Question

Let $f:R\to R$ be a function such that $f\left( x \right)={{x}^{3}}+{{x}^{2}}f'\left( 1 \right)+xf''\left( 2 \right)+f'''\left( 3 \right)$, $x\in R$. Find the value of $f\left( 2 \right)$?
(a) 8
(b) –2
(c) –4
(d) 30

Answer 1

Hint: We use the fact $f'''\left( 3 \right)$, $f''\left( 2 \right)$ and $f'\left( 1 \right)$ are constants as the function ‘f’ is dependent on the variable x to start the problem. We differentiate the function ‘f’ and substitute 1 in place of x after differentiating to get a relation between $f''\left( 2 \right)$ and $f'\left( 1 \right)$. We differentiate ‘f’ for the second time and substitute 2 in place of x to get another relation between $f''\left( 2 \right)$ and $f'\left( 1 \right)$. Using these two relations we calculate the values of $f''\left( 2 \right)$ and $f'\left( 1 \right)$. We differentiate ‘f’ for the third time and substitute 3 in place of x to get the value of $f'''\left( 3 \right)$. Using all these values we calculate the required value of $f\left( 2 \right)$.

Complete step-by-step answer:
Given that we have a function defined on $f:R\to R$ as $f\left( x \right)={{x}^{3}}+{{x}^{2}}f'\left( 1 \right)+xf''\left( 2 \right)+f'''\left( 3 \right)$. We need to find the value of $f\left( 2 \right)$.
We know that the values of $f'''\left( 3 \right)$, $f''\left( 2 \right)$ and $f'\left( 1 \right)$ are constants as the function ‘f’ is dependent on the variable x. So, this makes x as an independent variable and the derivatives of function ‘f’ must be dependent on x.
Now we have function $f\left( x \right)={{x}^{3}}+{{x}^{2}}f'\left( 1 \right)+xf''\left( 2 \right)+f'''\left( 3 \right)$.
Let differentiate the function ‘f’ on both sides with respect to x.
$\Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{d{{x}^{3}}}{dx}+\dfrac{d\left( {{x}^{2}}f'\left( 1 \right) \right)}{dx}+\dfrac{d\left( xf''\left( 2 \right) \right)}{dx}+\dfrac{d\left( f'''\left( 3 \right) \right)}{dx}$ ---(1).
We know that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$, $\dfrac{d}{dx}\left( af\left( x \right) \right)=a\dfrac{d}{dx}\left( f\left( x \right) \right)$ and $\dfrac{d}{dx}\left( a \right)=0$. We use this as equation (1).
$\Rightarrow f'\left( x \right)=3{{x}^{2}}+2xf'\left( 1 \right)+f''\left( 2 \right)+0$.
 $\Rightarrow f'\left( x \right)=3{{x}^{2}}+2xf'\left( 1 \right)+f''\left( 2 \right)$ ---(2).
Let us substitute 1 in place of x.
$\Rightarrow f'\left( 1 \right)=3{{\left( 1 \right)}^{2}}+2\left( 1 \right)f'\left( 1 \right)+f''\left( 2 \right)$.
$\Rightarrow f'\left( 1 \right)=3+2f'\left( 1 \right)+f''\left( 2 \right)$.
$\Rightarrow f'\left( 1 \right)-2f'\left( 1 \right)=3+f''\left( 2 \right)$.
$\Rightarrow -f'\left( 1 \right)=3+f''\left( 2 \right)$ ---(3).
Let us differentiate equation (2) with respect to x on both sides,
$\Rightarrow \dfrac{d\left( f'\left( x \right) \right)}{dx}=\dfrac{d\left( 3{{x}^{2}} \right)}{dx}+\dfrac{d\left( 2xf'\left( 1 \right) \right)}{dx}+\dfrac{d\left( f''\left( 2 \right) \right)}{dx}$.
$\Rightarrow f''\left( x \right)=6x+2f'\left( 1 \right)$ ---(4).
Let us substitute 2 in place of x in equation (4).
$\Rightarrow f''\left( 2 \right)=6\left( 2 \right)+2f'\left( 1 \right)$.
$\Rightarrow f''\left( 2 \right)=12+2f'\left( 1 \right)$ ---(5).
Let us substitute equation (5) in equation (3),
$\Rightarrow -f'\left( 1 \right)=3+12+2f'\left( 1 \right)$.
$\Rightarrow -f'\left( 1 \right)-2f'\left( 1 \right)=15$.
$\Rightarrow -3f'\left( 1 \right)=15$.
$\Rightarrow f'\left( 1 \right)=\dfrac{15}{-3}$.
$\Rightarrow f'\left( 1 \right)=-5$ ---(6).
Let us substitute the value of $f'\left( 1 \right)$ in equation (5).
$\Rightarrow f''\left( 2 \right)=12+2\left( -5 \right)$.
$\Rightarrow f''\left( 2 \right)=12-10$
$\Rightarrow f''\left( 2 \right)=2$ ---(7)
Let us differentiate equation (4) with respect to x on both sides,
$\Rightarrow \dfrac{d\left( f''\left( x \right) \right)}{dx}=\dfrac{d\left( 6x \right)}{dx}+\dfrac{d\left( 2f'\left( 1 \right) \right)}{dx}$.
$\Rightarrow f'''\left( x \right)=6+0$.
$\Rightarrow f'''\left( x \right)=6$.
Let us substitute 3 in place of x.
$\Rightarrow f'''\left( 3 \right)=6$ ---(8).
Let us the values obtained in equations (6), (7) and (8) in the function $f\left( x \right)$.
We get $f\left( x \right)={{x}^{3}}+{{x}^{2}}\left( -5 \right)+x\left( 2 \right)+6$.
$\Rightarrow f\left( x \right)={{x}^{3}}-5{{x}^{2}}+2x+6$.
Let us substitute 2 in place of x.
$\Rightarrow f\left( 2 \right)={{\left( 2 \right)}^{3}}-5{{\left( 2 \right)}^{2}}+2\left( 2 \right)+6$.
$\Rightarrow f\left( 2 \right)=8-5\left( 4 \right)+4+6$.
$\Rightarrow f\left( 2 \right)=18-20$.
$\Rightarrow f\left( 2 \right)=-2$.
We have found the value of $f\left( 2 \right)$ as –2.
∴ The value of $f\left( 2 \right)$ is –2.

So, the correct answer is “Option b”.

Note: We should not different $f'''\left( 3 \right)$, $f''\left( 2 \right)$ and $f'\left( 1 \right)$ again as they were constants. We always should check what are the independent and dependent variables involved in the function. We should not make any calculation mistakes while solving for the values of $f'''\left( 3 \right)$, $f''\left( 2 \right)$ and $f'\left( 1 \right)$. Similarly, we can expect problems to find the derivatives of function ‘f’.