Question

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

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

Hint:

Show \(f'(x)\le0\) on \((0,1]\), so f is decreasing and the extremes are at the end points.

Answer 1

The Correct Option is D

\(f'(x)=\dfrac{0.6(1+x)^{-0.4}(1+x^{0.6})-0.6(1+x)^{0.6}x^{-0.4}}{(1+x^{0.6})^2}\).
The numerator is \(0.6(1+x)^{-0.4}\left[1+x^{0.6}-(1+x)x^{-0.4}\right]=0.6(1+x)^{-0.4}\left[1-x^{-0.4}\right]\), which is \(\le0\) for \(0\lt x\le1\) since \(x^{-0.4}\ge1\).
So f is decreasing on [0,1]: maximum \(K=f(0)=1\), minimum \(k=f(1)=\frac{2^{0.6}}{2}=2^{-0.4}\).
Answer: (4) \((2^{-0.4}, 1)\)