# MATHEMATICS | Study Mode

Question 11
A

S + 2

B

S - 2

C

S2 + 2

D

S2 - 2

##### Explanation

$$S = { \sqrt{t^2-4t+4}}$$
$$S = { \sqrt{(t-2)(t-2)}}$$

S = t - 2

t = S + 2

Question 12
A

12

B

20

C

2

D

4

Question 13
A

42

B

63

C

18

D

21

##### Explanation

Let f(p) = 6p3 - p2 - 47p + 30

Then by the remainder theorem,

(p - 3): f(3) = remainder R,

i.e. f(3) = 6(3)3 - (3)2 - 47(3) + 30 = R

162 - 9 - 141 + 30 = R

192 - 150 = R

R = 42

Question 14
A

15

B

10

C

200/5

D

120/5

##### Explanation

P $$\propto$$ mu, p $$\propto \frac{1}{q}$$

p = muk ................ (1)

p = $$\frac{1}{q}k$$.... (2)

Combining (1) and (2), we get

P = $$\frac{mu}{q}k$$

4 = $$\frac{m \times u}{1}k$$

giving k = $$\frac{4}{6} = \frac{2}{3}$$

So, P = $$\frac{mu}{q} \times \frac{2}{3} = \frac{2mu}{3q}$$

Hence, P = $$\frac{2 \times 6 \times 4}{3 \times \frac{8}{5}}$$

P = $$\frac{2 \times 6 \times 4 \times 5}{3 \times 8}$$

p = 10

Question 15
A

s varies inversely as r2 and t.

B

s varies directly as r2 and t2 .

C

s varies directly as r and t.

D

s varies inversely as r and t2 .

Question 16
A
x > 4
B
x < 4
C
x > 4
D
x < 4

Question 17
A

y = x2 3x + 2

B

y = x2 x 1

C

y = x2 + x 2

D

y = x2 x 2

Question 18
A

2 < x < 3

B

-1 < x < 5

C

x < 1

D

x < 5

##### Explanation

Solve for x in |x-2|<3

| x − 2 | < 3

x − 2 < ± 3

Set up the positive portion of the ± solution.

x − 2 < 3

Move all terms not containing x to the right side

x < 5

Set up the negative portion of the ± solution.

x − 2 < − 3

Move all terms not containing x to the right side

x < − 3 + 2

x < − 1

The solution to the equation includes both the positive and negative portions of the solution.

− 1 < x < 5 (B)

Question 19
A
3
B
5
C
5
D
2

Question 20
A

3n+1n+1

B

3n+1n-1

C

3n-1n+1

D

1-3nn+1