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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
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B
12
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C
1
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D
none of these
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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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