QJEEM21Ma0f02: JEE Main 2021 Mathematics - Class 12 JEEM21Ma0f02 Solution: If the normal to the curve y(x) = at a point (a, b) is parallel …

Normal to the curve at point P(a, b) is parallel to the line x + 3y = $-$ 5.

m_normal = $- {1 \over 3}$

$\therefore$ m_tangent = 3 = ${{dy} \over {dx}}$

Given y(x) = $\int\limits_0^x {(2{t^2} - 15t + 10)dt}$

$\Rightarrow$ y'(x) = (2x² $-$ 15x + 10)

at point P(a, b)

3 = (2a² $-$ 15a + 10)

$\Rightarrow$ 2a² $-$ 15a + 7 = 0

$\Rightarrow$ 2a² $-$ 14a $-$ a + 7 = 0

$\Rightarrow$ 2a(a $-$ 7) $-$ 1 (a $-$ 7) = 0

a = ${1 \over 2}$ or 7,

given a > 1 $\therefore$ a = 7

As P(a, b) lies on curve

$\therefore$ $b = \int_0^a {(2{t^2} - 15t + 10)dt}$

$b = \int_0^7 {(2{t^2} - 15t + 10)dt}$

$6b = - 413$

$\therefore$ $|a + 6b|\, = 406$