Let $f(x)=x^m, m$ being a non-negative integer. The value of $m$ so that the equality $f^{\prime}(a+b)=f^{\prime}(a)+f^{\prime}(b)$ is valid for all $a$, $b 0$ is

Question

Let $f(x)=x^m, m$ being a non-negative integer. The value of $m$ so that the equality $f^{\prime}(a+b)=f^{\prime}(a)+f^{\prime}(b)$ is valid for all $a$, $b > 0$ is, 0, 1, 2, 3

MARKS App · Accepted Answer

Hint : Let $f(x)=x^m, m \geq 0, m \in I$ $\begin{aligned} & f^{\prime}(a+b)=f^{\prime}(a)+f^{\prime}(b), a, b > 0 \ & \Rightarrow m(a+b)^{m-1}=m a^{m-1}+m \cdot b^{m-1} \ & (a+b)^{m-1}=a^{m-1}+b^{m-1} \quad f^{\prime}(x)=m n^{m-1} \ & \text { for } m=0 \Rightarrow 0=0 \ & (a+b)^{-1}=a^{-1}+b^{-1} \end{aligned}$ Not satisfied for $m=1$ $\begin{aligned} & (a+b)^0=a^{1-1}+b^{1-1} \ & 1=2 \end{aligned}$ Not satisfied for $m=2$ $\begin{aligned} & (a+b)^{2-1}=a^{2-1}+b^{2-1} \ & a+b=a+b \text { (satisfied) } \ & \text { for } m=3 \ & (a+b)^{3-1}=a^{3-1}+b^{3-1} \end{aligned}$ Not satisfied

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