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 that 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 $-1$

We have $f^{\prime \prime }\left(x\right)=6(x-1)$
Integrating, we get
$f^{\prime }\left(x\right)=3(x-1{)}^2+c...(1)$

At $(2,1),y=3x-5$ is tangent to $y=f\left(x\right).$
Thus, $f^{\prime }\left(2\right)=3.$

From equation $\left(1\right)$,
$3=3(2-1{)}^2+c\Rightarrow c=0$
$\therefore f^{\prime }\left(x\right)=3(x-1{)}^2$
Integrating, we get
$f\left(x\right)=(x-1{)}^3+c^{\prime }.$

Since, the curve passes through $(2,1)$,
$1=(2-1{)}^3+c^{\prime }\Rightarrow c^{\prime }=0$
$\therefore f\left(x\right)=(x-1{)}^3$
$\therefore f\left(0\right)=-1$