First slide

Theory of equations

Question

If the roots of the given equation $(\cos p - 1) x^{2} + (\cos p) x + \sin p = 0$ are real, then

Moderate

A

$p \in (- π, 0)$
B

$p \in (- \frac{π}{2}, \frac{π}{2})$
C

$p \in (0, π)$
D

$p \in (0, 2 π)$

Solution

Given equation $(\cos p - 1) x^{2} + (\cos p) x + \sin p = 0$
Its discriminant $D \geq 0$ since roots are real
$\Rightarrow \cos^{2} p - 4 \cos p \sin p + 4 \sin p \geq 0 \Rightarrow \cos^{2} p - 4 \cos p \sin p + 4 \sin p \geq 0 \Rightarrow {(\cos p - 2 \sin p)}^{2} - 4 \sin^{2} p + 4 \sin p \geq 0 \Rightarrow {(\cos p - 2 \sin p)}^{2} + 4 \sin p (1 - \sin p) \geq 0 . . . . (i)$
$Now (1 - \sin p) \geq 0$ for all real p, $\sin p > 0 for$
$0 < p < π .$ Therefore $4 \sin p (1 - \sin p) \geq 0$ when
$0 < p < π or p \in (0, π)$

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