Polynomial Thrill

$A$ :If $f$ is polynomial,then make two cases. (i) If $f(x)$ = a constant $c$ , then the given condition is equivalent to $c = c^{2013}$ , which happens precisely for three values of c, namely $c = 0; 1;-1$ .Thus there are three constant functions with the given property.

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(ii) If $f(x)$ is a non-constant polynomial, then consider its range set $A$ ={f(x)|x belongs to $R)$ }. Now for all $a$ belonging to $A$ , we have by the given property $f(a) = a^{2013}$ . So the polynomial $f(x) = x^{2013}$ has all elements of A as its roots. Since there are infinitely many values in A (e.g. applying the intermediate value theorem because f is continuous), the polynomial $f(x)=x^{2013}$ has infinitely many roots and thus must be the zero polynomial, i.e., $f(x) = x^{2013}$ for all real number $x$ .

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To find infinitely many function with the given property, define $f(0) = 0; f(1) = 1;f(-1) = -1$ . For every other real number $x$ , arbitrarily define $f(x)$ to be 0; 1 or -1. It is easy to see that any such function satisfies the given property.

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$B:$ We have $3*p^2q + 3*p^2r + 2*q^+ 2*r^3=(q + r)(3p^2+ 2q^2+ 2r^2-2qr)=(q+r)(3(p^2+q^2+r^2)-(q^2+r^2-2qr)=(q+r)(3-(q+r)^2)=x(3-x^2)=3x-x^3$ ;where $x=q+r$ .Let's try now.

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$q^2+r^2<=p^2+q^2+r^2=1$ .So $-\sqrt{2}<=x<=\sqrt{2}$ .Now it is easy to see that in the given domain $-2<=f(x)<=2$ .

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Hence $a=\infty,b=4,L=2,S=-2$ .

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NOTE:BOTH OF THE QUESTIONS ARE TAKEN FROM PAST YEAR CMI PAPERS,SO NO CLAIM OF ORIGINALITY IS MADE