NIMCET 2024 — Mathematics PYQ

Step 1: Simplify the Exponential Equation

The given equation is:

$$5^{1+|\sin x|+|\sin x{|}^2+\cdots }=25$$

We know that $25$ can be written as $5^2$:

$$5^{1+|\sin x|+|\sin x{|}^2+\cdots }=5^2$$

Since the bases are identical on both sides, we can equate their exponents:

$$1+|\sin x|+|\sin x{|}^2+\cdots =2$$

Step 2: Identify and Evaluate the Infinite Geometric Series

The expression in the exponent is an infinite geometric progression (GP):

$$S=1+|\sin x|+|\sin x{|}^2+\cdots$$

Here, the first term ($a$) is $1$ and the common ratio ($r$) is $|\sin x|$.

For $x\in (-\pi ,\pi )$, except for $x=0$ where $\sin x=0$, the value of $|\sin x|$ satisfies:

0 < |\sin x| < 1

Since the common ratio |r| < 1, we can use the sum formula for an infinite geometric series ($S_{\infty }=\frac{a}{1-r}$):

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

Step 3: Solve for $|\sin x|$

Now, let's solve the simplified algebraic equation:

$$1=2(1-|\sin x|)$$

$$1=2-2|\sin x|$$

$$2|\sin x|=2-1$$

$$2|\sin x|=1$$

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

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

Removing the absolute value gives two sets of values for $\sin x$:

$$\sin x=\frac{1}{2}\text{or}\sin x=-\frac{1}{2}$$

Step 4: Find the Number of Solutions in the Interval $(-\pi ,\pi )$

We need to find the number of values of $x$ that satisfy these equations within the open interval $(-\pi ,\pi )$:

1. For $\sin x=\frac{1}{2}$ in $(-\pi ,\pi )$:

   Since $\sin x$ is positive in the first and second quadrants:

   $$x=\frac{\pi }{6},\frac{5\pi }{6}$$

   (Both solutions lie inside $(0,\pi )\subset (-\pi ,\pi )$)

2. For $\sin x=-\frac{1}{2}$ in $(-\pi ,\pi )$:

   Since $\sin x$ is negative in the third and fourth quadrants (measured clockwise for negative angles):

   $$x=-\frac{\pi }{6},-\frac{5\pi }{6}$$

   (Both solutions lie inside $(-\pi ,0)\subset (-\pi ,\pi )$)

Counting all valid values of $x$, we get:

$$x\in \left\{-\frac{5\pi }{6},-\frac{\pi }{6},\frac{\pi }{6},\frac{5\pi }{6}\right\}$$

There are exactly 4 distinct solutions.