For real values of x, if the expression $\frac{(a x - b) (d x - c)}{(b x - a) (c x - d)}$ assumes all real values then show that $(a^{2} - b^{2})$ and $(c^{2} - d^{2})$ must have the same sign. x not equal a/b and d/c

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Solution

y=given expression $\Rightarrow (a d - b c y) x^{2} + (a c + b d) (y - 1) x$
$+ (b c - a d y) = 0$
Since x is real $∴ Δ \geq 0$
${(a c + b d)}^{2} {(y - 1)}^{2} - 4 (a d - b c y) (b c - a d y) \geq 0$
$\forall y ϵ R$ or ${(a c - b d)}^{2} y^{2} + {2 (a^{2} d^{2} + b^{2} c^{2}) - {(a c + b d)}^{2}} y + {(a c - b d)}^{2} \geq 0 \forall y ϵ R$
Above expression is to be +ive and its first term is + ive. Hence $Δ < 0.$
$4 {2 (a^{2} d^{2} + b^{2} + c^{2}) - {(a c + b d)}^{2}}^{2}$
$- 4 {(a c - b d)}^{4} = - i v e$
Apply $L^{2} - M^{2} = (L + M) (L - M)$ and cancel 4.
or $[2 a^{2} d^{2} + 2 b^{2} c^{2} - {(a c + b d)}^{2} {(a c - b d)}^{2}] < 0$
or $[2 a^{2} d^{2} + 2 b^{2} c^{2} - 2 a^{2} c^{2} - 2 b^{2} d^{2}]$
$[2 a^{2} d^{2} + 2 b^{2} c^{2} - 4 a b c d] < 0$
Again cancel 2, the second factor is ${(a d - b c)}^{2}$ which is +ive and first factor is $a^{2} (d^{2} - c^{2}) + b^{2} (c^{2} - d^{2})$
or $(c^{2} - d^{2}) (b^{2} - a^{2}) = - (a^{2} - b^{2}) (c^{2} - d^{2})$
Hence the required condition is $- (a^{2} - b^{2}) (c^{2} - d^{2}) {(a d - b c)}^{2} < 0$
$- (a^{2} - b^{2}) (c^{2} - d^{2}) > 0$ i.e., +ive
Above will hold good if both $(a^{2} - b^{2}) a n d (c^{2} - d^{2})$ have the same sign i.e., either both +or both -ive

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