JEE MATHEMATICS MCQ: MAXIMA & MINIMA

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Analyzing +4 | -1

Let

f(x)=2 x+\tan ^{-1} x

and

g(x)=\log _{e}\left(\sqrt{1+x^{2}}+x\right), x \in[0,3]

. Then

View answer and explanation

Correct option: D

\max f(x) > \max g(x)

Option D (

\max f(x) > \max g(x)

) is correct because the maximum value of f(x) is greater than the maximum value of g(x) in the given interval.

More Questions from MAXIMA & MINIMA

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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}