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JEE Main 202227 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
27 June
Shift
Morning Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2022 · 27 June · Morning Shift · Mathematics · Limits

Let a be an integer such that _ x → 7 (18 - [1 - x])/([x - 3a])...

Let aa be an integer such that lim⁡x→718−[1−x][x−3a]\lim_{x \to 7}\dfrac{18 - \left[1 - x\right]}{\left[x - 3a\right]}exists,where[t]exists, where \left[t\right] is greatest integer ≤t\le t. Then aa is equal to

Options

  1. A−2- 2
  2. B66
  3. C−6- 6
  4. D−7- 7

Correct answer

C. −6- 6

Step-by-step solution

Since lim⁡x→718−[1−x][x−3a]\lim_{x \to 7}\dfrac{18 - \left[1 - x\right]}{\left[x - 3a\right]}exists(a∈핀)exists \left(a \in 핀\right)

∴L.H.L=R.H.L∴ L . H . L = R . H . L

⇒lim⁡h→018−(−6)[7−h−3a]=lim⁡h→018−(−7)[7+h−3a]⇒ \lim_{h \to 0}\dfrac{18 - \left(- 6\right)}{\left[7 - h - 3a\right]} = \lim_{h \to 0}\dfrac{18 - \left(- 7\right)}{\left[7 + h - 3a\right]}

⇒lim⁡h→024(7−3a)+[−h]=lim⁡h→025(7−3a)+[h]⇒ \lim_{h \to 0}\dfrac{24}{\left(7 - 3a\right) + \left[- h\right]} = \lim_{h \to 0}\dfrac{25}{\left(7 - 3a\right) + \left[h\right]}

⇒246−3a=257−3a⇒a=−6⇒ \dfrac{24}{6 - 3a} = \dfrac{25}{7 - 3a} ⇒ a = - 6

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