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The value of $$p$$ for which the function
$$f(x)=\displaystyle \left\{ \begin{array}{rl} \dfrac { (4^{ { x } }-1)^{ 3 } }{ \sin\dfrac { { x } }{ { p } } \log(1+\dfrac { { x }^{ 2 } }{ 3 } ) }  & { ;\quad x\neq 0 } \\ 12(\log  4)^{ 3 } & { ;\quad x=0\quad  } \end{array} \right. $$ is continuous at $${x}=0$$, is


A
4
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B
2
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C
3
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D
1
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Solution

The correct option is A $$4$$
We know that the function is continuous at $$x=0$$ if $$\displaystyle \lim _{  x\rightarrow0  }{ f(x) } $$ exists and is equal to $$f(0)$$.

Therefore, for continuity we have,
$$\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { { \left( { 4 }^{ x }-1 \right)  }^{ 3 } }{ \sin { \dfrac { x }{ p }  } \log { \left( 1+\dfrac { { x }^{ 2 } }{ 3 }  \right)  }  }  } =f\left( 0 \right) =12{ \left( \log { 4 }  \right)  }^{ 3 }$$

Since 
$$\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { { \left( { a }^{ x }-1 \right)  } }{ x }  } =\log { a }$$ for $$a\neq 0,a>1$$,

$$\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { \sin { x }  }{ x } =1 } $$, and

$$\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { \log { \left( 1+x \right)  }  }{ x } =1 } $$,

the limit on LHS can be rewritten as:

$$\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { { \left( \dfrac { { 4 }^{ x }-1 }{ x }  \right)  }^{ 3 }\cdot { x }^{ 3 } }{ \dfrac { \sin { \dfrac { x }{ p }  }  }{ \dfrac { x }{ p }  } \cdot \dfrac { x }{ p } \cdot \dfrac { \log { \left( 1+\dfrac { { x }^{ 2 } }{ 3 }  \right)  }  }{ \dfrac { { x }^{ 2 } }{ 3 }  } \cdot \dfrac { { x }^{ 2 } }{ 3 }  }  } $$

$$=\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { { \left( \log { 4 }  \right)  }^{ 3 }\cdot { x }^{ 3 } }{ 1\cdot \dfrac { x }{ p } \cdot 1\cdot \dfrac { { x }^{ 2 } }{ 3 }  }  }$$  (using standard limits stated above)

$$=3p{ \left( \log { 4 }  \right)  }^{ 3 }$$ .... (because $$\displaystyle \lim _{ x\rightarrow 0 }{ \dfrac { { x }^{ 3 } }{ { x }^{ 3 } } =1 } $$)

According to the condition for continuity, this must equal $$f\left( 0 \right) =12{ \left( \log { 4 }  \right)  }^{ 3 }$$

$$3p{ \left( \log { 4 }  \right)  }^{ 3 }=12{ \left( \log { 4 }  \right)  }^{ 3 }$$
$$\\ \Rightarrow 3p=12$$
$$\\ \Rightarrow p = 4$$

Mathematics

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