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Question

f:(0,)(0,),f(xf(y))=x2ya(aR) then

Number of solutions of 2f(x)=ex is

A
0.0
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B
1
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C
2
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D
3
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Solution

The correct option is D 3
Put y =1
f(xf(1))=x2
Let f(1)=K,f(Kx)=x2
f(x)=x2k2
f(1)=12k2=kk3=1
k =1
So f(x)=x2
f(x.y)2=x2ya(xy2)2=x2ya
x2y4=x2ya
a=4
For 2f(x)=ex
2x2=ex

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