Question

If the normal to the curve y(x) = `int_0^x(2t^2 - 15t + 10)dt` at a point (a, b) is parallel to the line x + 3y = –5, a > 1, then the value of |a + 6b| is equal to ______.

Options

• 405

• 406

• 407

• 408

Answer 1

If the normal to the curve y(x) = `int_0^x(2t^2 - 15t + 10)dt` at a point (a, b) is parallel to the line x + 3y = –5, a > 1, then the value of |a + 6b| is equal to 406.

Explanation:

Given, y(x) = `int_0^x(2t^2 - 15t + 10)dt`

Apply Leibnitz theorem

H(x) = `int_(f_2(x))^(f_1(x)) g(t)dt`

H'(x) = `g(f_1(x))f_1^'(x) - g(f_2(x))f_2^'(x)`

Using the Leibnitz theorem, we can write

y'(x) = `(2x^2 - 15x + 10)(d(x))/(dx) - (10)(d(0))/(dx)`

⇒ y'(x) = 2x² – 15x + 10

y'(x) at point P(a, b)

y'(a) = 2a² – 15a + 10  ...(i)

Normal is parallel to the line x + 3y = –5,

So slope of normal will be equal to `-1/3`

Let m_N = `-1/3`

m_r = 3;(m_N.m_r = –1)  ...(ii)

y'(a) = m_r

Now using equation (i) and (ii)

2a² – 15a + 10 = 3

⇒ (a – 7)(2a – 1) = 0

⇒ a = 7, `1/2`

So, a = 7, a = `1/2` rejected because a > 1  ...(iii)

For the value of b: b = y(a)

⇒ b = y(7)

⇒ b = `int_0^7(2t^2 - 15t + 10)dt`

⇒ b = `((2t^3)/3 - (15t^2)/2 + 10t)_0^7`

⇒ b = `(2/3)(7)^3 - 15/2(7)^2 + 10(7) - 0`

⇒ b = `((2)^2(7)^3 - (15)(7)^2(3) + (10)(7)(6))/6`

⇒ 6b = (4 × 343) – (15)(49)(3) + 420

⇒ 6b = 1372 – 2205 + 420

⇒ 6b = 1792 – 2205

⇒ 6b = –413  ...(iv)

Now using equation (iii) and (iv)

|a + 6b| = |7 – 413|

⇒ |a + 6b| = |–406|    ∵ |x| = `{(x; x ≥ 0),(-x; x < 0):}`

⇒ |a + 6b| = 406