Consider the polynomial f(x) = 1 + 2x + 3x2 + 4x3. Let s be the sum of all distinct real roots of f(x) and let t=

Linked Question 1

The function f '(x) is

increasing in (– t, t)

decreasing in (– t, t)

decreasing in $(- t, - \frac{1}{4})$ and increasing in $(- \frac{1}{4}, t)$

increasing in $(- t, - \frac{1}{4})$ and decreasing in $(- \frac{1}{4}, t)$

f (x) = 1 + 2x + 3x² + 4x³

g (x) = f ' (x) = 2 + 6x + 12x²

g ' (x) = 6 + 24x = 0 $⇒x= \frac{- 1}{4}$

g (x) is increasing in $(\frac{- 1}{4}, \infty)$ and decreasing in $(- \infty, \frac{- 1}{4})$

t = | s | = x₀ > 1

$\Rightarrow$ increasing in $(\frac{- 1}{4}, t)$ and decreasing in $(- t, \frac{- 1}{4})$

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Linked Question 2

The area bounded by the curve y = f(x) and the lines x = 0, y = 0 and x = t, lies in the interval

(9, 10)

$(0, \frac{21}{64})$

$(\frac{3}{4}, 3)$

$(\frac{21}{64}, \frac{11}{16})$

$A (t) = \int_{0}^{t} f (x) dx = t^{4} + t^{3} + t^{2} + t = t (\frac{1 - t^{4}}{1 - t})$

(1/2) = 15/16 & A(3/4) =3 $(\frac{175}{256})$

So A(t) $\in (\frac{3}{4}, 3)$

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Linked Question 3

The real number s lies in the interval

$(- \frac{3}{4}, - \frac{1}{2})$

$(- \frac{1}{4}, 0)$

$(- 11, - \frac{3}{4})$

$(0, \frac{1}{4})$

f(x) = 4x³ + 3x² + 2x + 1

f’(x) = 12x² + 6x + 2 is a always positive

f(0) = 1, f(–1/2) =1/4, f(–3/4) = $\frac{1}{2}$

so root $\in (\frac{- 3}{4}, \frac{- 1}{2}) ∵$ the equation have only one real root so s = $(\frac{- 3}{4}, \frac{- 1}{2}) and t k \in (\frac{1}{2}, \frac{3}{4})$

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