Determine the values of \[a,b,c\] for which the function given below is continuous at \[x=0\]
\[f\left( x \right)=\left\{ \begin{align}
& \dfrac{\sin \left( a+1 \right)x+\sin x}{x},\text{ for }x<0 \\
& c\text{ , for }x=0 \\
& \dfrac{{{\left( x+b{{x}^{2}} \right)}^{\dfrac{1}{2}}}-{{x}^{\dfrac{1}{2}}}}{b{{x}^{\dfrac{3}{2}}}}\text{ , for }x>0 \\
\end{align} \right.\]
(a) \['a'\] can take any value
(b) \[c=\dfrac{1}{2}\]
(c) \[a=\dfrac{-3}{2}\]
(d) \['b'\] can take any value
Answer
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Hint: We solve this problem by using the definition of a continuous function.
If a function \[f\left( x \right)\] is said to be continuous at a point \[x=a\] if and only if
\[\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=f\left( a \right)\]
By using the above equation we find the required limits for the given function to find the required values. Here, \[{{a}^{-}},{{a}^{+}}\] represents that the value of \['x'\] tends to \['a'\] from left side and right side of \['a'\] respectively.
Complete step by step answer:
We are given that the function is continuous at \[x=0\]
We know that if a function \[f\left( x \right)\] is said to be continuous at a point \[x=a\] if and only if
\[\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=f\left( a \right)\]
By using the above definition to given function we get
\[\displaystyle \lim_{x \to {{0}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{0}^{+}}}f\left( x \right)=f\left( 0 \right)......equation(i)\]
Now, let us assume that the first term as
\[\Rightarrow A=\displaystyle \lim_{x \to {{0}^{-}}}f\left( x \right)\]
We know that \[{{0}^{-}}\] represents that the value of \['x'\] tends to 0 from left side of 0 that is \[x<0\]
We are given that the value of function when \[x<0\] as
\[f\left( x \right)=\dfrac{\sin \left( a+1 \right)x+\sin x}{x}\]
By substituting this function in the above limit we get
\[\begin{align}
& \Rightarrow A=\displaystyle \lim_{x \to {{0}^{-}}}\left( \dfrac{\sin \left( a+1 \right)x+\sin x}{x} \right) \\
& \Rightarrow A=\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\sin \left( a+1 \right)x}{x}+\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\sin x}{x} \\
\end{align}\]
We know that the standard formula of limits that is
\[\displaystyle \lim_{x \to 0}\dfrac{\sin \left( nx \right)}{x}=n\]
By using this condition to above equation we get
\[\begin{align}
& \Rightarrow A=\left( a+1 \right)+1 \\
& \Rightarrow A=a+2 \\
\end{align}\]
Now, let us assume the second term in the equation (i) as
\[\Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}f\left( x \right)\]
We know that \[{{0}^{+}}\] represents that the value of \['x'\] tends to 0 from right side of 0 that is \[x>0\]
We are given that the value of function when \[x>0\] as
\[f\left( x \right)=\dfrac{{{\left( x+b{{x}^{2}} \right)}^{\dfrac{1}{2}}}-{{x}^{\dfrac{1}{2}}}}{b{{x}^{\dfrac{3}{2}}}}\]
By substituting this function in the above limit we get
\[\Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{{{\left( x+b{{x}^{2}} \right)}^{\dfrac{1}{2}}}-{{x}^{\dfrac{1}{2}}}}{b{{x}^{\dfrac{3}{2}}}}\]
Now, let us take the common term out from the numerator then we get
\[\begin{align}
& \Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{{{x}^{\dfrac{1}{2}}}{{\left( 1+bx \right)}^{\dfrac{1}{2}}}-{{x}^{\dfrac{1}{2}}}}{bx\left( {{x}^{\dfrac{1}{2}}} \right)} \right) \\
& \Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{{{\left( 1+bx \right)}^{\dfrac{1}{2}}}-1}{bx} \right) \\
\end{align}\]
Now, let us rationalise the numerator that is let us multiply and divide the above limit with \[\left( {{\left( 1+bx \right)}^{\dfrac{1}{2}}}+1 \right)\] then we get
\[\Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{{{\left( 1+bx \right)}^{\dfrac{1}{2}}}-1}{bx}\times \dfrac{{{\left( 1+bx \right)}^{\dfrac{1}{2}}}+1}{{{\left( 1+bx \right)}^{\dfrac{1}{2}}}+1} \right)\]
We know that the formula of algebra that is
\[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]
By using this formula to numerator of the above limit we get
\[\begin{align}
& \Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{\left( 1+bx \right)-1}{bx\left( {{\left( 1+bx \right)}^{\dfrac{1}{2}}}+1 \right)} \right) \\
& \Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{1}{{{\left( 1+bx \right)}^{\dfrac{1}{2}}}+1} \right) \\
\end{align}\]
Now, by expanding the limit we get
\[\begin{align}
& \Rightarrow B=\dfrac{1}{{{\left( 1+0 \right)}^{\dfrac{1}{2}}}+1} \\
& \Rightarrow B=\dfrac{1}{2} \\
\end{align}\]
Now, let us assume that the third term in equation (i) as
\[\Rightarrow C=f\left( 0 \right)\]
We are given that the value of \[f\left( x \right)\] at \[x=0\] as \['c'\]
By using the given function we get
\[\Rightarrow C=c\]
Now, by substituting the required values in the equation (i) we get
\[\Rightarrow a+2=\dfrac{1}{2}=c.......equation(ii)\]
Here, we can see that the limit of second term as \[B=\dfrac{1}{2}\] is independent of \['b'\]
So, we can say that \['b'\] can take any value because it has no effect on the limit.
