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 the tangent to the graph is $y=3x-5$, then the function is

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

Answer 1

Answer: (D)

$(x-1{)}^3$

$f^{\prime \prime }\left(x\right)=6(x-1)$
$\Rightarrow f^{\prime }\left(x\right)=3x^2-6x+k$...(1)
Slope of tangent at $(2,1)$ is $3$.
$\Rightarrow 3=3\times 2^2-6\times 2+k$
$\Rightarrow k=3$...(2)
From (1) and (2)
$f\left(x\right)=x^3-3x^2+3x+c$
Using $f\left(x\right)=y=1$ when $x=2$, we get
$1=2+c\Rightarrow c=-1$
Hence, the function is
$f\left(x\right)=x^3-3x^2+3x-1$
$\Rightarrow f\left(x\right)=(x-1{)}^3$