# Mathematics 2010 | Study Mode

Question 31
A
-1
B
14
C
4
D
1
##### Explanation
Given that y = -3 - 2x + X2

then, $$\frac{dy}{dx}$$ = -2 + 2x

At maximum value, $$\frac{dy}{dx}$$ = O

therefore, -2 + 2x

2x = 2

x = 2/2 = 1

Question 32
A
y + 4x + 11 = 0
B
y - 4x - 11 = 0
C
y + 4x - 11 = 0
D
y - 4x + 11 = 0
##### Explanation
By comparing y = mx + c

with y = -4x + 2,

the gradient of y = -4x + 2 is m1 = -4

let the gradient of the line parallel to the given line be m2,

then, m2 = m1 = -4

(condition for parallelism)

using, y - y1 = m2(x - x1)

Hence the equation of the parallel line is

y - 3 = -4(x-2)

y - 3 = -4 x + 8

y + 4x = 8 + 3

y + 4x = 11

y + 4x - 11 = 0

Question 33
A
$$\frac{5P - MX + 5}{M}$$
B
$$\frac{5P - MX - 5}{M}$$
C
$$\frac{5P + MX + 5}{M}$$
D
$$\frac{5P + MX - 5}{M}$$
##### Explanation
p = $$\frac{M}{5}$$(X + Q) + 1

P - 1 = $$\frac{M}{5}$$(X + Q)

$$\frac{5}{M}$$(p - 1) = X + Q

$$\frac{5}{M}$$(p - 1)- x = Q

Q = $$\frac{5(p -1) - Mx}{M}$$

= $$\frac{5p - 5 - Mx}{M}$$

= $$\frac{5p - Mx - 5}{M}$$

Question 34
A
2y + 3x
B
2y - 3x
C
3x + 2y
D
3x - 2y
##### Explanation
27x3 - 8y3 = (3x - 2y)3

But 9x2 + 6xy + 4y2 = (3x +2y)2

So, 27x3 - 8y3 = (3x - 2y)(3x - 2y)2

Hence the other factor is 3x - 2y

Question 35
A
$$\frac{x(x - 5)}{2(x + 2)}$$
B
$$\frac{x(x + 5)}{2(x + 2)}$$
C
$$\frac{x(x - 5)}{2(x - 2)}$$
D
$$\frac{x ^2 + 5}{2x + 4}$$
##### Explanation
$$\frac{x^3 + 3x^2 - 10x}{2x^2 - 8}$$ = $$\frac {x(x^2 + 3x - 10)}{2(x^2 - 4)}$$

= $$\frac {x(x^2 + 5x - 2x - 10)}{2(x + 2)(x - 2)}$$

= $$\frac {x(x - 2)(x + 5)}{2(x + 2)(x - 2)}$$

= $$\frac {x(x + 5)}{2(x + 2)}$$

Question 36
A
(-1, 3)
B
(3, 1)
C
(-3, 1)
D
(1, 3)
##### Explanation
x - y = 2 ...........(1)

x2 - y2 = 8 ........... (2)

x - 2 = y ............ (3)

Put y = x -2 in (2)

x2 - (x - 2)2 = 8

x2 - (x2 - 4x + 4) = 8

x2 - x2 + 4x - 4 = 8

4x = 8 + 4 = 12

x = $$\frac{12}{4}$$

= 3

from (3), y = 3 - 2 = 1

therefore, x = 3, y = 1

Question 37
A
6
B
12
C
3
D
5
##### Explanation
y = $$\alpha$$x ........(1)

y = kx ........(2)

When y = 3, x = 16,

(2) becomes 3 = k16 or 3 = k x 4

giving k = $$\frac{3}{4}$$

from (2), y = $$\frac{3}{4}$$x

When x = 64, y = $$\frac{3}{4}$$64

y = $$\frac{3}{4}$$ x 8

= 6

Question 38
A
4
B
5
C
1$$\frac{1}{4}$$
D
2$$\frac{1}{4}$$
##### Explanation
x $$\alpha$$ $$\frac{1}{y}$$ .........(1)

x = k x $$\frac{1}{y}$$ .........(2)

When x = 2$$\frac{1}{2}$$

= $$\frac{5}{2}$$, y = 2

(2) becomes $$\frac{5}{2}$$ = k x $$\frac{1}{2}$$

giving k = 5

from (2), x = $$\frac{5}{y}$$

so when y =4, x = $$\frac{5}{y}$$ = 1$$\frac{1}{4}$$

Question 39
A
x < $$\frac{3}{2}$$
B
x > $$\frac{3}{2}$$
C
x < -$$\frac{3}{2}$$
D
x > -$$\frac{3}{2}$$
##### Explanation
$$\frac{1}{2}$$x + $$\frac{1}{4}$$ > $$\frac{1}{3}$$x + $$\frac{1}{2}$$

Multiply through by through by the LCM of 2, 3 and 4

12 x $$\frac{1}{2}$$x + 12 x $$\frac{1}{4}$$ > 12 x $$\frac{1}{3}$$x + 12 x $$\frac{1}{2}$$

6x + 3 > 4x + 6

6x - 4x > 6 - 3

2x > 3

$$\frac{2x}{2}$$ > $$\frac{3}{2}$$

x > $$\frac{3}{2}$$

Question 40
A
-1 < x < 5
B
-1 < x $$\leq$$ 5
C
-1 $$\leq$$ x $$\leq$$ 6
D
-1 $$\leq$$ x < 6
##### Explanation
-6 $$\leq$$ 4 - 2x < 5 - x
split inequalities into two and solve each part as follows:

-6 $$\leq$$ 4 - 2x = -6 - 4 $$\leq$$ -2x

-10 $$\leq$$ -2x

$$\frac{-10}{-2}$$ $$\geq$$ $$\frac{-2x}{-2}$$

giving 5 $$\geq$$ x or x $$\leq$$ 5

4 - 2x < 5 - x

-2x + x < 5 - 4

-x < 1

$$\frac{-x}{-1}$$ > $$\frac{1}{-1}$$

giving x > -1 or -1 < x

Combining the two results, gives -1 < x $$\leq$$ 5

Try this quiz in in E-test/CBT Mode
switch to