Question 335 of 401HardMCQ
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 :
Solutions · 1
Official solution
x2−x−2=0
roots are 2 \& -1 \[ \Rightarrow \] \[\alpha \] = 2 (given \[\alpha\] is positive)
Now \[ \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {1 - \cos ({x^2} - x - 2)} } \over {(x - 2)}}\]
\[ = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {2{{\sin }^2}{{({x^2} - x - 2)} \over 2}} } \over {(x - 2)}}\]
\[ = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt 2 \sin \left( {{{(x - 2)(x + 1)} \over 2}} \right)} \over {(x - 2)}}\]
\[ = {3 \over {\sqrt 2 }}\] Answer the question to unlock