Question

If the tangent to the curve, y = x³ + ax – b at the point (1, –5) is perpendicular to the line –x + y + 4 = 0, then which of the following point lies on the curve?

विकल्प

• (–2, 2)

• (2, –2)

• (–2, 1)

• (2, –1)

Answer 1

(2, –2)

Explanation:

Given, curve is y = x³ + ax – b  ...(i)

Passes through point P(1, –5).

∴ –5 = 1 + a – b

`\implies` b – a = 6  ...(ii)

and slope of tangent at point P(1, –5) to the curve is,

m₁ = `dy/dx|_((1"," -5))`

= `[3x^2 + a]_((1"," -5))`

= a + 3

∵ The slope of tangent m₁ = a + 3 at point P(1, –5) is perpendicular to line –x + y + 4 = 0, whose slope is m₂ = 1.

∴ a + 3 = –1   ...[∵ m₁m₂ = –1]

`\implies` a = – 4  

Now, on substituting a = – 4 in equation (ii), we get b = 2

On putting a = – 4 and b = 2 in equation (i), we get

y = x³ – 4x – 2

Now, (2, –2) is the required point that lies on it.