# MATHEMATICS | Study Mode

Question 21
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 22
A

$\left[\begin{array}{cc}14& 8\\ 7& 7\end{array}\right]$

B

$\left[\begin{array}{cc}7& 7\\ 8& 14\end{array}\right]$

C

$\left[\begin{array}{cc}8& 14\\ 7& 7\end{array}\right]$

D

$\left[\begin{array}{cc}7& 7\\ 14& 8\end{array}\right]$

##### Explanation

Question 23
A

$\left[\begin{array}{cc}2& -32\\ -3& 52\end{array}\right]$

B

$\left[\begin{array}{cc}2& 32\\ -3& -52\end{array}\right]$

C

$\left[\begin{array}{cc}2& 32\\ -3& 52\end{array}\right]$

D

$\left[\begin{array}{cc}2& -32\\ -3& -52\end{array}\right]$

Question 24
A

40

B

45

C

15

D

30

Question 25
A

100

B

70

C

130

D

110

Question 26
A

42

B

66

C

12

D

30

Question 27
A

576

B

720

C

336

D

420

##### Explanation

Question 28
A

$$8\sqrt{2} cm$$

B

$$8\sqrt{3} cm$$

C

4 cm

D

8 cm

##### Explanation

Length of chord = $$2r \sin (\frac{\theta}{2})$$

= $$2 \times 8 \times \sin (\frac{90}{2})$$

= $$16 \times \frac{\sqrt{2}}{2}$$

= $$8\sqrt{2} cm$$

Question 29
A

16π cm2

B

32π cm2

C

4π cm2

D

8$\pi$cm2

##### Explanation

Diameter = 4$$\sqrt{3}$$ cm

radius = 2$$\sqrt{3}$$ cm

Area of major sector = $$\frac{\theta}{360} \times \pi r^{2}$$

$$\theta = 360 - 120 = 240°$$

= $$\frac{240}{360} \times \pi \times 12$$

= $$8\pi cm^{2}$$

Question 30
A

pair of parallel lines to PQ

B

perpendicular line to PQ

C

circle centre P

D

circle centre Q.

##### Explanation

The locus of points at a fixed distance from the point P is a circle with the given P at its centre.

The locus of points at a fixed distance from the point Q is a circle with the given point Q at its centre

The locus of points equidistant from two points P and Q i.e line PQ is the perpendicular bisector of the segment determined by the points

Hence, The locus of a point which is equidistant from the line PQ forms a perpendicular line to PQ

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