Question

Let \[f\left( x \right) = {x^n}\], \[n\] being a positive integer. Then what is the value of \[n\] for which the equality \[f'\left( {a + b} \right) = f'\left( a \right) + f'\left( b \right)\], \[f\] is valid for all \[a,b > 0\]?
A. \[0, 2\]
B. \[1, 3\]
C. \[3,4\]
D. None of these

Answer 1

Hint: First, differentiate the given function with respect to \[x\]. Then substitute \[a,b\] and \[a + b\] in the differential equation instead of \[x\] to get the various differential equation. In the end, substitute these three differential equations in the given equality to get the required answer.

Formula Used: \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]

Complete step by step solution:
The given function is \[f\left( x \right) = {x^n}\], \[n > 0\] and \[n\] satisfies the equality \[f'\left( {a + b} \right) = f'\left( a \right) + f'\left( b \right)\].

Let’s differentiate the given function with respect to \[x\].
\[f'\left( x \right) = n{x^{n - 1}}\]

Now substitute \[a,b\] and \[a + b\] in the above differential equation instead of \[x\].
\[f'\left( a \right) = n{\left( a \right)^{n - 1}}\]
\[f'\left( b \right) = n{\left( b \right)^{n - 1}}\]
\[f'\left( {a + b} \right) = n{\left( {a + b} \right)^{n - 1}}\]

Substitute the values of the above equations in the given inequality.
\[n{\left( {a + b} \right)^{n - 1}} = n{\left( a \right)^{n - 1}} + n{\left( b \right)^{n - 1}}\]
Cancel out the common factors from each side.
\[{\left( {a + b} \right)^{n - 1}} = {\left( a \right)^{n - 1}} + {\left( b \right)^{n - 1}}\] \[.....\left( 1 \right)\]

Now verify the values of \[n\].
Substitute \[n = 0\] in the equation \[\left( 1 \right)\].
\[{\left( {a + b} \right)^{0 - 1}} = {\left( a \right)^{0 - 1}} + {\left( b \right)^{0 - 1}}\]
\[ \Rightarrow \]\[\dfrac{1}{{a + b}} = \dfrac{1}{a} + \dfrac{1}{b}\]
\[ \Rightarrow \]\[\dfrac{1}{{a + b}} \ne \dfrac{{a + b}}{{ab}}\]
So, this value is incorrect.

Substitute \[n = 1\] in the equation \[\left( 1 \right)\].
\[{\left( {a + b} \right)^{1 - 1}} = {\left( a \right)^{1 - 1}} + {\left( b \right)^{1 - 1}}\]
\[ \Rightarrow \]\[{\left( {a + b} \right)^0} = {\left( a \right)^0} + {\left( b \right)^0}\]
\[ \Rightarrow \]\[1 = 1 + 1\]
\[ \Rightarrow \]\[1 \ne 2\]
So, this value is incorrect.

Substitute \[n = 2\] in the equation \[\left( 1 \right)\].
\[{\left( {a + b} \right)^{2 - 1}} = {\left( a \right)^{2 - 1}} + {\left( b \right)^{2 - 1}}\]
\[ \Rightarrow \]\[a + b = a + b\]
This is true.
So, this value is correct.

Substitute \[n = 3\] in the equation \[\left( 1 \right)\].
\[{\left( {a + b} \right)^{3 - 1}} = {\left( a \right)^{3 - 1}} + {\left( b \right)^{3 - 1}}\]
\[ \Rightarrow \]\[{\left( {a + b} \right)^2} = {\left( a \right)^2} + {\left( b \right)^2}\]
\[ \Rightarrow \]\[{a^2} + {b^2} + 2ab \ne {a^2} + {b^2}\]
So, this value is incorrect.

Substitute \[n = 4\] in the equation \[\left( 1 \right)\].
\[{\left( {a + b} \right)^{4 - 1}} = {\left( a \right)^{4 - 1}} + {\left( b \right)^{4 - 1}}\]
\[ \Rightarrow \]\[{\left( {a + b} \right)^3} = {\left( a \right)^3} + {\left( b \right)^3}\]
\[ \Rightarrow \]\[{a^3} + {b^3} + 3ab\left( {a + b} \right) \ne {a^3} + {b^3}\]
So, this value is incorrect.

Hence the correct option is D.

Note: Students often get confused with the formula of \[{\left( {a + b} \right)^3}\]. The correct formula is \[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)\].