Question

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

• A. 2
• B. 4
• C. 6
• D. infinite

Answer 1

Answer: (B)

We have, $3^{{\sec }^{2}x-1}\sqrt{9y^2-6y+2}\le 1$
$\Rightarrow 3^{{\sec }^{2}x}\sqrt{y^2-\frac{2}{3}y+\frac{2}{9}}\le 1$
$\Rightarrow 3^{{\sec }^{2}x}\sqrt{{\left(y-\frac{1}{3}\right)}^2+\frac{1}{9}}\le 1$
Now, ${\sec }^{2}x\ge 1\Rightarrow 3{\sec }^{2}x\ge 3$ and $\sqrt{{\left(y-\frac{1}{3}\right)}^2+\frac{1}{9}\ge \frac{1}{3}}$, 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.