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 tbe point $(2,1)$ and at that point the tangent to the curve is $y=3x-5$, then the function is:

• A. ${(x+1)}^3$
• B. ${(x-1)}^3$
• C. ${(x-1)}^2$
• D. ${(x+1)}^2$

Answer 1

The correct option is B ${(x-1)}^3$

Given $f^{\prime \prime }\left(x\right)=6(x-1)$
$\Rightarrow$ $f^{\prime }\left(x\right)=\frac{6{(x-1)}^2}{2}+c$
$\Rightarrow$ $3=3+c$ $\{\because f\left(x\right)=y=3x+5,f^{\prime }\left(x\right)=3\forall xϵR\}$
$\Rightarrow$ $c=0$
so $f^{\prime }\left(x\right)=3{(x-1)}^2$
$\Rightarrow$ $f\left(x\right)={(x-1)}^3+c_1$ as curve passes through (2, 1)
$\Rightarrow$ $1={(2-1)}^3+c_1$
$\Rightarrow$ $c_1=0$
$\therefore$ $f\left(x\right)={(x-1)}^3$