Question

Let \( f: \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \( f''(x)>0 \) for all \( x \in \mathbb{R} \) and \( f'(a-1) = 0 \), where \( a \) is a real number. Let \( g(x) = f(\tan^2 x - 2\tan x + a) \), \( 0<x<\frac{\pi}{2} \).
Consider the following two statements :
(I) \( g \) is increasing in \( (0, \frac{\pi}{4}) \)
(II) \( g \) is decreasing in \( (\frac{\pi}{4}, \frac{\pi}{2}) \)
Then,

• A. Only (II) is True
• B. Only (I) is True
• C. Both (I) and (II) are True
• D. Neither (I) nor (II) is True

Hint:

If \( f''(x)>0 \), then \( f'(x) \) is strictly increasing. Use the point where the derivative is zero to establish the sign of \( f' \) in other regions.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To determine whether a function \( g(x) \) is increasing or decreasing, we evaluate the sign of its first derivative \( g'(x) \).
A function \( g(x) \) is increasing if \( g'(x)>0 \) and decreasing if \( g'(x)<0 \).

Step 2: Key Formula or Approach:
Given \( g(x) = f(u) \), where \( u(x) = \tan^2 x - 2\tan x + a \).
Using the chain rule, \( g'(x) = f'(u) \cdot u'(x) \).

Step 3: Detailed Explanation:
First, find \( u'(x) \):
\[ u(x) = (\tan x - 1)^2 + a - 1 \]
\[ u'(x) = 2(\tan x - 1) \sec^2 x \]
Now, consider \( f'(u) \).
Given \( f''(x)>0 \), \( f'(x) \) is a strictly increasing function.
Since \( f'(a-1) = 0 \), and \( u(x) = (\tan x - 1)^2 + a - 1 \ge a - 1 \) for all \( x \), it follows that:
\( f'(u) \ge f'(a-1) \implies f'(u) \ge 0 \).
The derivative of \( g(x) \) is:
\[ g'(x) = f'(u) \cdot 2(\tan x - 1) \sec^2 x \]
Analyzing the interval \( (0, \frac{\pi}{4}) \):
In this range, \( 0<\tan x<1 \), so \( (\tan x - 1)<0 \).
Since \( f'(u) \ge 0 \) and \( \sec^2 x>0 \), we have \( g'(x) \le 0 \).
Thus, \( g \) is decreasing in \( (0, \frac{\pi}{4}) \). (Statement I is False)
Analyzing the interval \( (\frac{\pi}{4}, \frac{\pi}{2}) \):
In this range, \( \tan x>1 \), so \( (\tan x - 1)>0 \).
Thus, \( g'(x) \ge 0 \).
So, \( g \) is increasing in \( (\frac{\pi}{4}, \frac{\pi}{2}) \). (Statement II is False)

Step 4: Final Answer:
Both statements (I) and (II) are false.