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

L.H.L=f(2)=R.H.L Ltx→2--(x-2)x-2+a=f(2)=Ltx→2+(x-2)2-x+b ⇒−1+a=5=−1+ba=6, b=6(a, b)=(6, 6)

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

L.H.L=f(2)=R.H.L Ltx→2--(x-2)x-2+a=f(2)=Ltx→2+(x-2)2-x+b ⇒−1+a=5=−1+ba=6, b=6(a, b)=(6, 6)

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If $f (x) = \frac{| x - 2 |}{x - 2} + a$ for $x < 2$ or $\frac{| x - 2 |}{2 - x} + b = 5$ for $x > 2$ is continuous on $R$ then $(a, b) =$ –––––

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$(6, - 6)$

$(- 6, 6)$

$(- 6, - 6)$

$(6, 6)$

answer is C.

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Detailed Solution

$L . H . L = f (2) = R . H . L \underset{x \to 2^{-}}{L t} \frac{- (x - 2)}{x - 2} + a = f (2) = \underset{x \to 2^{+}}{L t} \frac{(x - 2)}{2 - x} + b$

$\Rightarrow - 1 + a = 5 = - 1 + b$

$a = 6, b = 6$

$(a, b) = (6, 6)$

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