The function f(x) = x3 – 6x2 + ax + b is such that f(2) = f(4) = 0. Consider two statements. (S1) : there exists x1, x2 ∈ (2,4), x1 < x2, such that f '(x1) = – 1 and f '(x2) = 0. (S2) : there exists x3, x4 ∈ (2,4), x3 < x4, such that f is decreasing in (2, x4), increasing in (x4, 4) and 2 f ' \left(\right. x_{3} \left.\right) = \sqrt{3} f \left(\right. x

Engineering

Mathematics

Inequalities

Question

The function f(x) = x³ – 6x² + ax + b is such that f(2) = f(4) = 0. Consider two statements.

(S₁) : there exists x₁, x₂ ∈ (2,4), x₁ < x₂, such that f '(x₁) = – 1 and f '(x₂) = 0.

(S₂) : there exists x₃, x₄ ∈ (2,4), x₃ < x₄, such that f is decreasing in (2, x₄), increasing in (x₄, 4) and $2 f' (x_{3}) = \sqrt{3} f (x_{4})$ then

(S₁) is true and (S₂) is false

both (S₁) and (S₂) are true

(S₁) is false and (S₂) is true

both (S₁) and (S₂) are false

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

f(2) = 0 ⇒ 2a + b = 16

f(4) = 0 ⇒ 4a + b = 32

⇒ a = 8, b = 0

f(x) = x³ – 6x² + 8x = x(x – 2)(x – 4)

f '(x) = 3x² – 12x + 8

$f' (x) = 0 \Rightarrow x = 2 \pm \frac{2}{\sqrt{3}}$

$\Rightarrow x_{2} = 2 + \frac{2}{\sqrt{3}} \in (2, 4)$

Now f '(x₁) = –1 ⇒ 3x₁² –12x₁ + 9 = 0 ⇒ x₁ = 1, 3 but x₁ $\in$ (2,4) ⇒ x = 3 (S₁ is true)

Also f(x) is decreasing in (2, x₄) & increasing in (x₄, 4)

⇒ f '(x₄) = 0

$\Rightarrow x_{2} = 2 + \frac{2}{\sqrt{3}} and 2 f' (x_{3}) = \sqrt{3} f (x_{4})$

So, $2 (3 x_{3}^{2} - 12 x_{3} + 8) = \sqrt{3} (2 + \frac{2}{\sqrt{3}}) (\frac{2}{\sqrt{3}}) (\frac{2}{\sqrt{3}} - 2)$

$X_{3} = \frac{8}{3}, \frac{4}{3}$ (S₂ is true)

But 2 < x₃ < x₄ so $x_{3} = \frac{8}{3}$

Please subscribe our Youtube channel to unlock this solution.