JEE Main 8 Jan 2020 Shift 1 Solved Paper

Let tan−1x=θ, θ∈(−

π

2

,

π

4

)∪(

π

4

,

π

2

)
f(x)=(sin‌θ+cos‌θ)2−1 =sin‌2‌θ=

2x

1+x2

Now,

dy

dx

=

1

2

d

dx

sin−1(

2x

1+x2

) =−

1

1+x2

,|x|>1
Since, we can integrate only in the continuous interval. So we have to take integral in two cases separtely namely for x < -1 and for x > 1.
⇒y={

−tan−1x+c1,	x>1
−tan−1+c2,	x<−1

So, c1=

π

2

as y(√3)=

π

6

But we cannot find c2 as we do not have any other additional information for x < -1. So, all of the given options may be correct as c2 is unknown so, it should be bonus.