Prime Optional Mathematics 10

Y X
B' 7

6
5
A' 4
3C
2
X'C' 1
–9 –8 –7 –6 –5 –4 –3 –2 –1 O 1 2 3 4 5 6 7 8 9
–1
–2 A
–3
–4
–5 B

Y'

Exercise 7.1

1. Answer the following questions.
i) Find r2or1, where r1 is reflection on x-axis and r2 is reflection on y-axis for a point
P(a, b)

ii) If T1 = <1F and T2 = <2 F , find T2oT1.
2 –1

iii) Find the image of P(a, b) under rotation about –90° followed by 180° with centre

origin.

iv) If E1[0, 3 ] and E2 [0, 2] are the two enlargements, find combination of enlargements.
2

v) Find the single transformation of F and G where F is reflection about y = – x and G is
rotation about +90° with centre origin.

2. Find the single transformation for the followings.
i) RrF1o2roG2Rww1,hhweerhreeerr1FeiRsis1roirsetafrlteeifocletnicoatniboonaubatob+uo9tu0yt°xa=-na–dxxirsa2aninsddroGRt2aistisiorronetfaalbeticootuniot na1b8ao0bu°otuw–ti9tyh0-a°cxewinsi.ttrhe
origin.
ii) centre
iii)

origin.

iv) ToR where T is translation about T = 2 and R is reflection about x + 2 = 0.
<F
–3

v) EoR where R is rotation about 180° with centre origin and E is enlargement about

E[0, 2].
3. Find the combined transformations of the following and find the image of a point P(–3,
5) with it.
i) RT1isisrroetfalteicotnioanbaobuotu–t9y0°=wxiathndcern2 tirserootraigtiionnanabdoFuits–r9e0fl°ewctiitohncaebnotruet origin.
ii) y = 0.

296 PRIME Opt. Maths Book - X

iii) F is rotation about + 90° with centre origin and G is rotation about –270° with centre

origin.
iv) P is reflection about y + 2 = 0 and T is rotation about + 270° with centre origin.

v) E is enlargement about E[(1, –2), 2] and T is translation about T = 2 .
<F
3

4. i) A point p is transferred to P'(–3, 7) under reflection about x – 1 = 0 followed by
rotation about –90° with centre origin. Find the co-ordinate of p.

ii) Image of a point A is A'(2, –7) under rotation about +90° with centre origin followed

by translation about T = –2 . Find the co-ordinate of A.
<F

3

iii) A point P(–2, 3) is transformed to P'(–6, –4) under enlargement E(o, k) followed by
rotation about –270°, find the scale factor ‘k'.

iv) A point M(–3, 2) is transformed to M'(1, –8) under a translation T followed by
reflection about y + 2 = 0, find the translation vector ‘T'.

v) A point P is transferred to P'(–3, 5) under reflection on x = 0 followed by enlargement
about E[(1, 2), –2], find the co-ordinate of the point P.

5. i) A → A'(5, 2) is obtained after the combined transformation of FOG where F is
reflection about x + 3 = 0 and G is rotation about –270° with centre origin. Find the
co-ordinate of A.

ii) A point p is transferred to P' (–3, –5) under a combined transformation of reflection

iii) iaAmbcoaougmet yobfi=na–enxdoabtnrjadencttsrfaAonrtsmolaAat'ti(oi–on3na, –ob9fo)ue,tnfiTlnad=rgte<h12meFe,cfnoit-nodarbdthoineuatctoeE-1oo[fr0dA, i.n32at]e of A. –2] gives
and E2[0,

iv) A point A(2, 3) is translated to A'(5, 5) under a combined transformation of T1 and

T2 = 2 , find the translation vector T1.
<F
–1

v) A point P(–3, 1) is transformed to P'(–4, 1) under a combined transformation of
reflection about x + y = 0 and translation T, find the translation vector T.

6. i) Find the image and plot in graph of DABC having vertices A(4, 5), B(1, 2) and C(5,
–3) under reflection about y + 2 = 0 followed by reflection about y = x.

ii) Find the image of DPQR under rotation about +90° followed by rotation about
–270° where the vertices are P(–1, 5), Q(–4, –1) and R(3, 3). Also plot the object and

images in graph.
iii) Transform a triangle having vertices A(–2, 1), B(1, –3), and C(3, 4) under the

translation T1 = 3 followed by T2 = 2 . Also show in graph.
<F <F
4 –1

iv) Transform a quadrilateral having vertices P(–2, 2), Q(–1, –3), R(2, –2) and S(1, 3)
uTnhdeevreartniceenslaorfgDeAmBeCntaraeboAu(1t ,E31[)0, B, 2(2] ,fo6l)loawndedC(b4y, E11)[.0F,i2n]d.
v) Also plot them in graph.
the image of DABC under
reflection about x - y = 0 followed by reflection about y-axis. Also plot the object and

images in graph.

7. i) Translate the figure having vertices O(0, 0), A(2, 0), B(3, 1) and C(1, 1) with a

PRIME Opt. Maths Book - X 297

translation vector T = 0 followed by reflection on the line x – 3 = 0. Also plot the
<F
2

object and image so obtained.
ii) Find the image of DABC having vertices A(5, 3), B(1, 0) and C(3, 5) under reflection

about y = 5 followed by translation about AB . Also plot the object and image in

graph.
iii) The vertices of triangle are M(2, 5), N(–1, 3) and P(4, 1). Find the image of DMNP

under rotation with 90° in clockwise direction followed by enlargement about E[(0,

0), 2]. Also plot them in graph.
iv) Transform a triangle having vertices P(2, 1), Q(5, 3) and R(7, –1) under rotation

about negative quarter turn with centre origin followed by reflection on x-axis. Also

plot them in graph.
v) Find the image of a quadrilateral having A(1, 1), B(2, 3), C(1, –2) and D(3, –1) under

an enlargement E[(–2, –1), 2] followed by translation about T = 2
object and image in graph. <F . Also plot the
3

8. i) Find the image of DPQR having vertices P(3, 4), Q(–2, 1) and R(4, 2) under rotation
about negative quarter turn followed by reflection on x-axis. Show the object and
image in graph. Also state the single transformation represented by them.
ii) Find the image of a quadrilateral having A(5, 2), B(–4, –3), C(–2, 3) and D(3, 2)
under reflection about y + 3 = 0 followed by rotation about positive quarter turn.

Also plot them in graph.
iii) A(2, 5), B(–1, 3) and C(4, 1) are the verticel of DABC, Find the co-ordinate of image
of DABC under rotation about +270° followed by enlargement about E[0, 2]. Also

plot them in graph.
iv) Find the image of DABC having vertices A(2, 0), B(3, 1) & C(1, 1) under the reflection
on x - y = 0 followed by enlargement about E[(–3, –4), 2]. Also plot them in graph.
v) The vertices of DPQR are P(7, 0), Q(4, –2) and R(5, –4). Find the co-ordinate of image

DPQR under reflection about y = 0 followed by translation about T = –3
of <F . Also
–1
plot them in graph.

