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

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Definite Integrals

(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 :

Answer: (A)

The orthocentre

(a, b)

of the triangle with vertices

(1, 2), (2, 3), (3, 1)

lies on the altitude relations that reduce to

x + y = 4

, so

a + b = 4

.

Using

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

I_{1} = \int_{a}^{b} x sin (4 x - x^{2}) d x

(noting

sin (4 (a + b - x) - (a + b - x)^{2}) = sin (4 x - x^{2})

when

a + b = 4

I_{1} = \int_{a}^{b} (4 - x) sin (4 x - x^{2}) d x

Adding this to the original form of

I_{1}

2 I_{1} = \int_{a}^{b} 4 sin (4 x - x^{2}) d x = 4 I_{2} \Rightarrow \frac{I _{1}}{I _{2}} = 2

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

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Single correct option