JEE MAIN - Mathematics (2021 - 27th August Morning Shift - No. 10)
$$\int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 44x + 484)}}dx} $$ is equal to :
6
8
5
10
Explanation
Let $$I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 44x + 484)}}dx} $$
$$I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 22)}}dx} $$ .... (1)
We know,
$$\int\limits_a^b {f(x)dx} = \int\limits_a^b {f(a + b - x)} \,dx$$ (king)
So, $$I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{{(22 - x)}^2} + {{\log }_e}{{(21 - (22 - x))}^2}}}} $$
$$I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{x^2} + {{\log }_e}{{(22 - x)}^2}}}dx} $$ .... (2)
(1) + (2)
$$2I = \int\limits_6^{16} {1.\,dx} = 10$$
I = 5
$$I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 22)}}dx} $$ .... (1)
We know,
$$\int\limits_a^b {f(x)dx} = \int\limits_a^b {f(a + b - x)} \,dx$$ (king)
So, $$I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{{(22 - x)}^2} + {{\log }_e}{{(21 - (22 - x))}^2}}}} $$
$$I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{x^2} + {{\log }_e}{{(22 - x)}^2}}}dx} $$ .... (2)
(1) + (2)
$$2I = \int\limits_6^{16} {1.\,dx} = 10$$
I = 5
Comments (0)
