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Question: The function defined by \(f(x) = \left\{ \begin{matrix} \left( x^{2} + e^{\frac{1}{2 - x}} \right)^{...

The function defined by f(x)={(x2+e12−x)−1,x≠2k,x=2 f(x) = \left\{ \begin{matrix} \left( x^{2} + e^{\frac{1}{2 - x}} \right)^{- 1} & , & x \neq 2 \\ k & , & x = 2 \end{matrix} \right.\ , is continuous from right at the point x = 2, then k is equal to

A

0

B

¼

C

–1/4

D

None of these

Answer

¼

Explanation

Solution

f(x)=[x2+e12−x]−1f(x) = \left\lbrack x^{2} + e^{\frac{1}{2 - x}} \right\rbrack^{- 1} and f(2)=kf(2) = k

If f(x)f(x) is continuous from right at x=2x = 2 then

lim⁡x→2+f(x)=f(2)=k\lim_{x \rightarrow 2^{+}}f(x) = f(2) = k

⇒ lim⁡x→2+[x2+e12−x]−1=k\lim_{x \rightarrow 2^{+}}\left\lbrack x^{2} + e^{\frac{1}{2 - x}} \right\rbrack^{- 1} = k ⇒ k=lim⁡h→0f(2+h)k = \lim_{h \rightarrow 0}f(2 + h)

⇒ k=lim⁡h→0[(2+h)2+e12−(2+h)]−1k = \lim_{h \rightarrow 0}\left\lbrack (2 + h)^{2} + e^{\frac{1}{2 - (2 + h)}} \right\rbrack^{- 1}

⇒ k=lim⁡h→0[4+h2+4h+e−1/h]−1k = \lim_{h \rightarrow 0}\left\lbrack 4 + h^{2} + 4h + e^{- 1/h} \right\rbrack^{- 1} ⇒ k=[4+0+0+e−∞]−1k = \lbrack 4 + 0 + 0 + e^{- \infty}\rbrack^{- 1}

⇒ k=14k = \frac{1}{4}.