If (a, b) be the orthocentre of the triangle whose vertices are (1,2),(2,3) and (3,1) , and I 1 =int a b x sin (4 x-x 2 ) d x, I 2 =int a b sin (4 x-x 2 ) d x , then 36 frac I 1 I 2 is equal to :

Mathematics · Definite Integration

JEE Main 2024 — 27 January, Shift 1 — Question 4

If $(a, b)$ be the orthocentre of the triangle whose vertices are $(1,2),(2,3)$ and $(3,1)$ , and $I_{1}=\int_{a}^{b} x \sin \left(4 x-x^{2}\right) d x, \quad I_{2}=\int_{a}^{b} \sin \left(4 x-x^{2}\right) d x$ , then $36 \frac{I_{1}}{I_{2}}$ is equal to :

Option A:
72
Correct
Option B:
88
Option C:
80
Option D:
66

Answer: A

Step-by-step solution

Equation of CE

$y-1=-(x-3)$

$x+y=4$

orthocentre lies on the line $x+y=4$

so, $a+b=4$

$I_{1}=\int_{a}^{b} x \sin (x(4-x)) d x$

Using king rule $\begin{gathered} I_{1}=\int_{a}^{b}(4-x) \sin (x(4-x)) d x \end{gathered}$ (i) + (ii) $2 \mathrm{I}_{1}=\int_{\mathrm{a}}^{\mathrm{b}} 4 \sin (x(4-\mathrm{x})) d x$

$2 \mathrm{I}_{1}=4 \mathrm{I}_{2}$

$\mathrm{I}_{1}=2 \mathrm{I}_{2}$

$\frac{I_{1}}{I_{2}}=2$

$\frac{36 \mathrm{I}_{1}}{\mathrm{I}_{2}}=72$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 1
Subject: Mathematics
Chapter: Definite Integration
Topic: Methods of solving definite integrals(kings rule,odd even)

If int 0^1 frac1√(3+x)+√(1+x) d x=a+b √(2)+c √(3) , where a, b, c are rational numbers, then 2 a+3 b-4 c is equal to :

Answer: A

Step-by-step solution

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