Let f(x)=2 x+ ^ -1 x and g(x)= _e (1+x^2+x ), x [0,3]. Then - Vidyadip

MCQ

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

A
There exists $\hat{ x } \in[0,3]$ such that $f ^{\prime}(\hat{ x }) < g ^{\prime}(\hat{ x })$
✓
$\max f(x) > \max g(x)$
C
There exist $0 < x_1 < x_2 < 3$ such that $f(x) < g(x)$, $\forall x \in\left( x _1, x _2\right)$
D
$\min f ^{\prime}( x )=1+\max g ^{\prime}( x )$

✓

Answer

Correct option: B.

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

b
$f ( x )=2 x +\tan ^{-1} x \text { and } g ( x )=\ln \left(\sqrt{1+x^2}+x\right)$

$\text { and } x \in[0,3]$

$g ^{\prime}( x )=\frac{1}{\sqrt{1+x^2}}$

Now, $0 \leq x \leq 3$

$0 \leq x^2 \leq 9$

$1 \leq 1+x^2 \leq 10$

So, $\quad 2+\frac{1}{10} \leq f^{\prime}(x) \leq 3$

$\frac{21}{10} \leq f^{\prime}(x) \leq 3$ and $\frac{1}{\sqrt{10}} \leq g^{\prime}(x) \leq 1$ option $(4)$ is incorrect

From above, $g ^{\prime}( x )< f ^{\prime}( x ) \forall x \in[0,3]$ Option $(1)$ is incorrect. $f ^{\prime}( x ) \& g ^{\prime}( x )$ both positive so $f ( x ) \& g ( x )$ both are increasing

So, $\quad \max \left( f ( x )\right.$ at $x =3$ is $6+\tan ^{-1} 3$

$\operatorname{Max}(g(x)$ at $x=3$ is $\ln (3+\sqrt{10})$

And $6+\tan ^{-1} 3 > \ln (3+\sqrt{10})$

Option $(2)$ is correct

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