MAH-CET 2026 — Mathematics PYQ

To find the total number of valid solutions $n$ within the given interval 0 < x < 2\pi, we must analyze the absolute value term $|\cot x|$ by splitting it into two distinct algebraic cases based on the quadrants.

Note that $x\ne \pi$ because $\cot x$ and $\frac{1}{\sin x}$ are undefined at $x=\pi$.

Case 1: When $\cot x\ge 0$

The cotangent function is positive or zero in the first quadrant (0 < x \leq \frac{\pi}{2}) and the third quadrant (\pi < x \leq \frac{3\pi}{2}).

In this domain, $|\cot x|=\cot x$. Substituting this back into the original equation:

$$\cot x=\cot x+\frac{1}{\sin x}$$

Subtracting $\cot x$ from both sides gives:

$$0=\frac{1}{\sin x}$$

This implies:

$$1=0$$

Since this statement is completely impossible, Case 1 yields no solutions.

Case 2: When \cot x < 0

The cotangent function is strictly negative in the second quadrant (\frac{\pi}{2} < x < \pi) and the fourth quadrant (\frac{3\pi}{2} < x < 2\pi).

In this domain, $|\cot x|=-\cot x$. Substituting this back into the original equation:

$$-\cot x=\cot x+\frac{1}{\sin x}$$

Rearrange the equation by moving $\cot x$ to one side:

$$-2\cot x=\frac{1}{\sin x}$$

Now, rewrite $\cot x$ in terms of sine and cosine:

$$-2\left(\frac{\cos x}{\sin x}\right)=\frac{1}{\sin x}$$

Since $\sin x\ne 0$ anywhere in our defined open sub-intervals, we can safely multiply both sides by $\sin x$:

$$-2\cos x=1$$

$$\cos x=-\frac{1}{2}$$

Step 3: Find values of $x$ satisfying $\cos x=-\frac{1}{2}$ under Case 2 constraints

The cosine function is negative in the second quadrant and the third quadrant. Let us check the specific principal solutions within our overarching boundary (0 < x < 2\pi):

1. In the Second Quadrant (\frac{\pi}{2} < x < \pi):

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

   Constraint check: In the second quadrant, $\cot \left(\frac{2\pi }{3}\right)$ is negative, which perfectly satisfies the condition for Case 2. Thus, $x=\frac{2\pi }{3}$ is a valid solution.

2. In the Third Quadrant (\pi < x < \frac{3\pi}{2}):

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

   Constraint check: In the third quadrant, $\cot \left(\frac{4\pi }{3}\right)$ is positive. However, Case 2 explicitly requires \cot x < 0. Therefore, $x=\frac{4\pi }{3}$ is rejected.

Conclusion

There is only $1$ single unique value ($x=\frac{2\pi }{3}$) that satisfies the original relationship. Therefore, the total number of solutions is $n=1$.

Hence, the correct option is (a) 1.