Question:
The function \(f(x)=xe^{x(1-x)}\),x∈R is
The function \(f(x)=xe^{x(1-x)}\),x∈R is
Updated On: Sep 24, 2024
- increasing in \((-\frac{1}{2} , 1)\)
- increasing in \((-\frac{1}{2} , 1)\)
- increasing in \((-1 , -\frac{1}{2})\)
- decreasing in \((-\frac{1}{2} , \frac{1}{2})\)
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The Correct Option is A
Solution and Explanation
\(f(x)=xe^{x(x−1)}\)
⇒ \(f′(x)=e^{x(1−x)}+x^2(−1)e^{x(1−x)}+x(1−x)e^{x(1−x)}\)
=−\(e^{x(1−x)}(2x^2−x−1)=−e^{x(1−x)}(x−1)(x+1/2) \)
\(f′(x)=0\) at \(x=−1/2\)\(,1\) and \(f′(x)>0 \space when −1/2<x<1\)
⇒ f(x) is increasing function 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
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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).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
- Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
- 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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