The max value of 4sin^2x+3cos^2x is Options are 4,3,7and 5?

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Answer 1 · 2018-08-04T02:53:15

Wring the given expression in terms of either #sin or cos# using the identity #sin^2x+cos^2x=1#.

#f=4sin^2x+3cos^2x#
#=>f=4sin^2x+3(1-sin^2x)#
#=>f=4sin^2x+3(1-sin^2x)#
#=>f=3+sin^2x#

We know that #3# is a constant and maximum value of #sinx=1#. We the result maximum value of the given expression can be

#f_max=3+(1)^2#
#=>f_max=4#

Answer 2 · 2018-08-04T08:22:55

5

#4 sin 2x + 3 cos 2x#

#= 5 ( 4/5 sin 2x + 3/5 cos 2x )#,

noting #sqrt ( 4^2 + 3^2 ) =sqrt 25 = 5#

# = 5 ( sin 2x cos alpha +cos 2x sin alpha), #

#cos alpha = 4/5 and sin alpha = 3/5#

#= 5 sin ( 2x + alpha ) in 5 [ - 1, 1 ] = [ - 5, 5 ]#.

So, the maximum = 5.