If the range of the function, f(x)=(x3-3 x2+2 x/2 x3+2 x2-4 x) is R- α, β, γ then (α+β+γ) equals (where R is the set of all real numbers)

Question

If the range of the function, f(x)=(x3-3 x2+2 x/2 x3+2 x2-4 x) is R- α, β, &gamma; then (α+β+&gamma;) equals (where R is the set of all real numbers) (A) (1/2) (B) (1/6) (C) (-1/2) (D) (-1/6)

Tardigrade Admin · Accepted Answer

f(x)=(x(x-1)(x-2)/2 x(x-1)(x+2)), Df: R- 0,1,-2 f ( x )=( x -2/2( x +2)), x ≠ 0,1 ∴ R f: R - (1/2), (-1/2), (-1/6) ∴ α+β+&gamma;=(1/2)-(1/2)-(1/6)=(-1/6)

Q. If the range of the function, $f (x) = \frac{x ^{3} - 3 x ^{2} + 2 x}{2 x ^{3} + 2 x ^{2} - 4 x}$ is $R - {α, β, γ}$ , then $(α + β + γ)$ equals (where $R$ is the set of all real numbers)

$\frac{1}{2}$

$\frac{1}{6}$

$\frac{- 1}{2}$

$\frac{- 1}{6}$

$f (x) = \frac{x ( x - 1 ) ( x - 2 )}{2 x ( x - 1 ) ( x + 2 )}, D_{f} : R - {0, 1, - 2}$
$f (x) = \frac{x - 2}{2 ( x + 2 )}, x \neq = 0, 1$
$∴ R_{f} : R - {\frac{1}{2}, \frac{- 1}{2}, \frac{- 1}{6}}$
$∴ α + β + γ = \frac{1}{2} - \frac{1}{2} - \frac{1}{6} = \frac{- 1}{6}$

Q. If the range of the function, f(x)=2x3+2x2−4xx3−3x2+2x​ is R−{α,β,γ}, then (α+β+γ) equals (where R is the set of all real numbers)

21​

61​

2−1​

6−1​