Manipal Entrance Test 2020 Math Paper

Given that,
Slope of the curve =‌

y−1

x2+x

⇒‌

dy

dx

=‌

y−1

x2+x

We can rewrite this as
⇒‌

dy

y−1

=‌

dx

x2+x

Integrate on both sides
⇒∫‌

dy

y−1

=∫‌

dx

x2+x

⇒∫‌

dy

y−1

=∫‌

dx

x(x+1)

⇒∫‌

dy

y−1

=∫(‌

1

x

−‌

1

x+1

)dx
⇒∫‌

dy

y−1

=∫‌

1

x

dx−∫‌

1

x+1

dx
Note that this integral is in the form of ∫‌

1

a

da=log‌a
Applying this formula, we get
⇒log(y−1)=log‌x−log(x+1)+log‌C
We know that ⇒log‌m+log‌n=log‌m‌n and
⇒log‌m−log‌n=log(‌

m

n

)
Using these formulae in the above expressions we get,
⇒log(y−1)=log‌Cx−log(x+1)

⇒log(y−1)=log(‌

Cx

x+1

)
Applying Anti-log on both sides we get,
⇒y−1=‌

Cx

x+1

‌‌‌‌....eq(1)
It is given that the curve passes through (1,0)
Substituting these values of x,y in the above equation we get,
⇒0−1=‌

C×1

1+1

⇒−1=‌

C

2

⇒C=−2

Substituting this value of C in eq(1)
⇒y−1=‌

−2x

x+1

⇒(y−1)(x+1)=−2x
⇒(y−1)(x+1)+2x=0
Hence, the required equation of the curve is ⇒(y−1)(x+1)+2x=0