Question

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

• A. $[0,\pi ]\cup \left\{2\pi \right\}$
• B. $(0,\pi ]$
• C. $[0,2\pi ]$
• D. $\left\{0\right\}\cup [\pi ,2\pi ]$

Answer 1

The correct option is A $[0,\pi ]\cup \left\{2\pi \right\}$

Given: $(\cos p-1)x^2+\left(\cos p\right)x+\sin p=0$

When $\cos p-1=0\Rightarrow \cos p=1$, we get
$\sin p=0$, so
$0\cdot x^2+x+0=0\Rightarrow x=0\therefore p=0,2\pi$

When $\cos p-1\ne 0\Rightarrow \cos p\ne 1$, we get
$(\cos p-1)x^2+\left(\cos p\right)x+\sin p=0$
For real roots
$D\ge 0\Rightarrow {\cos }^{2}p-4(\cos p-1)\sin p\ge 0\Rightarrow {\cos }^{2}p-4\cos p\sin p+4\sin p\ge 0\Rightarrow (\cos p-2\sin p{)}^2-4{\sin }^{2}p+4\sin p\ge 0\Rightarrow (\cos p-2\sin p{)}^2+4\sin p(1-\sin p)\ge 0$
As $(\cos p-2\sin p{)}^2\ge 0,1-\sin p\ge 0$, so the inequality holds when
$\sin p\ge 0\Rightarrow p\in (0,\pi ](\because \cos p\ne 1)$

Hence, the complete solution of $p$ is $[0,\pi ]\cup \left\{2\pi \right\}$