Question

Let $f\left(x\right)=x^2$ and $g\left(x\right)=3x+4$ . Then the value of $\frac{(f+g)\left(x\right)}{(f-g)\left(x\right)}$= when x = 1

• A. $\frac{4}{3}$
• B. $\frac{3}{4}$
• C. $\frac{-4}{3}$
• D. $\frac{-3}{4}$

Answer 1

The correct option is C $\frac{-4}{3}$

$(f+g)\left(x\right)=f\left(x\right)+g\left(x\right)=x^2+3x+4(f-g)\left(x\right)=f\left(x\right)-g\left(x\right)=x^2-3x-4\Rightarrow \frac{(f+x)\left(x\right)}{(f-g)\left(x\right)}=\frac{x^2+3x+4}{x^2-3x-4}$
At x = 1,
$\frac{x^2+3x+4}{x^2-3x-4}=\frac{1+3+4}{1-3-4}=\frac{8}{-6}=\frac{-4}{3}$