Question:
What is the maximum value of the function \(f(x) = \left(\frac{1}{x}\right)^{2x^2}\)?
फलन \(f(x) = \left(\frac{1}{x}\right)^{2x^2}\) का अधिकतम मान क्या है?
Options:
- \(e\)
- \(e^{2/e}\)
- \(e^{1/e}\)
- \(e^{-1/e}\)
- \(e\)
- \(e^{2/e}\)
- \(e^{1/e}\)
- \(e^{-1/e}\)
Correct Answer: c)
Explanation:
From the previous question, the maximum occurs at \(x = e^{-1/2}\).
Substitute \(x = e^{-1/2}\) into \(\ln y = -2x^2 \ln x\):
\[x^2 = (e^{-1/2})^2 = e^{-1} = \frac{1}{e}\]
\[\ln y = -2\left(\frac{1}{e}\right)\left(-\frac{1}{2}\right) = \frac{1}{e}\]
Taking exponential of both sides:
\[y = e^{1/e}\]
Hence, the maximum value of the function is \(e^{1/e}\).
Hence, the correct option is (c).
अधिकतम मान बिंदु \(x = e^{-1/2}\) पर प्राप्त होता है:
\[\ln y = -2(e^{-1})(-1/2) = \frac{1}{e} \implies y = e^{1/e}\]
अतः फलन का अधिकतम मान \(e^{1/e}\) है।
अतः सही विकल्प (c) है।