JEE Mains 02-Sep-2020 Shift 2 Solved Paper

Given equation of curve
y=(1+x)2y+cos2 (sin−1x )
at [x=0]
y=(1+0)2 y+cos2 (sin−10 )
y=1+1 [y=2]
So we have to find the normal at (0,2)
Now y=e2y‌ln(1+x) +cos2 (cos−1√1−x2 )
y=e2y‌ln(1+x)+ (√1−x2 )2
y=e2y‌ln(1+x)+ (1−x2 ) ......(1)
Now differentiate w.r.t. x
y′=e2y‌ln(1+x) [2y.(

1

1+x

)+ln(1+x).2y′] −2x
Put x=0 & y=2
y′=e2×2l n1 [2×2(

1

1+0

)+ln(1+0).2y′] −2×0
y′=e0[4+0]−0
y′=4= slope of tangent to the curve
so slope of normal to the curve =−

1

4

{m1m2=−1}
Hence equation of normal at (0,2) is
y−2=−

1

4

(x−0)
⇒4y−8=−x
⇒[x+4y=8]