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

Answer 1

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#

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Answer 2

Daniel Acosta

#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.

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Answer 3

Dylan Arnold

To find the maximum value of (4\sin^2(x) + 3\cos^2(x)), you can use the fact that (\sin^2(x) + \cos^2(x) = 1).

Given the expression (4\sin^2(x) + 3\cos^2(x)), you can rewrite it as (4(1 - \cos^2(x)) + 3\cos^2(x)).

Expanding, you get (4 - 4\cos^2(x) + 3\cos^2(x)), which simplifies to (4 - \cos^2(x)).

Since (\cos^2(x)) has a maximum value of (1), the maximum value of (4 - \cos^2(x)) is (4 - 1 = 3).

Therefore, the maximum value of (4\sin^2(x) + 3\cos^2(x)) is (3).

So, the correct option is (3).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.