If λ1>λ2>λ3>....>λn be the values of λ for which the equation x3−3x+λ=0 has all integral roots. If h (x) = x3−3x+|λ1|+|λ2|+....+|λn| for x∈[1,∞) and g(x) be the inverse of h(x) then, which of the following options is/are correct?

∫ 1 3 ( x 3 − 3 x ) d x + ∫ 2 22 g ( x ) d x = 64 , ∫ 1 3 ( x 3 − 3 x ) d x + ∫ 2 22 g ( x ) d x = 56 , λ 1 2 + λ 2 2 + ..... + λ n 2 = a number which is not the perfect square of an integer

If λ1>λ2>λ3>....>λn be the values of λ for which the equation x3−3x+λ=0 has all integral roots. If h (x) = x3−3x+|λ1|+|λ2|+....+|λn| for x∈[1,∞) and g(x) be the inverse of h(x) then, which of the following options is/are correct?

x 3 − 3 x + λ = 0 let roots = α , β , γ ∴ α , β , γ = 0. α β + β γ + γ α = − 3 , α β γ = − λ ∴ α 2 + β 2 + γ 2 = 6 ∵ α , β , γ ∈ I ∴ h ( x ) = x 3 − 3 x + 4 Required area = ∫ 2 5 h ( x ) d x = ∫ 2 5 ( x 3 − 3 x + 4 ) d x = 531 4 a n d ∫ 1 5 ( x 3 − 3 x ) d x = ∫ 2 22 g ( x ) d x ∫ 1 5 ( x 3 − 3 x + 4 ) d x + ∫ 2 22 g ( x ) d x − 8 = 66 − 2 − 8 = 56

∫ 1 3 ( x 3 − 3 x ) d x + ∫ 2 22 g ( x ) d x = 64

∫ 1 3 ( x 3 − 3 x ) d x + ∫ 2 22 g ( x ) d x = 56

Area bounded by g(x) and y-axis between limits y = 2 to y = 5 is 531 4 sq.units

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If $λ_{1} > λ_{2} > λ_{3} > .... > λ_{n}$ be the values of $λ$ for which the equation $x^{3} - 3 x + λ = 0$ has all integral roots. If h (x) = $x^{3} - 3 x + | λ_{1} | + | λ_{2} | + .... + | λ_{n} |$ for $x \in [1, \infty)$ and g(x) be the inverse of h(x) then, which of the following options is/are correct?

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$\int_{1}^{3} (x^{3} - 3 x) d x + \int_{2}^{22} g (x) d x = 64$

$\int_{1}^{3} (x^{3} - 3 x) d x + \int_{2}^{22} g (x) d x = 56$

Area bounded by g(x) and y-axis between limits y = 2 to y = 5 is $\frac{531}{4}$ sq.units

$λ_{1}^{2} + λ_{2}^{2} + ..... + λ_{n}^{2}$ = a number which is not the perfect square of an integer

answer is A, B, D.

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Detailed Solution

$x^{3} - 3 x + λ = 0$ let roots $= α, β, γ$
$∴ α, β, γ = 0. α β + β γ + γ α = - 3, α β γ = - λ$

$∴ α^{2} + β^{2} + γ^{2} = 6 ∵ α, β, γ \in I ∴ h (x) = x^{3} - 3 x + 4$
Required area
$= \int_{2}^{5} h (x) d x = \int_{2}^{5} (x^{3} - 3 x + 4) d x = \frac{531}{4} a n d \int_{1}^{5} (x^{3} - 3 x) d x = \int_{2}^{22} g (x) d x$
$\int_{1}^{5} (x^{3} - 3 x + 4) d x + \int_{2}^{22} g (x) d x - 8 = 66 - 2 - 8 = 56$