MCV4U_A_All_Answers verified to pass 2026/2027

To anyone deciding to take some high school credit while in university,
vectors grade 12 with ILC. herecalculus and
all units

1.
A)

B)

-4.9(0)^2+450=450m
-4.9(2)^2+450=430.4m
V=d/t
=(430.4-450)/(2-0)
=-9.8m/s

C)
i) 1 ≤ t ≤ 4
-4.9(1)2 + 450 = 445.1m
-4.9(4)2 + 450 = 371.6m
V=d/t
= (371.6 – 445.1) / (4-1)
= -24.5m/s

ii) 1 ≤ t ≤ 2
-4.9(1)2 + 450 = 445.1m
-4.9(2)^2+450=430.4m
V=d/t
= (430.4-445.1)/(2-1)
= -14.7m/s

iii) 1 ≤ t ≤ 1.5
-4.9(1)2 + 450 = 445.1m
-4.9(1.5)2 + 450 = 438.975m
V=d/t
= (438.975-455.1) / (1.5-1)
= -12.25m/s
D)
-4.9(1)2 + 450= 445.199m
-4.9(0.99) 2 + 450=445.10m
= (445.199 – 445.10)/ 0.01
= -9.751m/s

2.
A)
M=10.5-0.4t^2
0=10.5-0.4t^2
X=-5.123

,X=5.123

B)
10.5-0.4(0)^2= 10.5
10.5-0.4(1)^2= 10.1
(10.1 – 10.5) / 1-0
= 0.4g/s
C)

10.5-0.4(1.99)^2= 8.91596g
10.5-0.4(2)^2= 8.9g
= (-8.9 – 8.91569) / 2 – 1.99
= -1.596g/s

3.
A)

B)
2(4) ^2 =32m
2(7) ^2 =98m
= (98-32)/ (7-4)
=22m/s



C)
2(3.99)2 =31.8402m
2(4)2 =32m
= (32-31.8402)/(4-3.99)
= 15.98m/s

4.
A)
By using the formula f(x)= (f(x+h)-f(x))/h, we can solve for the instantaneous rate of change
with h being an extremely small number in which approaches zero thus giving us the ability
to use this equation to calculate the instantaneous rate of change of a given point without
having to uses small number is 0.01 or 0.001 thus making our lives easier.
b)
i)
lim┬(x→0) (x-3)/(2x^2-5)
=3/5
ii)
lim┬(x→2) (2x^2-7x+6)/(x-2)=(2x^2-4x)(-3x+6)/(x-2)=(2x(x-2)-3(x-2))/(x-2)=(2x-3)(x-2)/(x-2)
=
lim ┬(x→2) 2x-3=2(2)-3=1



5.

,A)
lim┬(h→0) ( -5(2+h) ^2+20(2+h)+1-(21))/h
=
lim┬(h→0) (-20-20h-5h^2+40+20h+1-21)/h
= -5h
=-5(0)
= 0
B)
([(4+h)^2-8(4+h)+15]-[4^2-8(4)+15])/h
=
lim ┬(h→0) ([16+8h+h^2-32-8h+15]-[-1])/h
=
lim ┬(h→0) h=0

6.
A)
The interval in which the rate of change is positive is
1<X<-1
The interval in which the rate of change is negative is
-1<X<1




B)
The rate of change is zero at the point
X=-1
X=1

C)
The local minimum is at
(1, -2)
The local maximum is at
(-1,2)

7.
A)
f(x) = 2x^3 – 7x^2 +4x + 1
f’(x) = 6x^2 – 14x +4
f’(0) = 6(0)^2 – 14(0) + 4
f’(0) = 4

f(x) = 2x^3 – 7x^2 +4x + 1
f’(x) = 6x^2 – 14x +4
f’(1) = 6(1)^2 -14(1) + 4
f’(1) = -4

, B)
At X = 0 the function is increasing since the instantaneous rate of change is positive in this
case 4.
At X = 1 the function is decreasing since the instantaneous rate of change is negative in this
case -4.

C)
Yes since there is a positive instantaneous rate of change at X = 0 and there is a negative
instantaneous rate of change at X = 1, therefore there has to be an arc or hump in the
function for this to be possible.
8.
A)
f(x) = x^2-2x+1
=((x +h)^2 – 2(x+h) + 1 –x2 – 2x +1))/ h
=(x^2 + 2xh + h^2 – 2x -2h + 1 – x^2 -2x +1)/h
= 2x + h – 2
= 2x -2

B)
f(x) = x^3 – 3x^2
((x+3) ^3 -3(x+h) ^2 –x^3 +3x^2)/h
= (3x^2 h+3xh^2 +h^3 – 6xh -3h^2)/h
= (h(3x^2 + 3xh +h^2 -6x – 3h)/h
=3x^2 +3xh +h^2 -6x -3h
= 3x^2 – 6x

C)
= (√(2x+2h) - √2x)/h X (√(2x+2h) + √2x /( √(2x +2h) + √2x)
= ((√2x+2h) ^2 – (√2x) ^2)/h (V2x+2h + √2x)
= (2x + 2h -2x)/h (√2x +2h + √2x)
= (2)/ (√ (2x+2h) + √2x)
= (1)/ √2x
D)
= ((4)/(x+h+1) – (4)/(x+1))/h
= (-4h)/h(x+h+1)(x+1)
=(-4)/(x+h+1)(x+1)
= (-4)/(x+1)^2

9.
If you were to find the value of the slope of the given values of x and calculate the different
of those point and proceed to calculate the different of the first differences you would see
that only the second difference would be the same thus confirming that the derivative of a
cubic function is a quadratic function.



10.