Equation of the line normal to function f(x) = (x - 8) $^{2 / 3}$ + 1 at P(0, 5) is

y = 3x - 5

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y = 3x + 5

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3y = x + 15

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3y = x - 5

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Solution

The correct option is B y = 3x + 5
f(x) = (x - 8) $^{2 / 3}$ + 1

f '(x) = $\frac{2}{3} \frac{1}{(x - 8)^{1 / 3}}$

At (0, 5) f '(x) = $\frac{2}{3} \times \frac{1}{(- 8)^{1 / 3}}$

= $- \frac{1}{3} =$ Slope of tangent

Solpe of normal = 3

$∴$ Equation of normal at (0, 5) is

y - 5 = +3 (x - 0)

$∴$ y = 3x + 5

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Equation of the line normal to function f(x) = (x - 8)2/3 + 1 at P(0, 5) is

The correct option is B y = 3x + 5f(x) = (x - 8)2/3 + 1 f '(x) = 231(x−8)1/3 At (0, 5) f '(x) = 23×1(−8)1/3 = −13= Slope of tangent Solpe of normal = 3 ∴ Equation of normal at (0, 5) is y - 5 = +3 (x - 0) ∴ y = 3x + 5

Equation of the line normal to function f(x) = (x - 8) $^{2 / 3}$ + 1 at P(0, 5) is