Now, from the equation (ii) we have the value of \['c'\]taking the variable and constant together that is
\[\Rightarrow c=\dfrac{1}{2}\]
Now, from equation (ii) we get the value of \['a'\] as
\[\begin{align}
& \Rightarrow a+2=\dfrac{1}{2} \\
& \Rightarrow a=\dfrac{-3}{2} \\
\end{align}\]
Therefore we can conclude that \[a=\dfrac{-3}{2},c=\dfrac{1}{2}\] and \['b'\] can take any value.
So, the correct answer is “Option b, c and d”.
Note: Students may do mistake in taking the value of \['b'\]
Here, we have the limit where \['b'\] is included as
\[\begin{align}
& \Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{1}{{{\left( 1+bx \right)}^{\dfrac{1}{2}}}+1} \right) \\
& \Rightarrow B=\dfrac{1}{{{\left( 1+0 \right)}^{\dfrac{1}{2}}}+1}=\dfrac{1}{2} \\
\end{align}\]
Here, we can see that the value of \['B'\] is independent of \['b'\] this means that \['b'\] can take any value in its domain.
But students may mistake and assume that \['b'\] has no value.
Having no value means it is not defined but here the value of \['b'\] is defined which means that \['b'\] has some value.
So, the correct answer will be \['b'\] can take any value.
If a function \[f\left( x \right)\] is said to be continuous at a point \[x=a\] if and only if
\[\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=f\left( a \right)\]
By using the above equation we find the required limits for the given function to find the required values. Here, \[{{a}^{-}},{{a}^{+}}\] represents that the value of \['x'\] tends to \['a'\] from left side and right side of \['a'\] respectively.
Complete step by step answer:
We are given that the function is continuous at \[x=0\]
We know that if a function \[f\left( x \right)\] is said to be continuous at a point \[x=a\] if and only if
\[\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=f\left( a \right)\]
By using the above definition to given function we get
\[\displaystyle \lim_{x \to {{0}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{0}^{+}}}f\left( x \right)=f\left( 0 \right)......equation(i)\]
Now, let us assume that the first term as
\[\Rightarrow A=\displaystyle \lim_{x \to {{0}^{-}}}f\left( x \right)\]
We know that \[{{0}^{-}}\] represents that the value of \['x'\] tends to 0 from left side of 0 that is \[x<0\]
We are given that the value of function when \[x<0\] as
\[f\left( x \right)=\dfrac{\sin \left( a+1 \right)x+\sin x}{x}\]
By substituting this function in the above limit we get
\[\begin{align}
& \Rightarrow A=\displaystyle \lim_{x \to {{0}^{-}}}\left( \dfrac{\sin \left( a+1 \right)x+\sin x}{x} \right) \\
& \Rightarrow A=\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\sin \left( a+1 \right)x}{x}+\displaystyle \lim_{x \to {{0}^{-}}}\dfrac{\sin x}{x} \\
\end{align}\]
We know that the standard formula of limits that is
\[\displaystyle \lim_{x \to 0}\dfrac{\sin \left( nx \right)}{x}=n\]
By using this condition to above equation we get
\[\begin{align}
& \Rightarrow A=\left( a+1 \right)+1 \\
& \Rightarrow A=a+2 \\
\end{align}\]
Now, let us assume the second term in the equation (i) as
\[\Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}f\left( x \right)\]
We know that \[{{0}^{+}}\] represents that the value of \['x'\] tends to 0 from right side of 0 that is \[x>0\]
We are given that the value of function when \[x>0\] as
\[f\left( x \right)=\dfrac{{{\left( x+b{{x}^{2}} \right)}^{\dfrac{1}{2}}}-{{x}^{\dfrac{1}{2}}}}{b{{x}^{\dfrac{3}{2}}}}\]
