JEE MAIN - Mathematics (2021 - 27th July Morning Shift - No. 20)
Let the domain of the function
$$f(x) = {\log _4}\left( {{{\log }_5}\left( {{{\log }_3}(18x - {x^2} - 77)} \right)} \right)$$ be (a, b). Then the value of the integral $$\int\limits_a^b {{{{{\sin }^3}x} \over {({{\sin }^3}x + {{\sin }^3}(a + b - x)}}} dx$$ is equal to _____________.
$$f(x) = {\log _4}\left( {{{\log }_5}\left( {{{\log }_3}(18x - {x^2} - 77)} \right)} \right)$$ be (a, b). Then the value of the integral $$\int\limits_a^b {{{{{\sin }^3}x} \over {({{\sin }^3}x + {{\sin }^3}(a + b - x)}}} dx$$ is equal to _____________.
Answer
1
Explanation
For domain
$${\log _5}\left( {{{\log }_3}(18x - {x^2} - 77)} \right) > 0$$
$${\log _3}(18x - {x^2} - 77) > 1$$
$$18x - {x^2} - 77 > 3$$
$${x^2} - 18x + 80 < 0$$
$$x \in (8,10)$$
$$\Rightarrow$$ a = 8 and b = 10
$$I = \int\limits_a^b {{{{{\sin }^3}x} \over {{{\sin }^3}x + {{\sin }^3}(a + b - x)}}} dx$$
$$I = \int\limits_a^b {{{{{\sin }^3}x(a + b - x)} \over {{{\sin }^3}x + {{\sin }^3}(a + b - x)}}} $$
$$2I = (b - a) \Rightarrow I = {{b - a} \over 2}$$ ($$\because$$ a = 8 and b = 10)
$$I = {{10 - 8} \over 2} = 1$$
$${\log _5}\left( {{{\log }_3}(18x - {x^2} - 77)} \right) > 0$$
$${\log _3}(18x - {x^2} - 77) > 1$$
$$18x - {x^2} - 77 > 3$$
$${x^2} - 18x + 80 < 0$$
$$x \in (8,10)$$
$$\Rightarrow$$ a = 8 and b = 10
$$I = \int\limits_a^b {{{{{\sin }^3}x} \over {{{\sin }^3}x + {{\sin }^3}(a + b - x)}}} dx$$
$$I = \int\limits_a^b {{{{{\sin }^3}x(a + b - x)} \over {{{\sin }^3}x + {{\sin }^3}(a + b - x)}}} $$
$$2I = (b - a) \Rightarrow I = {{b - a} \over 2}$$ ($$\because$$ a = 8 and b = 10)
$$I = {{10 - 8} \over 2} = 1$$
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