Question

Let f(x) is a cubic polynomial which has a local max at x = –1. If f(2) = 18, f(–1) = –1 and f'(x) has local min at x = 0, then... 

(a) the distance between (–1, 2) and (a, f(a)) where x = a is the point of local minima is 2√5

(b) f(x) is increasing for x ∈ [1, 2√5]

(c) f(x) has local minima at x = 1

(d) the value of f(0) is 5. 

Answer 1

Correct option (a, b, c)

Explanation:

Let f'(x) = (x + 1)(x – α)

⇒ f''(x) = a(x + 1) + a(x – α)

⇒ f'''(x) = a(2x + 1 – α)

Since f'(x) has a local minimum at x = 0,

f'(x) has a local minima at x = 0,

so f''(0) = 0 and f''(0) > 0

⇒ a(1 – a) = 0, a > 0 ⇒ a = 1

Thus, f'(x) = a(x + 1)(x – 1) = a(x² – 1)

⇒ f(x) = a(x³/3 – x) + k

As f(–1) = 2 and f(3) = 18, we get,

a( –1/3 + 1) + k = 2 and 6a + k = 18

a = 3, k = 0

Therefore, f(x) = x³ – 3x

⇒ f'(x) = 3x² – 3

⇒ f''(x) = 6x

Now, f'(x) = 0 gives x = 1, – 1.

As f''(x) = 6 > 0,

so f(x) has a local minimum at

x = 1.

Thus, a = 1 and f(0) = – 2.

Distance between (–1, 2) and (1, – 2)