Question

The least positive non integral solution of $\sin \pi (x^2+x)-\sin \pi x^2=0$ is

• A. rational
• B. irrational of the form $\sqrt{p}$
• C. irrational of the form $\frac{\sqrt{p}-1}{4}$, where p is an odd integer
• D. irrational of the form $\frac{\sqrt{p}-1}{4}$, where p is an even integer

Answer 1

The correct option is C irrational of the form $\frac{\sqrt{p}-1}{4}$, where p is an odd integer

given, $\sin \pi (x^2+x)-\sin \pi x^2=0$

The general solution is
$\pi (x^2+x)=n\pi +{(-1)}^n\pi x^2$

$(x^2+x)=n+{(-1)}^n{x}^2$

If $n$ is even i.e. $n=2m$

$(x^2+x)=2m+x^2$

$\Rightarrow x=2m\in I$
If $n$ is odd $n=k\left(say\right)$

$(x^2+x)=k-x^2$

$2x^2+x-k=0$

$\Rightarrow x=\frac{-1\pm \sqrt{1+8k}}{4}$

For least positive non-integral solution
$x=\frac{-1+\sqrt{1+8k}}{4}$

$x=\frac{-1+\sqrt{p}}{4}$ where p is an odd integer