Q107 / 225 · JEE Main 2004 · Mathematics

A function $y=f(x)$ has a 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 $y=3 x-5$, then the function is

  • A
    $(x-1)^2$
  • Correct answer
    B
    $(x-1)^3$
  • C
    $(x+1)^3$
  • D
    $(x+1)^2$

Solution

$f^{\prime \prime}(x)=6(x-1) \Rightarrow f^{\prime}(x)=3(x-1)^2+c$ and $f^{\prime}(2)=3 \Rightarrow c=0$ $\Rightarrow f(x)=(x-1)^3+k$ and $f(2)=1 \Rightarrow k=0$ $\Rightarrow f(x)=(x-1)^3$

Hence, the correct answer is option: 'B'.

Relevant Questions