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Question

The number of polynomial functions f of degree 1 satisfying f(x2) = (f (x)) 2 = f (f(x)) is


A

0

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B

1

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C

2

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D
Infinitely many
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Solution

The correct option is B

1


Let degree of f (x) be n.

deg of f(x2) = 2n

deg of ( f(x2)) = 2n

deg of f( f(x) = n2

n2 = 2n iff n = 0 or 2. But n ≥ 1

Degree of f(x) = 2

Since f(x2) = ( f(x2)), if the leading coefficient is a then the coefficient of x4 in the LHS is a and in the RHS is a2.

a = a2 Þ a = 1 ( a = 0 not admissible )

f (x) = x2 + bx + c

f (x2) = x4 + bx2 + c = (f(x))2 = (x2 + bx + c)2

== x4 + 2bx2 + b2x2 + 2cx2 + c2 + 2bcx

Thus Þ 0 = 2b ; b = 0

b2 + 2c = b Þ 2c = 0 Þ c = 0

The only function is f (x) = x2


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