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Functions: Number Solutions Interval Gif

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

Question
The number of solutions of [sin⁡x]+[x2π]+[2x5π]=9x10π[\sin x]+\left[\dfrac{x}{2\pi}\right]+\left[\dfrac{2x}{5\pi}\right] = \dfrac{9x}{10\pi} in the interval (30,40)(30,40) (where [⋅][\cdot] is GIF).
Solution
Answer: 1
Step 1: Rewrite using fractional parts Using [t]=t−{t}[t] = t-\{t\}:
[sin⁡x]=x2π+2x5π−[x2π]−[2x5π]={x2π}+{2x5π}[\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 {x2π}\left\{\dfrac{x}{2\pi}\right\} and {2x5π}\left\{\dfrac{2x}{5\pi}\right\} have periods 2π2\pi and 5π2\dfrac{5\pi}{2} respectively, so their sum has period lcm(2π,5π2)=10π\text{lcm}(2\pi, \frac{5\pi}{2}) = 10\pi. Step 3: Find solutions in (30,40)(30,40) For {x2π}={2x5π}=0\left\{\dfrac{x}{2\pi}\right\} = \left\{\dfrac{2x}{5\pi}\right\} = 0 (which gives [sin⁡x]=0[\sin x] = 0): x=10πnx = 10\pi n. In (30,40)(30,40): x=10π≈31.42x = 10\pi \approx 31.42 is the only such value. Verification: [sin⁡(10π)]+[5]+[4]=0+5+4=9=9(10π)10π[\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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