9. i) Find the image of DABC under a combination transformation of reflection on x-axis
and reflection on y-axis where the vertices are A(–3, 4), B(–1, 0) and C(1, 6) and
show the object and image in graph. Also find the single transformation.

ii) Find the image of DPQR under a combined transformation of rotation about +90°
and rotation about 180° where the vertices of DPQR are P(3, –2), Q(4, 4) and R(6,
0) and plot them in graph. Also write down the single transformation.

iii) Transformed DKLM having vertices k(–3, –2), L(–5, 0) & M(–6, –5) under a combined

transformation of translation about T1 = 3 and translation about T2 = –1 and
<F <F
1
2

shown in graph. Also find the single transformation.
iv) Transform a triangle having vertices A(–2, 3), B(3, 1) and C[1, –2] under a combined
3
enlargement of E1, [0, 2] and E2[0, – 2 ] and show in graph. Also write down the

single enlargement.
v) A(3, 4), B(2, 1) and C(4, 2) are the vertices of a triangle. Find the image of DABC

298 PRIME Opt. Maths Book - X

under a composite transformation of reflection on y-axis and reflection on y + 2 = 0.
Also plot the object and image in graph.

10. i) Find the image of DABC having vertices A(4, –1), B(1, 1) and c(2, 3) under oRr1iOgiRn2.
ii) AwlhseorpeloRt1 is reflection about yst-aatxeisthanedsiRn2gilse rotation about – 90° with centre
them in graph and transformation.
iii) Transform a quadrilateral having vertices P(1, 4), Q(5, 1), R(6, 3) and S(4, 5) under
iv)
v) TOR where T is translation about T = 2 and R is reflection about x = –1. Also plot
the object and image in graph. <F
0

The vertices of DABC are A(3, 1), B(5, 2) and C(6, 7). Find the image of DABC under
E1oE2 where E1[(0, 0), 2] and E2[(2, 0), 2] are the two also plot the object and image
in graph.
Find the image of an object having vertices A(–2, –3), B(–4, 4) and C(–6, 2) under
GoF where G is rotation about – 270° with centre origin and F is translation about
BC . Also plot the object and image in graph.
Transform a triangle having vertices P(2, 7), Q(–2, 3) and R(4, 0) under EOR where
R is reflection about x – y = 0 and E is an enlargement about E[(2, –5), 2]. Also plot
the object and image in graph.

11. PRIME more creative questions:
i) Find the equation of image of a straight line 2x + y = 8 under a translation with T =

<2F followed by reflection on x - axis.
1

ii) Find the image of a straight line 2x – y = 5 under a combined rotation of +90° with
centre origin and again rotation about positive quarter turn.

iii) Find the image of DABC having A(1, 2) B(3, –4) and C(6, 1) under a combined
rotation about +120° with centre origin and +60° with centre origin. Also plot them

in graph.
iv) Transform a triangle having vertices A(3, –2), B(5, –7) and C(6, 2) under an axis

rotation through 45° passes through the origin between the positive axes of followed
by the reflection axis which bisects the negative axis of x and positive axis of y. Also
plot the object and image in graph.
v) Transform a triangle having vertices A(–3, –1), B(–5, 4) and C(–1, 6) under rotation
about +90° with centre (1, –2) followed by rotation about 180° with centre (1, –2).
Also plot the object and image in graph.

Answer

1. Show to your teacher.
2. i) P'(–x, –y). It is rotation about 180°.

ii) P'(y, –x). It is rotation about –90°.
iii) P'(x, –y). It is reflection about x–axis.
iv) P'(– 2 – x, y – 3). It is reflection about x-axis.
v) P'(–2x, – 2y). It is enlargement about E[0, –2]

PRIME Opt. Maths Book - X 299

Answer

3. i) (x, –y), (–3, –5) ii) (y, x), (5, –3) iii) (–x, –y), (3, –5)
iv) (–4 –y, –x), (–9, 3) v) (2x + 1, 2y + 7), (–5, 17)

4. i) P(9, –3) ii) A(–10, –4) iii) k = 2 iv) T = 4 v) P(–3, 2)
iii) A(1, 3) <F
5. i) A(2, 11) ii) A(6, 5) 2

1 –1
iv) T1 = <F v) T2 = <F
3 –2

6. i) A''(–9, 4), B''(–6, 1), C''(–1, 5) ii) P''(1, –5), Q''(4, 1), R''(–3, –3)

iii) A''(3, 4), B''(6, 0), C''(8, 7) iv) P''(–8, 8), Q''(–4, –12), R''(8, –8)

v) A''(–3, 1), B''(–6, 2), C''(–1, 4)

7. i) O''(6, 2), A''(4, 2), B''(3, 3), C''(5, 3)
ii) A''(1, 4), B''(–3, 7), C''(–1, 2)
iii) M''(10, –4), N''(6, 2), P''(2, –8)
iv) P''(1, 2), Q''(3, 5), R''(–1, 7)
v) A''(6, 6), B''(8, 10), C''(6, 0), D''(10, 2)

8. i) P''(4, 3), Q''(1, –2), R''(2, 4). It is reflection on y = x.
ii) A''(8, 5), B''(3, –4), C''(9, –2), D''(8, 3). It is P(x, y) → P'(6 + y, x)
iii) A''(10, –4), B''(6, 2), C''(2, –8). It is P(x, y) → P'(2y, –2x)
iv) A''(3, 8), B''(5, 10), C''(5, 6). It is P(x, y) → P'(2y + 3, 2x + 4)
v) P''(4, –1), Q''(1, 1), R''(2, 3). It is P(x, y) → P'(x – 3, –y – 1)

9. i) A'(3, –4), B'(1, 0), C'(–1, –6); It is rotation about 180°.
ii) P'(–2, –3) Q'(4, –4), R'(0, –6); It is rotation about –90°.

iii) K'(–1, 1), L'(–3, 3), M'(–4, –2). It is translation about T = 2
<F
3

iv) A'(6, –9), B'(–9, 3), C'(–3, 6). It is enlargement about E[0, –3].
v) A'(–3, –8), B'(–2, –5), C'(–4, –2); It is rotation about 180° with centre (0, –2)

10. i) A'(1, –4), B'(–1, –1), C'(–3, –2); It is reflection on y = –x.
ii)
P'(–1, 4), Q'(–5, 1), R'(–6, 3), S'(–4, 5). It is reflection on y-axis.
iii)
iv) A'(8, 4), B'(16, 8), C'(20, –6); It is enlargement about E[( 4 , 0), 4]
v) 3
A'(5, –4), B(–2, –6) C'(0, –8). It is P(x, y) → P'(2 – y, x – 2)

P'(12, 9), Q'(4, 1), R'(–2, 13); It is P(x, y) → P'(2y – 2, 2x + 5)

11. i) 2x – y – 13 = 0 ii) 2x – y + 6 = 0
iii)A'(–1, –2), B'(–3, 4), C'(–6, –1) iv) A'(–3, 2), B'(–5, 7), C'(–6, –2)

300 PRIME Opt. Maths Book - X

7.2 Inversion Transformation

The transformation which is studied previously is a kind of geometry where image of an
object is calculated and transformed from one place to another place under any geometrical
conditions.

