Let a be largest real root and b be smallest real root of polynomial equation x^6-6 x^5+15 x^4-20 x^

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4-2.Quadratic Equations and Inequations

hard

Let $a$ be the largest real root and $b$ be the smallest real root of the polynomial equation $x^6-6 x^5+15 x^4-20 x^3+15 x^2-6 x+1=0$ Then $\frac{a^2+b^2}{a+b+1}$ is

A

$\frac{1}{2}$

B

$\frac{2}{3}$

C

$\frac{5}{4}$

D

$\frac{13}{7}$

(KVPY-2021)

Solution

(b)

$x ^6-6 x ^5+15 x ^4-20 x ^3+15 x ^2-6 x +1=0$ $( x -1)^6=0$

All roots are equal and are equal to 1 $\therefore a =1, b =1$

$\therefore \frac{a^2+b^2}{a+b+1}=\frac{2}{3}$

Standard 11

Mathematics

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