NIMCET 2024 — Mathematics PYQ

$\sin 4x=\frac{1}{2}$

For $\sin \theta =\frac{1}{2}$, the general solution is:

$4x=n\pi +(-1{)}^n\frac{\pi }{6}$

$x=\frac{n\pi }{4}+(-1{)}^n\frac{\pi }{24}$

However, to find the cardinality (number of solutions) in the interval $x\in (-9\pi ,3\pi )$, it is easier to look at the behavior of the sine function.

1. Total Range of the Interval:

   The width of the interval is $3\pi -(-9\pi )=12\pi$.

2. Number of Periods:

   The function is $\sin 4x$. The period of $\sin 4x$ is $\frac{2\pi }{4}=\frac{\pi }{2}$.

   Number of periods in $12\pi$ is:

   $\frac{12\pi }{\pi /2}=24$ periods.

3. Solutions per Period:

   In every single period (width $\frac{\pi }{2}$), the function $\sin 4x$ takes the value $\frac{1}{2}$ exactly twice.

4. Total Number of Solutions:

   $\text{Total solutions}=\text{Number of periods}\times \text{Solutions per period}$

   $\text{Total solutions}=24\times 2=48$

Therefore, the cardinality of set $C$ is $48$.

Correct Option: (B)