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Question Id : 11631 | Context :NIMCET 2024

Question

The number of solutions of ${5}^{1+|\sin x|+|\sin x{|}^2+\ldots}=25$ for $x\in(-\mathrm{\pi},\mathrm{\pi})$ is
🎥 Video solution / Text Solution of this question is given below:

Step 1: Recognize the series

The exponent is an infinite geometric series: $$ 1 + |\sin x| + |\sin x|^2 + |\sin x|^3 + \cdots $$

This is a geometric series with first term \( a = 1 \), common ratio \( r = |\sin x| \in [0,1] \), so: $$ \text{Sum} = \frac{1}{1 - |\sin x|} $$

Step 2: Rewrite the equation

$$ 5^{\frac{1}{1 - |\sin x|}} = 25 = 5^2 $$

Equating exponents: $$ \frac{1}{1 - |\sin x|} = 2 \Rightarrow 1 - |\sin x| = \frac{1}{2} \Rightarrow |\sin x| = \frac{1}{2} $$

Step 3: Solve for \( x \in (-\pi, \pi) \)

We want all \( x \in (-\pi, \pi) \) such that \( |\sin x| = \frac{1}{2} \)

So \( \sin x = \pm \frac{1}{2} \). Within \( (-\pi, \pi) \), the values of \( x \) satisfying this are:

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

✅ Final Answer: $\boxed{4}$ solutions

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