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प्रश्न
The points at which the tangent passes through the origin for the curve y = 4x3 – 2x5 are
विकल्प
(0, 0), (2, 1) and (–1, –2)
(0, 0), (2, 1) and (–2, –1)
(2, 0), (2, 1) and(–3, 1)
(0, 0), (1, 2) and (–1, –2)
उत्तर
(0, 0), (1, 2) and (–1, –2)
Explanation:
The equation of the given curve is y = `4x^3 – 2x^5`
`(dy)/(dx) = 12x^2 - 10x^4`
Therefore, the slope of the tangent at point (x, y) is `12x^2 - 10x^4`
The equation of the tangent at (x, y) is given by `Y - y = (12x^2 - 10x^4)(X - x)` .......(i)
When, the tangent passes through the origin (0, 0), then X = Y = 0
Therefore equation (i) reduce to `- y = (12x^2 - 10x^4) (- x)`
⇒ `y = 12x^3 - 10x^5`
Also, we have `y = 4x^3 - 2x^5`
`12x^3 - 10x^5 = 4x^3 - 2x^5`
∴ `12x^3 - 10x^5 = 4x^3 - 2x^5`
⇒ `8x^5 - 8x^3` = 0
⇒ `x^5 - x^3` = 0
⇒ `x^3(x^2 - 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
Hence, the require points are (0, 0), (1, 2) and (–1, –2).
