Verified Solution • 2025–26 Session moderate

QJEEM20Mae40b: JEE Main 2020 Mathematics - Class 12 JEEM20Mae40b Solution: If is positive root of the equation, p(x) = x 2 - x - 2 = 0,…

If α\alpha is positive root of the equation, p(x) = x2 - x - 2 = 0, then

lim⁡x→α+1−cos⁡(p(x))x+α−4\mathop {\lim }\limits_{x \to {\alpha ^ + }} {{\sqrt {1 - \cos \left( {p\left( x \right)} \right)} } \over {x + \alpha - 4}} is equal to :

Multiple Choice Options

  • A 12{1 \over \sqrt2}
  • B 12{1 \over 2}
  • C 32{3 \over \sqrt2}
  • D 32{3 \over 2}

Step-by-Step Verified Solution

x2−x−2=0{x^2} - x - 2 = 0

roots are 2 & −- 1 ⇒\Rightarrow α\alpha = 2 (given α\alpha is positive)

Now lim⁡x→2+1−cos⁡(x2−x−2)(x−2)\mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {1 - \cos ({x^2} - x - 2)} } \over {(x - 2)}}

=lim⁡x→2+2sin⁡2(x2−x−2)2(x−2)= \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {2{{\sin }^2}{{({x^2} - x - 2)} \over 2}} } \over {(x - 2)}}

=lim⁡x→2+2sin⁡((x−2)(x+1)2)(x−2)= \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt 2 \sin \left( {{{(x - 2)(x + 1)} \over 2}} \right)} \over {(x - 2)}}

=32= {3 \over {\sqrt 2 }}

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