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If $f(x)=x^{n}, n$ being 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
Options:
$f^{\prime}(\alpha+\beta)=f^{\prime}(\alpha)+f^{\prime}(\beta)$ for all $\alpha, \beta>0$ is
Solution:
2167 Upvotes
Verified Answer
The correct answers are:
2, 0
We have,
$\begin{aligned} f(x) &=x^{n} \\ f^{\prime}(x) &=n x^{n-1} \end{aligned}$
Now, $f^{\prime}(\alpha+\beta)=f^{\prime}(\alpha)+f^{\prime}(\beta)$
$\Rightarrow \quad n(\alpha+\beta)^{n-1}=n \alpha^{n-1}+n \beta^{n-1}$
$\Rightarrow(a+\beta)^{n-1}=a^{n-1}+\beta^{n-1}$
From options we see that $n=2$, satisfy the above equation.
$\therefore$
$$
n=2
$$
$\begin{aligned} f(x) &=x^{n} \\ f^{\prime}(x) &=n x^{n-1} \end{aligned}$
Now, $f^{\prime}(\alpha+\beta)=f^{\prime}(\alpha)+f^{\prime}(\beta)$
$\Rightarrow \quad n(\alpha+\beta)^{n-1}=n \alpha^{n-1}+n \beta^{n-1}$
$\Rightarrow(a+\beta)^{n-1}=a^{n-1}+\beta^{n-1}$
From options we see that $n=2$, satisfy the above equation.
$\therefore$
$$
n=2
$$
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