Question

The value of $\alpha$ so that the geometric mean of $x$ and $y$, where $x\ne y$ is $\frac{x^{\alpha +2}+y^{\alpha +2}}{x^{\alpha +1}+y^{\alpha +1}}$, is

• A. $-\frac{2}{3}$
• B. $-\frac{1}{4}$
• C. $-\frac{3}{2}$
• D. $\frac{7}{6}$

Answer 1

The correct option is C $-\frac{3}{2}$

The $G.M$ between $x$ and $y$
$=\sqrt{xy}$
Now,
$\frac{x^{\alpha +2}+y^{\alpha +2}}{x^{\alpha +1}+y^{\alpha +1}}=\sqrt{xy}\Rightarrow y\times \frac{{\left(\frac{x}{y}\right)}^{\alpha +2}+1}{{\left(\frac{x}{y}\right)}^{\alpha +1}+1}=\sqrt{xy}\Rightarrow \frac{{\left(\frac{x}{y}\right)}^{\alpha +2}+1}{{\left(\frac{x}{y}\right)}^{\alpha +1}+1}=\sqrt{\frac{x}{y}}$
Assuming $t=\frac{x}{y}$, we get
$\Rightarrow \frac{t^{\alpha +2}+1}{t^{\alpha +1}+1}=\sqrt{t}\Rightarrow t^{\alpha +2}+1-t^{\alpha +3/2}-t^{1/2}=0\Rightarrow (t^{\alpha +3/2}-1)(t^{1/2}-1)=0\Rightarrow t^{\alpha +3/2}=1\text{or}t=1\Rightarrow \alpha =-\frac{3}{2}\text{or}t=1$
When $t=1$, we get
$x=y$

Hence, $\alpha =-\frac{3}{2}$