Question

9. solve it step by step.
this is jee mains 2021 proble6​

Answer 1

Answer:

y+1=Y⇒dy=dY

x+2=X⇒ dx = dX

Y +Y/dX = XdY

dX X e X

e/x_ <= { |X|+c (3, 2)→ e €²¾/3 = (13/+c e + (n3-(n|X|> 0 en XI<(e2/3+ (3)

Let λ=(e2/3 + (n3)

|x +2|<e²

-e²<x+2<e* -e²-2<x<e^-2

в

a+B=4⇒la+ B| = 4