Question
The maximum value of $${\left( {\frac{1}{x}} \right)^{2{x^2}}}$$ is :
A.
$$e$$
B.
$$\root e \of e $$
C.
1
D.
none of these
Answer :
$$\root e \of e $$
Solution :
$$\eqalign{
& y = {\left( {\frac{1}{x}} \right)^{2{x^2}}} \Rightarrow \log \,y = 2{x^2}\log \frac{1}{x} \cr
& {\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}x,\,\,\frac{1}{y}.\frac{{dy}}{{dx}} = 4x\log \frac{1}{x} + 2{x^2}.\left( { - \frac{1}{x}} \right) \cr
& {\text{Then }}\frac{{dy}}{{dx}} = 0{\text{ if }} - 4x\log \,x - 2x = 0 \cr
& {\text{or,}}\,\,\,2\log \,x + 1 = 0 \cr
& \therefore \,\,x = {e^{ - \,\frac{1}{2}}} \cr
& {\text{Verify that }}\frac{{{d^2}y}}{{d{x^2}}} < 0{\text{ at }}x = {e^{ - \,\frac{1}{2}}} \cr} $$