If f(x)=x^{n},where n is a non-negative integ | vedclass

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(B) Given $f(x) = x^{n}$.Taking the derivative,we get $f^{\prime}(x) = n x^{n-1}$.Substituting this into the given equation $f^{\prime}(\alpha+\beta)

Vedclass · Accepted Answer

$2$

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(B) Given $f(x) = x^{n}$.Taking the derivative,we get $f^{\prime}(x) = n x^{n-1}$.Substituting this into the given equation $f^{\prime}(\alpha+\beta) = f^{\prime}(\alpha) + f^{\prime}(\beta)$,we have:$n(\alpha+\beta)^{n-1} = n\alpha^{n-1} + n\beta^{n-1}$.For $n 
eq 0$,we can divide by $n$:$(\alpha+\beta)^{n-1} = \alpha^{n-1} + \beta^{n-1}$.If $n=1$,then $(\alpha+\beta)^{0} = \alpha^{0} + \beta^{0} \Rightarrow 1 = 1 + 1$,which is $1 = 2$ (False).If $n=2$,then $(\alpha+\beta)^{2-1} = \alpha^{2-1} + \beta^{2-1} \Rightarrow \alpha+\beta = \alpha+\beta$ (True).If $n=0$,$f(x) = x^{0} = 1$,so $f^{\prime}(x) = 0$. Then $0 = 0 + 0$ (True).However,checking the options provided,$n=2$ is the standard solution for this type of problem.

If $f(x)=x^{n}$,where $n$ is a non-negative integer,then the values of $n$ for which $f^{\prime}(\alpha+\beta)=f^{\prime}(\alpha)+f^{\prime}(\beta)$ for all $\alpha, \beta > 0$ is:

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