Question:medium

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

Updated On: Apr 1, 2026
  • Only P and Q
  • Only P and R
  • Only Q and R
  • All P, Q and R
Show Solution

The Correct Option is D

Solution and Explanation

To determine which of the statements P, Q, and R are true, we need to analyze the function \( f(x) = 3^{(x^2 - 2)^3 + 4} \) and evaluate each statement step-by-step.

Step 1: Analyze Statement P

Statement P claims that \( x = 0 \) is a point of local minima of \( f \).

To find the point of local minima, we should find the first derivative \( f'(x) \) and check the sign changes around \( x = 0 \).

The derivative of \( f(x) \) is given by:

f'(x) = 3^{g(x)} \ln(3) \cdot g'(x), where \( g(x) = (x^2 - 2)^3 + 4 \).

\begin{align*} g'(x) &= 3(x^2 - 2)^2 \cdot 2x = 6x(x^2 - 2)^2. \end{align*}

At \( x = 0 \), \( g'(0) = 6 \cdot 0 \cdot (-2)^2 = 0 \), but to find if it is a minimum, check the second derivative \( f''(x) \):

f''(x) \propto g''(x) = 6(2(x^2 - 2) \cdot 2x + (x^2 - 2)^2 \cdot 2) = 12(x^2 - 2)(x) + 12x(x^2 - 2)^2.

Since at \( x = 0 \), \( g''(x) = 0 \), \( f''(0) \gt 0 \) indicates it is indeed a local minimum.

Step 2: Analyze Statement Q

Statement Q claims that \( x = \sqrt{2} \) is a point of inflection of \( f \).

A point of inflection occurs where the second derivative changes sign, so we calculate \( f''(x) \) and see if there’s a sign change at \( x = \sqrt{2} \).

g'(\sqrt{2}) = 6\sqrt{2}(0) = 0 and find \( g''(\sqrt{2}) \).

Now consider the pattern \( g''(x) \) for \( x \approx \sqrt{2} \). There is indeed a change in concavity, confirming it is a point of inflection.

Step 3: Analyze Statement R

Statement R claims \( f' \) is increasing for \( x \gt \sqrt{2} \).

For this, we need to ensure that the third derivative \( f'''(x) \) is positive for \( x \gt \sqrt{2} \).

Given \( g'(x) = 6x(x^2 - 2)^2 \), for \( x \gt \sqrt{2} \), since \( x^2 \gt 2 \), we conclude \( g'(x) \) is always increasing.

Thus, all conditions in P, Q, and R are satisfied.

The correct option is All P, Q, and R.

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