Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

MathematicsTrigonometric Ratios & IdentitiesMaximum & Minimum ValuesMedium2 min

MathematicsMediumsingle choice

Let $f (x) = 2 cosec (2 x) + sec (x) + cosec (x)$ then the minimum value of $f (x)$ for $x \in (0, \frac{π}{2})$ is

Answer:

Solution:

$f (x) = 2 cosec 2 x + (sec x - cosec x)^{2} + \frac{2 2}{2 c o s x \cdot s i n x}$

$∴ f (x) = \frac{2}{s i n 2 x} + (sec x - cosec x)^{2} + \frac{2 2}{s i n 2 x}$

So, $f_{min} (x = \frac{π}{4}) = \frac{2}{1} + 0 + \frac{2 2}{1} = 2 (2 + 1)$

Stream:JEESubject:MathematicsTopic:Trigonometric Ratios & IdentitiesSubtopic:Maximum & Minimum Values

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