JEE MAIN - Mathematics (2022 - 27th June Morning Shift - No. 4)
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 :
$$-$$6
$$-$$2
2
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
$$LHL = RHL$$
$${{25} \over {7 - 3a}} = {{24} \over {6 - 3a}}$$
$$ \Rightarrow {{25} \over {7 - 3a}} = {8 \over {2 - a}}$$
$$\therefore$$ $$a = - 6$$
Comments (0)
