Question:
If the minimum value of \(f(x)=\frac{5 x^2}{2}+\frac{\alpha}{x^5}, x>0\), is 14 , then the value of \(\alpha\) is equal to:
If the minimum value of \(f(x)=\frac{5 x^2}{2}+\frac{\alpha}{x^5}, x>0\), is 14 , then the value of \(\alpha\) is equal to:
Updated On: Jun 24, 2024
- 32
- 64
- 128
- 256
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The Correct Option is C
Solution and Explanation
The correct option will be (C): 128
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Concepts Used:
Continuity & Differentiability
Definition of Differentiability
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Definition of Continuity
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.