Minimum problem

Geometry Level 2

Find the minimum value of $4\cos ^{ 2 }{ x+9\sec ^{ 2 }{ x } }$ for real $x$ .

A number just less than 13 Exactly 13 Exactly 15 A number just more than 12 Exactly 12

1 solution

Prajwal Krishna
Nov 22, 2016

Let y = ( $4\cos ^{ 2 }{ x\quad +\quad 9\sec ^{ 2 }{ x } }$ )

$\frac{dy}{dx}$ = -8(cos x)(sin x)+18(sec x)(sec x)(tan x)

Now

8<18

sin x < tan x in [o, $\frac { \Pi }{ 2 }$ ]

cos x < (sec x)^2

Therefore , $\frac{dy}{dx}$ > 0 in [o, $\frac { \Pi }{ 2 }$ ]

Hence , minimum will occur at x=0

Hence y (minimum) = 13