If $f (x) = xe^{x(1-x)}$ , then f (x) is
- increasing on [-1/2, 1]
- decreasing on R
- increasing on R
- decreasing on [-1/2, 1]
The Correct Option is A
Solution and Explanation
The correct option is(A): increasing on [-1/2, 1]
\(f (x) = xe^{x (1 - x)}\)
\(\Rightarrow \, f'(x) = e^{x (1-x)} + (1 - 2x) x e^{x (1-x)}\)
\(= - e^{x (1-x) } (2x^2 - x - 1) = - e^{x (1 - x) } (2x + 1) (x - 1)\)
\(\therefore\) f (x) is increasing on [-1/2, 1]
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Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
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- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
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