Q63 / 160 · TS EAMCET 2020 (09 Sep Shift 1) · Mathematics

The equation $\sin^{4} x - (k + 3) \sin^{2} x - k - 4 = 0$ has a solution if

Solution

Given: The equation

$\sin^{4} x - (k + 3) \sin^{2} x - k - 4 = 0$ has a solution

It is a quadratic equation in $\sin^{2} x$

So, $\sin^{2} x = \frac{(k + 3) \pm \sqrt{{(k + 3)}^{2} + 4 (k + 4)}}{2}$

$= \frac{(k + 3) \pm \sqrt{{(k + 5)}^{2}}}{2}$

$= \frac{(k + 3) \pm (k + 5)}{2} = k + 4, - 2$

but $\sin^{2} x \in [0, 1]$

$\Rightarrow 0 \leq k + 4 \leq 1 \Rightarrow - 4 \leq k \leq - 3$ .

Hence, the correct answer is option: 'B'.