Here, we are going to discuss the transformation under a circle OQ P
called geometrical inversion or inversion circle.
O is the centre of a circle of radius ‘r'. Where P is transferred to Q
and vice versa.

i.e.,
• The point inside the circle is transferred to outside (P is inverse of Q )

• The point outside the circle is transferred to inside (Q is inverse of P)

• The point on the circumference of a circle is invariant point.

• As P moves further from centre ‘O', it's image ‘Q' moves closer to 0.

• The result for object point ‘P' and it's image ‘Q' in a circle is taken as, OP . OQ = r2

• Coordinate of image of a point under inversion circle having centre (a, b) and
radius ‘r' is,

p(x, y) → P'[(x – a, y – b). + (a, b)]

• If centre of circle becomes origin and radius ‘r', the image of P(x, y) will be
r2
P(x, y) → P'[(x, y). x2 + y2 ]

Theorem

Proof of OP . OP' = r2 in inversion circle where P' is the image of point P.

Proof : CQ P
Here O P' A
C be the inversion circle of centre O and radius OA = r.
P' is the image of point P under inversion.
Taking a point Q on the circumference

Then,
PQ will be the tangent

\ OQP = 90°
\ QP'O = 90°
Now,
In DQOP and DOQP'
\ QOP = \ QOP' (Common angle)
\ \ OQP = \ QP'O (both are 90°)
\ \ OPQ = \ OQP' (remaining angles)
\ DQOP ∼ DOQP' (by AAA axiom)

PRIME Opt. Maths Book - X 301

Here, the corresponding sides are proportional.
OQ = OP
or, OP' OQ

or, r = OP
OP' r

\ OP . OP' = r2

Way of finding inversion point of a point.

Y

A (h, k) P(x, y) P'(x', y')
Q R

X' O SM NX

Y'
Let P'(x', y') be the image of a point P(x, y) under inversion circle having centre A(h, k).
Draw,
PM^OX, P'N^OX, AR^PM and P'N,

AS^OX

Then,

AP = =

AQ = SM = OM – OS = x – h
AR = SN = ON – OS = x' – h
PQ = PM – QM = PM – AS =y–k
P'R = P'N – RN = P'N – AS = y' – k
Here, DAQP and DARP' are equiangular (i.e. similar)
Taking the ratios of similar triangles,

AR = P'R = AP'
AQ PQ AP

or, x' – h y' – k = AP' × AP
x–h = y–k AP AP

or, x' – h y' – k = [ a AP × AP' = r2]
x–h = y–k

Taking,

x' – h =
x–h

or, x' – h =

\ x' = +h

302 PRIME Opt. Maths Book - X

Similarly, +k
Taking,

y' – k
y–k =

\ y' =

\ P(x, y) → P'(x', y') =[ + h, r2 (y – k) + k]
(x–h)2 + (y–k)2

= [(x – h, y – k). + (h, k)]

If centre of the circle is the origin. + (0, 0)]
P(x, y) P'(x', y') = [(x – 0, y – 0) .

= [(x, y). r2 ]
x2 + y2

Worked out Examples

1. Find the image of a point P(1, 3) under a inversion circle having equation x2 + y2 = 25.
Solution :
The given equation of circle is,
x2 + y2 = 25
or, (x – 0)2 + (y – 0)2 = 52
\ centre (h, k) = (0, 0)
Radius (r) = 5 units.

Under inversion circle of centre origin and radius 5 units.

P(x, y) → P' <^x, yh . x2 r2 y2 F
+

` P(1, 3) → P' ;^x, yh . 52 E
12 + 32

→ P' :^1, 3h . 25 D
10

→ P' a 25 , 75 k
10 10

→ P'(2.5, 7.5)

2. Find the image of a point A(0, 4) under a circle of centre origin and radius 8 units.

Solution :
Under inversion circle of centre (0, 0) and radius 8 units.

P(x, y) → P' <^x, yh . x2 r2 y2 F
+

A(0, 4) → A' ;^0, 4h . 82 E
02 + 42

PRIME Opt. Maths Book - X 303

→ A'[(0, 4), 4]
→ A'(0, 16)

3. Find the image of a point A(2, 4) under a circle of centre (2, 1) and radius 6 units.
Solution :
Under inversion circle of centre (2, 1) and radius 6 units,

P(x, y) → P'

P(2, 4) → P'

→ P'[(0, 3) . 4 + (2, 1)]
→ P'(2, 13).

4. If a point P is transferred to P`(8, 0) under the inversion circle having centre origin and
radius find the co-ordinate of P.

Solution :
Under inversion circle of centre origin and radius 4 units.
r2
P(x, y) → P'[(x,y). x2 + y2 ]

→ P' [(x, y). 42 ]
22 + 02

→ P' [(4x, 4y)]

From given,
P(x, y) → P'(8, 0)
By equating the corresponding elements,

We get, and
4x = 8 4y = 0
\ x=2 y=0
\ Co-ordinate of P is (2, 0)

Exercise 7.2

1. Answer the following questions.
i) If Q is the image of a point P under inversion circle of radius 'r' and centre A. Write
down the result of AP.AQ.
ii) Write down the image of P(x, y) under inversion circle of centre origin and radius
'r'.
iii) Write down the image of A(x, y) under inversion circle of centre (a, b) and radius 'r'.
iv) If P' is the image of a point P under inversion circle of radius 4 units and centre O,
write down the result of OP.OP'.
v) What do your mean by inversion transformation.

