Let f(x) be a cubic polynomial with leading coefficient unity such that f (a) = b and f '(a) = f "(a) = 0. Suppose g(x) = f(x) – f(a) + (a – x) f '(x) + 3(x – a)2 for which conclusion of Rolle's theorem in[a, b] holds for x = 2 \in(a, b), then

Engineering

Mathematics

Rolles Theorem

Question

Let f(x) be a cubic polynomial with leading coefficient unity such that f (a) = b and f '(a) = f "(a) = 0. Suppose g(x) = f(x) – f(a) + (a – x) f '(x) + 3(x – a)² for which conclusion of Rolle's theorem in[a, b] holds for x = 2 $\in$ (a, b), then

$\int_{0}^{2} f (x) dx = 2 b$

f "(2) = 6

Number of roots of f(f(x)) = 0 is 2

a + 2b = 6

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$∵$ f '(x) = 0 has x = 0 as repeated roots.

$∴$ f '(x) = 3 (x – a)²

⇒ f(x) = (x – a)³ + b

$∴$ g(x) = (x – a)³ – 3 (x – a)³ + 3(x – a)²

= – 2 (x – a)³ + 3(x – a)²

g(a) = 0

$∴$ g(b) = 2 (b – a) – 3 = 0 ⇒ 2b = 2a + 3 and g'(2) = 0.

g'(x) = – 6 (x – a)² + 6 (x – a)

= – 6 (x – a) (x – a – 1)

⇒ g'(2) = 0 ⇒ 2 – a – 1 =0 ⇒ a = 1

$∴ b = \frac{5}{2}$

$∴$ f "(x) = 6 (x – a), f "(2) = 6 (2 – 1) = 6

$\int_{0}^{2} f (x) dx = \int_{0}^{2} ({(x - 1)}^{3} + \frac{5}{2}) dx = 5 = 2 b$

Clearly, f(x) is always increasing f(x) = 0 has only one real root ${(\frac{- 5}{2})}^{\frac{1}{3}} + 1 .$

$∴$ f (f(x)) = 0 ⇒ f(x) = ${(\frac{- 5}{2})}^{\frac{1}{3}} + 1$

will have only one solution.