UPSEE 2017 Solved Paper

Given,
y=2‌tan‌x−tan2x .......(i)
∴

dy

dx

=2sec2x−2‌tan‌x.sec2x
=2sec2x(1−tan‌x) .......(ii)
At point of maxima,

dy

dx

=0
2sec2x(1−tan‌x)=0 [From Eq. (ii)]
∴x=

π

4

,

π

2

[Here, x=

π

2

is not possible]
∴x=

π

4

[∵x∈[0,

π

2

] (given) ]
Now,

d2y

dx2

=4‌sec‌x.sec‌x.tan‌x(1−tan‌x) +2sec2x(0−sec2x)
=4sec2x‌tan‌x−4sec2xtan2x−2sec4x
∴

d2y

dx2

|x=

π

2

=4sec2

π

4

‌tan‌

π

4

−4sec2

π

4

tan2

π

4

−2sec4

π

4

=4(√2)2.1−4(√2)2.(1)2−2.(√2)4
=8−8−8
=−8 which is negative.
∴ At x=

π

4

, function y=2‌tan‌x−tan2x has maximum value.
∴ Maximum value of function at point x=

π

4

, will be
[y]x=

π

4

=1