Exams Joint Entrance Examination Mathematics Limits, Continuity and Differentiability Question 365 Question 365 of 401 Hard MCQ
Let a be an integer such that \[\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}\] exists, where [t] is greatest integer \[\le\] t. Then a is equal to :
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\[\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}\] exist \& \[a \in I\].
\[ = \mathop {\lim }\limits_{x \to 7} {{17 - [ - x]} \over {[x] - 3a}}\] exist
\[RHL = \mathop {\lim }\limits_{x \to {7^ + }} {{17 - [ - x]} \over {[x] - 3a}} = {{25} \over {7 - 3a}}\] \[\left[ {a \ne {7 \over 3}} \right]\]
\[LHL = \mathop {\lim }\limits_{x \to {7^ - }} {{17 - [ - x]} \over {[x] - 3a}} = {{24} \over {6 - 3a}}\] \[\left[ {a \ne 2} \right]\]
For limit to exist
L H L = R H L LHL = RHL L H L = R H L
\[{{25} \over {7 - 3a}} = {{24} \over {6 - 3a}}\]
\[ \Rightarrow {{25} \over {7 - 3a}} = {8 \over {2 - a}}\]
\[\therefore\] a = − 6 a = - 6 a = − 6 Answer the question to unlock