Question

The maximum value of $\left( \frac{1}{x} \right)^{2x^{2}}$is –

• A. 1
• B. e
• C. e^1/e
• D. None

Answer 1

Answer: (C)

e^1/e

Answer 2

y is maximum when log y = z is max.

\ z = 2x² log $\frac{1}{x}$ = – 2x² log x

$\frac{dz}{dx}$ = –2[x + 2x log x] = 0

= –2x(1 + 2 log x) = 0

\ log x = –$\frac{1}{2}$ or x = e^–1/2 =$\frac{1}{\sqrt{e}}$

$\frac{d^{2}z}{dx^{2}}$= – 2 $\left\lbrack (1 + 2\log x).1 + x.\frac{2}{x} \right\rbrack$

= –2 [0 + 2] = –4 = –ive

\ Max.

\ y = $\left( \frac{1}{x} \right)^{2x^{2}}$ = $(\sqrt{e})^{2.(1/e)} = e^{1/e}$.