Verified Solution • 2025–26 Session moderate

QJEEM05Ma650f: JEE Main 2005 Mathematics - Class 12 JEEM05Ma650f Solution: Let and be the distinct roots of , then is equal to

Let α\alpha and β\beta be the distinct roots of ax2+bx+c=0a{x^2} + bx + c = 0 , then

lim⁡x→α1−cos⁡(ax2+bx+c)(x−α)2\mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \left( {a{x^2} + bx + c} \right)} \over {{{\left( {x - \alpha } \right)}^2}}} is equal to

Multiple Choice Options

  • A 0
  • B a2(α−β)22{{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}
  • C −a2(α−β)22- {{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}
  • D (α−β)22{{{{\left( {\alpha - \beta } \right)}^2}} \over 2}

Step-by-Step Verified Solution

Given limit =lim⁡x→α1−cos⁡ a(x−α)(x−β)(x−α)2= \mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \,a\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over {{{\left( {x - \alpha } \right)}^2}}}

=lim⁡x→α2sin⁡2(a(x−α)(x−β)2)(x−α)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}}}

=lim⁡x→α2(x−α)2×sin⁡2(a(x−α)(x−β)2)a2(x−α)2(x−β)24×a2(x−α)2(x−β)24= \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}

=a2(α−β)22.= {{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}.

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