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

JEE · Math · previous-year question

  1. A.$${{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}$$correct
  2. B.0
  3. C.$$ - {{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}$$
  4. D.$${{{{\left( {\alpha - \beta } \right)}^2}} \over 2}$$

Answer

A. $${{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}$$

Explanation

Given limit $$ = \mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \,a\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over {{{\left( {x - \alpha } \right)}^2}}}$$ $$ = \mathop {\lim }\limits_{x \to \alpha } {{2{{\sin }^2}\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \right)} \over {{{\left( {x - \alpha } \right)}^2}}}$$ $$ = \mathop {\lim }\limits_{x \to \alpha } {2 \over {{{\left( {x - \alpha } \right)}^2}}} \times {{{{\sin }^2}\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \right)} \over {{{{a^2}{{\left( {x - \alpha } \right)}^2}{{\left( {x - \beta } \right)}^2}} \over 4}}} \times {{{a^2}{{\left( {x - \alpha } \right)}^2}{{\left( {x - \beta } \right)}^2}} \over 4}$$ $$ = {{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}.$$

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