304 PRIME Opt. Maths Book - X

2. Find the image of the following points under the inversion circle of the centre origin.
i) Point P(2, 0); equation of circle is x2 + y2 = 16
ii) Point A(1, 1); equation of circle is x2 + y2 = 4
iii) Point A(3, 4); radius = 10 units
iv) Point P(4, 2); radius = 10 units
v) Point P(6, 8); radius = 50 units

3. Find the image of the following points under the inversion circle of centre ‘Q' and radius ‘r'.
i) Point P(2, 4); Centre Q(6, 2); r = 10 units
ii) Point P(3, 6); Centre Q(3, 0); r = 18 units
iii) Point P(4, 6); Centre Q(2, 0); r = 80 units
iv) Point P(8, 5); equation of circle x2 + y2 – 2x – 8y – 83 = 0
v) Point P(0, 3); equation of circle x2 + y2 + 4x – 4y – 17 = 0

4. PRIME more creative questions.
i) If a point A(3, 4) is transformed to A'(12, 16) under inversion circle having center
origin and radius 'r', find the length of the radius.
ii) If a point P is transferred to P'(0, 16) under inversion circle having centre origin and
radius 8 units, find the co-ordinate of 'P'.
iii) If a point A(2, 4) is transferred to A'(2, 13) under inversion Circle of centre(2, 1) and
radius 'r', find the length of the radius.
iv) If a point P(0, 3) is transferred to P'(8, 7) under a inversion circle having centre (–2,
2) and radius 'r', find the value of 'r'.
v) Find the image of a triangle having vertices A(2, 0), B(–2, 2) and C(2, 2) under a
inversion circle having center origin and radius 4 units. Also plot the object and
image in graph.
vi) Find the image of DPQR having vertices P(0, 2), Q(2, –4) & R(4, 4) under a inversion
circle of equation x2 + y2 – 4x – 8y + 4 = 0. Also plot the object and image in graph.

Answer

1. Show to your teacher.

2. i) (8, 0) ii) (2, 2) iii) (12, 16) iv) (2, 1) v) (3, 4)
iv) (15, 6) v) (8, 7)
3. i) (4, 3) ii) (3, 3) iii) (6, 12) iv) 5 units

4. i) 10 units ii) (0, 4) iii) 6 units

v) A'(8, 0), B'(–4, 4), and C'(4, 4); graph

vi) P'(–2, 0), Q'(2, 2) and R'(10, 4)

PRIME Opt. Maths Book - X 305

7.3 Transformation using matrices

Let us taking an example : x + 2y = 5 and 3x – y = 1

The equations can be written in matrix form as, 1 2x 5
< F <F = <F
3 –1 y 1

For the variables the matrix so obtained is 1 2
< F
3
–1

Taking another example for image & object.
x = a and y = b
where x & y are the objects and a & b are the images where,
a = x and b = y
= x + 0.y = 0.x + y

a 10 x
The equation in matrix from are <bF = <0 1F <yF

10
Here 2 × 2 transformation matrix = < F
0 1

Above example conclude that formula for the transformation using 2 × 2 matrix is,

Image = (matrix) (object)

2 × 2 transformation matrix for different transformations

S.N. Transformations Derivation of equations 2 × 2 matrix
1. Reflection 10
i) on x - axis (x, y) → P' x <F
<F 0 –1
P(x, y) → P'(x, – y) –y
–1 0
→ =x + 0.yG <F
0.x – y
01
10 x
→ < F< F 01
0 –1 y <F
10
ii) on y-axis (x, y) → P' –x
p(x, y) → P'(–x, y) <F

y

→ =–0x.x++0y.yG

–1 0 x
→ < 0 F< F
1y

iii) on y = x (x, y) → P' y
P(x, y) → P'(y, x) <F
x

→ =0x.+x + yG
0.y

→ 0 1 x
< F <F
1 0y

306 PRIME Opt. Maths Book - X

iv) on y = – x (x, y) → P' –y 0 –1
P(x, y) → P'(–y, –x) <F <F
–x –1 0
0 –1
0.x – y <F
→ < F 10
–x + 0.y
01
0 –1 x <F
→ < F< F –1 0
–1 0y
–1 0
2. Rotation (with centre (x, y) → P' –y <F
i) origin) <F
0 –1
x k0
<F
About +90° 0.x – y 0k
P(x, y) → P'(–y, x) → < F
x + 0.y a
<F
→ <0 –1F <xF b
10 y used for addition

ii) About –90° (x, y) → P' y
P(x, y) → P'(y, –x) <F
–x

0.x + y
→ = G
–x + 0.y

01 x
→ < F< F
–1 0y

iii) About 180° (x, y) → P' –x
P(x, y) → P'(–x, –y) <F
–y

→ =–x + 0.yG
0.x – y

–1 0 x
→ < 0 F< F
–1 y

3. Enlargement about E[0, k] (x, y) → P' kx
P(x, y) → P'(kx, ky) <F
ky

→ =0k.xx++0k.yyG

k0 x
→ < F< F
0 ky

4. a (x, y) P' =x + aG
Translation about T = <F → y + b
b

P(x, y) → P'(x + a, y + b) ax
→ <F + <F
b y

PRIME Opt. Maths Book - X 307

Informations for transformation using matrix
i) For given image (x', y') = (ax + by, cx + dy)

x' = ax + by
y' = cx + dy

(x', y') = ax + by
= + G
cx
dy

ab x
= < F <F
c d y

i.e. Image = (matrix) (object) Y

a
ii) For 2 × 1 matrix M = <F
b (1,1)
(0,1)
Image = (matrix) + (object) X' (0, 0) (unit square)
X
ab
< F (1, 0)
iii) For 2 × 2 matrix M = c d

Image = (matrix) (object) Y'

iv) Co-ordinate of the vertices of unit square.
OABC is the unit square where the co-ordinate of the vertices are : O(0, 0), A(1, 0), B(1,
1), C(0, 1)

Way of finding the transformation represented by the 2 × 2 matrix:

1. Taking 2 × 2 matrix 12 for an object (x, y).
< F
1 –1

Using formula for matrix transformation:

Image = (matrix) (object)

(x', y') = 1 2x
< F< F
1
–1 y

= =11××xx++(2–1×)yyG

= =x + 2yG
x–y

Here, The transformation with equation,
x' = x + 2y and
y' = x – y

308 PRIME Opt. Maths Book - X

ii. Transformation for a 2 × 2 matrix 0 –1
< F
1 0

Using formula for matrix transformation:

Image = (matrix) (object)

(x', y') = 0 –1 x
< F< F
1
0y

= =0 × x + (–1) × yG
1× x +0× y

–y
= <xF

\ P(x, y) → P'(–y, x)
The transformation is rotation about +90° with centre origin.

Worked out Examples

1. Which transformation is represented by the matrix <02 02F ? Also find the image of A(3, –2)
with the matrix.

Solution :
Let, P(x, y) be the object point.

We have,

Image = (matrix)(object)

20 x
= < F< F
0 2y

= =2 × x + 0× yG
0× x + 2× y

2x
= <F
2y

\ P(x, y) → P'(2x, 2y)

It is enlargement about E[(0, 0), 2]

Again, for the point A[3, –2]

Image = (Matrix)(Object)

20 3
= < F< F
0 2 –2

2 × 3 + 0 (–2)
= = G
0× 3 + 2× (–2)

6
= <F
–4

\ A(3, –2) → P'(6, –4)

PRIME Opt. Maths Book - X 309

2. Find the image of a point P(–3, 2) using 2 × 2 matrix for reflection about y-axis.
Solution : Under reflection about y - axis,
P(x, y) → P'(–x, y)
–x
→< F
y

→ =–x + 0.yG
0.x + y

→ <–1 0F <xF
01 y

–1 0
\ 2 × 2 matrix is < 0 F
1

Again, for the point P(–3, 2)

Image = (Matrix)(Object)

–1 0 –3
= <0 F< F
12

=

3
= <F
2

\ A(–3, 2) → P'(3, 2)

3. Find 2 × 2 matrix represented by a transformation which gives x' = 2x – y and y' = x + 3y.

Also find the image of A(2, –5) using the matrix.

