Question

Number of solutions of the equation $\sin x+\cos x=x^2-2x+\sqrt{35}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $infinite$

Answer 1

The correct option is A $0$

LHS$=\sin x+\cos x$

We now that,

$-\sqrt{1+1}\le \sin x+\cos x\le \sqrt{1+1}$

$\therefore -\sqrt{2}\le \sin x+\cos x\le \sqrt{2}$

RHS$=x^2-2x+\sqrt{35}$

$\therefore =(x-1{)}^2+\sqrt{35}-1$

$\therefore$ RHS $\ge \sqrt{35}-1$

$\therefore$ RHS $\ge 4.9$ ………$\left(1\right)$

Also LHS, $-1.41\le \sin x+\cos x\le 1.41$

$\therefore -1.41\le LHS\le 1.41$ ……..$\left(2\right)$

$\therefore$ From $\left(1\right)$ and $\left(2\right)$

We can conclude that it has zero solution.