If f(x)=|x−2|x−2+a forx 2 is continuous on R then (a,b)=

Ltx→2−f(x)=f(2)Ltx→2−f(x)=|x−2|x−2+a=5=(−1)+a=5a=5+1a=6Ltx→2+f(x)=f(2)Ltx→2+|x−2|2−x+b=5−1+b=5⇒b=5+1=6

If f(x)=|x−2|x−2+a forx 2 is continuous on R then (a,b)=

Ltx→2−f(x)=f(2)Ltx→2−f(x)=|x−2|x−2+a=5=(−1)+a=5a=5+1a=6Ltx→2+f(x)=f(2)Ltx→2+|x−2|2−x+b=5−1+b=5⇒b=5+1=6

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If $f (x) = \{\begin{cases} \frac{| x - 2 |}{x - 2} + a f o r & x < 2 \\ 5 f o r & x = 2 \\ \frac{| x - 2 |}{x - 2} + b f o r & x > 2 \end{cases}$ is continuous on R then $(a, b) =$

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$(- 6, 6)$

$(6, - 6)$

$(- 6, - 6)$

$(6, 6)$

answer is D.

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Detailed Solution

$\begin{array}{l} \underset{x \to 2^{-}}{L t} f (x) = f (2) \\ \underset{x \to 2^{-}}{L t} f (x) = \frac{| x - 2 |}{x - 2} + a = 5 \\ = (- 1) + a = 5 \\ a = 5 + 1 \\ a = 6 \\ \underset{x \to 2^{+}}{L t} f (x) = f (2) \\ \underset{x \to 2^{+}}{L t} \frac{| x - 2 |}{2 - x} + b = 5 \\ - 1 + b = 5 \Rightarrow b = 5 + 1 = 6 \end{array}$