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

To solve this mathematics problem, we need to analyze the given function: \(f(x)=3^{(x^2-2)^3 + 4}\), where \(x \in \mathbb{R}\), and examine the given statements P, Q, and R. 

Statement P: x = 0 is a point of local minima of f

1. To determine if \(x = 0\) is a local minima, we need to find the first derivative of \(f(x)\): \(f'(x)\).
2. Let \(g(x) = (x^2 - 2)^3 + 4\). Then \(f(x) = 3^{g(x)}\).
3. Derivative of \(f(x)\) is \(f'(x) = \ln(3) \cdot 3^{g(x)} \cdot g'(x)\).
4. Find \(g'(x)\): \(g'(x) = 3((x^2 - 2)^2) \cdot 2x = 6x(x^2 - 2)^2\).
5. \(f'(x) = \ln(3) \cdot 3^{((x^2-2)^3 + 4)} \cdot 6x(x^2 - 2)^2\).
6. At \(x = 0\), \((x^2 - 2)^2 = 4\), thus \(f'(0) = \ln(3) \cdot 3^{4} \cdot 0 = 0\).
7. To verify if it is minima, check second derivative \(f''(x)\).
8. Evaluating second derivative at \(x = 0\), we find that the derivative test suggests a local minimum since \(f''(0) > 0\).

Statement Q: x = √2 is a point of inflection of f

1. A point of inflection occurs where the second derivative changes sign.
2. Under the form of \(g(x)\), the second derivative according to \(g'(x) = 6x(x^2 - 2)^2\) changes sign at \(x=\sqrt{2}\) as \((x^2-2)^2\) reaches zero.
3. Therefore, \(x = \sqrt{2}\) indeed is a point of inflection.

Statement R: f′ is increasing for x > √2

1. For this, analyze \(f'(x)\).
2. After \(x = \sqrt{2}\), \(f'(x)\) retains its positive increase due to the linear behavior of the multiplicative terms.
3. Therefore, \(f'\) is increasing for \(x > \sqrt{2}\).

Conclusion: All three statements P, Q, and R are true, thereby confirming the correct answer is: All P, Q, and R.

Answer 2

\(\begin{array}{l} f\left(x\right)=3^{\left(x^2-2\right)^3+4}, x\in R \end{array}\)
\(\begin{array}{l} f\left(x\right)=81.3^{\left(x^2-2\right)^3} \end{array}\)
\(\begin{array}{l} f’\left(x\right)=81.3^{\left(x^2-2\right)^3}\text{ln}2.3\left(x^2-2\right)2x\end{array}\)
\(\begin{array}{l} =\left(486\ \text{ln}2\right)\left(3^{\left(x^2-2\right)^3}\left(x^2-2\right)x\right)\end{array}\)

⇒ x = 0 is the local minima.
f′′(x) = (486 ln2)
\(\begin{array}{l} \begin{pmatrix}3^{\left(x^2-2\right)^3}\cdot\left(x^2-2\right) \\\left(5x^2-2+6x^2\text{ln}3\left(x^2-2\right)\right)\end{pmatrix}\end{array}\)
\(\begin{array}{l} f”\left(x\right)=0~~~~~x=\sqrt{2}\end{array}\)
\(\begin{array}{l} f”\left(\sqrt{2}^+\right)>0\end{array}\)
\(\begin{array}{l} \Rightarrow x=\sqrt{2}\end{array}\)
is point of inflection
\(\begin{array}{l} f”\left(x\right)>0~\forall~x>\sqrt{2}\end{array}\)
⇒ f′(x) is increasing for \(x > \sqrt2\)