Solution : A transformation gives image,
(x', y') = (2x – y, x + 3y)

or, (x', y') = 2x – y
<x + 3yF

2 –1 x
= <1 3 F <yF

2 –1
\ 2 × 2 matrix = <1 3 F

Again, for the point A(2, –5)
Image = (matrix)(object)

2 –1 2
= <1 3 F <–5F

=

= =4+5G
2 – 15

9
= <F
–13

\ A(2, –5) → A'(9, –13)

310 PRIME Opt. Maths Book - X

4. Find the image of triangle having vertices A(1, –3), B(–2, 1) and C(2, 4) by using the matrix

for rotation about +270° with centre origin. Also plot them in graph.
Solution : Under rotation about + 270°
P(x, y) → P'(y, –x)

y
→ <F
–x

→ =–0x.x++0y.yG

01 x
→ < F< F
–1 0y

01
\ 2 × 2 matrix = < F .
–1 0

Again, for the image of DABC.

(Image) = (matrix)(object)

= <0 1F < 1 –2 2F
–1 0 –3 1 4

=

–3 1 4
= < F
–1 2 –2

\ A(1, –3) → A'(–3, –1)
B(–2, 1) → B'(1, 2)
C(2, 4) → C'(4, –2)

5. Find the image of unit square <00 1 1 01F by using 2 × 2 matrix of a transformation which
0 1

gives an image (x', y') = (2y, x – y).

Solution :

0110
Unit square = < F
0 0 1 1

Under a transformation
(x', y') = (2y, x – y)

= < 2y F
x y
–

= =0.x + 2yG
x –y

02 x
= < F< F
1 –1 y

02
\ 2 × 2 matrix = < F
1 –1

PRIME Opt. Maths Book - X 311

Again, for the unit square,

Image = (matrix) (object)

0 2 0110
= < F< F
1 –1 0 0 1 1

=

002 2
= < F
0 1 0 –1

\ O(0, 0) → O'(0, 0)
A(1, 0) → A'(0, 1)
B(1, 1) → B'(2, 0)
C(0, 1) → C'(2, –1)

6. Find 2 × 2 transformation matrix which transform a unit square to a parallelogram

<00 1 3 42F . <a bF
3 7 c d

Solution : Let, 2 × 2 matrix be

0110
Object = unit square = < F
0 0 1 1

0132
Image = < F
0 3 7 4

We have,

Image = (matrix)(object)

0132 ab 0110
or, < F = < F< F
0 3 7 4 c d0 0 1 1

0132
or, < F =
0 3 7 4

or, 0 1 3 2 = =0 a a+b bG
< 3 7 F 0 c c+d d
0
4

By equating the corresponding elements,

a=1 b=2
c=3 d=4

12
< F
\ 2 × 2 matrix = 3 4

7. Find 2 × 2 transformation which transform A (1, –2) → A'(–2, –1) and B(3, 4) → B'(4, –3).

Also find the type of transformation.

ab
Solution : Let, 2 × 2 transformation matrix be < F
Which gives, c d

A(1, –2) → A'(–2, –1)
B(3, 4) → B'(4, –3)

We have,

Image = (matrix)(object)

312 PRIME Opt. Maths Book - X

–2 4 a b 1 3
or, < F = < F< F
–1 –3 c d –2 4

–2 4
or, < F =
–1 –3

or, –2 4 = =a–2b 3a + 4bG
< F c–2d 3c + 4d
–1
–3

By equation the corresponding elements,

a – 2b = –2 ⇒ a = 2b – 2 ...... (i)

3a + 4b = 4 ⇒ a= 4 – 4b ...... (ii)
3

c – 2d = –1 ⇒ c = 2d – 1 ...... (iii)

3c + 4d = –3 ⇒ c= –3 – 4d ...... (iv)
3

Solving equation (i) and (ii)

2b – 2 = 4 – 4b
3

or, 6b – 6 = 4 – 4b

or, 6b + 4b = 10
\ b=1

Putting the value of ‘b' in equation (i)
a=2×1–2=0

Again, solving equation (iii) and (iv)

2d – 1 = –3–4d
3

or, 6d – 3 = – 3 – 4d
or, 6d + 4d = 0
\ d= 0
Putting the value of ‘d' in equation (iii)

C = 2 × 0 – 1 = –1

01
\ 2 × 2 matrix is < F
–1 0

Again, Taking an object A(x, y)

Image = (matrix) (object)

01 x
= < F< F
–1 0y

= =–01××xx++10××yyG

y
= <F
–x

\ A(x, y) → A'(y, –x)

It is rotation about –90° with centre origin.

PRIME Opt. Maths Book - X 313

8. A transformation gives image of an object A(3, –4) → A'(4, –3). Find 2 × 2 transformation
matrix represented by it. Also find the image of point P(5, –2)

Solution: The transformation gives.
A(3, –4) → A'(4, –3)

It is equivalent to,

P(x, y) → P'(–y, –x)

→ P' <–yF
–x

0.x – y
→ < F
–x + 0.y

0 –1 x
→ < F< F
–1 0y

0 –1
\ 2 × 2 transformation matrix is < F.
–1 0

Again, For the point (5, –2)

Image = (matrix) (object)

0 –1 5
= < F< F
–1 0 –2

=

= <2 F
–5

\ P(5, 2) → P'(2, –5)

314 PRIME Opt. Maths Book - X

Exercise 7.3

1. Find 2 × 2 transformation matrix represented by the following transformations.
i) Reflection about y = 0.
ii) Reflection about y = –x.
iii) Rotation about –90° with centre origin.

iv) Enlargement about E[(0, 0), –3]
v) Rotation about –270° with centre origin.

2. Which transformations defined by the following matrices?

–1 0 –2 0 0 –1
i) < F ii) <F iii) <F
0 1 0 –2 –1 0

–1 0 0 –1
iv) < 0 F v) <F
–1 10

3. i) Find the matrix represented by translation about T = <3F .
2

ii) For a transformation which gives A(3, –2) → A'(2, –3).
iii) For a transformation which gives A(a, b) → A'(–b, a).
iv) For a transformation which gives x' = x + 2y and y' = 3x – y.
v) For a transformation which gives x' = 2x and y' = 2y – x.