By substituting this function in the above limit we get
\[\Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{{{\left( x+b{{x}^{2}} \right)}^{\dfrac{1}{2}}}-{{x}^{\dfrac{1}{2}}}}{b{{x}^{\dfrac{3}{2}}}}\]
Now, let us take the common term out from the numerator then we get
\[\begin{align}
& \Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{{{x}^{\dfrac{1}{2}}}{{\left( 1+bx \right)}^{\dfrac{1}{2}}}-{{x}^{\dfrac{1}{2}}}}{bx\left( {{x}^{\dfrac{1}{2}}} \right)} \right) \\
& \Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{{{\left( 1+bx \right)}^{\dfrac{1}{2}}}-1}{bx} \right) \\
\end{align}\]
Now, let us rationalise the numerator that is let us multiply and divide the above limit with \[\left( {{\left( 1+bx \right)}^{\dfrac{1}{2}}}+1 \right)\] then we get
\[\Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{{{\left( 1+bx \right)}^{\dfrac{1}{2}}}-1}{bx}\times \dfrac{{{\left( 1+bx \right)}^{\dfrac{1}{2}}}+1}{{{\left( 1+bx \right)}^{\dfrac{1}{2}}}+1} \right)\]
We know that the formula of algebra that is
\[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]
By using this formula to numerator of the above limit we get
\[\begin{align}
& \Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{\left( 1+bx \right)-1}{bx\left( {{\left( 1+bx \right)}^{\dfrac{1}{2}}}+1 \right)} \right) \\
& \Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{1}{{{\left( 1+bx \right)}^{\dfrac{1}{2}}}+1} \right) \\
\end{align}\]
Now, by expanding the limit we get
\[\begin{align}
& \Rightarrow B=\dfrac{1}{{{\left( 1+0 \right)}^{\dfrac{1}{2}}}+1} \\
& \Rightarrow B=\dfrac{1}{2} \\
\end{align}\]
Now, let us assume that the third term in equation (i) as
\[\Rightarrow C=f\left( 0 \right)\]
We are given that the value of \[f\left( x \right)\] at \[x=0\] as \['c'\]
By using the given function we get
\[\Rightarrow C=c\]
Now, by substituting the required values in the equation (i) we get
\[\Rightarrow a+2=\dfrac{1}{2}=c.......equation(ii)\]
Here, we can see that the limit of second term as \[B=\dfrac{1}{2}\] is independent of \['b'\]
So, we can say that \['b'\] can take any value because it has no effect on the limit.
Now, from the equation (ii) we have the value of \['c'\]taking the variable and constant together that is
\[\Rightarrow c=\dfrac{1}{2}\]
Now, from equation (ii) we get the value of \['a'\] as
\[\begin{align}
& \Rightarrow a+2=\dfrac{1}{2} \\
& \Rightarrow a=\dfrac{-3}{2} \\
\end{align}\]
Therefore we can conclude that \[a=\dfrac{-3}{2},c=\dfrac{1}{2}\] and \['b'\] can take any value.
So, the correct answer is “Option b, c and d”.
Note: Students may do mistake in taking the value of \['b'\]
Here, we have the limit where \['b'\] is included as
\[\begin{align}
& \Rightarrow B=\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{1}{{{\left( 1+bx \right)}^{\dfrac{1}{2}}}+1} \right) \\
& \Rightarrow B=\dfrac{1}{{{\left( 1+0 \right)}^{\dfrac{1}{2}}}+1}=\dfrac{1}{2} \\
\end{align}\]
Here, we can see that the value of \['B'\] is independent of \['b'\] this means that \['b'\] can take any value in its domain.
But students may mistake and assume that \['b'\] has no value.
Having no value means it is not defined but here the value of \['b'\] is defined which means that \['b'\] has some value.
So, the correct answer will be \['b'\] can take any value.
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