Sol. It is given that f(x) is increasing for all x ∈ (– ∞ , 1) ∪ (2, 3) and decreasing for all x ∈ (1, 2) ∪ (3, ∞ ). Therefore, f ′ (x) > 0 for all x ∈ (– ∞ , 1) ∪ (2, 3) and, f ′ (x) < 0 for all x ∈ (1, 2) ∪ (3, ∞ )

The changes in signs of f ′ (x) for different values of x are shown in fig.

We find that f ′ (x) is changing its sign from positive to negative in the neighbourhood of x = 1. Therefore, x = 1 is a point of local maximum. Similarly, x = 3 is a point of local maximum and x = 2 is a point of local minimum. Thus, x = 1, 2, 3 are roots of f ′ (x). So, let

f ′ (x) = λ (x –1) (x –2) (x –3)

⇒ f ′ (x) = λ (x 3 –6x 2 + 11x –6)

⇒ f(x) = λ + c

Now,

f(0) = 1

⇒ 1 = c

Hence, f(x) = λ + 1

= A (x 4 –8x 3 + 22x 2 – 24 x) + 1, where A =