Question

If $f\left(x\right)=x^3+x^2{f}^{\prime }\left(1\right)+x{f}^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(3\right).$ Then $f\left(2\right)$ is

• A. -2
• B. 1
• C. 30
• D. 7

Answer 1

The correct option is A -2

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

on differentiating both sides

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

again differentiating both sides

$f^{\prime \prime }\left(x\right)=6x+2f^{\prime }\left(1\right)$

again differentiating

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

$\begin{array}{cc}\therefore f^{\prime }\left(1\right)& =3(1{)}^2+2f^{\prime }(1)+f^{\prime \prime }(2)\\ f^{\prime }\left(1\right)& =-3-f^{\prime \prime }\left(2\right)\\ f^{\prime \prime }\left(2\right)& =6\left(2\right)+2f^{\prime }\left(1\right)\\ & =12+2(-3-f^{\prime \prime }(2\left)\right)\end{array}$

$f^{\prime \prime }\left(2\right)=12-6-2f^{\prime \prime }\left(2\right)$

$3f^{\prime \prime }\left(2\right)=6$

$f^{\prime \prime }\left(2\right)=2$

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

$=-3-2$

$f^{\prime }\left(1\right)=-5$

$f^{-11}\left(3\right)=6$

on Puting

$\begin{array}{cc}f\left(2\right)& =(2{)}^3+(2{)}^2{f}^{\prime }\left(1\right)+2f^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(3\right)\\ & =8+4(-5)+2\left(2\right)+6\\ & =8-20+4+6\\ & =18-20\\ f\left(2\right)& =-2\\ \text{option}& \text{A is correct}\end{array}$