JEE Mains · Maths · STD 12 - 7.2 definite integral
If \((a, b)\) be the orthocentre of the triangle whose vertices are \((1,2),(2,3)\) and \((3,1)\), and \(I_1=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{x} \sin \left(4 \mathrm{x}-\mathrm{x}^2\right) \mathrm{dx}, \mathrm{I}_2=\int_{\mathrm{a}}^{\mathrm{b}} \sin \left(4 \mathrm{x}-\mathrm{x}^2\right) \mathrm{dx}\) , then \(36 \frac{\mathrm{I}_1}{\mathrm{I}_2}\) is equal to :
- A \(72\)
- B \(88\)
- C \(80\)
- D \(66\)
Answer & Solution
Correct Answer
(A) \(72\)
Step-by-step Solution
Detailed explanation
Equation of \(\mathrm{CE}\) \( y-1=-(x-3)\) \( x+y=4\) orthocentre lies on the line \(x+y=4\) \( \text { so, } a+b=4 \) \( I_1=\int_a^b x \sin (x(4-x)) d x\) \(.....(i)\) Using king rule \( I_1=\int_a^b(4-x) \sin (x(4-x)) d x \)\(......(ii)\) \(text { (i) }+ \text { (ii) }\)…
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