f(x) is a quadratic polynomial satisfying the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $f(x) = ax^2 + bx + c$. The given equation is $f(x) + f(1/x) = f(x)f(1/x)$,which can be rewritten as $f(x)f(1/x) - f(x) - f(1/x) + 1 = 1$,or $

Vedclass · Accepted Answer

$(-\infty, 1]$

Vedclass · Answer

(C) Let $f(x) = ax^2 + bx + c$. The given equation is $f(x) + f(1/x) = f(x)f(1/x)$,which can be rewritten as $f(x)f(1/x) - f(x) - f(1/x) + 1 = 1$,or $(f(x) - 1)(f(1/x) - 1) = 1$. Let $g(x) = f(x) - 1$. Then $g(x)g(1/x) = 1$. Since $f(x)$ is a quadratic polynomial,$g(x)$ is also a quadratic polynomial. Let $g(x) = ax^2 + bx + c'$. Then $(ax^2 + bx + c')(a/x^2 + b/x + c') = 1$. Comparing coefficients,we find $g(x) = x^2$ or $g(x) = 1/x^2$ (not a polynomial) or $g(x) = -x^2$ or $g(x) = 1$. Given $f(-1) = 0$,we have $g(-1) = f(-1) - 1 = -1$. If $g(x) = x^2$,$g(-1) = 1 
eq -1$. If $g(x) = -x^2$,$g(-1) = -1$. Thus $f(x) - 1 = -x^2$,which means $f(x) = 1 - x^2$. The range of $f(x) = 1 - x^2$ is $(-\infty, 1]$.

$f(x)$ is a quadratic polynomial satisfying the condition $f(x) + f\left(\frac{1}{x}\right) = f(x) f\left(\frac{1}{x}\right)$. If $f(-1) = 0$,then the range of $f$ is

Similar Questions

The range of the real valued function $f(x) = \sqrt{\frac{x^2+2x+8}{x^2+2x+4}}$ is

If $f:[2,3] \rightarrow R$ is defined by $f(x)=x^3+3x-2$,then the range of $f(x)$ is contained in the interval

The domain and range of $y(x) = \cos x - 3$ are respectively

The range of the function $f(x)=-\sqrt{5-6x-x^2}$ is

The domain of $f(x) = \sqrt{\left(\frac{1}{\sqrt{x}} - \sqrt{x+1}\right)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module