JEE MAIN - Mathematics (2024 - 27th January Morning Shift - No. 5)
If $(a, b)$ be the orthocentre of the triangle whose vertices are $(1,2),(2,3)$ and $(3,1)$, and $\mathrm{I}_1=\int\limits_{\mathrm{a}}^{\mathrm{b}} x \sin \left(4 x-x^2\right) \mathrm{d} x, \mathrm{I}_2=\int\limits_{\mathrm{a}}^{\mathrm{b}} \sin \left(4 x-x^2\right) \mathrm{d} x$, then $36 \frac{\mathrm{I}_1}{\mathrm{I}_2}$ is equal to :
80
72
66
88
Explanation
Equation of CE
$$\begin{aligned} & y-1=-(x-3) \\ & x+y=4 \end{aligned}$$
_27th_January_Morning_Shift_en_5_1.png)
orthocentre lies on the line $$x+y=4$$
so, $$a+b=4$$
$$I_1=\int_\limits a^b x \sin (x(4-x)) d x\quad$$ ..... (i)
Using king rule
$$I_1=\int_\limits a^b(4-x) \sin (x(4-x)) d x\quad$$ .... (ii)
$$\begin{aligned} & \text { (i) }+ \text { (ii) } \\ & 2 \mathrm{I}_1=\int_\limits{\mathrm{a}}^{\mathrm{b}} 4 \sin (\mathrm{x}(4-\mathrm{x})) \mathrm{dx} \\ & 2 \mathrm{I}_1=4 \mathrm{I}_2 \\ & \mathrm{I}_1=2 \mathrm{I}_2 \\ & \frac{\mathrm{I}_1}{\mathrm{I}_2}=2 \\ & \frac{36 \mathrm{I}_1}{\mathrm{I}_2}=72 \end{aligned}$$
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