4. i) Find the image of a point P(–2, 3) under 2 × 2 matrix for reflection about x-axis.
ii) Find the image of a point A(2, –4) under 2 × 2 matrix for rotation about –270° with

centre origin..

iii) Find the image of a point P(3, –1) under a matrix for translation about T = –2
<3 F.

iv) Find the image of a point M(4, –1) under 2 × 2 matrix for enlargement about E[(0,

0), 3].

v) Which transformation is represented by the matrix 0 1 ? Also find the image of a
point A(5, –2) with the matrix. < F
1
0

5. i) Find the image of DABC having vertices A(–1, –2), B(–3, 4) and C(–7, 2) under a

matrix <1 2F . Also plot the object and image in graph.
–1 –2

ii) Find the image of DPQR having vertices P(1, –3), Q(3, 3) and R(5, –1) under a matrix
for reflection on x – y = 0. Also show them in graph.

iii) Find the image of a triangle having vertices A(–2, –3), B(–1, 3) and C(3, –1) under a
2 × 2 matrix for enlargement about E[0, 2]. Also plot the object and image in graph.

iv) Find the image of DKLM having vertices k(–1, –2), L(3, 2) and M(–2, 5) under a
matrix for rotation about –270° with centre origin. Also plot them in graph.

v) Find the image of an object having vertices A(3, 0), B(1, 3) and C(4, 7) under a
transformation which gives x' = 2x + y and y' = x – 3y. Also plot them in graph.

PRIME Opt. Maths Book - X 315

6. i) Find the image of a unit square 0 1 1 0 under a 2 × 2 matrix of 1 2
the object and image in graph. < 0 1 F < F . Also plot
0 3 –1
1

ii) Find the image of a unit square under a matrix represented by reflection about x – y

= 0.

iii) Find the image of a unit square under 2 × 2 transformation matrix for enlargement

about E[0, –2].

0 1 0 –1 under a matrix 1 –1 & find the vertices
iv) Transform a parallelogram < –2 1 F < F
0 –2
of image. 3 3

0 –2 1 3 under a matrix –2 3
v) Transform a parallelogram < 3 –1 F < F . Also plot them
in graph. 0 –4
–4 3

7. i) Find 2 × 2 transformation matrix which transform a unit square to a parallelogram

having vertices 0 0 1 1 . Which transformation is it?
< 1 1 F
0
0

ii) Find 2 × 2 transformation matrix which transform the unit square to a parallelogram

0132
having vertices < F .
0 3 7 4

iii) Find 2 × 2 transformation matrix which transform the parallelogram of 0 1 0 –1
to the unit square. < –2 1 F
0
3

iv) Find 2 × 2 transformation matrix which gives A(3, –2) → A'(–8, 0) and B(–1, 4) →
B'(6, 10).

v) Find 2 × 2 transformation matrix which transform P(2, 3) and Q(–3, 1) to P'(3, 8)
and Q'(–10, –1)

8. PRIME more creative questions.
i) Transform a triangle having vertices A(1, 2), B(2, 1) and C(3, 4) under a inverse

25
matrix of < F .
1 3
ii) Transform a unit square under a 2 × 2 matrix for reflection about y-axis followed by
rotation about –90° with centre origin.
iii) Transform a triangle having vertices P(1, –5), Q(–2, 2) and R(4, 1) under a combined

1 2 and 2 1
matrix of < F < F .
–1 1 –1
–2

iv) The image of triangle ABC having vertices A(–2, 1), B(1, –3) and C(2, 4) are A'(–5,
–3), B'(0, –1) and C'(10, 8) under a 2 × 2 matrix. Find the inverse of the matrix so
formed.

v) The image of DABC under a matrix 3 1 are A'(7, 6), B'(–7, –5) and C'(9, 5). Find
the vertices of DABC. < F
2
1

316 PRIME Opt. Maths Book - X

9. Project work
i. Prove that the rotation about +90° with centre origin followed by ratation about 180° is

the reflection abot y-axis by calculating 2 × 2 matrix.
ii. Prepare a chart of the formula of different transformations and 2 × 2 matrix represented

by them in chart paper and present the report in your classroom.

Answer

10 0 –1 01 –3 0 0 –1
1. i) <F ii) <F iii) <F iv) <F v) <F
0 –1 –1 0 –1 0 0 –3 10

2. i) Reflection on y-axis ii) Enlargement about E[O, –2]
iii) Reflection on y = –x iv) Rotation about 180° with centre origin.
v) Rotation about + 90° with centre origin

3 0 –1 0 –1 12 20
3. i) <F ii) <F iii) <F iv) <F v) <F
2 –1 0 10 3 –1 –1 2

4. i) (–2, –3) ii) (4, 2) iii) (1, 2)
iv) (12, 3) v) It is reflection about y = x. (2, 5)

5. i) (–5, 5), (5, –5), (–3, 3) ii) (–3, 1), (3, 3), (–1, 5)

iii) (–4, –6), (–2, 6), (6, –2) iv) (2, –1), (–2, 3), (–5, –2)

v) (6, 3), (5, –8), (15, –17)

013 2 0011 0 –2 –2 0
6. i) <F ii) <F iii) <F
0 3 2 –1 0110 0 0 –2 –2

0 3 –1 –4
iv) <F v)
0 –8 3 11

7. i) 0 1 , It is reflection about y = x. ii) 12
< F <F
1 34
0

32 –2 1 3 –1
iii) <F iv) <F v) <F
11 23 12

0011
8. i) (–7, 3), (1, 0), (–11, 5) ii) <F
iii) (–9, –18), (2, 4), (2, 4) 0110

1 –1
iv) <F v) (1, 4), (–2, –1), (4, –3)
–2 3

PRIME Opt. Maths Book - X 317

Transformation

Unit Test Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. Which transformation gives congruent image of an object and which one gives only

similar?

2. a) Find the image of a point A(–3, 2) under reflection about x-axis followed by rotation
about +270° with centre origin.

b) If a point A(a, b) is transferred to A'(–5, –1) under FOG where F is reflection about x
+ 2 = 0 and G is rotation about + 90° with centre origin. Find the value of ‘a' and ‘b'.

c) Find the image of a point P(2, –2) under inversion circle of centre origin and radius
4 units.

3. a) Find the image of triangle having vertices A(3, 3), B(–2, –1) and C(–6, –3) under a

combined transformation of translation about T = 3 and rotation about 180° with
<F
2

centre origin. Also plot the object and image in graph.
b) Find 2 × 2 transformation matrix which transform a parallelogram having

0 2 –1 –3
< 2 F to the unit square.
0 –1 1

4. Transform a quadrilateral having vertices A(–2, 2), B(–1, –3) and C(2, –4) and D(3, 4)

under an enlargement about E[0, –2] followed by translation about T = 4
them in graph. <F . Also plot
3

318 PRIME Opt. Maths Book - X

Unit 8 Statistics

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions – 1 2 – 3 10 12
Weight – 2 8 –

K = Knowledge, U = Understanding, A = Application, HA = Higher ability, TQ = Total

Questions, TM = Total Marks

Objectives : At the end of the lesson
• Students are able to find the quartile deviation, mean deviation and

standard deviation using continuous series.
• Students are able to analyses the data using measure of dispersion and

coefficient.
• Students are able to find the scatteredness of the observations from their

central values.
• Students are able to know which dispersion is better to analyses the data.

