The equation sin ^4 x-(k+3) sin ^2 x-k-4=0 ha | vedclass

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

Question

(B) Given the equation: $\sin ^4 x-(k+3) \sin ^2 x-k-4=0$. This is a quadratic equation in terms of $\sin ^2 x$. Using the quadratic formula,$\sin ^2

Vedclass · Accepted Answer

$-4 \leq k \leq -3$

Vedclass · Answer

(B) Given the equation: $\sin ^4 x-(k+3) \sin ^2 x-k-4=0$. This is a quadratic equation in terms of $\sin ^2 x$. Using the quadratic formula,$\sin ^2 x = \frac{(k+3) \pm \sqrt{(k+3)^2 + 4(k+4)}}{2}$. Simplifying the discriminant: $(k+3)^2 + 4k + 16 = k^2 + 6k + 9 + 4k + 16 = k^2 + 10k + 25 = (k+5)^2$. So,$\sin ^2 x = \frac{(k+3) \pm (k+5)}{2}$. This gives two possible values: $\sin ^2 x = \frac{2k+8}{2} = k+4$ or $\sin ^2 x = \frac{-2}{2} = -1$. Since $\sin ^2 x$ cannot be negative,we must have $\sin ^2 x = k+4$. Given that $0 \leq \sin ^2 x \leq 1$,we have $0 \leq k+4 \leq 1$. Subtracting $4$ from all parts,we get $-4 \leq k \leq -3$.

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

Similar Questions

If $\alpha$ and $\beta$ are different values of $x$ satisfying $a \cos x + b \sin x = c,$ then $\tan \left( \frac{\alpha + \beta}{2} \right) = $

Let $S = \left\{ \theta \in [-\pi, \pi] - \left\{ \pm \frac{\pi}{2} \right\} : \sin \theta \tan \theta + \tan \theta = \sin 2 \theta \right\}$. If $T = \sum_{\theta \in S} \cos 2 \theta$,then $T + n(S)$ is equal to:

In a $\triangle ABC$, $\tan A$ and $\tan B$ are the roots of the equation $pq(x^{2}+1) = r^{2}x$. Then, $\triangle ABC$ is:

If in $\Delta ABC,$ $a = 6, b = 3$ and $\cos(A - B) = \frac{4}{5},$ then its area will be ..... $square \, unit.$

If $\alpha$ and $\beta$ are acute angles such that $\alpha+\beta$ and $\alpha-\beta$ satisfy the equation $\tan^2\theta - 4\tan\theta + 1 = 0$, then $\alpha$ and $\beta$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module