Question

Find k such that the given function is continuous \[f\left( x \right)=\left\{ \begin{matrix}
  & \dfrac{\sqrt{3}\sin x+\cos x}{x+\dfrac{\pi }{6}},&x\ne \dfrac{\pi }{6} \\
 & k,&x=-\dfrac{\pi }{6} \\
\end{matrix} \right\}\] at \[x= \dfrac{\pi }{6}\].

Answer 1

Hint: In this type of question we have to use the concept of continuity of a function along with some values as well as rules of trigonometric functions. We know that if a function \[f\left( x \right)\] is continuous at \[x=a\] then \[\displaystyle \lim_{x \to a}f\left( x \right)=f\left( a \right)\]. Also here we have to use one of the rule of continuity that is \[\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1\]. Also we have to consider some values of the trigonometric functions that are \[\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}\And \cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}\]. Also we have to use the trigonometric formula, \[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\].

Complete step-by-step solution:
Now, here we have to find the value of k such that the given function \[f\left( x \right)\] is continuous at \[x=\dfrac{\pi }{6}\] and \[f\left( x \right)=\left\{ \begin{matrix}
  & \dfrac{\sqrt{3}\sin x+\cos x}{x+\dfrac{\pi }{6}},&x\ne \dfrac{\pi }{6} \\
 & k,&x=-\dfrac{\pi }{6} \\
\end{matrix} \right\}\] at \[x=\dfrac{\pi }{6}\]
As \[f\left( x \right)\] is continuous at \[x=\dfrac{\pi }{6}\] we can write
\[\Rightarrow \displaystyle \lim_{x \to \dfrac{\pi }{6}}f\left( x \right)=f\left( \dfrac{\pi }{6} \right)\]
Also, we have given that, \[f\left( x \right)=k\] at \[x=-\dfrac{\pi }{6}\]
Thus we can write,
\[\Rightarrow f\left( -\dfrac{\pi }{6} \right)=k\]
Now, if we directly substitute the value of \[x\] in the function \[f\left( x \right)=\dfrac{\sqrt{3}\sin x+\cos x}{x+\dfrac{\pi }{6}}\] then the denominator of \[f\left( x \right)\] becomes zero.
Thus let us consider.
\[\Rightarrow \displaystyle \lim_{x \to \text{ }-\text{ }\dfrac{\pi }{6}}f\left( x \right)=k\]
\[\Rightarrow \displaystyle \lim_{x \to \text{ }-\text{ }\dfrac{\pi }{6}}\dfrac{\sqrt{3}\sin x+\cos x}{x+\dfrac{\pi }{6}}=k\]
\[\Rightarrow \displaystyle \lim_{x \to \text{ }-\text{ }\dfrac{\pi }{6}}\dfrac{2\left( \dfrac{\sqrt{3}}{2}\sin x+\dfrac{1}{2}\cos x \right)}{x+\dfrac{\pi }{6}}=k\]
Now, by using \[\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}\And \cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}\] we get,
\[\Rightarrow \displaystyle \lim_{x \to \text{ }-\text{ }\dfrac{\pi }{6}}\dfrac{2\left( \sin x\cos \dfrac{\pi }{6}+\cos x\sin \dfrac{\pi }{6} \right)}{\left( x+\dfrac{\pi }{6} \right)}=k\]
Also we know that, \[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\]
\[\Rightarrow \displaystyle \lim_{x \to \text{ }-\text{ }\dfrac{\pi }{6}}\dfrac{2\sin \left( x+\dfrac{\pi }{6} \right)}{\left( x+\dfrac{\pi }{6} \right)}=k\]
Now, put \[\left( x+\dfrac{\pi }{6} \right)=t\] then as \[x \to -\dfrac{\pi }{6}\] we get \[t\to 0\].
Hence, by using the above substitution we can write
\[\Rightarrow \displaystyle \lim_{t\to \text{ 0}}\dfrac{2\sin t}{t}=k\]
As 2 is a constant we get,
\[\Rightarrow 2\displaystyle \lim_{t\to \text{ 0}}\dfrac{\sin t}{t}=k\]
By using the rule of continuity that is \[\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1\] we get,
\[\Rightarrow 2=k\]
Hence, for the given \[f\left( x \right)\] the value of k is 2.

Note: In this type of question interpretation of the question is important. Students have to take care in calculation of the limit and using the appropriate formulae of limit and continuity as well as of trigonometry. Also students have to recall the values of trigonometric sine and cosine function for the angles like \[\dfrac{\pi }{6}\], \[\dfrac{\pi }{3}\], \[\dfrac{\pi }{4}\],……etc.