ISI2017-MMA-24

Question

The number of polynomial function $f$ of degree $\geq$ 1 satisfying $f(x^{2})=(f(x))^{2}=f(f(x))$ for all real $x$, is

• A. $0$
• B. $1$
• C. $2$
• D. infinitely many

Comments

Option B only one polynomial function.

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Explain

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Explained in the answer section.

Answer 1

Consider the equality $(f(x))^2 = f(f(x)).$

Let $f(x) = a_0 x^n + a_1 x^{(n-1)} + a_2 x^{(n-2)} + a_3 x^{(n-3)} +\ldots+ a_n$

Note that the degree of $(f(x))^2 = 2n$ and the degree of $f(f(x)) = n^2$

Since $(f(x))^2 = f(f(x)),$ we have $2n = n^2$ .
$\implies n= 2.$

So, $f(x) = a_0 x^2 + a_1 x + a_2 $

Now, we have the equality $f(x^2) = (f(x))^2$

Now, $f(x^2) = a_0 x^4 + a_1 x^2 +a_2$ and
$(f(x))^2 = a_0^2 x^4 + 2 a_0a_1 x^3 + (2 a_0a_2 + a_1^2 ) x^2 + 2 a_1a_2 x + a_2^2.$

Now we have,

$a_0 x^4 + a_1 x^2 +a_2  =  a_0^2 x^4 + 2 a_0a_1 x^3 + (2 a_0a_2 + a_1^2 ) x^2 + 2 a_1a_2 x + a_2^2\quad \to (1)$

Now the coefficient of $x^3$ in L.H.S. is $0$ and the coefficient of $x^3$ in R.H.S. is $2a_0a_1.$

So, $2 a_0a_1 = 0.$

Hence, $a_1 = 0$ $($because $a_0$ can not be $0$ since it is the leading coefficient$)$

Now we have by putting $a_1 = 0$ in equation $(1)$

$a_0 x^4 + a_2  =  a_0^2 x^4 + 2 a_0a_2 x^2 + a_2^2 \quad \to(2)$

Similarly, by comparing the coefficients of $x^2$ on both sides we get $a_2 = 0.$

So, equation $(2)$ becomes

$a_0 x^4 = a_0^2 x^4$
$\implies a_0 = a_0^2$
$\implies a_0 = 1.$

So, the required polynomial function is $f(x) = x^2$

Hence, there is only one polynomial which $x^2$

So, option B is correct.

Comments

I have doubt in this question please explain

Answer 2

Let f(x)=xⁿ

f(x²)=x²ⁿ and (f(x))² =x²ⁿ

f(f(x))=f(xⁿ)=(xⁿ)ⁿ=x^{n^2} , according to question:

f(x²)= (f(x))² =f(f(x))

i.e. x^{n^2} = x²ⁿ or n²=2n or n=0,2

In the question it is given , degree must be >=1, therefore n=2 is the only answer. So x² is the only polynomial that could satisfy the given conditions.

Answer 3

Assume, $f(x) = x^2$

then, $f(x^2) $ will be $\left(x^2\right)^2 = x^4$

$\qquad \left(f(x)\right)^2 $ will be $\left(x^2\right)^2 = x^4$

$\qquad f\left(f(x)\right) $ will be $f(x^2) = $$ \left(x^2\right)^2 = x^4$

$∴\text{There are only 1 solutions}$

Correct answer is option B) 1

Answer 4

Two function are said to be equal iff

1. Both range are equal 
2. Both domain is equal 
3. Both co- domain is equal .

So domain of f(f(x))= f(x^2)

Means f(x) =x^2 

So answer is b