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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Write the value of b which $\text{f(x)}=\begin{cases}5\text{x}-4,&0<\text{x}\leq1\\4\text{x}^2+3\text{bx},&1<\text{x}<2\end{cases}$ is continuous at x = 1.
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Answer

Given, $\text{f(x)}=\begin{cases}5\text{x}-4,&0<\text{x}\leq1\\4\text{x}^2+3\text{bx},&1<\text{x}<2\end{cases}$
If f(x) is continuous at x = 1, then
$\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x})=\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x})=\text{f}(1)\ ...(\text{i})$
Now,
$\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x})=\lim\limits_{{\text{h}}\rightarrow0}\text{f}(1-\text{h})\\=\lim\limits_{{\text{h}}\rightarrow0}5(1-\text{h})-4=5-4=1$
$\lim\limits_{{\text{x}}\rightarrow1^+}\text{f(x})=\lim\limits_{{\text{h}}\rightarrow0}\text{f}(1+\text{h})\\=\lim\limits_{{\text{h}}\rightarrow0}4(1+\text{h})^2+3\text{b}(1+\text{h})=4+3\text{b}$
Also,
$\text{f}(1)=5(1)-4=1$
$=\lim\limits_{{\text{x}}\rightarrow1^-}\text{f(x)}=\lim\limits_{{\text{x}}\rightarrow1^+}\text{f(x)}=\text{f}(1)$ [From eq. (i)]
$\Rightarrow1=4+3\text{b}=1$
$\Rightarrow1=4+3\text{b}$
$\Rightarrow-3=3\text{b}$
$\Rightarrow\text{b}=-1$
Thus, for b = -1, the function f(x) is continuous at x = 1.

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