QJEEM18Maf3f9: JEE Main 2018 Mathematics - Class 12 JEEM18Maf3f9 Solution: Let f(x) = { ( x − 1 ) 1 2 − x , x > 1 , x ≠ 2 k , x = 2…

Question

Verified Solution • 2025–26 Session moderate

Let f(x) =

{\begin{matrix} {(x - 1)}^{\frac{1}{2 - x}}, & x > 1, x \neq 2 \\ k & , x = 2 \end{matrix}

Thevaue of k for which f s continuous at x = 2 is :

Accepted Answer

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}

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