Question

The value of \[p\] for which the function \[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{{{{\left( {{4^x} - 1} \right)}^3}}}{{\sin \left( {\dfrac{x}{p}} \right)\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}};{\text{ }}x \ne 0} \\
  {{\text{ }}12{{\left( {\log 4} \right)}^3}{\text{ }};{\text{ }}x = 0}
\end{array}} \right.\] is continuous at \[x = 0\], is
(a) 4
(b) 2
 (c) 3
 (d) 1

Answer 1

Hint: Here, we need to find the value of \[p\]. A function is continuous at \[x = a\], if \[f\left( a \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right)\]. We will use the given information for the point at which the function is continuous. Then, we will rewrite the expression and use limit equations to simplify the expression. Finally, we will solve the equation to find the required value of \[p\].

Formula Used:
 A function is continuous at \[x = a\], if \[f\left( a \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right)\].
Limit equation: If \[x\] is approaching 0, then \[\mathop {\lim }\limits_{x \to 0} \dfrac{{{a^x} - 1}}{x}\] is equal to \[\log a\].
Limit equation: If \[x\] is approaching 0, then \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x}\] is equal to 1.
Limit equation: If \[x\] is approaching 0, then \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + x} \right)}}{x}\] is equal to 1.

Complete step-by-step answer:
A function is continuous at \[x = a\], if \[f\left( a \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right)\].
The given function is continuous at \[x = 0\].
Therefore, we get
\[f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} f\left( x \right)\]
Now, since \[x\] is approaching 0, it is not equal to 0.
Therefore, substituting \[f\left( x \right) = \dfrac{{{{\left( {{4^x} - 1} \right)}^3}}}{{\sin \left( {\dfrac{x}{p}} \right)\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}\] in the equation, we get
\[ \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {{4^x} - 1} \right)}^3}}}{{\sin \left( {\dfrac{x}{p}} \right)\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}\]
We will simplify the expression on the right hand side using limit equations.
Dividing the numerator and denominator of the fraction by \[{x^3}\], we get
\[ \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{{{\left( {{4^x} - 1} \right)}^3}}}{{{x^3}}}}}{{\dfrac{{\sin \left( {\dfrac{x}{p}} \right)\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{{x^3}}}}}\]
Rewriting the equation, we get
\[\begin{array}{l} \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {\dfrac{{{4^x} - 1}}{x}} \right)}^3}}}{{\dfrac{{\sin \left( {\dfrac{x}{p}} \right)\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{x \cdot {x^2}}}}}\\ \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {\dfrac{{{4^x} - 1}}{x}} \right)}^3}}}{{\dfrac{{\sin \left( {\dfrac{x}{p}} \right)}}{x} \times \dfrac{{\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{{x^2}}}}}\end{array}\]
Writing the limit before the terms, we get
\[ \Rightarrow f\left( 0 \right) = \dfrac{{{{\left( {\mathop {\lim }\limits_{x \to 0} \dfrac{{{4^x} - 1}}{x}} \right)}^3}}}{{\left[ {\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {\dfrac{x}{p}} \right)}}{x}} \right]\left[ {\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{{x^2}}}} \right]}}\]
Now, we will use limit equations to simplify the equation further.
Limit equation: If \[x\] is approaching 0, then \[\mathop {\lim }\limits_{x \to 0} \dfrac{{{a^x} - 1}}{x}\] is equal to \[\log a\].
Substituting \[a = 4\] in the limit equation, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{4^x} - 1}}{x} = \log 4\]
Substituting \[\mathop {\lim }\limits_{x \to 0} \dfrac{{{4^x} - 1}}{x} = \log 4\] in the equation \[f\left( 0 \right) = \dfrac{{{{\left( {\mathop {\lim }\limits_{x \to 0} \dfrac{{{4^x} - 1}}{x}} \right)}^3}}}{{\left[ {\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {\dfrac{x}{p}} \right)}}{x}} \right]\left[ {\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{{x^2}}}} \right]}}\], we get
\[ \Rightarrow f\left( 0 \right) = \dfrac{{{{\left( {\log 4} \right)}^3}}}{{\left[ {\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {\dfrac{x}{p}} \right)}}{x}} \right]\left[ {\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{{x^2}}}} \right]}}\]
Dividing and multiplying the denominator of the first parentheses by \[p\], we get
