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Q.
The range of $\frac{1}{3 sin \theta + 4 cos \theta + 2}$ is
NTA AbhyasNTA Abhyas 2022
Solution:
We know that range of the expression $asinx+bcosx$ is $\left[- \sqrt{a^{2} + b^{2}} , \sqrt{a^{2} + b^{2}}\right]$ i.e.,
$-\sqrt{a^{2} + b^{2}}\leq asinx+bcosx\leq \sqrt{a^{2} + b^{2}}$
So, $-\sqrt{3^{2} + 4^{2}}\leq 3sinx+4cosx\leq \sqrt{3^{2} + 4^{2}}$
$\Rightarrow -5\leq 3sinx+4cosx\leq 5$
$\Rightarrow -5+2\leq 3sinx+4cosx+2\leq 5+2$
$\Rightarrow -3\leq 3sinx+4cosx+2\leq 7$
$\Rightarrow -\frac{1}{3}\geq \frac{1}{3 sin x + 4 cos x + 2}\geq \frac{1}{7}$
So, range of $\frac{1}{3 \sin x+4 \cos x+2}$ is $\left(-\infty, \quad-\frac{1}{3}\right] \cup\left[\frac{1}{7}, \infty\right)$.