A) \[f(x)=x{{e}^{x}}^{(1-x)},\]
B) \[\left[ 1\frac{1}{2},1 \right]\]
C) ?e
D) e
Correct Answer: B
Solution :
We have,\[f(x)=x{{e}^{-x}}\] On differentiating both sides w.r.t. x, we get \[f'(x)={{e}^{-x}}(1-x)\] Put\[f'(x)=0\]\[\Rightarrow \]x = 1 Now,\[f(0)=0\]and\[\underset{x\to \infty }{\mathop{\lim }}\,f(x)=\underset{x\to \infty }{\mathop{\lim }}\,x{{e}^{-x}}\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\frac{x}{{{e}^{x}}}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{x}{{{e}^{x}}}=0\] Hence, the greatest value of f (x) is\[\frac{x}{e}.\]
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