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Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

The number of solutions of

[\sin x]+\left[\dfrac{x}{2\pi}\right]+\left[\dfrac{2x}{5\pi}\right] = \dfrac{9x}{10\pi}

in the interval

(30,40)

(where

[\cdot]

is GIF).

Solution

Answer: 1

Step 1: Rewrite using fractional parts Using

[t] = t-\{t\}

[\sin x] = \frac{x}{2\pi}+\frac{2x}{5\pi} - \left[\frac{x}{2\pi}\right] - \left[\frac{2x}{5\pi}\right] = \left\{\frac{x}{2\pi}\right\}+\left\{\frac{2x}{5\pi}\right\}

Step 2: Identify periodicity Both

\left\{\dfrac{x}{2\pi}\right\}

and

\left\{\dfrac{2x}{5\pi}\right\}

have periods

2\pi

and

\dfrac{5\pi}{2}

respectively, so their sum has period

\text{lcm}(2\pi, \frac{5\pi}{2}) = 10\pi

. Step 3: Find solutions in

(30,40)

For

\left\{\dfrac{x}{2\pi}\right\} = \left\{\dfrac{2x}{5\pi}\right\} = 0

(which gives

[\sin x] = 0

x = 10\pi n

. In

(30,40)

x = 10\pi \approx 31.42

is the only such value. Verification:

[\sin(10\pi)]+[5]+[4] = 0+5+4 = 9 = \dfrac{9(10\pi)}{10\pi}

. ✓ Number of solutions = 1. Answer: 1

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