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

To solve this problem, we need to compare the maximum values of the functions \( f(x) \) and \( g(x) \) over the interval \( x \in [0, 3] \).

First, let's analyze the function \( f(x) = 2x + \tan^{-1}x \).

1. The derivative of \( f(x) \) is given by: \(f'(x) = 2 + \frac{1}{1 + x^2}\)
2. This derivative is always positive for \( x \geq 0 \), which means \( f(x) \) is an increasing function on the given interval.
3. Thus, the maximum value of \( f(x) \) on \([0, 3]\) occurs at \( x = 3 \).
4. Substituting \( x = 3 \) into \( f(x) \), we get: \(f(3) = 2 \times 3 + \tan^{-1}(3)\) \(= 6 + \tan^{-1}(3)\)

Now, let's examine the function \( g(x) = \log_e(\sqrt{1+x^2} + x) \).

1. The derivative of \( g(x) \) can be calculated using the chain rule: \(g'(x) = \frac{d}{dx} \left( \log_e(\sqrt{1+x^2} + x) \right)\) \(= \frac{1}{\sqrt{1+x^2} + x} \cdot \left(\frac{x}{\sqrt{1+x^2}} + 1\right)\)
2. Simplifying gives: \(g'(x) = \frac{x + \sqrt{1+x^2}}{(\sqrt{1+x^2}+x)\sqrt{1+x^2}}\) \(= \frac{1}{\sqrt{1+x^2}}\)
3. This function is also positive and decreasing, reaching its maximum at \( x = 0 \).
4. Therefore, at \( x = 0 \): \(g(0) = \log_e(\sqrt{1+0} + 0) = \log_e(1) = 0\)

Comparing the maximum values:

• The maximum of \( f(x) \) is \(6 + \tan^{-1}(3)\).
• The maximum of \( g(x) \) is 0.

Thus, \(\max f(x) > \max g(x)\), i.e., \(6 + \tan^{-1}(3) > 0\), which is true.

Therefore, the correct option is: \( \max f(x) > \max g(x) \).