Question

If f(x)=$\frac{x^1}{2}$, g(x)=$\frac{x^1}{3}$ and h(x)=$\frac{x^2}{3}$. Find $\frac{(f+g)\left(x\right)}{(f+h)\left(x\right)}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$(f+g)\left(x\right)isf\left(x\right)+g\left(x\right)$
We want to find $\frac{(f+g)\left(x\right)}{(f+h)\left(x\right)}$.

$\Rightarrow \frac{(f+g)\left(x\right)}{(f+h)\left(x\right)}$ = $\frac{f\left(x\right)+g\left(x\right)}{f\left(x\right)+h\left(x\right)}$

$=\frac{x^{\frac{1}{2}}+x^{\frac{1}{3}}}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$
We see all the options are in the $\frac{x^1}{n}$ or $\frac{1}{x^{\frac{1}{n}}}$. This means we have to find a common factor or simplify the expression. For this, we can either guess it by looking at the options or we will replace x with $y^6$. $y^6$, because we don't want any irrational terms, 6 is the L.C.M of denominator of powers.
$\Rightarrow \frac{x^{\frac{1}{2}}+x^{\frac{1}{3}}}{x^{\frac{1}{2}}+x^{\frac{2}{3}}}$ = $\frac{y^3+y^2}{y^3+y^4}$, where x=$y^6$ or y=$x^{\frac{1}{6}}$
$=\frac{(y^3+y^2)}{y^2+y^{3}xy}$

$=\frac{1}{y}$

$=\frac{1}{x^6}$