Question

For real values x, if the expression $\frac{(ax-b)(dx-c)}{(bx-a)(cx-d)}$ assumes all real values then $(a^2-b^2)$ and $(c^2-d^2)$ must have the same sign.

• A. True
• B. False

Answer 1

The correct option is A True

y= given expression
$\Rightarrow (ad-bcy)x^2+(ac+bd)(y-1)x+(bc-ady)=0$
Since x is real $\therefore \Delta \ge 0$
${(ac+bd)}^2{(y-1)}^2-4(Ad-bcy)(bc-ady)\ge 0\forall yϵR$
or ${(ac-bd)}^2{y}^2+2\left\{2\right(a^2{d}^2+b^2{c}^2)-{(ac+bd)}^2\}y+{(ac-bd)}^2\ge 0\forall yϵR$
Above expression is to be $+ive$ and its first term is $+ive$. Hence $\Delta <0$.
$4{\left\{2\right(a^2{d}^2+b^2{c}^2)-{(ac-bd)}^2\}}^2-4{(ac-bd)}^4=-ive$
Apply $L^2-M^2=(L+M)(L-M)$ and cance $4$.
or $[2a^2{d}^2+2b^2{c}^2-{(ac+bd)}^2-{(ac-bd)}^2]$
$[2a^2{d}^2+2b^2{c}^2-{(ac+bd)}^2-{(ac-bd)}^2]<0$
or $[2a^2{d}^2+2b^2{c}^2-2a^2{c}^2-2b^2{d}^2]$
$[2a^2{d}^2+2b^2{c}^2-4abcd]<0$
Again cancel $2,$ the second factor is ${(ad-bc)}^2$ which $+ive$ and first factor is
$a^2(d^2-c^2)+b^2(c^2-d^2)$
or $(c^2-d^2)(b^2-a^2)=-(a^2-b^2)(c^2-d^2)>0$ i.e., $+ive$
Above will hold good if both $(a^2-b^2)$ and $(c^2-d^2)$ have the same sign i.e., either both $+ive$ or both $-ive$.