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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=k⇒k3=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