JEE MATHEMATICS MCQ: MAXIMA & MINIMA

1 Hard

1 of 1

Analyzing +4 | -1

The maximum value of

f\left( x \right) = \frac{x}{{4 + x + {x^2}}}

on [-1,1] is

View answer and explanation

Correct option: C

\frac{1}{6}

Option C (

\frac{1}{6}

) is correct because it represents the maximum value of the function on the given interval, found by evaluating the function at critical points and endpoints.

More Questions from MAXIMA & MINIMA

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Let

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

. Let

m(b)

be the minimum value of

f(x)

(that is least value). As

b

varies, the range of

m(b)

is Let

f: R \rightarrow R

be a function defined by

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

. Then, which of the following is NOT true? The equation

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

possesses Let the function

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

-6 have a maxima for some value of

x<0

and a minima for some value of

x>0

. Then, the set of all values of

p

is The least value of the

f\left( x \right)

given by

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

in the interval

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

, is If the total maximum value of the function

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}

, then

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

is equal to