A-Maths嘢

1a) dy/dx = 5(ln 2x + cos x)4 d(ln 2x + cos x)/dx

= 5(ln 2x + cos x)4 (1/x - sin x)

b) dy/dx = x2 d[√(1 - 3x + x2)]/dx + 2x√(1 - 3x + x2)

= x2/[2√(1 - 3x + x2)] d(1 - 3x + x2)/dx + 2x√(1 - 3x + x2)

= x2(-3 + 2x)/[2√(1 - 3x + x2)] + 2x√(1 - 3x + x2)

= [(-3x2 + 2x3) + 4x(1 - 3x + x2)]/[2√(1 - 3x + x2)]

= (-3x2 + 2x3 + 4x - 12x2 + 4x3)/[2√(1 - 3x + x2)]

= (6x3 - 15x2 + 4x)/[2√(1 - 3x + x2)]

2a) f'(x) = 4x(1 - 2x) - 4x2

= -12x2 + 4x

f"(x) = -24x + 4

For f'(x) = 0, x = 0 or 1/3

With f"(0) > 0 and f"(1/3) < 0, x = 0 and x = 1/3 gives a local min. and max. resp.

b) Min. = (0, 0), max. (1/3, 2/27)

Another x-intercept at (1/2, 0)

Point of inflexion at (1/6, 1/54)

4) We take the improper integral:

∫(x = 6 → ∞) dx/x

= [ln x] (x = 6 → ∞)

which yields an infinite value since ln x tends to ∞ as x → ∞.

5) (13x - 4)/(6x2 - x - 2)

= (13x - 4)[(3x - 2)(2x + 1)]

= 2/(3x - 2) + 3/(2x + 1)

So ∫(13x - 4)dx/(6x2 - x - 2)

= ∫[2/(3x - 2) + 3/(2x + 1)]dx

= (2/3) ln |3x - 2| + (3/2) ln |2x + 1| + C