
If we have a function $f\left( x \right) = {x^n}$, n being a non negative integer , then the value of n for which $f'\left( {\alpha + \beta } \right) = f'\left( \alpha \right) + f'\left( \beta \right)$ for all $\alpha ,\beta > 0$ is
$
{\text{A}}{\text{. 1}} \\
{\text{B}}{\text{. 2}} \\
{\text{C}}{\text{. 0}} \\
{\text{D}}{\text{. 5}} \\
$
Answer
616.5k+ views
Hint:- Here we have a given function and we have a result on its differentiation and we have to find the value of n for which condition is satisfied. We have to first differentiate the given function and turn it into format so that we can find the value of n.
Complete step-by-step solution -
We have
$f\left( x \right) = {x^n}$
On differentiating it , we get
$f'\left( x \right) = n{x^{n - 1}}$
So when we replace x by $\alpha + \beta $ we get,
$f'\left( {\alpha + \beta } \right) = n{\left( {\alpha + \beta } \right)^{n - 1}}$$ \ldots \ldots \left( i \right)$
And when we replace x by $\alpha $ we get,
$f'\left( \alpha \right) = n{\alpha ^{n - 1}}$$ \ldots \ldots \left( {ii} \right)$
And at last when we replace x by $\beta $ we get,
$f'\left( \beta \right) = n{\beta ^{n - 1}}$$ \ldots \ldots \left( {iii} \right)$
We have given in the question
$f'\left( {\alpha + \beta } \right) = f'\left( \alpha \right) + f'\left( \beta \right)$
Using equation (i) , (ii) and (iii) we get,
$n{\left( {\alpha + \beta } \right)^{n - 1}} = n{\alpha ^{n - 1}} + n{\beta ^{n - 1}}$
On cancel out , we get
${\left( {\alpha + \beta } \right)^{n - 1}} = {\alpha ^{n - 1}} + {\beta ^{n - 1}}$
Now we will check first option , put n = 1,
1 = 1 +1 , which is not correct hence option A is wrong.
Now we will check second option , put n = 2 we get
$\left( {\alpha + \beta } \right) = \alpha + \beta $
Hence option B is the correct option.
As we know this is a single correct question so no need to check option C and option D.
Note:- Whenever we get this type of question the key concept of solving is we have to check option wise which option is satisfying the condition given in question. And one thing is clear that we have been asked about the first derivative of function so first we have to differentiate then use the value of n to get in the format required.
Complete step-by-step solution -
We have
$f\left( x \right) = {x^n}$
On differentiating it , we get
$f'\left( x \right) = n{x^{n - 1}}$
So when we replace x by $\alpha + \beta $ we get,
$f'\left( {\alpha + \beta } \right) = n{\left( {\alpha + \beta } \right)^{n - 1}}$$ \ldots \ldots \left( i \right)$
And when we replace x by $\alpha $ we get,
$f'\left( \alpha \right) = n{\alpha ^{n - 1}}$$ \ldots \ldots \left( {ii} \right)$
And at last when we replace x by $\beta $ we get,
$f'\left( \beta \right) = n{\beta ^{n - 1}}$$ \ldots \ldots \left( {iii} \right)$
We have given in the question
$f'\left( {\alpha + \beta } \right) = f'\left( \alpha \right) + f'\left( \beta \right)$
Using equation (i) , (ii) and (iii) we get,
$n{\left( {\alpha + \beta } \right)^{n - 1}} = n{\alpha ^{n - 1}} + n{\beta ^{n - 1}}$
On cancel out , we get
${\left( {\alpha + \beta } \right)^{n - 1}} = {\alpha ^{n - 1}} + {\beta ^{n - 1}}$
Now we will check first option , put n = 1,
1 = 1 +1 , which is not correct hence option A is wrong.
Now we will check second option , put n = 2 we get
$\left( {\alpha + \beta } \right) = \alpha + \beta $
Hence option B is the correct option.
As we know this is a single correct question so no need to check option C and option D.
Note:- Whenever we get this type of question the key concept of solving is we have to check option wise which option is satisfying the condition given in question. And one thing is clear that we have been asked about the first derivative of function so first we have to differentiate then use the value of n to get in the format required.
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