If the roots of the given equation (cos p - 1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the quadratic equation $(\cos p - 1)x^2 + (\cos p)x + \sin p = 0$.For the roots to be real,the discriminant $D$ must be greater than or equa

Vedclass · Accepted Answer

$p \in (0, \pi)$

Vedclass · Answer

(C) Given the quadratic equation $(\cos p - 1)x^2 + (\cos p)x + \sin p = 0$.For the roots to be real,the discriminant $D$ must be greater than or equal to zero $(D \ge 0)$.$D = b^2 - 4ac = (\cos p)^2 - 4(\cos p - 1)(\sin p) \ge 0$.Expanding this,we get $\cos^2 p - 4\sin p \cos p + 4\sin p \ge 0$.This can be rewritten as $(\cos p - 2\sin p)^2 - 4\sin^2 p + 4\sin p \ge 0$.$(\cos p - 2\sin p)^2 + 4\sin p(1 - \sin p) \ge 0$.Since $(\cos p - 2\sin p)^2 \ge 0$ for all real $p$,the inequality holds if $4\sin p(1 - \sin p) \ge 0$.Since $(1 - \sin p) \ge 0$ for all real $p$,we require $\sin p \ge 0$.In the interval $(0, 2\pi)$,$\sin p \ge 0$ holds for $p \in (0, \pi)$.

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

Similar Questions

Find the quadratic equation whose roots are reciprocals of the roots of the equation $3x^{2}-20x+17=0$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $(x+y)^{2} = 3136$
$II.$ $y+2513 = 2569$

Solve the given two equations and select the correct answer from the given options.
$I.$ $6x^2 + 77x + 121 = 0$
$II.$ $y^2 + 9y - 22 = 0$

If the roots of the equation $ax^2 + x + b = 0$ are real,then the roots of the equation $x^2 - 4\sqrt{ab}x + 1 = 0$ will be

Let $x_1, x_2, x_3 \in R - \{0\}$,$x_1 + x_2 + x_3 \neq 0$ and $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{1}{x_1 + x_2 + x_3}$. Then $\frac{1}{x_1^n + x_2^n + x_3^n} = \frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{x_3^n}$ holds good for:

Vedclass Test Series

Exam Paper Generator

Online Exam Module