Question

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

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

Answer 1

Answer: (B)

${(x-1)}^3$

Given, $f^{{}^{\prime \prime }}\left(x\right)=6(x-1)$
On integrating both sides, we get
$f^{{}^{\prime }}\left(x\right)=\frac{6{(x-1)}^2}{2}+C$
$\Rightarrow f^{{}^{\prime }}\left(x\right)=3{(x-1)}^2+C$ ....(i)
At the point $(2,1)$ the tangent of graph is
$y=3x-5$
Slope of tangent $=3$
$\Rightarrow f^{{}^{\prime }}\left(2\right)=3$
$\Rightarrow 3{(2-1)}^2+c=3$
$\Rightarrow c=0$
From equation (i), we have

$f^{{}^{\prime }}\left(x\right)=3{(x-1)}^2$
$\Rightarrow f\left(x\right)=\frac{3{(x-1)}^3}{3}+k$ ....(ii)
Sine, graph passes through $(2,1)$.
Thus $1={(2-1)}^2+k$
$\Rightarrow k=0$
Equation of function is $f\left(x\right)={(x-1)}^3$.