Question

Given that $\frac{d}{dx}f\left(x\right)=f\text{'}\left(x\right).$ The relationship $f\text{'}(a+b)=f\text{'}\left(a\right)+f\text{'}\left(b\right)$ is valid if $f\left(x\right)$ is equal to

• A. $x$
• B. $x^2$
• C. $x^3$
• D. $x^4$

Answer 1

The correct option is B

$x^2$

Explanation for the correct option:

Step 1. Find the value of $f\left(x\right)$

Given that $\frac{d}{dx}f\left(x\right)=f\text{'}\left(x\right).$

and $f\text{'}(a+b)=f\text{'}\left(a\right)+f\text{'}\left(b\right)$ should be valid

Check the options and verify the condition.

Step 2. Check Option (B) $f\left(x\right)=x^2$

$\Rightarrow$ $f\text{'}\left(x\right)=2x$

$\Rightarrow$ $f\text{'}\left(a\right)=2a$

$\Rightarrow$ $f\text{'}\left(b\right)=2b$

Now,

$\begin{array}{rcl}f\text{'}(a+b)& =& 2(a+b)\\ & =& 2a+2b\\ & =& f\text{'}\left(a\right)+f\text{'}\left(b\right)\end{array}$

Explanation for the Incorrect option:

Step 3. Check Option (A) $f\left(x\right)=x$

$\Rightarrow$$f\text{'}\left(x\right)=1$

Step 4. Check Option (C) $f\left(x\right)=x^3$

$\Rightarrow$$f\text{'}\left(x\right)=3x^2$

$\Rightarrow$$f\text{'}\left(a\right)=3a^2$

$\Rightarrow$$f\text{'}\left(b\right)=3b^2$

Now,

$\begin{array}{rcl}f\text{'}\left(a\right)+f\text{'}\left(b\right)& =& 3a^2+3b^2\\ & \ne & f\text{'}(a+b)\end{array}$

Step 5. Check Option (D)$f\left(x\right)=x^4$

$\Rightarrow$$f\text{'}\left(x\right)=4x^3$

$\Rightarrow$$f\text{'}\left(a\right)=4a^3$

$\Rightarrow$$f\text{'}\left(b\right)=4b^3$

Now,

$\begin{array}{rcl}f\text{'}\left(a\right)+f\text{'}\left(b\right)& =& 4a^3+4b^3\\ & \ne & f\text{'}(a+b)\end{array}$

$\therefore$The value of $f\left(x\right)$ is ${\mathit{x}}^{\mathbf{2}}$

Hence, the correct option is (B).