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प्रश्न
Find points on the curve given by y = x3 − 6x2 + x + 3, where the tangents are parallel to the line y = x + 5.
उत्तर
Equation of the curve is y = x3 − 6x2 + x + 3
Differentiating w.r.t. x, we get
`("d"y)/("d"x)` = 3x2 – 12x + 1 ...(i)
Slope of the line y = x + 5 is 1.
Since the tangent is parallel to y = x + 5, their slopes are equal.
∴ Slope of the tangent = `("d"y)/("d"x)` = 1
∴ 3x2 – 12x + 1 = 1 ...[From (i)]
∴ 3x2 – 12x = 0
∴ 3x(x – 4) = 0
∴ x = 0 or x = 4
When x = 0, y = (0)3 – 6(0)2 + 0 + 3 = 3
When x = 4, y = (4)3 – 6(4)2 + 4 + 3 = – 25
∴ The points on the given curve are (0, 3) and (4, – 25).
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