Q181 / 270 · NDA 2016 (Phase 1) · Mathematics

Consider the function $\mathrm{f}(\mathrm{x})=(\mathrm{x}-1)^2(\mathrm{x}+1)(\mathrm{x}-2)^3$. Let $f(x)$ and $g(x)$ be twice differentiable functions on $[0,2]$ satisfying $\mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{g}^{\prime \prime}(\mathrm{x}), \mathrm{f}^{\prime}(1)=4, \mathrm{~g}^{\prime}(1)=6, f(2)=3$ and $g(2)=9$. Then what is $f(x)-g(x)$ at $x=4$ equal to ?

  • Correct answer
    A
    –10
  • B
    –6
  • C
    –4
  • D
    2

Solution

$f^{\prime \prime}(x)=g^{\prime \prime}(x)$ integrate b|s w.r.t. (x) $f^{\prime}(x)+C_1=g^{\prime}(x)+C_2$ ...(1) Putting $x=1$ $\begin{aligned} & f^{\prime}(1)+C_1=g^{\prime}(1)+C_2 \\ & \Rightarrow C_1-C_2=g^{\prime}(1)-f^{\prime}(1) \\ & \Rightarrow C_1-C_2=6-4=2 \end{aligned}$ again integrating equation (1) $f(x)+C_1 x+a_1=g(x)+C_2 x+a_2$ ...(3) Putting $x=2$ $\begin{aligned} & f(2)+C_1 \times 2+a_1=g(2)+2 C_2+a_2 \\ & \Rightarrow a_1-a_2=g(2)+2 C_2-2 C_1-f(2) \\ & \quad=9-3-2\left(C_1-C_2\right) \\ & \quad=6-2 \times 2=2 \\ & \Rightarrow \mathrm{a}_1-\mathrm{a}_2=2 \end{aligned}$ Rearranging equation (3) again, we get $f(x)-g(x)=-x\left(C_1-C_2\right)-\left(a_1-a_2\right)$ Putting $x=4$, we get $\begin{aligned} {[f(x)-g(x)]_{x=4} } & =-4\left(C_1-C_2\right)-\left(a_1-a_2\right) \\ & =-4 \times(2)-2 \\ {[f(x)-g(x)]_{x=4} } & =-10 \end{aligned}$

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

Relevant Questions