If the function f(x) = xe -x , x ∈ R attains its maximum value β at x = α then (α, β) =
If the function f(x) = xe -x , x ∈ R attains its maximum value β at x = α then (α, β) =
\((2,\frac{1}{e})\)
\((1, \frac{1}{e})\)
\((2,\frac{-1}{e})\)
\((\frac{1}{e}, 1)\)
The Correct Option is B
Solution and Explanation
We are given the function \( f(x) = x e^{-x} \) and need to find the point \( (\alpha, \beta) \) where it attains its maximum value.
1. First, find the derivative \( f'(x) \):
Using product rule, let \( u = x \), \( v = e^{-x} \):
\[ f'(x) = u'v + uv' = e^{-x} - x e^{-x} \]
2. Set derivative equal to zero:
\[ e^{-x} - x e^{-x} = 0 \Rightarrow e^{-x}(1 - x) = 0 \]
Since \( e^{-x} \neq 0 \), we get \( x = 1 \).
3. This gives the critical point. The function decreases for large \( x \), so \( x = 1 \) is a maximum.
4. Find the maximum value:
\[ f(1) = 1 \cdot e^{-1} = \frac{1}{e} \]
Thus, the required point is \( (1, \frac{1}{e}) \).
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.