Mathematics

# If $g(1)=g(2)$ then the value of  $\int _{ 1 }^{ 2 }{ { \left[ f\{ g(x)\} \right] }^{ -1 } } f'\{ g(x)\} g'(x)dx\quad is$

$0$

##### SOLUTION
$\because g(1)=g(2)\\ \int _{ 1 }^{ 2 }{ { [f\{ g(x)\} ] }^{ -1 } } f^{ ' }\{ g(x)\} g^{ ' }\left( x \right) dx\\ \int _{ 1 }^{ 2 }{ \cfrac { 1 }{ f\{ g(x)\} } } f^{ ' }\{ g(x)\} g^{ ' }\left( x \right) dx\quad [\because \int { \cfrac { 1 }{ x } dx } =\log { x } ]\\ \therefore { =[\log { f(g(x)) } ] }_{ 1 }^{ 2 }\\ \Rightarrow \log { f(g(2)) } -\log { f(g(1)) } \\ \Rightarrow \log { f(t) } -\log { f(t) } \quad [let\quad g(1)=g(2)=t]\\ =0$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$