If be the orthocentre of the triangle whose vertices are (1, 2), (2, 3) and (3, 1) and then is equal to

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Solution

Q.
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 : [2024]

1 66

2 88

3 72

4 80

Ans.
(3)

Let the vertices of the triangle are $A (3, 1), B (1, 2)$ and $C (2, 3)$ .

$Slope of B C = \frac{3 - 2}{2 - 1} = 1$

Since, AD is perpendicular to BC.

Slope of AD = $- 1$

Equation of line $A D$ is

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

Since, $(a, b)$ lies on AD.

$∴$ $b - 1 = - 1 (a - 3) \Rightarrow a + b = 4$

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

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

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

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

$\Rightarrow 2 I_{1} = 4 I_{2} \Rightarrow \frac{I_{1}}{I_{2}} = 2$ $∴$ $\frac{36 I_{1}}{I_{2}} = 36 \times 2 = 72$

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