Let a be an integer such that lim x → 7 18 - 1 - x x - 3 a exists, where t is greatest integer ≤ t . Then a is equal to

Since lim x → 7 18 - 1 - x x - 3 a exists a ∈ 𝕀 ∴ L . H . L = R . H . L ⇒ lim h → 0 18 - - 6 7 - h - 3 a = lim h → 0 18 - - 7 7 + h - 3 a ⇒ lim h → 0 24 7 - 3 a + - h = lim h → 0 25 7 - 3 a + h ⇒ 24 6 - 3 a = 25 7 - 3 a ⇒ a = - 6

Mathematics - Limits Question with Solution | TestHub

MathematicsLimitsGeneral ( Limit existence)Medium2 minPYQ_2022

MathematicsMediumsingle choice

Let $a$ be an integer such that $\lim_{x \to 7} \frac{18 - [1 - x]}{[x - 3 a]}$ exists, where $[t]$ is greatest integer $\leq t$ . Then $a$ is equal to

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Answer:

Solution:

Since $\lim_{x \to 7} \frac{18 - [1 - x]}{[x - 3 a]}$ exists $(a \in 𝕀)$

$∴ L . H . L = R . H . L$

$\Rightarrow \lim_{h \to 0} \frac{18 - (- 6)}{[7 - h - 3 a]} = \lim_{h \to 0} \frac{18 - (- 7)}{[7 + h - 3 a]}$

$\Rightarrow \lim_{h \to 0} \frac{24}{(7 - 3 a) + [- h]} = \lim_{h \to 0} \frac{25}{(7 - 3 a) + [h]}$

$\Rightarrow \frac{24}{6 - 3 a} = \frac{25}{7 - 3 a}$ $\Rightarrow a = - 6$

Stream:JEESubject:MathematicsTopic:LimitsSubtopic:General ( Limit existence)

⏱ 2mℹ️ Source: PYQ_2022

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