Question

# Let $$\displaystyle { I }_{ 1 }=\int _{ \sec ^{ 2 }{ z } }^{ 2-\tan ^{ 2 }{ z } }{ xf\left( x\left( 3-x \right) \right) dx }$$ and $$\displaystyle { I }_{ 2 }=\int _{ \sec ^{ 2 }{ z } }^{ 2-\tan ^{ 2 }{ z } }{ f\left( x\left( 3-x \right) \right) dx }$$, where $$f$$ is a continuous function and $$z$$ is any real number, $$\displaystyle \frac { { I }_{ 1 } }{ { I }_{ 2 } } =$$

A
32
B
12
C
1
D
none of these

Solution

## The correct option is B $$\displaystyle \frac { 3 }{ 2 }$$We have, $$\displaystyle { I }_{ 1 }=\int _{ \sec ^{ 2 }{ z } }^{ 2-\tan ^{ 2 }{ z } }{ xf\left( x\left( 3-x \right) \right) dx }$$$$\displaystyle =\int _{ \sec ^{ 2 }{ z } }^{ 2-\tan ^{ 2 }{ z } }{ \left( 3-x \right) f\left( \left( 3-x \right) \left( 3-\left( 3-x \right) \right) \right) } dx$$   $$\displaystyle \left[ \because \int _{ a }^{ b }{ f\left( x \right)dx } =\int _{ a }^{ b }{ f\left( a+b-x \right) dx } \right]$$$$\displaystyle =\int _{ \sec ^{ 2 }{ z } }^{ 2-\tan ^{ 2 }{ z } }{ \left( 3-x \right) f\left( x\left( 3-x \right) \right) dx }$$$$\displaystyle =3\int _{ \sec ^{ 2 }{ z } }^{ 2-\tan ^{ 2 }{ z } }{ f\left( x\left( 3-x \right) \right) dx } -\int _{ \sec ^{ 2 }{ z } }^{ 2-\tan ^{ 2 }{ z } }{ xf\left( x\left( 3-x \right) \right) } dx$$$$=3{ I }_{ 2 }-{ I }_{ 1 }$$$$\displaystyle \therefore 2{ I }_{ 1 }=3{ I }_{ 2 }\Rightarrow \frac { { I }_{ 1 } }{ { I }_{ 2 } } =\frac { 3 }{ 2 }$$Mathematics

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