A) 2
B) 3
C) 5
D) 4
Correct Answer: B
Solution :
\[f(xy)=f(x)\,\,f(y)\forall x,\,\,y\,\in \,\,R\,and\,f(0)\,\,\ne \,\,0\] put \[x=y=0\] \[\Rightarrow \,\,\,f\left( 0 \right)={{[f\left( 0 \right)]}^{2}}\] \[\Rightarrow \,\,\,\,f\left( 0 \right)=1~~~~~~~~~~~~\] put \[y=0\,\,\,\Rightarrow \,\,f(0)=f(x)\,f(0)\] \[\Rightarrow \]\[f\left( x \right)=1\] given that \[\frac{\text{dy}}{\text{dx}}\text{=f(x)}\] \[\therefore \,\,\,\,\frac{dy}{dx}\,\,=\,\,1\,\,\,\,\,\Rightarrow \,\,y=x+k\] given that \[y\left( 0 \right)=1\] \[\therefore \,\,\,k=1\] hence \[y=x+1\] \[y\left( \frac{1}{4} \right)+y\left( \frac{3}{4} \right)=\left( \frac{1}{4}+1 \right)+\left( \frac{3}{4}+1 \right)\,\,=\,\,3\]You need to login to perform this action.
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