Determine the values of a, b, c for which the function f(x)= (a+1)x+ x , &for x 0 is continuous at x = 0.

The given function can be rewritten as, f(x)= (a+1)x+ x , &for x 0 (x)= (a+1)x+ x , &for x 0 We observe (LHL at x=0)= _ x 0^- f(x)= _ h 0 f(0-h)= _ h 0 f(-h) = _ h 0 [ - (a+1)h- (-h) h ]= _ h 0 [ - (a+1)h h - h ] =-(a-1) _ h 0 [ - (a+1)h (a+1)h ]- _ h 0 h =-a-1 (RHL at x=0)= _ x 0^+ f(x)= _ h 0 f(0+h)= _ h 0 f(h) = _ h 0 ( 1+bh -1 bh )= _ h 0 ( bh bh( 1+bh +1) )= _ h 0 (1 1+bh +1 )=1 2 And, f(0) = c If f(x) is continuous at x = 0, then = _ x 0^- f(x )= _ x 0^+ f(x )=f(0) -a-1=1 2 =c -a-1=1 2 and c=1 2 =-3 2 ,c=1 2 Now, 1+bx -1 bx exists only if bx 0 0. - 0

Determine the values of a, b, c for which the function f(x)= (a+1…

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Question

Determine the values of a, b, c for which the function
$\text{f}\text{(x)}=\begin{cases}\frac{\sin\text{(a}+1)\text{x}+\sin\text{x}}{\text{x}}, &\text{for}\text{ x}<0,&\\\text{ c},&\text{for x}=0\\\frac{\sqrt{\text{x}+\text{bx}^2}-\sqrt{\text{x}}}{\text{bx}^\frac{3}{2}},&\text{for x}>0\end{cases}$ is continuous at x = 0.

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Answer

The given function can be rewritten as,
$\text{f}\text{(x)}=\begin{cases}\frac{\sin\text{(a}+1)\text{x}+\sin\text{x}}{\text{x}}, &\text{for}\text{ x}<0,&\\\text{ c},&\text{for x}=0\\\frac{\sqrt{\text{x}+\text{bx}^2}-\sqrt{\text{x}}}{\text{bx}^\frac{3}{2}},&\text{for x}>0\end{cases}$
$\Rightarrow\text{f}\text{(x)}=\begin{cases}\frac{\sin\text{(a}+1)\text{x}+\sin\text{x}}{\text{x}}, &\text{for}\text{ x}<0,&\\\text{ c},&\text{for x}=0\\\frac{\sqrt{1+\text{bx}}-1}{\text{bx}},&\text{for x}>0\end{cases}$
We observe
$\text{(LHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\Big[\frac{-\sin(\text{a}+1)\text{h}-\sin(-\text{h})}{\text{h}}\Big]=\lim\limits_{\text{h} \rightarrow 0}\Big[\frac{-\sin(\text{a}+1)\text{h}}{\text{h}}-\frac{\sin\text{h}}{\text{h}}\Big]$
$=-(\text{a}-1)\lim\limits_{\text{h} \rightarrow 0}\Big[\frac{-\sin(\text{a}+1)\text{h}}{(\text{a}+1)\text{h}}\Big]-\lim\limits_{\text{h} \rightarrow 0}\frac{\sin\text{h}}{\text{h}}=-\text{a}-1$
$\text{(RHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\sqrt{1+\text{bh}}-1}{\text{bh}}\Big)=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\text{bh}}{\text{bh}(\sqrt{1+\text{bh}}+1)}\Big)=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{1}{\sqrt{1+\text{bh}}+1}\Big)=\frac{1}{2}$
And, f(0) = c
If f(x) is continuous at x = 0, then
$=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x )}=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x )}=\text{f}(0)$
$\Rightarrow-\text{a}-1=\frac{1}{2}=\text{c}$
$\Rightarrow-\text{a}-1=\frac{1}{2}$ and $\text{c}=\frac{1}{2}$
$\Rightarrow\text{a}=\frac{-3}{2},\text{c}=\frac{1}{2}$
Now, $\frac{\sqrt{1+\text{bx}}-1}{\text{bx}}$ exists only if $\text{bx}\neq0\Rightarrow\text{b}\neq0.$
$\therefore\text{b}\in\text{R}-\{0\}$

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