JEE Main 2019 — Application Of Derivatives Question with Solution

Slope of the tangent to the curve y = x³ + ax – b at point (1, –5)

m₁ = $${\frac{dy}{dx}|}_{\left(1,-5\right)}$$ = 3x² + a = 3 + a

Slope of the line –x + y + 4 = 0,

m₂ = 1

As line and tangent to the curve are perpendicular to each other,

$$\therefore$$ m₁ $$\times$$ m₂ = -1

$$\Rightarrow$$ (3 + a) $$\times$$ 1 = -1

$$\Rightarrow$$ a = - 4

$$\therefore$$ Curve becomes y = x³ - 4x – b

This curve goes through (1, –5)

$$\therefore$$ -5 = 1 - 4 - b

$$\Rightarrow$$ b = 2

So curve is y = x³ - 4x – 2

By checking each options you can see,

(2, –2) lies on the curve.