MHT CET 2018 Math Solved Paper

(a) Line y = 4x - 5 → slope of line m = 4 ...(i)
curve y2=ax3 + b
∴ differentiating w.r.t. ‘x’
2y

dy

dx

=3ax2

dy

dx

=

3ax2

2y

= slope of tangent
∴

dy

dx

|(2,3)=

3a×4

2×3

= 2a
from(i) and (ii), we get
4 = 2a ⇒ a = 2
Since, (2,3) is a point on the curve: y2=ax3 + b.
∴ (3)2=2(2)3 + b ⇒ b = - 7
∴ 7a + 2b = 7 x 2 + 2 (-7) = 0