Materials
• Marks ledger of the students of terminal examination.
• Chart of the number of students of each classes according to their scoring.
• Chart of the pass out students after SEE and their grading analysis.
• Chart of different types of measures of central tendency and measure of

dispersions.
• Chart of the formula of mean, median, quartiles and measure of dispersions

etc.

PRIME Opt. Maths Book - X 319

8.1 Measure of dispersion:

Enjoy the recall:

1. Arithmetic mean of the continuous frequency distribution.
/fx
i) x = N , where x is mid - value of the class.

ii) x = A + /fd , where deviation d = x – A.
N

iii) x=A+ /fd' # h , where step - deviation d' = x–A
N h

2. Median of the observations of the continuous frequency distribution.
• Construct cumulative frequency table.
• Median class lies in, (Finding position of median class)

= Size of ` N th item
2
j

= Just greater C.F. than N .
2

= Corresponding class

Then,

N – C.F.
F
Median (Md) = L+ 2 ×h

3. Quartile deviation, (Semi inter quartile range)
1
Q.D. = 2 ^Q3 – Q1h

Coefficient of Q.D = Q3 – Q1
Q3 + Q1

4. Mean deviation,
/f
M.D. ^Xh = x–x ( from mean )
N

M.D. ^Mdh = /f x –Md ( from median )
N

Coefficient of M.D.(x) = M.D. ( from mean )
x

Coefficient of M.D. ^Mdh = M.D. ( from median )
Md

5. Standard deviation:

• S.D. (σ) = /f^x – xh2 (Using mean)
N

• S.D. (σ) = (Shortcut method)

• S.D. (σ) = (Step Deviation method)

• S.D. (σ) = h ×

320 PRIME Opt. Maths Book - X

Where,

d = xas–suAm, de'd=mxe–hanA S.D.
A = x

• Coefficiemt of S.D. = S.D.
• x
• Coefficient of variation (C.V.) = × 100 %

Variance = σ2

6. Syllabus for grade ‘X'.
• Quartile deviation.
• Mean deviation.
• Standard deviation.
• For all the measures of dispersion , only continuous frequency distribution have
to be used.

Quartile Deviation

Quartile deviation is defined as the half of the range of the
quartile of the given observations. Hence it is also called
the semi - inter quartile range.

QH3er–eQ, Q1.1TahnednQq3uaarretitlheedeexvtiraetimonescqanuabretielexsporef sthseedoibnseforrvmatuiolansaswhere range of the quartiles is

Q.D. = 1 ^Q3 – Q1h
2
Q3 – Q1
Coefficient of Q.D. = Q3 + Q1

Steps for the calculation of Quartile deviation.
• Construct a cumulative frequency table.
• Finding of first quartile (Q1) and third quartile (Q3) as the calculation of median using
N th 3N th
their classes obtained by finding size of ` 4 item and ` 4 item respectively.
j j

Where,

N – c.f. 3N – c.f
f f
Q1 = L + 4 ×h Q3 = L + 4 #h

• Calculation of quartile deviation and it's coefficient using formula,
1
Q.D. = 2 ^Q3 – Q1h

Coefficient of Q.D. = Q3 – Q1
Q3 + Q1
• Inter - quartile range can be calculated as,
Inter - quartile range = (Q3 – Q1).
Note:
• Only first quartile and third quartile are involved in the quartile deviation.
• Due to its cause, quartile deviation is not the good measurement of dispersion in

practice.

PRIME Opt. Maths Book - X 321

Worked out Examples

1. If first and third quartiles are 25 and 40 respectively of a continuous frequency table. Find

the quartile deviation and its coefficient.

Solution:

FTihrisrtdqquuaarrtitliele((QQ1)3)==4205
Then,

Q.D. = 1 ^Q3 – Q1h
2

Q.D. = 1 ^40 – 25h
2

= 1 # 15
2

= 7.5

Coefficient of Q.D. = Q3 – Q1
Q3 + Q1

= 40 – 25
40 + 25

= 15
65

= 0.230

2. If quartile deviation and it's coefficient of a continuous frequency distribution are 20 and

0.5 respectively, find the first and third quartile.

Solution:

Q.D. = 20
Coefficient of Q.D. = 0.5

We have,

Q.D. = 1 ^Q3 – Q1h
2

or, 20 = 1 ^Q3 – Q1h
2

or, Q3 = 40 + Q1 ..........(i)
Again, of
or, Coefficient Q1 Q.D. = Q3 – Q1
Q3 – Q3 + Q1
Q3 + Q1
0.5 =

or, 1 = Q3 – Q1
2 Q3 + Q1

or, Q3+Q1= 2Q3 – 2Q1
or, 3Q1 = Q3
or, 3Q1 = 40 + Q1 [ a From equation (i) ]

or, 2Q1 = 40
\ Q1 = 20
Putting the value of Q1 in equation (i),
Q3 = 40 + 20 = 60
\ First quartile = 20
Second quartile = 60

322 PRIME Opt. Maths Book - X

3. Find the semi - interquartile range and it's coefficient of

Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60
10 2
f 3 5 8 12
Solution:

Class f c.f.
3 3
0 - 10 5 3+5=8
10 - 20 8 8 + 8 = 16
20 - 30 12 16 + 12 = 28
30 - 40 10 28 + 10 38
40 - 50 2 38 + 2 = 40
50 - 60

Here, Q1 class lies in.

= Size of ` N th item
4
j

= Size of ` 40 th item
4
j

= Size of 10th item

16 is just greater than 10 in c.f.
\ Q1 class is 20 - 30.
Then,

N – c.f.
f
Q1 =L+ 4 #h

= 20 + 10 – 8 # 10
8

= 20 + 20
8

= 22.5

Q3 class lies in, 3N th
4
= Size of ` j item

= Size of 30th item

38 is the just greater than 30 in c.f.
\ Q3 class is (40 - 50).

