Can you please checkout this solution.

1 vote

If two real polynomials $f(x)$ and $g(x)$ of degrees $m\: (\geq 2)$ and $n\: (\geq 1)$ respectively, satisfy

$$f(x^2+1)=f(x)g(x),$$

for every $x \in \mathbb{R}$, then

- $f$ has exactly one real root $x_0$ such that $f’(x_0) \neq 0$
- $f$ has exactly one real root $x_0$ such that $f’(x_0) = 0$
- $f$ has $m$ distinct real roots
- $f$ has no real root

1 vote

$\underline{\mathbf{Answer:D}}$

$\underline{\mathbf{Solution:}}$

Let's say, $\mathrm x$ is a root then $x\mathrm{^2+1}$ should also be a root.

Thus, if $\mathrm {x_1}$ is the largest real root, then we should have $\mathrm{x_1 \ge x_1^2+1}$

$\mathrm{\Rightarrow x_1^2 -x_1 + 1 \le 0}$

But,

$\mathrm{x_1^2-x_1+1 = (x_1-\frac{1}{2})^2 + \frac{3}{4} > 0}$

which is a contradiction.

$\therefore \mathbf f$ has no real roots.

$\therefore \mathbf D$ is the correct answer.

$\underline{\mathbf{Solution:}}$

Let's say, $\mathrm x$ is a root then $x\mathrm{^2+1}$ should also be a root.

Thus, if $\mathrm {x_1}$ is the largest real root, then we should have $\mathrm{x_1 \ge x_1^2+1}$

$\mathrm{\Rightarrow x_1^2 -x_1 + 1 \le 0}$

But,

$\mathrm{x_1^2-x_1+1 = (x_1-\frac{1}{2})^2 + \frac{3}{4} > 0}$

which is a contradiction.

$\therefore \mathbf f$ has no real roots.

$\therefore \mathbf D$ is the correct answer.

1

@`JEET

The answer is correct. But your proof is fallacious as you show for the case only when $x_1$ is the largest real root. But you need to show what if $x_1$ is the lowest real root! I meant what if $x_1< {x_{1}}^2+1$.

0 votes

f($x^{_{2}}+1$)=f(x)g(x)

when x=0; f(1)=f(0)g(0)

if k,q are the constant terms of polynomials f,g then

f(1)=kq

hence f(x)=k$x^{_{q-1}}$+k$x^{_{q-2}}$+.......+k

power of x terms in f($x^{_{2}}+1$) are always even numbers. hence f(x) and g(x) should have x terms with even powers.

f(x) is an polynomial with even degree m>=2 and all-one polynomial. hence **f(x) has no real distinct roots.**

**ANSWER: D) no distinct real roots.**