Question

A function $y=f\left(x\right)$ has a second-order derivative $f^{\prime \prime }\left(x\right)=6(x-1)$. If its graph passes through the point $(2,1)$ and at the point tangent to the graph is $y=3x-5$, then the value of $f\left(0\right)$ is

• A. $1$
• B. $-1$
• C. $2$
• D. $0$

Answer 1

The correct option is B $-1$

$f^{\prime \prime }\left(x\right)=6x-6$
$\int f^{\prime \prime }\left(x\right)dx=\int (6x-6)dx$
$\Rightarrow f^{\prime }\left(x\right)=3x^2-6x+C$
Since, it passes through (2,1) and tangent is $y=3x-5$
$\Rightarrow C=3$
So, $f^{\prime }\left(x\right)=3x^2-6x+3$
$\int f^{\prime }\left(x\right)dx=\int (3x^2-6x+3)dx$
$\Rightarrow f\left(x\right)=x^3-3x^2+3x+C_1$
Since, it passes through $(2,1)$
$\Rightarrow C_1=-1$
Hence, $f\left(x\right)=x^3-3x^2+3x-1$
$f\left(0\right)=-1$