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Find the values of a and b such that the function f defined by

$f(x)=\left\{\begin{array}{ll} \frac{x-4}{|x-4|}+a, & \text { if } x<4 \\ a+b & , \text { if } x=4 \\ \frac{x-4}{|x-4|}+b, & \text { if } x>4 \end{array}\right.$

is a continuous function at x = 4.

Answers (2)

Given,

$f(x)=\left\{\begin{array}{ll} \frac{x-4}{|x-4|}+a, & \text { if } x<4 \\ a+b & , \text { if } x=4 \\ \frac{x-4}{|x-4|}+b, & \text { if } x>4 \end{array}\right.$ …(1)

We need to find the value of a & b such that f(x) is continuous at x = 4.

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$

Where h is a very small number very close to 0 (h→0)

Now, let’s assume that f(x) is continuous at x = 4.

$\therefore \lim _{h \rightarrow 0} \mathrm{f}(4-\mathrm{h})=\lim _{h \rightarrow 0} \mathrm{f}(4+\mathrm{h})=\mathrm{f}(4)$

to find a & b, we have to pick out a combination so that we get a or b in our equation.

In this question first we take LHL = f(4)

$\therefore \lim _{h \rightarrow 0} f(4-h)=f(4)$

$\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{4-h-4}{|4-h-4|}+a\right\}=a+b$ {using equation 1}

$\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{-h}{|-h|}+a\right\}=a+b$

$\because$ h > 0 as defined in theory above.

$\therefore|-h|=h$

$\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{-h}{h}+a\right\}=a+b$

$\Rightarrow \lim _{h \rightarrow 0}\{a-1\}=a+b$

$\Rightarrow$ a - 1 = a + b

$\therefore$ b = -1

Now, taking other combination,
RHL = f(4)

$\lim _{h \rightarrow 0} \mathrm{f}(4+\mathrm{h})=\mathrm{f}(4)$

$\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{4+h-4}{|4+h-4|}+b\right\}=a+b$ {using equation 1}

$\underset{h \rightarrow 0}{\lim }\left\{\frac{h}{|h|}+b\right\}=a+b$

$\because$ h > 0 as defined in theory above.

$\therefore$ |h| = h

$\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{h}{h}+b\right\}=a+b$

$\Rightarrow \lim _{h \rightarrow 0}\{b+1\}=a+b$

⇒ b + 1 = a + b

∴ a = 1

Hence,

a = 1 and b = -1

Posted by

infoexpert21

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Posted by

infoexpert22

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