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Question

Find the values of $$a$$ and $$b$$ such that the function defined by 
$$f(x) =$$ $$\begin{cases} 5,\ if\ x \le 2 \\ ax + b,\ if\ 2 < x < 10 \\ 21,\ if\ x \ge 10\end{cases}$$ is a continuous function.


Solution

Given definition of $$f$$ is 
$$f(x) = \begin{cases} 5, \,\, if \, x \le 2 \\ ax + b, \,\,  if \, 2 < x < 10 \\ 21, \,\,  if \, x \ge 10\end{cases}$$             .....(1)

Also, given $$f(x)$$ is a continuous function.

So, $$f(x)$$ is continuous at all points .

So, $$f(x)$$ is continuous at $$x=2$$

$$\Rightarrow \displaystyle \lim _{ x\rightarrow 2^{ - } }{ f(x) } =\lim _{ x\rightarrow 2^{ + } }{ f(x) } =f(2)$$     ....(2)

Now, $$RHL =\displaystyle \lim_{x\rightarrow 2^+}f(x)$$

$$=\displaystyle \lim_{h\rightarrow 0}f(2+h)$$

$$=\displaystyle \lim_{h\rightarrow 0} a(2+h)+b$$

$$\Rightarrow RHL=2a+b$$

Also, $$f(2)=5$$ 

Substituting these values in (2), we get
$$2a+b=5$$                 .......(3)

Also, $$f(x) $$ is continuous at $$x=10$$

$$\therefore\displaystyle \lim _{ x\rightarrow 10^{ - } }{ f(x) } =\lim _{ x\rightarrow 10^{ + } }{ f(x) } =f(10)$$                      ......(4)

Now, $$LHL=\displaystyle \lim_{x\rightarrow 10^{-}}f(x)$$

$$=\displaystyle \lim_{h\rightarrow 0} f(10-h)$$

$$=\displaystyle \lim_{h\rightarrow 0} a(10-h)+b$$

$$=10a+b$$

Also, by (1), $$f(10)=21$$

Substituting these values in eq (4), we get

$$10a+b=21$$            .....(5)

Solving eq $$(3)$$ from eqn $$(5)$$, we get

$$8a=16$$

$$\Rightarrow a=2$$ 

Put this value in $$(3)$$, we get

$$2(2)+b=5$$

$$\Rightarrow b=1$$

Mathematics
RS Agarwal
Standard XII

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