If the function defined by⨍ x=x2+e12-x-1 ,⨍ or x≠2k,⨍ or x=2 is right continuous at x=2, then k=

fx=x2+e12-x-1;x≠2k :x=2is right continuous at x=2Then k=f2=limx→2+f(x)k=limx→2+ 1x2+e12-x=14Option (3) is correct

If the function defined by⨍ x=x2+e12-x-1 ,⨍ or x≠2k,⨍ or x=2 is right continuous at x=2, then k=

fx=x2+e12-x-1;x≠2k :x=2is right continuous at x=2Then k=f2=limx→2+f(x)k=limx→2+ 1x2+e12-x=14Option (3) is correct

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If the function defined by
$⨍ (x) = \{\begin{matrix} \{\begin{cases} {(x^{2} + e^{\frac{1}{2 - x}})}^{- 1} \end{cases} & , & ⨍ or x≠2 \\ k & , & ⨍ or x=2 \end{matrix}$
is right continuous at $x = 2,$ then $k =$

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$- \frac{1}{4}$

$\frac{1}{4}$

answer is C.

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Detailed Solution

$f (x) = \{\begin{matrix} {(x^{2} + e^{\frac{1}{2 - x}})}^{- 1}; x \neq 2 \\ k : x = 2 \end{matrix}$ is right continuous at $x = 2$

Then $k = f (2) = \lim_{x \to 2^{+}} f (x)$

$k = \lim_{x \to 2^{+}} \frac{1}{x^{2} + e^{\frac{1}{2 - x}}} = \frac{1}{4}$

Option (3) is correct

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If the function defined by⨍ x=x2+e12-x-1 ,⨍ or x≠2k,⨍ or x=2 is right continuous at x=2, then k=

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If the function defined by
$⨍ (x) = \{\begin{matrix} \{\begin{cases} {(x^{2} + e^{\frac{1}{2 - x}})}^{- 1} \end{cases} & , & ⨍ or x≠2 \\ k & , & ⨍ or x=2 \end{matrix}$
is right continuous at $x = 2,$ then $k =$