P:R→R,P(0)=0,P(x)>x2∀x≠0 P(x) will be polynomial of even degree. P(x)=a0x2n+a1x2n−1+.......+a2n−2x2+a2n−1x P”(0)=
1
2
a2n−2=
1
4
P(x)=a0x2n+a1x2n−1+.......
1
4
x2+a2n−1x>x2∀x≠0 P(x)=a0x2n+a1x2n−1+.......−
3
4
x2+a2n−1x>0∀x≠0 Which is not always true Method II: P(x)>x2 P(x)=x2+f(x)f(x)>0∀x∈R0 ∵ P(0)=0⇒f(0)=0 Now, P”(x) = 2 + f”(x) f”(0) = negative ⇒ f(x) is concave down so f(x) can’t be +ve always ⇒ which is contradiction