MCQ
The equation ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$ is solvable for
- A$ - \frac{1}{2} \le \alpha \le \frac{1}{2}$
- B$ - 3 \le \alpha \le 1$
- ✓$ - \frac{3}{2} \le \alpha \le \frac{1}{2}$
- D$ - 1 \le \alpha \le 1$
==> ${({\sin ^2}x + {\cos ^2}x)^2} - 2{\sin ^2}x{\cos ^2}x + \sin 2x + \alpha = 0$
==> ${\sin ^2}2x - 2\sin 2x - 2 - 2\alpha = 0$
Let $sin 2x = y$. Then the given equation becomes
${y^2} - 2y - 2(1 + \alpha ) = 0$,
where $ - 1 \le y \le 1$, $({\rm{ }} - 1 \le \sin 2x \le 1)$
For real, discriminant
$ \ge 0$$ \Rightarrow $$3 + 2\alpha \ge 0$
$ \Rightarrow $ $\alpha \ge - \frac{3}{2}$
Also $ - 1 \le y \le 1 \Rightarrow - 1 \le 1 - \sqrt {3 + 2\alpha } \,\, \le 1$
$ \Rightarrow $ $3 + 2\alpha \le 4 \Rightarrow \alpha \le \frac{1}{2}$.
Thus $ - \frac{3}{2} \le \alpha \le \frac{1}{2}$.
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Assertion $(A)$ : The circle ${x^2} + {y^2} = 1$ has exactly two tangents parallel to the $x$ - axis
Reason $(R)$ : $\frac{{dy}}{{dx}} = 0$ on the circle exactly at the point $(0, \pm 1)$.
Of these statements