The point(s) at each of which the tangents to the curve $y = x^{3} - 3 x^{2} - 7 x + 6$ cut off on the positive semi axis $O X$ a line segment half that on the negative semi axis $O Y$ , then the co-ordinates of the point(s) is/are give by:

(- 1, 9)

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(3, - 15)

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(1, - 3)

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none

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Solution

The correct option is B $(3, - 15)$
$y = x^{3} - 3 x^{2} - 7 x + 6 \frac{d y}{d x} = 3 x^{2} - 6 x - 7$
Let coordinate of X be $(a, 0)$ then coordinate of Y is $(0, - 2 a)$
Therefore slope of XY is $\frac{0 + 2 a}{a - 0} = 2$
Equating slope $3 x^{2} - 6 x - 7 = 2 3 x^{2} - 6 x - 9 = 0$
Solving we get
$x = - 1, y = 9 x = 3, y = - 15$
Hence, option 'B' is correct.

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Similar questions

y = x^{3} - 3 x^{2} - 7 x + 6

y = x^{3} - 9 x^{2} + 15 x + 3

y = x^{3} - 3 x^{2} - 9 x + 7

x -

(1, 2)

\frac{1}{3}

x = a {cos}^{3} t, y = a {sin}^{3} t

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