Question

$\sin x+\cos x=y^2-y+a$ will have no solution in $x$ and $y$ if $a$ belongs to

• A. $(0,\sqrt{3})$
• B. $(-\sqrt{3},0)$
• C. $(-\infty ,-\sqrt{3})$
• D. $(\sqrt{2}+\frac{1}{4},\infty )$

Answer 1

Answer: (D)

$(\sqrt{2}+\frac{1}{4},\infty )$

$\sin x+\cos x=y^2-y+a$
As $-\sqrt{2}\le \sin x+\cos x\le \sqrt{2}$
Therefore,
$y^2-y+a\ge \sqrt{2}$ or $y^2-y+a\le -\sqrt{2}$
$y^2-y+a-\sqrt{2}\ge 0$ or $y^2-y+a+\sqrt{2}\le 0$
$1-4(a-\sqrt{2})\le 0$ or $1-4(a+\sqrt{2})\ge 0$
$a\ge \frac{1}{4}+\sqrt{2}$ or $a\le \frac{1}{4}-\sqrt{2}$