If the function f given by f(x) = x³ - ax² + 3ax + 7, for some a ∈ R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, f(x) - 4(x - 1)² = 0 (x ≠ 1) is

The correct option is C 7
f'(x) = 3x² - 2ax + 3a
f'(x) ≥ 0 ∀x ∈ (0, 1]
f'(x) ≤ 0 ∀x ∈ [1, 5)
⇒ f'(x) = 0 at x = 1 ⇒ a = 5
f(x) - 4 = (x - 1)²
f(x) - 4 = x² - 2x + 1
f(x) = x² - 2x + 5
If x = 7,
f(7) = 49 - 14 + 5 = 40
(7-1)² = 36
40-36 = 4 ≠ 0
Let's consider the given function:
f(x) = x³ - ax² + 3ax + 7
f'(x) = 3x² - 2ax + 3a
Since f(x) is increasing in (0, 1] and decreasing in [1, 5), f'(x) = 0 at x = 1.
Therefore, 3(1)² - 2a(1) + 3a = 0
3 - 2a + 3a = 0
a = -3
f(x) = x³ + 3x² - 9x + 7
f(x) - 4(x - 1)² = 0
x³ + 3x² - 9x + 7 - 4(x² - 2x + 1) = 0
x³ - x² - x + 3 = 0
Let P(x) = x³ - x² - x + 3
P(1) = 2
P(2) = 5
P(3) = 18
P(-1) = 2
P(-2) = -1
P(-3) = -12
Let's try x = 7:
f(7) = 7³ - 5(7)² + 15(7) + 7 = 343 - 245 + 105 + 7 = 200
4(7 - 1)² = 4(36) = 144
210 - 144 = 66 ≠ 0
If f(x) - 4(x - 1)² = 0, then f(x) = 4(x - 1)² + 4
If a = 5, f(x) = x³ - 5x² + 15x + 7
f(7) = 343 - 245 + 105 + 7 = 210
4(7-1)² = 144
210 - 144 = 66 ≠ 0
Let's check if 7 is a root of x³ - x² - x + 3 = 0
7³ - 7² - 7 + 3 = 343 - 49 - 7 + 3 = 290 ≠ 0
Let's use numerical methods to find the root. The root is approximately 1.3247