AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

It is given that,
α=

lim

x→0

‌

x⋅2x−x

1−cos‌x

Apply L hospital rule,
α=

lim

x→0

‌

2x+x⋅2x‌log‌2−1

sin‌x

Again, apply L hospital rule,
α=

lim

x→0

‌

2x‌log‌x+2x‌log‌2+x⋅2x(log‌2)2

cos‌x

α=log‌2+log‌2
α=2‌log‌2‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I)
And
β=

lim

x→0

‌

x⋅2x−x

√1+x2−√1−x2

Apply L hospital rule,

β=

lim

x→0

‌

2x+x⋅2x‌log‌2−1

(‌ 2x 2√1+x2 −‌ 2x 2√1−x2 )

Again, apply L hospital rule,

β=

lim

x→0

‌

2x‌log‌2+2x‌log‌2+x⋅2x(log‌2)2

left.√1+x2−‌ 2x 2√1+x2 )+(√1−x2+‌ 2x 2√1−x2 )

β=‌

2‌log‌2

(1−0)+(1+0)

=‌

2‌log‌2

2

β=log‌2‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II)
From equation
(I) and (II), α=2β