Question

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`

Answer 2

5

Explanation:

`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.