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Mathematics Question on limits and derivatives

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

A

0

B

1

C

2

D

3

Answer

0

Explanation

Solution

The correct answer is/are option(s) :
(A): 0
(C): 2

f(x)=xm⇒f′(x)=mxm−1f(x)=x^m\Rightarrow f'(x)=mx^{m-1}

f′(a+b)=f′(a)+f′(b)f'(a+b)=f'(a)+f'(b)

⇒n(a+b)m−1=am−1+bm−1\Rightarrow n(a+b)^{m-1}=a^{m-1}+b^{m-1}

Also, for m=0,f(x)=1m=0, f(x)=1

So, f′(x)=0;f′(a+b)=0f'(x)=0;f'(a+b)=0

So,f′(a+b)=f′(a)+f′(b)So, f'(a+b)=f'(a)+f'(b)

Hence, there are two values of m. which are 0 and 2.