Question

A function $y=f(x)$ has second order derivative $f^{\prime \prime }(x)=6(x-1)$ . 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)

We have, $f^{\prime \prime }(x)=6x-6$ On integrating, we get $f^{\prime }(x)=3x^2-6x+c$ At $x=2,$ $f^{\prime }(2)=3$
$\Rightarrow$ $12-12+c=3$
$\Rightarrow$ $c=3$
$\therefore$ $f^{\prime }(x)=3x^2-6x+3$
Again integrating we get $f(x)=x^3-3x^2+3x+d$
when $x=2,y=1$ $f(2)=1={(2)}^3-3{(2)}^2+3(2)+d$
$\Rightarrow$ $d=-1$
$\therefore$ $f(x)=x^3-3x^2+3x-1$
$={(x-1)}^3$