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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Trigonometric Equations

BITSAT 2020 · Mathematics · Trigonometric Equations

The least positive non-integral solution of the equation π(x^ 2...

The least positive non-integral solution of the equation sin⁡π(x2+x)=sin⁡πx2\sin \pi\left(x^{2}+x\right)=\sin \pi x^{2} is

Options

  1. Arational
  2. Birrational of the form p\sqrt{\mathrm{p}}
  3. Cirrational of the form p−14,\frac{\sqrt{\mathrm{p}}-1}{4}, where p\mathrm{p} is an odd integer
  4. Dirrational of the form p+14,\frac{\sqrt{\mathrm{p}}+1}{4}, where p\mathrm{p} is an even integer

Correct answer

A. rational

Step-by-step solution

We have, sin⁡π(x2+x)=sin⁡πx2⇒π(\mathrmx2+x)=nπ+(−1)nπ\mathrmx2∴\sin \pi\left(x^{2}+x\right)=\sin \pi x^{2}\Rightarrow \quad \pi\left(\mathrmx^{2}+\mathrm{x}\right)=\mathrm{n} \pi+(-1)^{\mathrm{n}} \pi \mathrmx^{2}\therefore Either \mathrmx2+x=2m+\mathrmx2⇒x=2m∈I\mathrmx^{2}+\mathrm{x}=2 \mathrm{ m}+\mathrmx^{2} \Rightarrow \mathrm{x}=2 \mathrm{ m} \in \mathrm{I} or \mathrmx2+\quad \mathrmx^{2}+\mathrm{x}=\mathrm{k}−\mathrmx2,-\mathrmx^{2}, where k\mathrm{k} is an odd integer ⇒2\mathrmx2+x−k=0⇒x=−1±1+8k4\Rightarrow \quad 2 \mathrmx^{2}+\mathrm{x}-\mathrm{k}=0 \Rightarrow \mathrm{x}=\frac{-1 \pm \sqrt{1+8 \mathrm{k}}}{4} For least positive non-integral solution is x=12,x=\frac{1}{2}, when k=1k=1
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