If the normal to the curve $y (x) = \int_{0}^{x} (2 t^{2} - 15 t + 10) d t$ at a point $(a, b)$ is parallel to the line $x + 3 y = - 5, a > 1$ then the value of $| a + 6 b |$ is equal to

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Solution

$y^{'} (x) = (2 x^{2} - 15 x + 10)$
at point $P (a, b)$ slope of normal is $- 1 / 3$
$∴ 3 = (2 a^{2} - 15 a + 10) \Rightarrow 2 a^{2} - 15 a + 7 = 0 \Rightarrow 2 a^{2} - 14 a - a + 7 = 0 \Rightarrow 1 a (a - 7) - 1 (a - 7) = 0 a = \frac{1}{2} or 7, given a > 1 ∴ a = 7 also, P lies on curve ∴ b = \int_{0}^{a} (2 t^{2} - 15 t + 10) d t b = \int_{0}^{7} (2 t^{2} - 15 t + 10) d t 6 b = - 413 ∴ | a + 6 b | = 406$

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Similar questions

y (x) = \int_{0}^{x} (2 t^{2} - 15 t + 10) d t

(a, b)

x + 3 y = - 5, a > 1

| a + 6 b |

y = x^{3} + ax + b

y = x^{3} + ax + b

a x + b y + c = 0

x^{3} y = 1

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