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Definite Integration question

2021 · 16 Mar · Shift 1 · Q39
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Definite Integration question

2021 · 16 Mar · Shift 1 · Q39

JEE MainMathematicsDefinite IntegrationNumerical+4 / −1
If the normal to the curve y(x) = ∫0x(2t2−15t+10)dt\int\limits_0^x {(2{t^2} - 15t + 10)dt}0∫x​(2t2−15t+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 ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 406

  1. Find the curve explicitly

Given

y(x)=∫0x(2t2−15t+10) dty(x)=\int_0^x (2t^2-15t+10)\,dty(x)=∫0x​(2t2−15t+10)dt

Integrating,

y(x)=[2t33−15t22+10t]0x=2x33−15x22+10xy(x)=\left[\frac{2t^3}{3}-\frac{15t^2}{2}+10t\right]_0^x =\frac{2x^3}{3}-\frac{15x^2}{2}+10xy(x)=[32t3​−215t2​+10t]0x​=32x3​−215x2​+10x

So at the point (a,b)(a,b)(a,b),

b=y(a)=2a33−15a22+10ab=y(a)=\frac{2a^3}{3}-\frac{15a^2}{2}+10ab=y(a)=32a3​−215a2​+10a
  1. Use the condition on the normal

The line

x+3y=−5x+3y=-5x+3y=−5

can be written as

y=−13x−53y=-\frac{1}{3}x-\frac{5}{3}y=−31​x−35​

so its slope is

−13-\frac{1}{3}−31​

Since the normal to the curve is parallel to this line, the slope of the normal is also

−13-\frac{1}{3}−31​

Hence the slope of the tangent is the negative reciprocal:

mtangent=3m_{\text{tangent}}=3mtangent​=3
  1. Differentiate the curve

By the Fundamental Theorem of Calculus,

y′(x)=2x2−15x+10y'(x)=2x^2-15x+10y′(x)=2x2−15x+10

At x=ax=ax=a,

y′(a)=2a2−15a+10y'(a)=2a^2-15a+10y′(a)=2a2−15a+10

Since the tangent slope is 333,

2a2−15a+10=32a^2-15a+10=32a2−15a+10=3 2a2−15a+7=02a^2-15a+7=02a2−15a+7=0

Solve:

2a2−14a−a+7=02a^2-14a-a+7=02a2−14a−a+7=0 (2a−1)(a−7)=0(2a-1)(a-7)=0(2a−1)(a−7)=0

So,

a=12ora=7a=\frac12 \quad \text{or} \quad a=7a=21​ora=7

Given a>1a>1a>1, we get

a=7a=7a=7
  1. Find bbb

Substitute a=7a=7a=7 into y(x)y(x)y(x):

b=2(7)33−15(7)22+10(7)b=\frac{2(7)^3}{3}-\frac{15(7)^2}{2}+10(7)b=32(7)3​−215(7)2​+10(7) b=6863−7352+70b=\frac{686}{3}-\frac{735}{2}+70b=3686​−2735​+70

Take LCM 666:

b=1372−2205+4206b=\frac{1372-2205+420}{6}b=61372−2205+420​ b=−4136b=\frac{-413}{6}b=6−413​
  1. Compute ∣a+6b∣|a+6b|∣a+6b∣
a+6b=7+6(−4136)=7−413=−406a+6b=7+6\left(-\frac{413}{6}\right)=7-413=-406a+6b=7+6(−6413​)=7−413=−406

Therefore,

∣a+6b∣=406|a+6b|=406∣a+6b∣=406
  1. Comparison with stored answer

Derived answer is 406406406, which matches the stored correct answer.

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