Let f(x)=2 x+\tan ^{-1} x and g(x)=\log _e (\sqrt{1+x^2}+x), {\x} \in[0,3]. Then [1-Feb-2023 Shift 1]

Correct Answer is : maxf(x)>maxg(x)

There exists {\x} \in[0,3] such that {\f}^{'}( {\x})< {\g}^{'}( {\x})

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

There exist 0 < {\x}_1 < {\x}_2 < 3 such that {\f}( {\x}) < {\g} (x), \forall {\x} \in ( {\x}_1, {\x}_2)

\min {\f}^{'}( {\x})=1+\max {\g}^{'}( {\x})

JEE Mains 1-Feb-2023 Shift 1 Solved Paper

Section: Mathematics

Question : 78 of 90

Marks: +1, -0

Let f(x)=2x+tan−1x and g(x)=loge(√1+x2+x), x∈[0,3]. Then

[1-Feb-2023 Shift 1]

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