JEE MAIN - Mathematics (2018 - 15th April Evening Slot - No. 21)
Let f(x) = $$\left\{ {\matrix{
{{{\left( {x - 1} \right)}^{{1 \over {2 - x}}}},} & {x > 1,x \ne 2} \cr
{k\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {,x = 2} \cr
} } \right.$$
Thevaue of k for which f s continuous at x = 2 is :
Thevaue of k for which f s continuous at x = 2 is :
1
e
e-1
e-2
Explanation
Since f(x) is continuous at x = 2.
$$\therefore\,\,\,$$$$\mathop {\lim }\limits_{x \to 2} $$ f(x) = f(2)
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 2} $$ $${\left( {x - 1} \right)^{{1 \over {2 - x}}}}$$ = k ($${{1^\infty }}$$ form)
$$ \therefore $$ $${{e^l}}$$ = k
Where $$l$$ = $$\mathop {\lim }\limits_{x \to 2} $$(x $$-$$ $$l$$ $$-$$ 1) $$ \times $$ $${1 \over {2 - x}}$$ = $$\mathop {\lim }\limits_{x \to 2} {{x - 2} \over {2 - x}}$$
= $$\mathop {\lim }\limits_{x \to 2} \left( {{{x - 2} \over {x - 2}}} \right)$$
$$ \Rightarrow $$ k = e$$-$$1
$$\therefore\,\,\,$$$$\mathop {\lim }\limits_{x \to 2} $$ f(x) = f(2)
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 2} $$ $${\left( {x - 1} \right)^{{1 \over {2 - x}}}}$$ = k ($${{1^\infty }}$$ form)
$$ \therefore $$ $${{e^l}}$$ = k
Where $$l$$ = $$\mathop {\lim }\limits_{x \to 2} $$(x $$-$$ $$l$$ $$-$$ 1) $$ \times $$ $${1 \over {2 - x}}$$ = $$\mathop {\lim }\limits_{x \to 2} {{x - 2} \over {2 - x}}$$
= $$\mathop {\lim }\limits_{x \to 2} \left( {{{x - 2} \over {x - 2}}} \right)$$
$$ \Rightarrow $$ k = e$$-$$1
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