[JEE Advanced 1990] If $(\cos p-1)x^2+(\cos p)x+\sin p=0$ has real roots in $x$, then $p$ can lie in
If $(\cos p-1)x^2+(\cos p)x+\sin p=0$ has real roots in $x$, then $p$ can lie in
(a) $(0,2\pi)$
(b) $(-\pi,0)$
(c) $(-\frac\pi2,\frac\pi2)$
(d) $(0,\pi)$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 1mo ago
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Correct answer: (d) $(0,\pi)$
Real roots need discriminant $\cos^2p-4(\cos p-1)\sin p\ge0$. Analysing on $(0,\pi)$ (where $\sin p>0$ and $\cos p-1\le0$) the condition holds, giving $p\in(0,\pi)$.
JEE Advanced 1990 · Trigonometry — verified solution by the Vidaara Team.
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