Let a be an integer such that limx - 718-1-x-x-3a- exists- where -t- is greatest integer - t- Then a is equal to -

Lim_x → 718 [1 x][x 3a] exists & a ∈ I =lim_x → 717 [ x][x] 3a nbsp; exists RHL =lim_x → 7^+17 [ x][x] 3a=257 3a [a ≠73] LHL =lim_x → 7^ 17 [ x][x] 3a= ...

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Standard XI

Mathematics

Standard Derivative

Let a be an i...

Question

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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Solution

The correct option is B $- 6$
$lim x \to 7 \frac{18 - [1 - x]}{[x - 3 a]}$ exists $& a \in I$
$= lim x \to 7 \frac{17 - [- x]}{[x] - 3 a}$ exists
RHL $= lim x \to 7^{+} \frac{17 - [- x]}{[x] - 3 a} = \frac{25}{7 - 3 a} [a \neq \frac{7}{3}]$
LHL $= lim x \to 7^{-} \frac{17 - [- x]}{[x] - 3 a} = \frac{24}{6 - 3 a} [a \neq 2]$
For limit to exist
LHL = RHL
$\frac{25}{7 - 3 a} = \frac{24}{6 - 3 a}$
$\Rightarrow \frac{25}{7 - 3 a} = \frac{8}{2 - a}$
$∴ a = - 6$

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Let a be an integer such that limx→718−[1−x][x−3a] exists, where [t] is greatest integer ≤t. Then a is equal to :

The correct option is B −6limx→718−[1−x][x−3a] exists & a∈I =limx→717−[−x][x]−3a exists RHL =limx→7+17−[−x][x]−3a=257−3a [a≠73] LHL =limx→7−17−[−x][x]−3a=246−3a [a≠2] For limit to exist LHL = RHL 257−3a=246−3a ⇒257−3a=82−a ∴a=−6

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 :