Question:
The function \(f(x) = \frac{x^2}{e^x}\) is monotonically increasing if:
फलन \(f(x) = \frac{x^2}{e^x}\) एकदिष्ट वर्धमान है यदि:
Options:
- \(x < 0\) only
- \(x > 2\) only
- \(0 < x < 2\)
- \(x \in (-\infty, 0) \cup (2, \infty)\)
- केवल \(x < 0\)
- केवल \(x > 2\)
- \(0 < x < 2\)
- \(x \in (-\infty, 0) \cup (2, \infty)\)
Correct Answer: c)
Explanation:
A function is monotonically increasing where \(f'(x) > 0\).
Differentiating \(f(x) = x^2 e^{-x}\):
\[f'(x) = 2x e^{-x} – x^2 e^{-x} = x(2 – x)e^{-x}\]
Since \(e^{-x} > 0\) for all real \(x\):
\[f'(x) > 0 \iff x(2 – x) > 0 \iff x(x – 2) < 0 \iff 0 < x < 2\]
Hence, the correct option is (c).
फलन एकदिष्ट वर्धमान होता है जहाँ \(f'(x) > 0\) हो।
\(f(x) = x^2 e^{-x}\) का अवकलन करने पर:
\[f'(x) = 2x e^{-x} – x^2 e^{-x} = x(2 – x)e^{-x}\]
चूँकि सभी वास्तविक \(x\) के लिए \(e^{-x} > 0\) होता है:
\[f'(x) > 0 \iff x(2 – x) > 0 \iff 0 < x < 2\]
अतः सही विकल्प (c) है।