Test Index

Quadratic Equations

Section: Mathematics

© examsnet.com

Question : 18 of 51

Marks: +1, -0

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)

f(-1)=0

, then the range of

f

[AP EAPCET 22 May 2025 S2]

$[1, \infty)$
$[-1,1]$
$(-\infty, 1]$
$\mathbb{R}$

Solution:

f(x)=a x^{2}+b x+c

substitute

f(x)

into the given equation

\;a x^{2}+b x+c+a\left(\frac{1}{x}\right)^{2}+b\left(\frac{1}{x}\right)+c

\;= (a x^{2}+b x+c)\left(a\left(\frac{1}{x}\right)^{2}+b\left(\frac{1}{x}\right)+c\right)

\;\Rightarrow a x^{2}+b x+c+\frac{a}{x^{2}}+\frac{b}{x}+c

\;= (a x^{2}+b x+c)\left(\frac{a}{x^{2}}+\frac{b}{x}+c\right)

\;\Rightarrow a x^{4}+b x^{3}+c x^{2}+a+b x+c x^{2}

\;= (a x^{2}+b x+c)(a+b x+c x^{2})

\;\Rightarrow a x^{4}+b x^{3}+2c x^{2}+b x+a

\;=a^{2} x^{2}+a b x^{3}+a c x^{4}+a b x+b^{2} x^{2}+b c x^{3}+a c+b c x+c^{2} x^{2}

Compare coefficients :

a=a c

,

b\;=a b+b c

2c\;=a^{2}+b^{2}+c^{2},\;b=a b+b c,\;a=a c

From

a=a c

, we have

c=1

or

a=0

, if

a=0

then

f(x)

is not quadratic, so

c=1

from

b=a b+b c

, we have

b=a b+b

, so

a b=0

, since

a \neq 0,\;b=0

From

2 c=a^{2}+b^{2}+c^{2}

, we have

2=a^{2}+0+1

So,

\;\;a^{2}=1

and

a= \pm 1

thus,

f(x)=x^{2}+1

or

f(x)=-x^{2}+1

Since,

f(-1)=0

, we have

f(x)=-x^{2}+1

\;\Rightarrow \;\;f(x)=-x^{2}+1

\;\Rightarrow \;\text{ Range }=(-\infty, 1]

© examsnet.com

Go to Question:

Prev Question

Next Question