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.
$\Rightarrow f(x)=\frac{(1+x{)}^{3/5}}{\left(1+x^{3/5}\right)}$
$\Rightarrow f^{\prime }(x)=\frac{\left(1+x^{3/5}\right)\cdot \frac{3}{5}(1-x{)}^{-2/5}-(1+x{)}^{3/5}\cdot \frac{3}{5}{x}^{-2/5}}{{\left(1+x^{3/5}\right)}^2}$
$\Rightarrow f^{\prime }(x)=\frac{3}{5}\left[\frac{\left(1+x^{3/5}\right)x^{2/5}-(1+x)}{x^{2/5}(1+x{)}^{2/5}{\left(1+x^{3/3}\right)}^2}\right]$
$\Rightarrow f^{\prime }(x)=\frac{3}{5}\left[\frac{x^{2/5}+x-1-x}{x^{2/5}(1+x{)}^{2/5}{\left(1+x^{3/3}\right)}^2}\right]$
$\Rightarrow f^{\prime }(x)=0$ for $x=1,-1\therefore f(-1)=0,f(1)=\frac{(2{)}^{3/5}}{2}$
$\Rightarrow f^{\prime }(x)=0$ for $x=1,-1$
$\therefore f(-1)=0,f(1)=\frac{(2{)}^{3/5}}{2}$
Also $f(0)=1\Rightarrow \mathrm{In}[0,1],$
$f(x)$ has minimum value $=f(1)=(2{)}^{-2/5}=(2{)}^{0.4}=k$
and maximum value $=f(0)=1=k$
$\therefore (k,k)=(2^{-04},1)$