Question

If $x, y \in[0,10]$, then the number of solutions $(x, y)$ of the inequation $3^{\sec ^{2} x-1} \sqrt{9 y^{2}-6 y+2} \leq 1$ is

• A. 2
• B. 4
• C. 6
• D. infinite

Answer 1

Answer: (B)

We have, $3^{\sec ^{2} x-1} \sqrt{9 y^{2}-6 y+2} \leq 1$
$\Rightarrow 3^{\sec ^{2} x} \sqrt{y^{2}-\frac{2}{3} y+\frac{2}{9}} \leq 1$
$\Rightarrow 3^{\sec ^{2} x} \sqrt{\left(y-\frac{1}{3}\right)^{2}+\frac{1}{9}} \leq 1$
Now, $\sec ^{2} x \geq 1 \Rightarrow 3 \sec ^{2} x \geq 3$ and $\sqrt{\left(y-\frac{1}{3}\right)^{2}+\frac{1}{9} \geq \frac{1}{3}}$, so we
must have $\sec ^{2} x=1$ and $y-\frac{1}{3}=0$
$\Rightarrow x=0, \pi, 2 \pi, 3 \pi $ and $y=\frac{1}{3}$
$\therefore $ There are $4$ solutions.