Application of DerivativeHard
Question
The point at which the tangent to the curve y = x3 + 5 is perpendicular to the line x + 3y = 2 are-
Options
A.(6, 1), (-1, 4)
B.(6, 1) (4, -1)
C.(1, 6), (1, 4)
D.(1, 6), (-1, 4)
Solution
Let point (x1, y1)
y = x3 + 5
= 3x2
= 3 x21
It is ⊥ to x + 3y - 2 = 0
So 3 x21 × -
= -1
x1 = ±1
x1 = 1, y1 = 6
x1 = -1, y1 = 4
(1, 6) and (-1, 4)
y = x3 + 5
= 3x2= 3 x21It is ⊥ to x + 3y - 2 = 0
So 3 x21 × -
= -1x1 = ±1
x1 = 1, y1 = 6
x1 = -1, y1 = 4
(1, 6) and (-1, 4)
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