Question

If $R\to R$ is a function which satisfies $f\left(x\right)=x^3+x^2{f}^{\prime }\left(3\right)+x{f}^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(1\right).$ Then find $f\left(1\right)=?$

• A. $\frac{15}{7}$
• B. $\frac{11}{7}$
• C. $2$
• D. $\frac{16}{7}$

Answer 1

The correct option is D $\frac{16}{7}$

Given $f\left(x\right)=x^3+x^{2}f\left(3\right)+x{f}^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(1\right).$ --------(1)

On putting $x=1$ in (1); we get $f\left(1\right)=1+f^{\prime }\left(3\right)+f^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(1\right)$

Differentiating both sides $f\left(1\right)$ ; we get $f^{\prime }\left(x\right)={3x}^2+2x{f}^{\prime }\left(3\right)+f^{\prime \prime }\left(2\right)$ --------(2)

Again differentiating both sides $f^{\prime \prime }\left(x\right)=6x+2f^{\prime }\left(3\right)$ -------(3)

and differentiating again w.r.t $x$, we get

$f^{\prime \prime \prime }\left(x\right)=6$

$\therefore f^{\prime \prime }\left(1\right)=6\Rightarrow f\left(0\right)=6(\because f^{\prime \prime \prime }\left(1\right)=f\left(0\right))$

Putting $x=2$ in (3); we get

$f^{\prime \prime }\left(2\right)=12+2f^{\prime }\left(3\right)$ --------(4)

and purring $x=3$ in (2); we get

$f^{\prime }\left(3\right)=27+6f^{\prime }\left(3\right)+f^{\prime \prime }\left(2\right)$ -------(5)

Solving (4) and (5); we get

$f\left(3\right)=-\frac{39}{3},f\left(3\right)=\frac{6}{7}$

$\therefore f\left(x\right)=x^3-\frac{39}{7}{x}^2+\frac{6}{7}x+6$

$=\frac{1}{7}(7x^3-39x^2+6x+42)$

Putting $x=2$, we get:

$f\left(1\right)=\frac{1}{7}(7-39+6+42)=\frac{16}{7}.$