MHTCET 2016 Math Solved Paper

(a) Since, 6y = x3 + 2
and Δy = 8 Δx
After differentiating on both sides of eq. (i) w.r.t. x, we get

6dy

dx

=3x2 ⇒

dy

dx

=

1

2

x2
As, Δy=

dy

dx

Δx ⇒ 8Δx=

1

2

x2Δx
So, x2 = 16 ⇒ x = ± 4
When x = 4, then 6y = (4)3 + 2
So, 6y = 66 ⇒ y = 11
Hence, required point is (4,11).