Let k and K be the minimum and the maximum values of the function… - Vidyadip

MCQ

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

✓
$(2^{-0·4}, 1)$
B
$(2^{-0.4}, 2^{0.6})$
C
$(2^{-0·6}, 1)$
D
$( 1, 2^{0.6})$

✓

Answer

Correct option: A.

$(2^{-0·4}, 1)$

a
${\rm{ Let }}f(x) = \frac{{{{(1 + x)}^{\frac{3}{5}}}}}{{1 + {x^{\frac{3}{5}}}}}{\rm{\,\, and }}\,\,x \in [0,1]$

$ \Rightarrow f'(x) = \frac{{\left( {1 + {x^{\frac{3}{5}}}} \right)\frac{3}{5}{{(1 + x)}^{ - \frac{2}{5}}} - \frac{3}{5}{{(1 + x)}^{\frac{3}{5}}}\left( {{x^{\frac{{ - 2}}{5}}}} \right)}}{{{{\left( {1 + {x^{\frac{3}{5}}}} \right)}^2}}}$

$ = \frac{3}{5}\left. {\left[ {\left( {1 + {x^{\frac{3}{5}}}} \right){{\left( {1 + x} \right)}^{ - \frac{2}{5}}}} \right. - {{\left( {1 + x} \right)}^{\frac{3}{5}}}{x^{\frac{{ - 2}}{5}}}} \right]$

$ = \frac{3}{5}\left[ {\frac{{1 + {x^{\frac{3}{5}}}}}{{{{\left( {1 + x} \right)}^{\frac{2}{5}}}}} - \frac{{{{(1 + x)}^{\frac{3}{5}}}}}{{{x^{\frac{2}{5}}}}}} \right]$

$ = \frac{{{x^{\frac{2}{5}}} + x - x}}{{{x^{\frac{2}{5}}}{{\left( {1 + x} \right)}^{\frac{2}{5}}}}} = \frac{{{x^{\frac{2}{5}}} - 1}}{{{x^{\frac{2}{5}}}{{\left( {1 + x} \right)}^{\frac{2}{5}}}}} < 0$

Alos, $f(0) = 1$

$ \Rightarrow f\left( x \right) \in \left[ {{2^{ - 04}},1} \right]$

$f(1) = {2^{ - 0,4}}$

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