Question:
Consider the function \(f(x) = \left(\frac{1}{x}\right)^{2x^2}\), where \(x > 0\). At what value of \(x\) does the function attain its maximum value?
फलन \(f(x) = \left(\frac{1}{x}\right)^{2x^2}\), जहाँ \(x > 0\) पर विचार करें। \(x\) के किस मान पर फलन अधिकतम मान प्राप्त करता है?
Options:
- \(e\)
- \(\sqrt{e}\)
- \(\frac{1}{\sqrt{e}}\)
- \(\frac{1}{e}\)
- \(e\)
- \(\sqrt{e}\)
- \(\frac{1}{\sqrt{e}}\)
- \(\frac{1}{e}\)
Correct Answer: c)
Explanation:
Let \(y = (1/x)^{2x^2} = x^{-2x^2}\).
Taking natural log of both sides:
\[\ln y = -2x^2 \ln x\]
Differentiating with respect to \(x\):
\[\frac{1}{y}\frac{dy}{dx} = -4x\ln x – 2x^2\left(\frac{1}{x}\right) = -4x\ln x – 2x = -2x(2\ln x + 1)\]
For stationary points, set \(\frac{dy}{dx} = 0\):
Since \(x > 0\), \(2\ln x + 1 = 0 \implies \ln x = -\frac{1}{2} \implies x = e^{-1/2} = \frac{1}{\sqrt{e}}\]
The second derivative confirms this is a local maximum.
Hence, the correct option is (c).
माना \(y = x^{-2x^2}\)। दोनों पक्षों का लघुगणक लेने पर:
\[\ln y = -2x^2 \ln x\]
अवकलन करने पर:
\[\frac{1}{y}\frac{dy}{dx} = -2x(2\ln x + 1) = 0 \implies \ln x = -1/2 \implies x = e^{-1/2} = \frac{1}{\sqrt{e}}\]
अतः फलन \(x = \frac{1}{\sqrt{e}}\) पर अधिकतम मान प्राप्त करता है।
अतः सही विकल्प (c) है।