Question

If a, b are the roots of quadratic equation 6x² – 6x + 1 = 0 then the value of

$\frac{1}{2}$ (a + ba + ca² + da³) + $\frac{1}{2}$ (a + bb + cb²+ db³) is –

• A. $\frac{12d + 6c + 4b + a}{12}$
• B. 12a + 6b + 4c + 9d
• C. $\frac{1}{12}$ (12a + 6b + 4c + 3d)
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{12}$ (12a + 6b + 4c + 3d)

Answer 2

a + b = 1

ab = 1/6 Now expression

= $\frac{1}{2}$ (a + a) + $\frac{b}{2}$ (a + b) + $\frac{c}{2}$ (a² + b²) + $\frac{d}{2}$ (a³ + b³)

= a + $\frac{b}{2}$ (1) + $\frac{c}{2}$ [(a + b)² – 2ab] + $\frac{d}{2}$ [(a + b)³ – 3ab

(a + b)]

= a + $\frac{b}{2}$ + $\frac{c}{2}$ [1 – 2 $\left( \frac{1}{6} \right)$] + $\frac{d}{2}$ [1 – $\frac{3}{6}$ (1)]

= a + $\frac{b}{2}$+ $\frac{2c}{6}$ + $\frac{d}{2}$ $\left( \frac{1}{2} \right)$

= $\frac{12a + 6b + 4c + 3d}{12}$