JEE MAIN - Mathematics (2020 - 5th September Morning Slot - No. 1)
If $$\alpha $$ is positive root of the equation, p(x) = x2 - x - 2 = 0, then
$$\mathop {\lim }\limits_{x \to {\alpha ^ + }} {{\sqrt {1 - \cos \left( {p\left( x \right)} \right)} } \over {x + \alpha - 4}}$$ is equal to :
$$\mathop {\lim }\limits_{x \to {\alpha ^ + }} {{\sqrt {1 - \cos \left( {p\left( x \right)} \right)} } \over {x + \alpha - 4}}$$ is equal to :
$${1 \over \sqrt2}$$
$${1 \over 2}$$
$${3 \over \sqrt2}$$
$${3 \over 2}$$
Explanation
$${x^2} - x - 2 = 0$$
roots are 2 & $$-$$1 $$ \Rightarrow $$ $$\alpha $$ = 2 (given $$\alpha$$ is positive)
Now $$ \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {1 - \cos ({x^2} - x - 2)} } \over {(x - 2)}}$$
$$ = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {2{{\sin }^2}{{({x^2} - x - 2)} \over 2}} } \over {(x - 2)}}$$
$$ = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt 2 \sin \left( {{{(x - 2)(x + 1)} \over 2}} \right)} \over {(x - 2)}}$$
$$ = {3 \over {\sqrt 2 }}$$
roots are 2 & $$-$$1 $$ \Rightarrow $$ $$\alpha $$ = 2 (given $$\alpha$$ is positive)
Now $$ \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {1 - \cos ({x^2} - x - 2)} } \over {(x - 2)}}$$
$$ = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {2{{\sin }^2}{{({x^2} - x - 2)} \over 2}} } \over {(x - 2)}}$$
$$ = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt 2 \sin \left( {{{(x - 2)(x + 1)} \over 2}} \right)} \over {(x - 2)}}$$
$$ = {3 \over {\sqrt 2 }}$$
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