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JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Evening)

  • question_answer
    Let f: \[\left[ 0,\text{ }1 \right]\,\,\to \,\,R\] be such that\[f\left( xy \right)=f\left( x \right).f\left( y \right)\], for all \[x,\,\,y\,\,\in \,\,\,[0,\,\,\,1]\] and\[f(0)\,\,\ne \,\,0\]. If \[y\,\,=\,\,y(x)\]satisfies the differential equation, \[\frac{dy}{dx}\,\,=\,\,f(x)\,\]with \[y\left( 0 \right)\,\,=\,\,1\], then \[y\left( \frac{1}{4} \right)+y\left( \frac{3}{4} \right)\]is equal to: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

    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 \[y=x+1\] \[y\left( \frac{1}{4} \right)+y\left( \frac{3}{4} \right)=\left( \frac{1}{4}+1 \right)+\left( \frac{3}{4}+1 \right)\,\,=\,\,3\]


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