The equation sin^4 x + cos^4 x + sin 2x + alp | vedclass

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

Question

(C) Given the equation $\sin^4 x + \cos^4 x + \sin 2x + \alpha = 0$.We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x =

Vedclass · Accepted Answer

$-\frac{3}{2} \le \alpha \le \frac{1}{2}$

Vedclass · Answer

(C) Given the equation $\sin^4 x + \cos^4 x + \sin 2x + \alpha = 0$.We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2 2x$.Substituting this into the equation:$1 - \frac{1}{2}\sin^2 2x + \sin 2x + \alpha = 0$.Let $y = \sin 2x$,where $-1 \le y \le 1$.The equation becomes $1 - \frac{1}{2}y^2 + y + \alpha = 0$,which simplifies to $y^2 - 2y - 2(1 + \alpha) = 0$.For $y$ to have a real solution,the discriminant $D \ge 0$:$D = (-2)^2 - 4(1)(-2(1 + \alpha)) = 4 + 8(1 + \alpha) = 12 + 8\alpha \ge 0 \Rightarrow \alpha \ge -\frac{3}{2}$.The roots are $y = \frac{2 \pm \sqrt{4 + 8(1 + \alpha)}}{2} = 1 \pm \sqrt{1 + 2(1 + \alpha)} = 1 \pm \sqrt{3 + 2\alpha}$.Since $-1 \le y \le 1$,we must have $-1 \le 1 - \sqrt{3 + 2\alpha} \le 1$ (the root $1 + \sqrt{3 + 2\alpha}$ will always be $\ge 1$).$-1 \le 1 - \sqrt{3 + 2\alpha}$ $\Rightarrow \sqrt{3 + 2\alpha} \le 2$ $\Rightarrow 3 + 2\alpha \le 4$ $\Rightarrow \alpha \le \frac{1}{2}$.Thus,the range is $-\frac{3}{2} \le \alpha \le \frac{1}{2}$.

The equation $\sin^4 x + \cos^4 x + \sin 2x + \alpha = 0$ is solvable for

Similar Questions

In a triangle,if $r_1 = 2r_2 = 3r_3$,then $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is equal to

If $\alpha, \beta$ and $\gamma$ are the lengths of the altitudes of a $\triangle ABC$ with area $\Delta$,then $\frac{\Delta^2}{R^2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right)$ is equal to

In a triangle $ABC$,if $\sin \frac{A}{2} = \frac{1}{4} \sqrt{\frac{3}{5}}$,$a = 2$,$c = 5$ and $b$ is an integer,then the area (in sq. units) of triangle $ABC$ is

If in a triangle $ABC$,angle $C$ is $45^o$,then $(1 + \cot A)(1 + \cot B) = $

If $\Delta$ denotes the area of triangle $ABC$,then $(b \sin C + c \sin B)(b \cos C + c \cos B) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module