Question

Number of integral value(s) of m for the equation $\sin x-\sqrt{3}\cos x=\frac{4m-6}{4-m}$ has solutions $xϵ[0,2\pi ]$ is

Answer 1

$-2\le (\sin x-\sqrt{3}\cos x)\le 2[Note:-\sqrt{a^2+b^2}\le (a\sin x+b\cos x)\le \sqrt{a^2+b^2}$

$So-2\le \frac{4m-6}{4-m}\le 2$

$-1\le \frac{2m-3}{4-m}\le 1$

$Sowehavetwoconditions\frac{2m-3}{4-m}\le 1......\left(a\right)and\frac{2m-3}{4-m}\ge -1.........\left(b\right)$

$from\left(a\right)\frac{3m-7}{m-4}\ge 0m\in (-\infty ,\frac{7}{3}]\cup (4,\infty )$

$from\left(b\right)\frac{m+1}{m-4}\le 0m\in [-1,\frac{7}{3}]$

$SotheIntersectionofboththesolutionis$

$m\in [-1,\frac{7}{3}]$

$SointergralValuesofm=\{-1,0,1,2\}$

$Soweget4values$