Question

If the equation $(\cos p-1)x^2+\left(\cos p\right)x+\sin p=0$ in $x$ has real roots, then the set of values of $p$ is

• A. $[0,2\pi ]$
• B. $[-\pi ,0]$
• C. $[-\frac{\pi }{2},\frac{\pi }{2}]$
• D. $[0,\pi ]$

Answer 1

The correct option is D $[0,\pi ]$

Discriminant of given equation is,

$D={\cos }^{2}p-4\sin p(\cos p-1)$

$D={\cos }^{2}p+4\sin p(1-\cos p)$

For equation to posses real solution, $D\ge 0$

now $-1\le \cos p\le 1\Rightarrow 1-\cos p\ge 0$

thus for $D\ge 0$ only we need is, $\sin p\ge 0$

$\Rightarrow p\in [0,\pi ]$