Question

Let 
\(\begin{array}{l} f\left(x\right)=3^{\left(x^2-2\right)^3+4},x\in \mathbb{R}.\end{array}\)
Then which of the following statements are true? 
P : x = 0 is a point of local minima of f
Q : x = √2 is a point of inflection of f 
R : f ′ is increasing for x > √2

• A. Only P and Q
• B. Only P and R
• C. Only Q and R
• D. All P, Q and R

Answer 1

The Correct Option is D

Let's analyze each statement to determine their truth value concerning the given function \( f(x) = 3^{(x^2-2)^3 + 4} \).

Step-by-Step Analysis:

1. Calculate the first derivative \( f'(x) \):
   The function is of the form \( f(x) = 3^g(x) \), where \( g(x) = (x^2-2)^3 + 4 \). Using the chain rule, the derivative of \( f(x) \) is given by: \(f'(x) = \ln(3) \cdot 3^{g(x)} \cdot g'(x)\)
   Here, \(g(x) = (x^2-2)^3 + 4\) and \(g'(x) = 3(x^2-2)^2 \cdot 2x = 6x(x^2-2)^2\).
2. Checking statement P (Local Minima at \(x=0\)):
   At \(x=0\), \(g(x) = (-2)^3 + 4 = -8 + 4 = -4\), thus, \(f(x) = 3^{-4}\).
   The first derivative is \( f'(0) = \ln(3) \cdot 3^{-4} \cdot 6 \cdot 0 \cdot (-2)^2 = 0 \), indicating a critical point.
   To determine the nature of this critical point, calculate \( f''(x) \). At \( x = 0 \), \( f''(0) > 0 \), which confirms a local minima, hence statement P is true.
3. Checking statement Q (Point of Inflection at \(x=\sqrt{2}\)):
   At this point, \( g(x) = 0^3 + 4 = 4 \), leading to \(f(x) = 3^4\).
   Here, \( g'(x) = 0 \), hence \( f'(x) = 0 \). Now, check \( f''(x) \). Here, \( f''(\sqrt{2}) = 0 \) and \( f'''\) changes sign, indicating a point of inflection. Thus, statement Q is true.
4. Checking statement R (\( f' \) is increasing for \( x > \sqrt{2} \)):
   Compute \( f''(x) \) and simplify. The sign of \( f''(x) \) indicates the nature of \( f'(x) \). For \( x > \sqrt{2} \), \( g''(x) > 0 \) making \( f'(x) \) an increasing function. Hence, statement R is true.

Therefore, after evaluating each statement, we conclude that all statements P, Q, and R are correct.

Conclusion:

The correct answer is: All P, Q and R.