Let a be an integer such that limx→718−[1−x][x−3a] exists, where [t] is the greater integer ≤t . Then ‘a’ equal to

R.H L= limh→018−[1−(7+h)][7+h]−3a=257−3a L.H.C limh→018−[1−(7−h)][7−h]−3a=82−a L.H.C=R.H.L ⇒a=−6

Let a be an integer such that limx→718−[1−x][x−3a] exists, where [t] is the greater integer ≤t . Then ‘a’ equal to

R.H L= limh→018−[1−(7+h)][7+h]−3a=257−3a L.H.C limh→018−[1−(7−h)][7−h]−3a=82−a L.H.C=R.H.L ⇒a=−6

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Let a be an integer such that $\lim_{x \to 7} \frac{18 - [1 - x]}{[x - 3 a]}$ exists, where [t] is the greater integer $\leq t$ . Then ‘a’ equal to

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$- 2$

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$6$

answer is A.

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R.H L= $\lim_{h \to 0} \frac{18 - [1 - (7 + h)]}{[7 + h] - 3 a} = \frac{25}{7 - 3 a}$
L.H.C $\lim_{h \to 0} \frac{18 - [1 - (7 - h)]}{[7 - h] - 3 a} = \frac{8}{2 - a}$

$L . H . C = R . H . L \Rightarrow a = - 6$

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Let a be an integer such that limx→718−[1−x][x−3a] exists, where [t] is the greater integer ≤t . Then ‘a’ equal to

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Let a be an integer such that $\lim_{x \to 7} \frac{18 - [1 - x]}{[x - 3 a]}$ exists, where [t] is the greater integer $\leq t$ . Then ‘a’ equal to