JEE MAIN - Mathematics (2021 - 16th March Morning Shift - No. 17)
If the normal to the curve y(x) = $$\int\limits_0^x {(2{t^2} - 15t + 10)dt} $$ at a point (a, b) is parallel to the line x + 3y = $$-$$5, a > 1, then the value of | a + 6b | is equal to ___________.
Answer
406
Explanation
Normal to the curve at point P(a, b) is parallel to the line x + 3y = $$-$$5.
mnormal = $$ - {1 \over 3}$$
$$ \therefore $$ mtangent = 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
$$ \therefore $$ $$b = \int_0^a {(2{t^2} - 15t + 10)dt} $$
$$b = \int_0^7 {(2{t^2} - 15t + 10)dt} $$
$$6b = - 413$$
$$ \therefore $$ $$|a + 6b|\, = 406$$
mnormal = $$ - {1 \over 3}$$
$$ \therefore $$ mtangent = 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
$$ \therefore $$ $$b = \int_0^a {(2{t^2} - 15t + 10)dt} $$
$$b = \int_0^7 {(2{t^2} - 15t + 10)dt} $$
$$6b = - 413$$
$$ \therefore $$ $$|a + 6b|\, = 406$$
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