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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 $$b = 4$$

Mathematics
NCERT
Standard XI

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