Question

The total number of ordered pair $(x,y)$ satisfying $|x|+|y|=4,\sin \left(\frac{\pi x^2}{3}\right)=1$ is equal to :

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (B)

$3$

Given that,

$|x|+|y|=4......\left(1\right)$

And

$\sin \left(\frac{\pi x^2}{3}\right)=1......\left(2\right)$

So,

$\sin \left(\frac{\pi x^2}{3}\right)=\sin \left(\frac{\pi }{2}\right)\therefore \sin \theta =1$

Now,

$\frac{\pi x^2}{3}=2n\pi +\frac{\pi }{2}......\left(3\right)$

It is positive, so, $n\in I$

$n\ge 0$

$n=0,1,2,3,4......$

$\frac{\pi x^2}{3}=\frac{\pi }{2}$

$x^2=\frac{3}{2}$

$x=\pm \sqrt{\frac{3}{2}}$

Now,

$|x|+|y|=4$

$|\pm \frac{\sqrt{3}}{2}|+|y|=4$

$y=4-\frac{\sqrt{3}}{2}$

So, put $n=1$, in equation (3) and we get.

$\frac{\pi x^2}{3}=2\pi +\frac{\pi }{2}$

$x^2=\frac{15}{2}$

$x=\sqrt[\pm ]{\pm \frac{15}{2}}$

Now,

$|x|+|y|=4$

$|\pm \sqrt{\frac{15}{2}}|+|y|=4$

$y=4-\frac{\sqrt{15}}{2}$

Put

$n=2,in\left(3\right)$

$\frac{\pi x^2}{3}=4\pi +\frac{\pi }{2}$

$\frac{\pi x^2}{3}=\frac{9\pi }{2}$

$x^2=\frac{27}{2}$

$x=\pm 3\frac{\sqrt{3}}{\sqrt{2}}$

$\text{So,}$

$|\text{x}|\text{+}|\text{y}|\text{=4}$

$|\pm \frac{3\sqrt{3}}{2}|+|y|=4$

$y=4-\frac{3\sqrt{3}}{2}$

Hence, the total number of ordered pair $(x,y)$ is $3$.

Hence, this is the answer.