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 :
JEE · Math · previous-year question
- A.$$-$$6correct
- B.$$-$$2
- C.2
- D.6
Answer
A. $$-$$6
Explanation
$$\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}$$ 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 $$LHL = RHL$$ $${{25} \over {7 - 3a}} = {{24} \over {6 - 3a}}$$ $$ \Rightarrow {{25} \over {7 - 3a}} = {8 \over {2 - a}}$$ $$\therefore$$ $$a = - 6$$
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