Question

If ${\log }_{m+1}(x+\sqrt{m+1}y)+{\log }_{m+1}(x-\sqrt{m+1}y)=1$
Where $m>0$;
Then

• A. $|x|-|y|\ge \sqrt{m}$
• B. $|x|-|y|\ge 0$
• C. $|x|-|y|\le -\sqrt{m}$
• D. $|x|-|y|<-\sqrt{m}$

Answer 1

Answer: (A), (B)

The correct options are
A $|x|-|y|\ge \sqrt{m}$
B $|x|-|y|\ge 0$

$x+\sqrt{m+1}y>0$ .....(i)

and

$x-\sqrt{m+1}y>0$ .... (ii)

Also, from the given equation,

$(x+\sqrt{m+1}\cdot y)\cdot (x-\sqrt{m+1}\cdot y)=m+1$
$\Rightarrow x^2-(m+1)y^2=m+1$ ....(iii)

From (i) and (ii),
$x-\sqrt{m+1}|y|>0$, ie $x>\sqrt{m+1}|y|$

$\text{Case1 : y > 0}$,

Set $p=x-y$. Thus, $x=p+y$

Eqn (iii) $\Rightarrow$

$(p+y{)}^2-(m+1)y^2=m+1$

$\Rightarrow$

$y^2\left(m\right)+y(-2p)+(m+1-p^2)=0$

For $y$ to be Real, Discriminant $(b^2-4ac)$ should be non-negative.
$4p^2-4\left(m\right)(m+1-p^2)\ge 0$
$\Rightarrow p^2\ge m$

So $p\ge \sqrt{m}$ or $p\le -\sqrt{m}$

But, $x>\sqrt{m+1}|y|>0$

$\Rightarrow$

$x>|y|$ .....(Since, $m>0$)

$\Rightarrow |x|-|y|>0$

Hence $p\ge m$ (Reject $p<-\sqrt{m})$
i.e. $|x|-|y|\ge \sqrt{m}$ in case $1$

$\text{Case2 : y < 0}$,

Let $z=-y$

So,
${\log }_{m+1}(x-\sqrt{m+1}z)+{\log }_{m+1}(x+\sqrt{m+1}z)=1$ which, from symmetry, leads us to,
$|x|-|z|\ge \sqrt{m}$

So, $|x|-|y|\ge \sqrt{m}$
i.e $|x|-|y|\ge \sqrt{m}$ (In case $2$).

$\text{Case 3: y = 0}$

$\Rightarrow {\log }_{m+1}\left(x\right)+{\log }_{m+1}\left(x\right)=1$ ie $x^2=m+1$
ie $|x|=\sqrt{m+1}$

So $|x|-|y|=\sqrt{m+1}-0=\sqrt{m+1}$
$\Rightarrow |x|-|y|\ge \sqrt{m}$ is satisfied.

So in all Possible Cases,
$\text{|}x|-|y|\ge \sqrt{m}.$

Further, $\sqrt{PositiveRealQuantity}$ is always non negative.

Hence, we select option $B$ in addition to option $A$.