4sin2x+9cosec2x find out the minimum value of this function
Anik Garg , 8 Years ago
Grade 12th pass
3 Answers
ayush
Last Activity: 8 Years ago
Let sin^2x=k>=0.Thus given expression by am-gm inequality will have lowest value when sin^2x=cosec^2 x. Thus min value is 13.
jfjyf
Last Activity: 5 Years ago
Let write equation as:
4(sin2x + cosec2x) + 5cosec2x
Applying AM-GM inequality in sin2x + cosec2x
(sin2x + cosec2x)/2 >= (sin2x.cosec2x)1/2
=> (sin2x + cosec2x) >= 2
So, minimum of (sin2x + cosec2x) is 2
which implies sin2x = cosec2x = 1.
Minimum of
4(sin2x + cosec2x) + 5cosec2x = 4*2 + 5*1 = 13
ANUBHAV SHARMA
Last Activity: 3 Years ago
Y=4sin^2x+9cosec^2xSolutionBy making whole square4sin^2x+9cosec^2x+12-12(2sinx-3cosecx)^2+12(2sinx-3cosecx)^2>=0(2sinx-3cosecx)^2+12>=12Hence the min value of y is 12Y=[12,infinity)
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