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
B
α22(αβ)2
C
0
D
α22(αβ)2

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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