Question

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

Answer 1

The Correct Option is C

The correct answer is (C) : \(\max f(x)>\max g(x)\)
\(f'(x)=2+\frac{1}{1+x^2},\ g'(x)=\frac{1}{\sqrt{1+x^2}}\)
Both does not have critical values
\(f(0)=0,f(3)=6+\tan^{-1}(3)\)
\(g(0)=0,g(3)=\log(3+\sqrt{10})\)
Let h(x) = f(x) - g(x)
\(h'(x) > 0∀x∈(0,3)\)
∴ h(x) is increasing function

Answer 2

Computing derivatives: \[ g'(x) = \frac{1}{\sqrt{1 + x^2} + x} \] Since \( g(x) \) is always increasing in \( [0,3] \), \[ \min g'(x) = \frac{1}{\sqrt{10} + 3}, \quad \max g'(x) = \frac{1}{\sqrt{1} + 0} = 1 \] For \( f(x) \), \[ f(x) = 2x + \tan^{-1} x \] Since \( f(x) \) is increasing, \[ \max f(x) = 6 + \tan^{-1} 3 \] For \( g(x) \), \[ \max g(x) = \ln (3 + \sqrt{10}) \] \[ 6 + \tan^{-1} 3 > \ln (3 + \sqrt{10}) \]