Question

Determine the values of $a,b,c$ for which the function given below is continuous at $x=0$

& \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{aligned} \right.$$ (a) $$'a'$$ can take any value (b) $$c=\dfrac{1}{2}$$ (c) $$a=\dfrac{-3}{2}$$ (d) $$'b'$$ can take any value

Answer 1

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

& \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{aligned}$$ We know that the standard formula of limits that is $$\displaystyle \lim_{x \to 0}\dfrac{\sin \left( nx \right)}{x}=n$$ By using this condition to above equation we get $$\begin{aligned} & \Rightarrow A=\left( a+1 \right)+1 \\\ & \Rightarrow A=a+2 \\\ \end{aligned}$$ Now, let us assume the second term in the equation (i) as $$\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{aligned} & \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{aligned}$$ Now, let us rationalise the numerator that is let us multiply and divide the above limit with $$\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{aligned} & \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{aligned}$$ Now, by expanding the limit we get $$\begin{aligned} & \Rightarrow B=\dfrac{1}{{{\left( 1+0 \right)}^{\dfrac{1}{2}}}+1} \\\ & \Rightarrow B=\dfrac{1}{2} \\\ \end{aligned}$$ Now, let us assume that the third term in equation (i) as $$\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{aligned} & \Rightarrow a+2=\dfrac{1}{2} \\\ & \Rightarrow a=\dfrac{-3}{2} \\\ \end{aligned}$$ Therefore we can conclude that $$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{aligned} & \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{aligned}$$ Here, we can see that the value of $$'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.