For the curve y = 4x³ - 2x⁵ find all the points at which the tangent passes through the origin

Solution:

The equation of the given curve is y = 4x³ - 2x⁵

Therefore,

dy/dx = 12x² - 10x⁴

Hence, the slope of the tangent at the point (x, y) is 12x² - 10x⁴

Thus, the equation of the tangent at (x, y) is given by,

Y - y = (12x² - 10x⁴)(X - x)

When the tangent passes through the origin (0, 0),

X = Y = 0.

Therefore,

- y = (12x² - 10x⁴)(- x)

y = 12x³ - 10x⁵

Also, we have y = 4x³ - 2x⁵

Hence,

∴ 12x³ - 10x⁵ = 4x³ - 2x⁵

⇒ 8x⁵ - 8x³ = 0

⇒ x⁵ - x³ = 0

⇒ x³ (x² - 1) = 0

⇒ x = 0, ± 1

When, x = 0,

⇒ y = 4 (0)³ - 2 (0)⁵ = 0

When, x = 1,

⇒ y = 4 (1)³ - 2 (1)⁵ = 2

When, x = - 1,

⇒ y = 4 (- 1)³ - 2 (- 1)⁵ = - 2

Thus, the points are (0, 0) , (1, 2) and (- 1, - 2)

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 18

For the curve y = 4x³ - 2x⁵ find all the points at which the tangent passes through the origin

Summary:

For the given curve y = 4x³ - 2x⁵, the points at which the tangent passes through the origin are (0, 0) , (1, 2) and (- 1, - 2)