First slide

Extreme values of trigonometric functions in trigonometry

Question

If $0 \leq x \leq \frac{π}{3}$ then range of $f (x) = \sec (\frac{π}{6} - x) + \sec (\frac{π}{6} + x)$ is

Moderate

A

$(\frac{4}{\sqrt{3}}, \infty)$
B

$[\frac{4}{\sqrt{3}}, \infty)$
C

$(0, \frac{4}{\sqrt{3}}]$
D

$(0, \frac{4}{\sqrt{3}})$

Solution

if a,b>0
Using A.M. $\geq$ G.M., we get
$\begin{array}{l} \frac{1}{a} + \frac{1}{b} \geq \frac{2}{\sqrt{ab}} \\ \Rightarrow f (x) \geq \frac{2}{\sqrt{\cos (\frac{π}{6} - x) \cos (\frac{π}{6} + x)}} \\ = \frac{2}{\sqrt{\cos^{2} \frac{π}{6} - \sin^{2} x}} \\ = \frac{2}{\sqrt{\frac{3}{4} - \frac{1 - \cos 2 x}{2}}} \\ = \frac{2}{\sqrt{\frac{1}{4} + \frac{\cos 2 x}{2}}} \end{array}$
$Now for 0 \leq x \leq \frac{π}{3}, \frac{- 1}{2} \leq \cos 2 x \leq 1$
$\begin{array}{l} \Rightarrow 0 \leq \sqrt{\frac{1}{4} + \frac{\cos 2 x}{2}} \leq \frac{\sqrt{3}}{2} \\ \Rightarrow f (x) \geq \frac{4}{\sqrt{3}} \end{array}$
Since 'f is continuous, range of $f^{'} is [\frac{4}{\sqrt{3}}, \infty) .$

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