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

$\left(2^{-0.4}, 2^{0.6}\right)$

JEE MATHEMATICS MCQ: MAXIMA & MINIMA

1 Easy

1 of 1

Analyzing +4 | -1

Let

\mathrm{k}

and

\mathrm{K}

be the minimum and the maximum values of the function

\mathrm{f}(x)=\frac{(1+x)^{0.6}}{1+x^{0.6}}

[0,1]

respectively, then the ordered pair

(\mathrm{k}, \mathrm{K})

is equal to :

View answer and explanation

Correct option: A

\left(2^{-0.4}, 1\right)

Option A (

\left(2^{-0.4}, 1\right)

) is correct. The function is decreasing in the interval [0,1], so the maximum value is at x=0 and the minimum value is at x=1.

Practice All Chapter MCQs →

f(x)=\left(1+b^2\right) x^2+2 b x+1

m(b)

f(x)

b

m(b)

f: R \rightarrow R

f(x)=(x-3)^{n_1}(x-5)^{n_2}, n_1, n_2 \in N

|2 a x-3|+|a x+1|+|5-a x|=\frac{1}{2}

f(x)=2 x^{3}+(2 p-7) x^{2}+3(2 p-9) x

x<0

x>0

p

f\left( x \right)

f\left( x \right) = {\tan ^{ - 1}}x - \frac{1}{2}{\log _e}x

\left[ {1/\sqrt 3 ,\;\sqrt 3 } \right]

f(x)=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^{2} x}, x \in\left(0, \frac{\pi}{2}\right),

\frac{k}{e}

\left(\frac{k}{e}\right)^{8}+\frac{k^{8}}{e^{5}}+k^{8}