The value of c and k that make the functionf(x)=x+2c, x≤23cx+k, 2<x<13x−2k, x≥1 continuous on (−∞,∞) are respectively

at x=2 LHL=limx→2-x+2c=2+2c RHL=limx→2+3cx+k=6c+k since function is continuous at x=2 then LHL=RHL ⇒2+2c=6c+k⇒4c+k=2 at x=1 LHL=limx→1-3cx+k=3c+k RHL=limx→1+3x-2k=3-2k since function is continuous at x=1 then LHL=RHL ⇒3c+k=3-2k⇒3c+3k=3⇒c+k=1 3c=1⇒c=13,k=23

The value of c and k that make the functionf(x)=x+2c, x≤23cx+k, 2<x<13x−2k, x≥1 continuous on (−∞,∞) are respectively

at x=2 LHL=limx→2-x+2c=2+2c RHL=limx→2+3cx+k=6c+k since function is continuous at x=2 then LHL=RHL ⇒2+2c=6c+k⇒4c+k=2 at x=1 LHL=limx→1-3cx+k=3c+k RHL=limx→1+3x-2k=3-2k since function is continuous at x=1 then LHL=RHL ⇒3c+k=3-2k⇒3c+3k=3⇒c+k=1 3c=1⇒c=13,k=23

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The value of c and k that make the function
$f (x) =$ $\{\begin{cases} \begin{matrix} x + 2 c, x \leq 2 \\ 3 c x + k, 2 < x < 1 \\ 3 x - 2 k, x \geq 1 \end{matrix} \end{cases}$
continuous on $(- \infty, \infty)$ are respectively

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$\frac{1}{3}, \frac{2}{3}$

$\frac{1}{3}, \frac{- 2}{3}$

$\frac{- 1}{3}, \frac{2}{3}$

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answer is A.

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Detailed Solution

$a t x = 2 L H L = \lim_{x \to 2^{-}} x + 2 c = 2 + 2 c R H L = \lim_{x \to 2^{+}} 3 c x + k = 6 c + k \sin c e f u n c t i o n i s c o n t i n u o u s a t x = 2 t h e n L H L = R H L \Rightarrow 2 + 2 c = 6 c + k \Rightarrow 4 c + k = 2 a t x = 1 L H L = \lim_{x \to 1^{-}} 3 c x + k = 3 c + k R H L = \lim_{x \to 1^{+}} 3 x - 2 k = 3 - 2 k \sin c e f u n c t i o n i s c o n t i n u o u s a t x = 1 t h e n L H L = R H L \Rightarrow 3 c + k = 3 - 2 k \Rightarrow 3 c + 3 k = 3 \Rightarrow c + k = 1 3 c = 1 \Rightarrow c = \frac{1}{3}, k = \frac{2}{3}$