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 $f\left(2\right)$ is

• A. $2$
• B. $1$
• C. $3$
• D. $4$

Answer 1

The correct option is B $1$

We have,

$f^{\prime \prime }\left(x\right)=6x-6$

Now, integrating we have,

$f^{\prime }\left(x\right)=3x^2-6x+c$.......(1) [Where $c$ is the integrating constant]

Again integrating we have,

$f\left(x\right)=x^3-3x^2+cx+d$........(2) [Where $d$ is also integrating constant]

According to the problem $f\left(x\right)$ and $y=3x-5$ have the same gradient at $(2,1)$.

Then

$f^{\prime }\left(x\right){|}_{(3,2)}=\frac{dy}{dx}{|}_{(2,1)}$

or, $12-12+c=3$

or, $c=3$.

Then $f\left(x\right)=x^3-3x^2+3x+d$ passes through $(2,1)$.

or, $1=8-12+6+d$

or, $d=1$.

$\therefore f\left(x\right)=(x-1{)}^3$

So, $f\left(x\right)=(2-1{)}^3=1$.