Let $f : \mathbb{R} ightarrow \mathbb{R}$ 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( an^2 x - 2 an x + a),\ 0

Question

Let $f : \mathbb{R} ightarrow \mathbb{R}$ 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 $g(x) = f(	an^2 x - 2 	an x + a),\ 0 < x < \frac{\pi}{2}$.

Consider the following two statements:

(I) g is increasing in $\left(0, \frac{\pi}{4}ight)$

(II) g is decreasing in $\left(\frac{\pi}{4}, \frac{\pi}{2}ight)$

Then,

Accepted Answer

B. 1. **Why**: Complete the square inside $f$ so the argument is centred at $a-1$, matching the given condition $f'(a-1)=0$. $$ g(x)=f\big(( an x-1)^2+(a-1)\big) $$ 2. **Why**: Differentiate using the chain rule. $$ g'(x)=f'\big(( an x-1)^2+a-1\big)\cdot 2( an x-1)\sec^2x $$ 3. **Why**: Since $f''(x)>0$ everywhere, $f'$ is a **strictly increasing** function, so we can compare $f'$ at different points just by comparing their arguments. 4. **Why**: On $0f'(a-1)=0 $$ Also $( an x-1)<0$ here, so $$ g'(x)=(+)(-)(+)<0 $$ So $g$ is **decreasing** on $\left(0,\frac{\pi}{4} ight)$ — statement **(I)** is **False**. 5. **Why**: A similar argument on $\left(\frac{\pi}{4},\frac{\pi}{2} ight)$, where $ an x>1$, flips the sign of $( an x-1)$ to positive while $f'(\cdot)>0$ still holds, giving $g'(x)>0$. So $g$ is **increasing** there — statement **(II)** is **False**. Hence, neither (I) nor (II) is true — the answer is **Option B**.