Question

The value(s) of 'a' for which the system of equations $\sin x\cos x2y=(a^2-1{)}^2+1$ & $\cos x\sin 2y=a+1$ has a solution is/are equal to

• A. $1$
• B. $-1$
• C. $-2$
• D. None of these

Answer 1

Answer: (B)

$-1$

Given system of equations
$\sin x\cos 2y=(a^2-1{)}^2+1$
& $\cos x\sin 2y=a+1$
Using fact $\left|\sin x\cos 2y\right|\le 1$ and $\left|\cos x\sin 2y\right|\le 1$
$\therefore \left|\right(a^2-1{)}^2+1|\le 1$ and $|a+1|\le 1$
$\Rightarrow (a^2-1{)}^2+1\le 1$ and $-1\le a+1\le 1$
$\Rightarrow (a^2-1{)}^2\le 0$ and $-2\le a\le 0$
But $(a^2-1{)}^2\ge 0$
$\Rightarrow a^2=1$ and $-2\le a\le 0$
$\therefore a=-1$
Hence the given equations are valid for $a=-1$