Exams Joint Entrance Examination Mathematics Limits, Continuity and Differentiability Question 373 Question 373 of 401 Medium MCQ
The set of all values of a a a for which \[\mathop {\lim }\limits_{x \to a} ([x - 5] - [2x + 2]) = 0\], where [\[\alpha\]] denotes the greatest integer less than or equal to \[\alpha\] is equal to
Solutions · 1 Post your solution Official solution
\[\mathop {\lim }\limits_{x \to a} \left( {[x - 5] - [2x + 2]} \right) = 0\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {[x] - 5 - [2x] - 2} \right) = 0\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to a} [x] - [2x] - 7 = 0\]
Case 1 :
If \[a \in \left[ {n,n + {1 \over 2}} \right)\]
then \[\mathop {\lim }\limits_{x \to a} \left( {[x] - [2x] - 7} \right) = 0\]
\[ \Rightarrow n - 2n - 7 = 0\]
\[ \Rightarrow n = - 7\]
\[\therefore\] \[a \in \left[ { - 7, - 6.5} \right)\]
Case 2 :
If \[a \in \left[ {n + {1 \over 2},n + 1} \right)\]
then \[\mathop {\lim }\limits_{x \to a} \left( {[x] - [2x] - 7} \right) = 0\]
\[ \Rightarrow n - (2n + 1) - 7 = 0\]
\[ \Rightarrow - n - 8 = 0\]
\[ \Rightarrow n = - 8\]
\[\therefore\] \[a \in \left[ { - 7.5, - 7} \right)\]
\[R.H.L = \mathop {\lim }\limits_{x \to - {{7.5}^ + }} \left( {\left[ x \right] - \left[ {2x} \right] - 7} \right)\]
\[ = - 8 - \left( { - 15} \right) - 7\]
= − 8 + 15 − 7 = 0 = - 8 + 15 - 7 = 0 = − 8 + 15 − 7 = 0
\[\eqalign{
\& L.H.L = \mathop {\lim }\limits_{x \to - {{7.5}^ - }} \left( {\left[ x \right] - \left[ {2x} \right] - 7} \right) \cr \\
\& = - 8 - \left( { - 16} \right) - 7 \cr \\
\& = - 8 + 16 - 7 = 1 \cr} \]
\[ \therefore \] R.H.L \[ \ne \] L.H.L
\[ \therefore \] At x = -7.5, limit does not exists.
\[\therefore\] \[a \in \left( { - 7.5, - 6.5} \right)\] Answer the question to unlock