MCQ
If $(2\cos x - 1)(3 + 2\cos x) = 0,\,0 \le x \le 2\pi $, then $x = $
- A$\frac{\pi }{3}$
- ✓$\frac{\pi }{3},\frac{{5\pi }}{3}$
- C$\frac{\pi }{2},\frac{{5\pi }}{3},{\cos ^{ - 1}}\left( { - \frac{3}{2}} \right)$
- D$\frac{{5\pi }}{3}$
Then $\cos x = \frac{1}{2}{\rm{ as}}\,{\rm{ }}\cos x \ne \frac{{ - 3}}{2}$
$ \Rightarrow $ $x = 2n\pi \pm \frac{\pi }{3};\,\,\left\{ \begin{array}{l}{\rm{ \,for \,\,}}n = 0,\,\,x = \frac{\pi }{3},\,\frac{{5\pi }}{3}\\{\rm{ \,for\,\, }}n = 1,\,\,x = \frac{{5\pi }}{3}\end{array} \right\}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$L _1: x \sqrt{2}+ y -1=0$ and $L _2: x \sqrt{2}- y +1=0$
For a fixed constant $\lambda$, let $C$ be the locus of a point $P$ such that the product of the distance of $P$ from $L_1$ and the distance of $P$ from $L_2$ is $\lambda^2$. The line $y=2 x+1$ meets $C$ at two points $R$ and $S$, where the distance between $R$ and $S$ is $\sqrt{270}$.
Let the perpendicular bisector of $RS$ meet $C$ at two distinct points $R ^{\prime}$ and $S ^{\prime}$. Let $D$ be the square of the distance between $R ^{\prime}$ and $S ^{\prime}$.
($1$) The value of $\lambda^2$ is
($2$) The value of $D$ is