Question

The equation $4{\sin }^{2}x+4\sin x+a^2-3=0$ possesses a solution if 'a' belongs to the interval.

• A. $(-1,3)$
• B. $(-3,1)$
• C. $[-2,2]$
• D. $R-(-2,2)$

Answer 1

The correct option is C $[-2,2]$

We are given, the equation

$4{\sin }^{2}x+4\sin x+a^2-3=0$

Here, $A=4,B=4$ & $C=a^2-3$

now, Discriminant $D=\sqrt{B^2-4AC}$

$\therefore D^2=B^2-4AC;D\ge 0$

here solution is possible so,

$D^2\ge 0$

$\therefore B^2-4AC\ge 0$

$\therefore (4{)}^2-4(4\left)\right(a^2-3)\ge 0$

$\therefore 16-16(a^3-3)\ge 0$

$\therefore 16(1-a^3+3)\ge 0$

$\therefore 1-a^2+3\ge 0$

$\therefore a^2-4\le 0$

$\therefore (a-2)(a+2)\le 0$

Now, From the value of a we get

the condition $B^2\ge 4AC$

For that $a\in [-2,2]$ So that the condition will be true.