Question

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

• A. $4$
• B. $-4$
• C. $0$
• D. $-2$

Answer 1

The correct option is B

$-4$

Explanation for the correct option.

Step 1: Integrate $f\text{'}\text{'}\left(x\right)=g\text{'}\text{'}\left(x\right)$.

$f\text{'}\text{'}\left(x\right)=g\text{'}\text{'}\left(x\right)$

Integrating with respect to $x$, we get

$f\text{'}\left(x\right)=g\text{'}\left(x\right)+c$

Now, for $x=1$

$\begin{array}{rcl}f\text{'}\left(1\right)& =& g\text{'}\left(1\right)+c\\ \Rightarrow 4& =& 6+c\\ \Rightarrow c& =& -2\end{array}$

Step 2: Integrate $f\text{'}\left(x\right)=g\text{'}\left(x\right)+c$.

Now, $f\text{'}\left(x\right)=g\text{'}\left(x\right)+c$ can be written as $f\text{'}\left(x\right)=g\text{'}\left(x\right)-2$

Integrating with respect to $x$, we get

$f\left(x\right)=g\left(x\right)-2x+c_1$

Now, for $x=2$

$$\begin{gathered} f\left(2\right)=g\left(2\right)-2\left(2\right)+c_1 \\ \Rightarrow 3=9-4+c_1 \\ \Rightarrow c_1=-2 \end{gathered}$$

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

Step 3: Find the value of $f\left(1\right)-g\left(1\right)$.

$\begin{array}{rcl}f\left(1\right)-g\left(1\right)& =& -2\left(1\right)-2\\ & =& -4\end{array}$

Hence, option B is correct.