Electronics and Communication Engineering (ECE) Exam > Electronics and Communication Engineering (ECE) Questions > The maximum value of ( 1/x)xisa)eb)e-1/ec)(1/...
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The maximum value of ( 1/x)x is
- a)e
- b)e-1/e
- c)(1/e)e
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The maximum value of ( 1/x)xisa)eb)e-1/ec)(1/e)ed)none of theseCorrect...
y=(1/x)^x
ln y = x ln (1/x)
ln y = x [ -ln x]
ln y = - x ln x
(1/y) dy/dx = -ln (x) - x 1/x = -ln x -1
dy/dx = y [ -ln x - 1]
dy/dx = (1/x)^x [ -ln x - 1]
set dy/dx =0
-ln x -1 = 0
ln x =-1
x=e^-1 =1/e
The maximum or minimum value is 1/e
d^2y/dx^2 = dy/dx (-ln x -1) + y (-1/x)
d^2y/dx^2 = (1/x)^x (-ln x-1) + (1/x)^x (-1/x)
At x=1/e, d^2y/dx^2 = -3.927 < 0
Therefore, f(x) has a maximum at x=1/e
The maximum value is f(1/e) = [1/(1/e)]^(1/e) = e^(1/e)
Most Upvoted Answer
The maximum value of ( 1/x)xisa)eb)e-1/ec)(1/e)ed)none of theseCorrect...
Understanding the Expression
To find the maximum value of the expression \( \left(\frac{1}{x}\right)^x \), we can rewrite it in a more manageable form:
- Let \( y = \left(\frac{1}{x}\right)^x = x^{-x} \)
Taking the Natural Logarithm
To simplify the maximization process, we take the natural logarithm:
- \( \ln y = -x \ln x \)
Now, we need to maximize \( \ln y \), which implies we can differentiate it with respect to \( x \):
Finding the Derivative
- Differentiate:
\( \frac{d}{dx}(\ln y) = -\ln x - 1 \)
Setting the derivative to zero for maximization:
- \( -\ln x - 1 = 0 \)
- \( \ln x = -1 \)
- Exponentiating both sides gives:
\( x = e^{-1} = \frac{1}{e} \)
Second Derivative Test
To confirm it's a maximum, we take the second derivative:
- \( \frac{d^2}{dx^2}(\ln y) = -\frac{1}{x} \)
Since this is negative for all \( x > 0 \), we conclude that we have a maximum.
Maximum Value Calculation
Now substituting \( x = \frac{1}{e} \) back into the expression:
- \( y = \left(\frac{1}{\frac{1}{e}}\right)^{\frac{1}{e}} = e^{\frac{1}{e}} \)
Thus, the maximum value of \( \left(\frac{1}{x}\right)^x \) occurs at \( e^{\frac{1}{e}} \).
Final Conclusion
The correct answer is indeed:
- Option B: \( e^{\frac{1}{e}} \)
To find the maximum value of the expression \( \left(\frac{1}{x}\right)^x \), we can rewrite it in a more manageable form:
- Let \( y = \left(\frac{1}{x}\right)^x = x^{-x} \)
Taking the Natural Logarithm
To simplify the maximization process, we take the natural logarithm:
- \( \ln y = -x \ln x \)
Now, we need to maximize \( \ln y \), which implies we can differentiate it with respect to \( x \):
Finding the Derivative
- Differentiate:
\( \frac{d}{dx}(\ln y) = -\ln x - 1 \)
Setting the derivative to zero for maximization:
- \( -\ln x - 1 = 0 \)
- \( \ln x = -1 \)
- Exponentiating both sides gives:
\( x = e^{-1} = \frac{1}{e} \)
Second Derivative Test
To confirm it's a maximum, we take the second derivative:
- \( \frac{d^2}{dx^2}(\ln y) = -\frac{1}{x} \)
Since this is negative for all \( x > 0 \), we conclude that we have a maximum.
Maximum Value Calculation
Now substituting \( x = \frac{1}{e} \) back into the expression:
- \( y = \left(\frac{1}{\frac{1}{e}}\right)^{\frac{1}{e}} = e^{\frac{1}{e}} \)
Thus, the maximum value of \( \left(\frac{1}{x}\right)^x \) occurs at \( e^{\frac{1}{e}} \).
Final Conclusion
The correct answer is indeed:
- Option B: \( e^{\frac{1}{e}} \)
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