Let $k$ and $K$ be the minimum and the maximum values of the function $f \left(x\right)=\frac{\left(1+x\right)^{0.6}}{1+x^{0.6}}$ in $[0, 1]$ respectively, then the ordered pair $(k, K)$ is equal to :

• A. $\left(1, 2^{0.6}\right)$
• B. $\left(2^{-0.4}, 2^{0.6}\right)$
• C. $\left(2^{-0.6}, 1\right)$
• D. $\left(2^{-0.4}, 1\right)$

Answer 1

Answer: (D)

$f(x)=\frac{(1+x)^{0.6}}{1+x^{0.6}}, x \in[0,1] ;(k, k)=?$; where $k=f(x)$ min and $k=f(x)$ max.
$\Rightarrow f(x)=\frac{(1+x)^{3 / 5}}{\left(1+x^{3 / 5}\right)}$
$\Rightarrow f'(x)=\frac{\left(1+x^{3 / 5}\right) \cdot \frac{3}{5}(1-x)^{-2 / 5}-(1+x)^{3 / 5} \cdot \frac{3}{5} x^{-2 / 5}}{\left(1+x^{3 / 5}\right)^{2}}$
$\Rightarrow f'(x)=\frac{3}{5}\left[\frac{\left(1+x^{3 / 5}\right) x^{2 / 5}-(1+x)}{x^{2 / 5}(1+x)^{2 / 5}\left(1+x^{3 / 3}\right)^{2}}\right]$
$\Rightarrow f'(x)=\frac{3}{5}\left[\frac{x^{2 / 5}+x-1-x}{x^{2 / 5}(1+x)^{2 / 5}\left(1+x^{3 / 3}\right)^{2}}\right]$
$\Rightarrow f'(x)=0$ for $x=1,-1 \therefore f(-1)=0, f(1)=\frac{(2)^{3 / 5}}{2}$
$\Rightarrow f'(x)=0$ for $x=1,-1$
$\therefore f(-1)=0, f(1)=\frac{(2)^{3 / 5}}{2}$
Also $f(0)=1 \Rightarrow \operatorname{In}[0,1],$
$f(x)$ has minimum value $=f(1)=(2)^{-2 / 5}=(2)^{0.4}=k$
and maximum value $=f(0)=1=k$
$\therefore (k, k)=(2^{-04}, 1)$