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

Option 1 -

Option 2 -

Option 3 -

Option 4 -

2 Views | Posted 2 months ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

$\underset{x \to 7}{l i m} \frac{18 - [1 - x]}{[x - 3 a]}$

exist & $a \in I .$

$= \underset{x \to 7}{l i m} \frac{17 - [- x]}{[x] - 3 a}$

exist

RHL = $\underset{x \to 7^{+}}{l i m} \frac{17 - [- x]}{[x] - 3 a} = \frac{25}{7 - 3 a} [a \neq \frac{7}{3}]$

$L H L = \underset{x \to 7^{-}}{l i m} \frac{17 - [- x]}{[x] - 3 a} = \frac{24}{6 - 3 a} [a \neq 2]$

LHL = RHL

$\Rightarrow \frac{25}{7 - 3 a} = \frac{8}{2 - a}$

$∴ a = - 6$

<math> <mrow> <munder> <mrow> <mi>l</mi> <mi>i</mi> <mi>m</mi> </mrow> <mrow> <mi>x</mi> <mo>→</mo> <mn>7</mn> </mrow> </munder> <mfrac> <mrow> <mn>1</mn> <mn>8</mn> <mo>−</mo> <mrow> <mo>[</mo> <mrow> <mn>1</mn> <mo>−</mo> <mi>x</mi> </mrow> <mo>]</mo> </mrow> </mrow> <mrow> <mrow> <mo>[</mo> <mrow> <mi>x</mi> <mo>−</mo> <mn>3</mn> <mi>a</mi> </mrow> <mo>]</mo> </mrow> </mrow> </mfrac> </mrow> </math> exist  & <math> <mrow> <mi>a</mi> <mo>∈</mo> <mi>I</mi> <mo>.</mo> </mrow> </math>   <math> <mrow> <mo>=</mo> <munder> <mrow> <mi>l</mi> <mi>i</mi> <mi>m</mi> </mrow> <mrow> <mi>x</mi> <mo>→</mo> <mn>7</mn> </mrow> </munder> <mfrac> <mrow> <mn>1</mn> <mn>7</mn> <mo>−</mo> <mrow> <mo>[</mo> <mrow> <mo>−</mo> <mi>x</mi> </mrow> <mo>]</mo> </mrow> </mrow> <mrow> <mrow> <mo>[</mo> <mrow> <mi>x</mi> </mrow> <mo>]</mo> </mrow> <mo>−</mo> <mn>3</mn> <mi>a</mi> </mrow> </mfrac> </mrow> </math>      existRHL = <math> <mrow> <munder> <mrow> <mi>l</mi> <mi>i</mi> <mi>m</mi> </mrow> <mrow> <mi>x</mi> <mo>→</mo> <msup> <mrow> <mn>7</mn> </mrow> <mrow> <mo>+</mo> </mrow> </msup> </mrow> </munder> <mfrac> <mrow> <mn>1</mn> <mn>7</mn> <mo>−</mo> <mrow> <mo>[</mo> <mrow> <mo>−</mo> <mi>x</mi> </mrow> <mo>]</mo> </mrow> </mrow> <mrow> <mrow> <mo>[</mo> <mrow> <mi>x</mi> </mrow> <mo>]</mo> </mrow> <mo>−</mo> <mn>3</mn> <mi>a</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mn>5</mn> </mrow> <mrow> <mn>7</mn> <mo>−</mo> <mn>3</mn> <mi>a</mi> </mrow> </mfrac> <mrow> <mo>[</mo> <mrow> <mi>a</mi> <mo>≠</mo> <mfrac> <mrow> <mn>7</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> <mo>]</mo> </mrow> </mrow> </math>   <math> <mrow> <mi>L</mi> <mi>H</mi> <mi>L</mi> <mo>=</mo> <munder> <mrow> <mi>l</mi> <mi>i</mi> <mi>m</mi> </mrow> <mrow> <mi>x</mi> <mo>→</mo> <msup> <mrow> <mn>7</mn> </mrow> <mrow> <mo>−</mo> </mrow> </msup> </mrow> </munder> <mfrac> <mrow> <mn>1</mn> <mn>7</mn> <mo>−</mo> <mrow> <mo>[</mo> <mrow> <mo>−</mo> <mi>x</mi> </mrow> <mo>]</mo> </mrow> </mrow> <mrow> <mrow> <mo>[</mo> <mrow> <mi>x</mi> </mrow> <mo>]</mo> </mrow> <mo>−</mo> <mn>3</mn> <mi>a</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mn>4</mn> </mrow> <mrow> <mn>6</mn> <mo>−</mo> <mn>3</mn> <mi>a</mi> </mrow> </mfrac> <mrow> <mo>[</mo> <mrow> <mi>a</mi> <mo>≠</mo> <mn>2</mn> </mrow> <mo>]</mo> </mrow> </mrow> </math> LHL = RHL <math> <mrow> <mo>⇒</mo> <mfrac> <mrow> <mn>2</mn> <mn>5</mn> </mrow> <mrow> <mn>7</mn> <mo>−</mo> <mn>3</mn> <mi>a</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>8</mn> </mrow> <mrow> <mn>2</mn> <mo>−</mo> <mi>a</mi> </mrow> </mfrac> </mrow> </math> <math> <mrow> <mo>∴</mo> <mi>a</mi> <mo>=</mo> <mo>−</mo> <mn>6</mn> </mrow> </math>

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

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Let a be an integer such that l i m x → 7 1 8 − [ 1 − x ] [ x − 3 a ] exists, where [t] is greatest integer ≤ t. Then a is equal to:

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