\[ \Rightarrow f\left( 0 \right) = \dfrac{{{{\left( {\log 4} \right)}^3}}}{{\left[ {\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {\dfrac{x}{p}} \right)}}{{x \times \dfrac{p}{p}}}} \right]\left[ {\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{{x^2}}}} \right]}}\]
Rewriting the expression, we get
\[ \Rightarrow f\left( 0 \right) = \dfrac{{{{\left( {\log 4} \right)}^3}}}{{\left[ {\dfrac{1}{p} \times \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {\dfrac{x}{p}} \right)}}{{\dfrac{x}{p}}}} \right]\left[ {\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{{x^2}}}} \right]}}\]
Limit equation: If \[x\] is approaching 0, then \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x}\] is equal to 1.
Substituting \[\dfrac{x}{p}\] for \[x\] in the limit equation, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \dfrac{x}{p}}}{{\dfrac{x}{p}}} = 1\]
Substituting \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \dfrac{x}{p}}}{{\dfrac{x}{p}}} = 1\] in the equation \[f\left( 0 \right) = \dfrac{{{{\left( {\log 4} \right)}^3}}}{{\left[ {\dfrac{1}{p} \times \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {\dfrac{x}{p}} \right)}}{{\dfrac{x}{p}}}} \right]\left[ {\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{{x^2}}}} \right]}}\], we get
\[ \Rightarrow f\left( 0 \right) = \dfrac{{{{\left( {\log 4} \right)}^3}}}{{\left[ {\dfrac{1}{p} \times 1} \right]\left[ {\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{{x^2}}}} \right]}}\]
Multiplying the terms in the expression, we get
\[ \Rightarrow f\left( 0 \right) = \dfrac{{{{\left( {\log 4} \right)}^3}}}{{\dfrac{1}{p}\left[ {\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{{x^2}}}} \right]}}\]
Dividing and multiplying the denominator of the parentheses by 3, we get
\[ \Rightarrow f\left( 0 \right) = \dfrac{{{{\left( {\log 4} \right)}^3}}}{{\dfrac{1}{p}\left[ {\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{{x^2} \times \dfrac{3}{3}}}} \right]}}\]
Rewriting the expression, we get
\[ \Rightarrow f\left( 0 \right) = \dfrac{{{{\left( {\log 4} \right)}^3}}}{{\dfrac{1}{p}\left[ {\dfrac{1}{3}\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{\dfrac{{{x^2}}}{3}}}} \right]}}\]
Limit equation: If \[x\] is approaching 0, then \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + x} \right)}}{x}\] is equal to 1.
Substituting \[\dfrac{{{x^2}}}{3}\] for \[x\] in the limit equation, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{\dfrac{{{x^2}}}{3}}} = 1\]
Substituting \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{\dfrac{{{x^2}}}{3}}} = 1\] in the equation \[f\left( 0 \right) = \dfrac{{{{\left( {\log 4} \right)}^3}}}{{\dfrac{1}{p}\left[ {\dfrac{1}{3}\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}{{\dfrac{{{x^2}}}{3}}}} \right]}}\], we get
\[ \Rightarrow f\left( 0 \right) = \dfrac{{{{\left( {\log 4} \right)}^3}}}{{\dfrac{1}{p}\left[ {\dfrac{1}{3} \times 1} \right]}}\]
Multiplying the terms in the denominator, we get
\[ \Rightarrow f\left( 0 \right) = \dfrac{{{{\left( {\log 4} \right)}^3}}}{{\dfrac{1}{p}\left[ {\dfrac{1}{3}} \right]}}\]
\[ \Rightarrow f\left( 0 \right) = \dfrac{{{{\left( {\log 4} \right)}^3}}}{{\dfrac{1}{{3p}}}}\]
Rewriting the expression, we get
\[ \Rightarrow f\left( 0 \right) = 3p{\left( {\log 4} \right)^3}\]
It is given that if \[x = 0\], then \[f\left( x \right) = 12{\left( {\log 4} \right)^3}\].
Therefore, we get
\[ \Rightarrow f\left( 0 \right) = 12{\left( {\log 4} \right)^3}\]
Comparing the equations \[f\left( 0 \right) = 3p{\left( {\log 4} \right)^3}\] and \[f\left( 0 \right) = 12{\left( {\log 4} \right)^3}\], we get
\[ \Rightarrow 3p{\left( {\log 4} \right)^3} = 12{\left( {\log 4} \right)^3}\]
Dividing both sides by \[3{\left( {\log 4} \right)^3}\], we get
\[ \Rightarrow \dfrac{{3p{{\left( {\log 4} \right)}^3}}}{{3{{\left( {\log 4} \right)}^3}}} = \dfrac{{12{{\left( {\log 4} \right)}^3}}}{{3{{\left( {\log 4} \right)}^3}}}\]
Thus, we get
\[\therefore p=4\]

Therefore, we get the value of \[p\] as 4. The correct option is option (a).

Note: We simplified the right hand side expression of the equation \[f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {{4^x} - 1} \right)}^3}}}{{\sin \left( {\dfrac{x}{p}} \right)\log \left( {1 + \dfrac{{{x^2}}}{3}} \right)}}\] using limit equations. This is because we put \[x = 0\] without simplifying using limit equations, then the equation will become of the form \[\dfrac{0}{0}\]. This should be avoided by using limit equations.