Question

The point at which the tangent to the curve y = x³ + 5 is perpendicular to the line x + 3y = 2 are
1. (1, 6), (-1, 4)
2. (1, 6), (1, 4)
3. (6, 1), (4, 1)
4. (6, 1), (-1, 4)

Answer 1

Correct Answer - Option 1 : (1, 6), (-1, 4)

Explanation:

Tangent to the curve y = x³ + 5

\({m_1} = \frac{{dy}}{{dx}} = 3{x^2}\)      ---(i)

Line equation 3y = 2 – x

⇒ \(y = - \frac{1}{3}x + 2\)

\({m_2} = \frac{{dy}}{{dx}} = \frac{{ - 1}}{3}\)    ---(ii)

Now,

m₁ × m₂ = -1 (for lines to be perpendicular to (ii)]

\(3{x^2} \times \left( {\frac{{ - 1}}{3}} \right) = - 1\)

x² = 1

x = ± 1

Now,

y = 1 + 5

y = 6

y = -1 + 5 = 4

∴ Points at which these lines are perpendicular are (1, 6), (-1, 4)