Question

If the given function\[f\left( x \right)\] is continuous at \[x = 0\] then find the value of \[a + 2b\].
\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{{\sin \left( {a + 2} \right)x + \sin x}}{x},x < 0} \\
  {b{\text{ ,}}x = 0} \\
  {\dfrac{{{{\left( {x + 3{x^2}} \right)}^{\dfrac{1}{3}}} - {x^{\dfrac{1}{3}}}}}{{{x^{\dfrac{4}{3}}}}}{\text{ ,x > 0}}}
\end{array}} \right.\]
A. \[ - 2\]
B. \[1\]
C. \[0\]
D. \[ - 1\]

Answer 1

Hint: In this question, for determining the value of \[b\], we have to equate the left-hand limit of a function to the right-hand limit of a function. After getting, the value of \[b\], we can easily calculate the value of \[b\] and also the value of \[a + 2b\].


Formula Used: If \[f\left( x \right)\] is continuous at \[x = 0\] then \[\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = f\left( 0 \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\]
Also, \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\] and \[\cos \left( 0 \right) = 1\]

Complete step by step answer:
 We know that \[f\left( x \right)\] is continuous at \[x = 0\].
\[\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = f\left( 0 \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\]
Thus, we can say that
\[\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = b = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\]
\[b = \mathop {\lim }\limits_{h \to 0} f\left( {0 + h} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {h + 3{h^2}} \right)}^{\dfrac{1}{3}}} - {h^{\dfrac{1}{3}}}}}{{{h^{\dfrac{4}{3}}}}}\]
\[b = \mathop {\lim }\limits_{h \to 0} \dfrac{{{h^{\dfrac{1}{3}}}\left[ {{{\left( {1 + 3h} \right)}^{\dfrac{1}{3}}} - 1} \right]}}{{{h^{\dfrac{4}{3}}}}}\]
\[b = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {{{\left( {1 + 3h} \right)}^{\dfrac{1}{3}}} - 1} \right]}}{h}\]
By simplifying, we get
\[b = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{3}\left[ {{{\left( {1 + 3h} \right)}^{\dfrac{{ - 2}}{3}}} \times 3} \right]\]
\[b = \dfrac{1}{3}\left[ {{{\left( {1 + 0} \right)}^{\dfrac{{ - 2}}{3}}} \times 3} \right]\]
\[b = \dfrac{1}{3}\left[ 3 \right]\]
\[b = 1\]
Now we can have
\[
  \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = 1 \\
   \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {a + 2} \right)\left( x \right) + \sin \left( x \right)}}{x} \\
   \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {a + 2} \right)\left( x \right) + \sin \left( x \right)}}{x} = 1 \\
 \]
Here, we will use the trigonometric identity such as \[\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\]
Thus, we get \[\sin \left( {ax + 2x} \right) = \sin \left( {ax} \right)\cos \left( {2x} \right) + \cos \left( {ax} \right)\sin \left( {2x} \right)\]
By simplifying, we get
\[
   \Rightarrow \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin \left( {ax} \right)\cos \left( {2x} \right) + \cos \left( {ax} \right)\sin \left( {2x} \right) + \sin \left( x \right)}}{x}} \right) = 1 \\
   \Rightarrow \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin \left( {ax} \right)\cos \left( {2x} \right)}}{x} + \dfrac{{\cos \left( {ax} \right)\sin \left( {2x} \right)}}{x} + \dfrac{{\sin \left( x \right)}}{x}} \right) = 1 \\
   \Rightarrow \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{a\sin \left( {ax} \right)\cos \left( {2x} \right)}}{{ax}} + \dfrac{{2\cos \left( {ax} \right)\sin \left( {2x} \right)}}{{2x}} + \dfrac{{\sin \left( x \right)}}{x}} \right) = 1 \\
   \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{a\sin \left( {ax} \right)\cos \left( {2x} \right)}}{{ax}} + \mathop {\lim }\limits_{x \to 0} \dfrac{{2\cos \left( {ax} \right)\sin \left( {2x} \right)}}{{2x}} + \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( x \right)}}{x} = 1 \\
 \]
But we know that \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\]
So, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} a\dfrac{{\sin \left( {ax} \right)}}{{ax}}\mathop {\lim }\limits_{x \to 0} \cos \left( {2x} \right) + \mathop {\lim }\limits_{x \to 0} 2\dfrac{{\sin \left( {2x} \right)}}{{2x}}\mathop {\lim }\limits_{x \to 0} \cos \left( {ax} \right) + \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( x \right)}}{x} = 1\]
\[
   \Rightarrow a\left( 1 \right)\cos \left( 0 \right) + 2\left( 1 \right)\cos \left( 0 \right) + \left( 1 \right) = 1 \\
   \Rightarrow a\left( 1 \right)\left( 1 \right) + 2\left( 1 \right)\left( 1 \right) + \left( 1 \right) = 1 \\
 \]
By simplifying further, we get
\[
   \Rightarrow a + 2 + 1 = 1 \\
   \Rightarrow a + 3 = 1 \\
   \Rightarrow a = 1 - 3 \\
   \Rightarrow a = - 2 \\
 \]
Thus, we get
\[
  a + 2b = - 2 + 2\left( 1 \right) \\
   \Rightarrow a + 2b = 0 \\
 \]
The value of \[a + 2b\]is \[0\].

Therefore, the option (C) is correct.

Additional Information: A function \[f\left( x \right)\]is continuous at a point \[x = b\], in its domain if and only if the below three conditions are fulfilled:
1) \[f\left( b \right)\]exists (That means the value of \[f\left( a \right)\]is finite)
2) \[\mathop {\lim }\limits_{x \to b} f\left( x \right)\]exists (That means the right-hand limit is equal to the left-hand limit, and both are finite in this case.)
3) \[\mathop {\lim }\limits_{x \to b} f\left( x \right) = f\left( b \right)\]
If any of the above three continuous-function criteria fails, the function is seen to be discontinuous at that moment. A limit is a value at which a function approaches the output for the given input values. This is the fundamental key point of calculus and analysis.

Note: Many students make mistakes in solving the limit of a function and further part. This is the only way, through which we can solve the example in the simplest way. In this question, the key point is the function is continuous which means the right-hand limit is equal to the left-hand limit of a given function.