Let f(x)=xm, m being a non-negative integer. The value of m so that the equality f'(a+b)=f'(a)+f'(b) is valid for all a,b>0 is
- 0
- 1
- 2
- 3
The Correct Option is A, C
Solution and Explanation
Given: Let f(x) = x^m
Then, the derivative is: $f'(x) = m x^{m - 1}$
We are given the condition: $f'(a + b) = f'(a) + f'(b)$
Substitute the derivative expression: $m(a + b)^{m - 1} = m a^{m - 1} + m b^{m - 1}$
Divide both sides by $m$ (assuming $m \ne 0$): $(a + b)^{m - 1} = a^{m - 1} + b^{m - 1}$
We now solve for values of $m$ such that the equation holds for all $a, b$.
Try m = 0: Then, $f(x) = 1$ and $f'(x) = 0$ So, $f'(a + b) = 0 = f'(a) + f'(b)$ ✅ Holds true
Try m = 2: Then, LHS: $(a + b)^{2 - 1} = a + b$ RHS: $a^{2 - 1} + b^{2 - 1} = a + b$ ✅ Holds true
These are the only values that satisfy the condition in general.
Final Answer: (A): 0 and (C): 2
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Concepts Used:
Limits And Derivatives
Mathematically, a limit is explained as a value that a function approaches as the input, and it produces some value. Limits are essential in calculus and mathematical analysis and are used to define derivatives, integrals, and continuity.
Limits Formula:
Derivatives of a Function:
A derivative is referred to the instantaneous rate of change of a quantity with response to the other. It helps to look into the moment-by-moment nature of an amount. The derivative of a function is shown in the below-given formula.
Properties of Derivatives:
Read More: Limits and Derivatives