JEE Main 2015 (Online) 11th AprilQuestion 1
Let $k$ and $K$ be the minimum and the maximum values of the function $f(x)=\frac{(1+x)^{0.6}}{1+x^{0.6}}$ in $[0,1]$, respectively, then the ordered pair $(k,K)$ is equal to :
Differentiation and its Application
Solution
Concept. Show $f'\le0$ on $[0,1]$; then the extremes are at the end points. Formulas used. (1) Quotient rule: $\left(\frac uv\right)'=\frac{u'v-uv'}{v^2}$Step 1 - Derivative. By (1), $$f'(x)=\frac{0.6(1+x)^{-0.4}\left(1+x^{0.6}\right)-0.6(1+x)^{0.6}x^{-0.4}}{(1+x^{0.6})^2}.$$ Take out $0.6(1+x)^{-0.4}$. Since $(1+x)x^{-0.4}=x^{-0.4}+x^{0.6}$, the rest is $1-x^{-0.4}$: $$f'(x)=\frac{0.6(1+x)^{-0.4}\left(1-x^{-0.4}\right)}{(1+x^{0.6})^2}.$$ Step 2 - Sign. For $0<x\le1$, $x^{-0.4}\ge1$, so $f'\le0$: $f$ decreases on $[0,1]$. Step 3 - Values. $K=f(0)=1$ and $k=f(1)=\frac{2^{0.6}}{2}=2^{-0.4}$. Answer: (A)