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Question

Let $$\alpha $$ and $$\beta$$ be the distinct roots of $$a{ x }^{ 2 }+bx+c=0$$, then $$\displaystyle\lim _{ x\rightarrow \alpha  }{ \dfrac { 1-\cos { \left( a{ x }^{ 2 }+bx+c \right)  }  }{ { \left( x-\alpha  \right)  }^{ 2 } }  } $$ is equal to


A
12(αβ)2
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B
α22(αβ)2
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C
0
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D
α22(αβ)2
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Solution

The correct option is B $$\dfrac { { \alpha }^{ 2 } }{ 2 } { \left( \alpha -\beta \right) }^{ 2 }$$
We have, $$\displaystyle\lim _{ x\rightarrow \alpha  }{ \dfrac { 1-\cos { \left( a{ x }^{ 2 }+bx+c \right)  }  }{ { \left( x-\alpha  \right)  }^{ 2 } }  }$$

$$=\displaystyle\lim _{ x\rightarrow \alpha  }{ \dfrac { 2\sin ^{ 2 }{ \left( \dfrac { a{ x }^{ 2 }+bx+c }{ 2 }  \right)  }  }{ { \left( x-\alpha  \right)  }^{ 2 } }  } $$
[Since, $$\alpha $$ and $$\beta$$ are the roots of $$a{ x }^{ 2 }+bx+c=0$$, so it can be written as $$a\left( x-\alpha  \right) \left( x-\beta  \right) =0$$]

$$=\displaystyle\lim _{ x\rightarrow \alpha  }{ \dfrac { 2\sin ^{ 2 }{ \left( \dfrac { a\left( x-\alpha  \right) \left( x-\beta  \right)  }{ 2 }  \right)  }  }{ { \left( x-\alpha  \right)  }^{ 2 } }  } $$

$$=\displaystyle\lim _{ x\rightarrow \alpha  }{ \dfrac { 2\sin ^{ 2 }{ \left( \dfrac { a }{ 2 } \left( x-\alpha  \right) \left( x-\beta  \right)  \right)  } { \left( \dfrac { a }{ 2 }  \right)  }^{ 2 }{ \left( x-\beta  \right)  }^{ 2 } }{ { \left[ \left( \dfrac { a }{ 2 }  \right) \left( x-\alpha  \right) \left( x-\beta  \right)  \right]  }^{ 2 } }  } $$

$$=\displaystyle\lim _{ x\rightarrow \alpha  }{ 2{ \left( \dfrac { a }{ 2 }  \right)  }^{ 2 }{ \left( x-\beta  \right)  }^{ 2 } } $$

$$=\displaystyle\lim _{ x\rightarrow \alpha  }{ \dfrac { { a }^{ 2 } }{ 2 } \cdot { \left( x-\beta  \right)  }^{ 2 } } =\dfrac { { a }^{ 2 } }{ 2 } { \left( \alpha -\beta  \right)  }^{ 2 }$$

Mathematics

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