Question

The equation $(\cos p-1)x^2+\left(\cos p\right)x+\sin p=0$, in the variable $x$ has a real root. Then $p$ can take value in the interaval

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

Answer 1

Answer: (A)

$(0,\pi )$

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

For equation to have real roots,

${\cos }^{2}p-4(\cos p-1)\sin p\ge 0\Rightarrow {\cos }^{2}p\ge 4(\cos p-1)\sin p$

$\Rightarrow {\cos }^{2}p\ge -8{\sin }^2\frac{p}{2}\sin p$

For $p=\frac{3\pi }{2}$ or $-\frac{\pi }{2}$, $R.H.S>1$ but $L.H.S<1$

$\therefore$ the choices (B),(C) and (D) are ruled out. the correct alternative is (A)