Let f(x)=x^m, 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

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Let f(x)=x^m, 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

MARKS Content Team · Accepted Answer

Correct answer: (A) 0; (C) 2. Hint : Let f(x)=x^m, m ≥ 0, m ∈ I f^′(a+b)=f^′(a)+f^′(b), a, b > 0 ⇒ m(a+b)^m-1=m a^m-1+m · b^m-1 (a+b)^m-1=a^m-1+b^m-1 f^′(x)=m n^m-1 for m=0 ⇒ 0=0 (a+b)^-1=a^-1+b^-1 Not satisfied for m=1 (a+b)^0=a^1-1+b^1-1 1=2 Not satisfied for m=2 (a+b)^2-1=a^2-1+b^2-1 a+b=a+b (satisfied) for m=3 (a+b)^3-1=a^3-1+b^3-1 Not satisfied

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

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