Given as the slope of tangent to the curve y = x3 + ax + b at (1, – 6)
Let us find the slope of tangent
Differentiate with respect to x
Slope of tangent to the curve y = x3 + ax + b at (1, -6) is
Given as line is x – y + 5 = 0
y = x + 5 is the form of equation of a straight line y = mx + c, here is the slope of the line.
Therefore the slope of the line is y = 1 × x + 5
Therefore the Slope is 1. … (2)
Given, the point (1, – 6) lie on the tangent, so
x = 1 & y = – 6 satisfies the equation, y = x3 + ax + b
– 6 = 13 + a × 1 + b
⇒ – 6 = 1 + a + b
⇒ a + b = – 7 … (3)
Here, the tangent is parallel to the line, from (1) & (2)
Thus, 3 + a = 1
⇒ a = – 2
From the equation (3)
a + b = – 7
⇒ – 2 + b = – 7
⇒ b = – 5
Therefore the value is a = – 2 & b = – 5
