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Q.
The range of the function $ y = \frac{1}{2 - \sin^3 \, x}$ for real x is :
Relations and Functions
Solution:
We know
$- 1 \le \sin \, x \, \le 1$
Cubing both sides, we get
$- 1 \le \, \sin^3 \, x \le 1 $
Multiply by - ve, we get
$1 \ge - \sin^3 x \ge -1 $
Add 2 on each side
$3 \ge 2 - \sin^3 x \ge 1$
Take reciprocal on each side, we get
$\frac{1}{3} \le \frac{1}{2 - \sin^{3} x} \le1$
$ \Rightarrow \frac{1}{3} \le y \le1 \left(\because y = \frac{1}{2- \sin^{3}x}\right) $