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JEE Main 202127 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
27 July
Shift
Morning Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2021 · 27 July · Morning Shift · Mathematics · Functions

Let the domain of the function f (x) = _ 4 ( _ 5 ( _ 3 (18x - x ^ 2...

Let the domain of the function ff\left(x\right) = \log_{4}\left(\log_{5}\left(\log_{3}\left(18x - {x}^{2} - 77\right)\right)\right)be(a,b)be \left(a, b\right). Then the value of the integral ∫absin⁡3x(sin⁡3x+sin⁡3(a+b−x)){∫}_{a}^{b}\dfrac{\sin^{3}x}{\left(\sin^{3}x + \sin^{3}\left(a + b - x\right)\right)} is equal to _____.

Correct answer

0

Step-by-step solution

For domain log⁡5(log⁡3(18x−x2−77))>0\log_{5}\left(\log_{3}\left(18x - {x}^{2} - 77\right)\right) > 0

log⁡3(18x−x2−77)>1\log_{3}\left(18x - {x}^{2} - 77\right) > 1

⇒18x−x2−77>3⇒ 18x - {x}^{2} - 77 > 3

⇒x2−18x+80<0⇒ {x}^{2} - 18x + 80 < 0

⇒x∈(8,10)⇒ x \in \left(8, 10\right)

⇒a=8⇒ a = 8 and b=10b = 10

I=∫absin⁡3xsin⁡3x+sin⁡3(a+b−x)dxI = {∫}_{a}^{b}\dfrac{\sin^{3}x}{\sin^{3}x + \sin^{3}\left(a + b - x\right)}dx

Apply ∫abf(x)dx=∫abf(a+b−x)dx{∫}_{a}^{b}f\left(x\right)dx = {∫}_{a}^{b}f\left(a + b - x\right)dx

I=∫absin⁡3(a+b−x)sin⁡3x+sin⁡3(a+b−x)dxI = {∫}_{a}^{b}\dfrac{\sin^{3}\left(a + b - x\right)}{\sin^{3}x + \sin^{3}\left(a + b - x\right)}dx

⇒2I=(b−a)⇒I=b−a2(∵a=8 and b=10)⇒ 2I = \left(b - a\right) ⇒ I = \dfrac{b - a}{2}\left(∵ a = 8 \text{ and } b = 10\right)

⇒I=10−82=1⇒ I = \dfrac{10 - 8}{2} = 1

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