IIT JEE Advanced 2010 Paper 1

We have

1

4cos2θ+1+ 3 2 sin2θ

⇒

1

2[1+cos2θ]+1+ 3 2 sin2θ

which lies between

1

2

to

11

2

. Therefore, the maximum value is 2.
The minimum value of 1+4cos2θ+3sinθcosθ is as follows:
1 +

4(1+cos2θ)

2

+

3

2

sin2θ
= 1 + 2 + 2 cos 2θ +

3

2

sin 2θ
= 3 + 2 cos 2θ +

3

2

2θ
= 3−√4+

9

4

= 3−

5

2

=

1

2

So, the maximum value of the given expression or

1

4cos2θ+1+ 3 2 sin2θ

is 2.