The function f(x) = x^3 - 6x^2 + ax + b is such that f(2) = f(4) = 0.

Question

The function $f(x) = {x^3} - 6{x^2} + ax + b$ is such that $f(2) = f(4) = 0$. Consider two statements :

Statement 1 : there exists x1, x2 $$\in(2, 4), x1 < x2, such that f'(x1) = -1 and f'(x2) = 0.

Statement 2 : there exists x3, x4 \[\in$$ (2, 4), x3 < x4, such that f is decreasing in (2, x4), increasing in (x4, 4) and $$2f'({x_3}) = \sqrt 3 f({x_4})$$.

Then

Accepted Answer

$f(x) = {x^3} - 6{x^2} + ax + b$ $f(2) = 8 - 24 + 2a + b = 0$ $2a + b = 16$ .... (1) $f(4) = 64 - 96 + 4a + b = 0$ $4a + b = 32$ .... (2) Solving (1) and (2) a = 8, b = 0 $f(x) = {x^3} - 6{x^2} + 8x$ $f'(x) = 3{x^2} - 12x + 8$ $f''(x) = 6x - 12$ $$\Rightarrow$$ f'(x) is $$ \uparrow $$ for x > 2, and f'(x) is $$ \downarrow $$ for x < 2 $f'(2) = 12 - 24 + 8 = - 4$ $f'(4) = 48 - 48 + 8 = 8$ $f'(x) = 3{x^2} - 12x + 8$ vertex (2, -4) f'(2) = -4, f'(4) = 8, f'(3) = 27 - 36 + 8 f'(x1) = -1, then x1 = 3 f'(x2) = 0 Again f'(x) < 0 for x $$\in$$ (2, x4) f'(x) > 0 for x $$\in$$ (x4, 4) x4 $$\in$$ (3, 4) f(x) = x3 - 6x2 + 8x f(3) = 27 - 54 + 24 = -3 f(4) = 64 - 96 + 32 = 0 For x4(3, 4) f(x4) < -3$$\sqrt 3 $$ and f'(x3) > -4 2f'(x3) > -8 So, 2f'(x3) = $$\sqrt 3 $$ f(x4) Correct Ans. (a).

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