JEE MAIN - Mathematics (2020 - 3rd September Morning Slot - No. 3)
Let [t] denote the greatest integer
$$ \le $$ t. If for some
$$\lambda $$ $$ \in $$ R - {1, 0}, $$\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}} \right|$$ = L, then L is equal to :
$$\lambda $$ $$ \in $$ R - {1, 0}, $$\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}} \right|$$ = L, then L is equal to :
1
2
0
$${1 \over 2}$$
Explanation
Here $$\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + [x]}}} \right| = L$$
Here L.H.L. $$\mathop {\lim }\limits_{h \to 0^-} \left| {{{1 + h + h} \over {\lambda + h - 1}}} \right| = \left| {{1 \over {\lambda - 1}}} \right|$$
R.H.L. = $$\mathop {\lim }\limits_{h \to 0^+} \left| {{{1 - h + h} \over {\lambda + h + 0}}} \right| = \left| {{1 \over \lambda }} \right|$$
$$ \because $$ Limit exists. Hence L.H.L. = R.H.L.
$$ \Rightarrow $$ $$\left| {\lambda - 1} \right| = \left| \lambda \right|$$
$$ \Rightarrow $$ $$\lambda = {1 \over 2}$$
$$ \therefore $$ L = $${1 \over {\left| \lambda \right|}}$$ = 2
Here L.H.L. $$\mathop {\lim }\limits_{h \to 0^-} \left| {{{1 + h + h} \over {\lambda + h - 1}}} \right| = \left| {{1 \over {\lambda - 1}}} \right|$$
R.H.L. = $$\mathop {\lim }\limits_{h \to 0^+} \left| {{{1 - h + h} \over {\lambda + h + 0}}} \right| = \left| {{1 \over \lambda }} \right|$$
$$ \because $$ Limit exists. Hence L.H.L. = R.H.L.
$$ \Rightarrow $$ $$\left| {\lambda - 1} \right| = \left| \lambda \right|$$
$$ \Rightarrow $$ $$\lambda = {1 \over 2}$$
$$ \therefore $$ L = $${1 \over {\left| \lambda \right|}}$$ = 2
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