JEE Mains Model Paper 1

We have
y =
=

−cos2x+ 1 4 sin22x

1− 1 4 sin22x

=

−4cos2x+1−cos22x

4−1+cos22x

=

1−4cos2x−cos22x

3+cos22x

⇒ (1 + y) cos22x + 4cos2x + 3y - 1 = 0
Since cos2x is real 16 - 4(3y - 1) (1 + y) > 0
or 3y2+ 2y - 5 ≤ 0
or (3y + 5) (y - 1) ≤ 0
⇒ −

5

3

≤ y ≤ 1
But y = 1 implies cos2x = -1 i.e. x =

π

2

which is not permissible