Trigonometrical Equations 3 Question 6
7. The equation $(\cos p-1) x^{2}+(\cos p) x+\sin p=0$ in the variable $x$, has real roots. Then, $p$ can take any value in the interval
(1990, 2M)
(a) $(0,2 \pi)$
(b) $(-\pi, 0)$
(c) $-\frac{\pi}{2}, \frac{\pi}{2}$
(d) $(0, \pi)$
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Answer:
Correct Answer: 7. (d)
Solution:
- Since, the given quadratic equation
$ (\cos p-1) x^{2}+(\cos p) x+\sin p=0 $
has real roots.
$\therefore$ Discriminant, $\cos ^{2} p-4 \sin p(\cos p-1) \geq 0$
$\Rightarrow \quad(\cos p-2 \sin p)^{2}-4 \sin ^{2} p+4 \sin p \geq 0$
$\Rightarrow \quad(\cos p-2 \sin p)^{2}+4 \sin p(1-\sin p) \geq 0$
$\because \quad 4 \sin p(1-\sin p)>0$ for $0<p<\pi$
and $\quad(\cos p-2 \sin p)^{2} \geq 0$
Thus, $(\cos p-2 \sin p)^{2}+4 \sin p(1-\sin p) \geq 0$
for
$0<p<\pi$.
Hence, the equation has real roots for $0<p<\pi$.
