Maths Applications of Derivatives

Since p (x) has relative extreme at x=1 and 2
so p' (x)=0 at x=1 and 2
⇒ p' (x)=A (x-1) (x-2)
⇒ p (x)=∫A (x²-3x+2)dx
p (x)=A (x³/3 - 3x²/2 + 2x)+C
P (1)=8
From (1)
8=A (1/3-3/2+2)+C
⇒ 8=5A/6+C ⇒ 48=5A+6C
P (2)=4
⇒ 4=A (8/3-6+4)+C
⇒ 4=-2A/3+C ⇒ 12=-2A+3C
From 3 and 4, C=-12
So P (0)=C=-12

√3|a+b|+|a-b| ≤

2√ (3+1)=4.

S.D.=√ (Σ (x-a)²/n - (Σ (x-a)/n)²) = √ (a-1).

f' (x) = 12sin³xcosx+30sin²xcosx+12sinxcosx = 3sin2x (2sin²x+5sinx+2) = 3sin2x (2sinx+1) (sinx+2).
In [-π/6, π/2], sinx+2>0. 2sinx+1>0 except at x=-π/6. sin2x>0 for x∈ (0, π/2), <0 for x (-/6,0).
So f' (x)<0 on (-/6,0) (decreasing) and f' (x)>0 on (0, π/2) (increasing).

f(3)=f(4) ⇒ α=12
f'(x) = (x²-12)/(x(x²+12))
∴ f'(c)=0 ⇒ c=√12
∴ f''(c) = 1/12

0.50
 y = x²-3x+2, x+y=a, x-y=b
x?=2 x?=1
y? = 4-6+2 = 0 y?=0
(2,0)
(1,0)
b=2
a=1
∴ a/b = 1/2 = 0.5

f(x) = (3x²+ax-2-a)e?
f'(x) = (3x²+ax-2-a)e? + e?(6x+a)
= e?(3x²+(a+6)x-2)
∴ x=1 is a critical point
∴ f'(1)=0
∴ 3+a+6-2=0
a = -7
∴ f'(x) = e?(3x²-x-2)
= e?(3x+2)(x-1)

∴ maxima at x = -2/3
∴ minima at x = 1

e? y'x? + 4x³e? + 2y' / (2√ (y+1) = 0 at (1,0)
y' + 4 + y' = 0 ⇒ y' = -2
equation of tangent at (1,0) is 2x + y - 2 = 0
So option (C) is correct.