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$$2(2)+b=5$$$$\Rightarrow b=1$$MathematicsRS AgarwalStandard XII

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