Question

If $f\left(x\right)$ and $g\left(x\right)$ are twice differentiable functions on $(0,3)$ satisfying $f^{\prime \prime }\left(x\right)=g^{\prime \prime },f^{\prime }\left(1\right)=4,g^{\prime }\left(1\right)=6,f\left(2\right)=3,g\left(2\right)=9$, then $f\left(1\right)-g\left(1\right)$ is:

• A. $4$
• B. $-4$
• C. $0$
• D. $-2$

Answer 1

The correct option is B $-4$

According to question,

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

Integrating w.r.t. $x$, we get

$f^{\prime }\left(x\right)=g^{\prime }\left(x\right)+C_1$

Put $x=1\Rightarrow f^{\prime }\left(1\right)=g^{\prime }\left(1\right)+C_1$

$\Rightarrow 4=6+C_1$

$\therefore C_1=2$

$\therefore f^{\prime }\left(x\right)=g^{\prime }\left(x\right)-2$

Again, integrating w.r.t. $x$, we get

$f\left(x\right)=g\left(x\right)-2x+C_2$

$\Rightarrow 3=9-4+C_2\Rightarrow C_2=-2$

$\therefore f\left(x\right)=g\left(x\right)-2x-2$

Put $x=1$, we get

$f\left(1\right)-g\left(1\right)=-2\left(1\right)-2=-4$