The maximum value of x–x is
Answer & explanation
Correct answer: option 2
f′(x) = x−x(−logx−1)
= −x−x(logx+1)
f′(x) = 0 when x = $\frac{1}{e}$
f′(x) > 0 when 0 < x < $\frac{1}{e}$
f′(x) < 0 when x > $\frac{1}{e}$
f(x) is maximum at x = $\frac{1}{e}$
$f(\frac{1}{e})=(e)^{\frac{1}{e}}$