Manipal Entrance Test 2015 Math Paper

Let y=tan−1(

√1+x2−1

x

)
Putting x = tan θ, we get
y=tan−1(

sec‌θ−1

tan‌θ

) =tan−1(tan‌

θ

2

) =

1

2

tan−1x
On differentiating both sides w.r.t. x, we get

dy

dx

=

1

2(1+x2)

and z=tan−1(

2x√1−x2

1−2x2

)
Putting x=sin‌θ, we get
z=tan−1(

2‌sin‌θ‌cos‌θ

cos‌2‌θ

) =tan−1(tan‌2‌θ)
⇒z=2θ=2sin−1x
On differentiating both sides w.r.t. x, we get

dz

dx

=

2

√1−x2

Thus,

dy

dz

=

dy∕dx

dz∕dx

=

1

4(1+x2)

√1−x2
∴[

dy

dz

]x=0=

1

4