Question

# 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
B
2
C
3
D
1

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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