Question

# Suppose $$f(x) =$$ $$\displaystyle \begin{cases}a+bx, & \text{ } x<1 \\4, & \text{ } x= 1 \\b-ax & \text{ } x>1\end{cases}$$ and if $$\displaystyle \lim_{x\rightarrow 1}f\left ( x \right )=f\left ( 1 \right )$$, then what is the possible value of $$a,b$$?

Solution

## We have,$$f(x) = \displaystyle \begin{cases}a+bx, & \text{ } x<1 \\4, & \text{ } x= 1 \\b-ax & \text{ } x>1\end{cases}$$$$\displaystyle \lim_{x\rightarrow 1^{-}}f\left ( x \right )=\lim_{x\rightarrow 1}\left ( a+bx \right )=a+b$$$$\displaystyle \lim_{x\rightarrow 1^{+}}f\left ( x \right )=\lim_{x\rightarrow 1}\left ( b-ax \right )=b-a$$Also  $$\displaystyle f\left ( 1 \right )=4$$It is given that $$\displaystyle \lim_{x\rightarrow 1}f\left ( x \right )=f\left ( 1 \right )\cdot$$$$\displaystyle \therefore$$$$\displaystyle \lim_{x\rightarrow 1^{-}}f\left ( x \right )=\lim_{x\rightarrow 1^{+}}f\left ( x \right )=\lim_{x\rightarrow 1}f\left ( x \right )=f\left ( 1 \right )$$$$\displaystyle \Rightarrow a+b=4$$ and $$b - a = 4$$Solving these two equations we obtain $$a = 0$$ and $$b = 4$$MathematicsNCERTStandard XI

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