Minimum positive values of x and y such that x + y = 2 π and sec x + sec y = 2 2

sec ( 2 x + y ) ≤ 2 s e c x + s e c y , i.e. sec x + sec y ≥ 2 sec 4 π = 2 2 , with equality at x = y = 4 π . The values asked for are positive, so x = y = 4 π .

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MathematicsTrigonometric Ratios & IdentitiesMaximum & Minimum ValuesMedium2 minQB

MathematicsMediummultiple choice

Minimum positive values of $x$ and $y$ such that $x + y = \frac{π}{2}$ and $sec x + sec y = 22$

Options:(select one or more)

Answer:

A, B

Solution:

$sec (\frac{x + y}{2}) \leq \frac{s e c x + s e c y}{2}$ , i.e. $sec x + sec y \geq 2 sec \frac{π}{4} = 22$ , with equality at $x = y = \frac{π}{4}$ . The values asked for are positive, so $x = y = \frac{π}{4}$ .

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

⏱ 2mℹ️ Source: QB

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