Question

If the solution of $ax+\frac{y}{b}=a^2-b^2$ and $\frac{x}{a}-by=1+b^4$ satisfies $k=b^{2}x+\frac{a}{b}y$ where a, b ≠ 0, then k equals

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is A 0

$ax+\frac{y}{b}=a^2-b^2.....\left(i\right)$
$\frac{x}{a}-by=1+b^4.....\left(ii\right)$
Multiplying (i) by $b^2$, and adding with (ii) we get
$a{b}^{2}x+by=a^2{b}^2-b^4+1+b^4$
$\Rightarrow a{b}^{2}x+\frac{x}{a}=a^2{b}^2+1$
$\Rightarrow (a^2{b}^2+1)\frac{x}{a}=(a^2{b}^2+1)$
$\Rightarrow \frac{x}{a}=1$
$\Rightarrow y=-b^3$

Now, substituting x = a and $y=-b^{3}ink=b^{2}x+\frac{a}{b}y,$ we get
$k=b^2\left(a\right)+\frac{a}{b}(-b^3)$
$\Rightarrow k=a{b}^2-a{b}^2=0$
Hence, the correct answer is option (1).