To prove that denominator has discriminate 0.

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F(x)= $\frac{a_1x^2+b_1x+c_1}{a_2x^2+b_2x+c_2}$

Given: (i)$a_1,a_2,b_1,b_2,c_1,c_2$ are all real.

(ii) $a_1,a_2 \not=$0

(iii) $(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)=\ (c_1a_2-c_2a_1)^2$

(iv) $a_1b_2-a_2b_1\not=$0

(v) Range of the function is R-{$\lambda$}

To prove : (1) $b_1^2-4a_1c_1\ge$0

(2) $\lambda=\frac{a_1}{a_2}$

(3) $b_2^2-4a_2c_2=0$


As (iii) 3rd condition is of atleast 1 common root, and by (iv) we can say both roots cannot be equal. As the constants are real so if any roots are imaginary then both roots would be common which is not possible .Hence roots are real therefore (1)(as D$\ge$0) is proven . But i am not able to prove (3).Any hints would be appreciated.

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Statement (3) is simply not true. Counterexample:

$f(x)=\frac{x^2}{x^2+bx}\ \ {\rm for\ }b\neq 0$

Then: (i): all coefficients are real

(ii): leading coefficients non-zero

(iii): $(b)(0)=(-0)^2$: true.

(iv): $(1)(b)\neq (1)(0)$: true.

(v): The function simplifies to $\frac{x}{x+b}=1-\frac{b}{x+b}$, and so its range is clearly $\Bbb R\setminus\{1\}$.

However, (3) is false, since $b^2-(1)(0)\neq 0$ whenever $b\neq 0$.