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Question

Let $$f\left( xy \right) =f\left( x \right) \cdot f\left( y \right) $$ for all $$x,y\in R$$. If $$f^{ ' }\left( 1 \right) =2$$ and $$f\left( 4 \right) =4$$, then $$f^{ ' }\left( 4 \right) $$ equal to


A
4
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B
1
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C
12
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D
8
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Solution

The correct option is D $$8$$
We have $$f\left( x,y \right) =f\left( x \right) f\left( y \right) $$ for all $$x,y\in R$$.
Putting $$x=y=1$$, we get
$$f\left( 1 \right) =f\left( 1 \right) f\left( 1 \right) $$
$$\Rightarrow f\left( 1 \right) \left[ 1-f\left( 1 \right)  \right] =0$$
$$\Rightarrow f\left( 1 \right) =1$$             $$\left[ \because f\left( 1 \right) \neq 0 \right] $$
Now, $$f^{ ' }\left( 1 \right) =2$$
$$\Rightarrow \displaystyle\lim _{ h\rightarrow 0 }{ \dfrac { f\left( 1+h \right) -f\left( 1 \right)  }{ h }  } =2$$
$$\Rightarrow f\left( 1 \right) \displaystyle\lim _{ h\rightarrow 0 }{ \dfrac { f\left( h \right) -1 }{ h }  } =2$$
$$\Rightarrow \displaystyle\lim _{ h\rightarrow 0 }{ \dfrac { f\left( 1 \right) f\left( h \right) -f\left( 1 \right)  }{ h }  } =2$$          [using $$f\left( 1 \right) =1$$]
$$\Rightarrow \displaystyle\lim _{ h\rightarrow 0 }{ \dfrac { f\left( h \right) -1 }{ h }  } =2$$                    ....(i)
Now, $$f^{ ' }\left( 4 \right) =\displaystyle\lim _{ h\rightarrow 0 }{ \dfrac { f\left( 4+h \right) -f\left( 4 \right)  }{ h }  } $$
$$=\displaystyle\lim _{ h\rightarrow 0 }{ \dfrac { f\left( 4 \right) \cdot f\left( h \right) -f\left( 4 \right)  }{ h }  } $$
$$=\left\{ \displaystyle\lim _{ h\rightarrow 0 }{ \dfrac { f\left( h \right) -1 }{ h }  }  \right\} \cdot f\left( 4 \right) =2f\left( 4 \right) $$      ....{from (i)}
$$=2\times 4=8$$

Mathematics

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