Then,

3N – c.f.
f #h
Q3 =L+ 4

= 40 + 30 – 28 # 10
10

= 42

PRIME Opt. Maths Book - X 323

Again,

Semi - interquartile range: (Quartile deviation)
1
= 2 ^Q3 – Q1h

= 1 ^42 – 22.5h
2

= 1 # 19.5
2

= 9.75

Coefficient of Semi-interquartile range (Coeff. of Q.D.)
Q3 – Q1
= Q3 + Q1

= 42 – 22.5
42 + 22.5

= 19.5
64.5

= 0.302

4. Find the inter - quartile range of :

Mid - value 6 18 30 42 54
14 10 6
f 8 12

Solution:

For the class interval,

Correction factor = 18 – 6 =6
2

Then, Lower limit = 6 – 6 = 0
Upper limit = 6 + 6 = 12

Class f c.f
8 8
0 - 12 12 8 + 12 = 20
12 - 24 14 20 + 14 = 34
24 - 36 10 34 + 10 = 44
36 - 48 6 44 + 6 = 50
48 - 60

N = 50

Now,

Q1 Class lies in,
N th
= Size of 4 item

= Size of 12.5th item.

20 is just greater than 12 . 5 in c.f.
\ Q1 class is (12 - 24) .

324 PRIME Opt. Maths Book - X

Then,

N – c.f.
f
Q1 =L+ 4 #h

= 12 + 12 . 5 – 8 # 12
12

= 16.5

Again,

Q3 Class lies in,
= Size of
` 3N th item
4
j

= Size of 37.5th item

44 is the just of greater than 37.5 in c.f.
\ Q3 class is (36 - 48).

Then, 3N – c.f.
Q3 f #h
=L+ 4
\
= 36 + 37 . 5 – 34 × 12
10

= 36 + 4.2

= 40.2 = Q3 – Q1
Inter - quartile range = 40.2 – 16.5
= 23.7

Exercise 8.1

1. Compute the followings from a continuous frequency distribution.
i) What is quartile deviation? Write down its calculating formula.
ii) If first and third quartiles are 5 and 25 respectively. Find the value of quartile
deviation and it's coefficient.
iii) If inter - quartile range and third quartile are respectively 20 and 30, find the first
quartile. Also find the quartile deviation and it's coefficient.
iv) If quartile deviation and coefficient of quartile deviation are respectively 10 and
0.25, find the value of first quartile and third quartile. Also find the inter - quartile
range.
v) If quartile deviation and first quartile are respectively 10 and 20, find the third
quartile. Also find the coefficient of quartile deviation.
vi) If coefficient of quartile deviation and third quartile are respectively 0.5 and 30, find
the first quartile. Also find semi - inter - quartile range.

2. Find the quartile deviation and it's coefficient from the followings.

i) Marks 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100
14
f 6 12 18 10

PRIME Opt. Maths Book - X 325

ii) Age 0 - 12 12 - 24 24 - 36 36 - 48 48 - 60 60 - 72 72 - 84

f 3 6 10 14 9 5 3

iii) Class 0 - 15 15 - 30 30 - 45 45 - 60 60 - 75 75 - 90

f 3 5 6 12 10 4

iv) Marks 0-5 5 - 10 10 - 15 15 - 20 20 - 25

f 35525

v) Class 0-8 0 - 16 0 - 24 0 - 32 0 - 40

f 7 17 25 33 36

3. PRIME more creative questions:
i) Find quartile deviation and its coefficient of.

Class 10 - 19 20 - 29 30 - 39 40 - 49 50 - 59
8 2
f 5 6 11

ii) Find quartile deviation and its coefficient of.

Mid - value 4 12 20 28 36
f 4 7982

iii) Find quartile deviation and its coefficient of.

Mid - value 8 20 32 44 56 68
f 2 5 6 12 10 5

iv) Find the quartile deviation by constructing frequency distribution table
taking a suitable class interval.
12, 48, 42, 7, 6, 34, 28, 30, 20, 24, 22, 37, 43, 17, 11, 18, 23, 26, 37, 32,

19, 29, 44, 33, 26, 5, 9, 13, 14, 31.
v) If first quatile of the observations is 32, find the quartile deviation.

Marks 2 – 14 14 – 26 26 – 38 38 – 50 50 – 62 62 – 74
f 2 5 6 m 10 5

1. i) Show to your teacher. Answer iii) 10, 10, 0.5

iv) 30, 50, 20 v) 40, 0.33 ii) 10, 0.67
vi) 10, 10
2. i) 21.5 ii) 12.9, 0.313 iii) 15.5, 0.307 iv) 6.5, 0.48 v) 8.2, 0.46
iii) 12, 0.36
3. i) 8.75, 0.263 ii) 7, 0.33 iv) 10 v) 12

326 PRIME Opt. Maths Book - X

8.2 Mean deviation:

Mean deviation is defined as the mean of the deviation of
the observations from the central values like arithmetic
mean, median and mode.

Here,
rxeasnpdecMtivd ealrye. the mean and median of the observations and deviation d = x – x and d = x – Md

Then formula of mean deviation can be written as,
/f
• M.D. from mean = x–x
N

• M.D. from median = /f x – Md
N

• Coefficient of M.D. (X) = M.D.
x

• Coefficient of M.D. (Md) = M.D.
Md

Steps for the calculation of mean deviation.

i) Mean deviation from mean. /fx
cNalled
• Finding of mean (x) of continuous frequency distribution. x = modulous).
• Finding absolute value of the deviation x – x (always positive
• Finding the f x – x and it's sum.
• Finding M.D. and it's coefficient as the formula.
/f x – x
M.D. (X) = N

Coefficient of M.D. (X) = M.D.
x

Model of the table

Class f Mid - value(x) fx x x–x f x–x

N = ∑fx = ∑f x – x =

ii) Mean deviation from median: class.
• Finding the value of median (Md) by calculating size of median
N
– c.f
M.D. = L + 2 f #h

• Finding absolute value of x – Md which is modulus (always positive)
• Finding f x – Md and it's sum.
• Finding mean deviation and it's coefficient using formula,
/f
M.D. = x – Md
N
M.D.
Coefficient of M.D. (Md) = Md

PRIME Opt. Maths Book - X 327

Model of the table

Class f cf Mid - value(x) Md x – Md f x – Md

N = ∑f x – Md =

Worked out Examples

1. Find mean deviation from mean and coefficient from the given data.

Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
9 8
f 5 7 11

Solution:

Marks f Mid - value (x) fx x x–x f x–x
22 110
0 - 10 5 5 25 12 84
2 22
10 - 20 7 15 105 8 72
20 - 30 11 25 275 27 18 144

30 - 40 9 35 315 ∑ x – x = 432

40 - 50 8 45 360

N = 40 ∑fx = 1080

Here, /fx
N
Mean (x) =

= 1080
40

= 27

Again, = /f x–x
M.D. (X) N

= 432
40

= 10.8 = M.D.
Coefficient of M.D. (X) x

= 10.8
27

= 0.4

328 PRIME Opt. Maths Book - X

2. Find mean deviation from median and it's coefficient of:

Class 0-8 8 - 16 16 - 24 24 - 32 32 - 40
11 8
f 7 9 15
Solution:

Class f c.f. Md mid- value (x) x – Md f x – Md
16.8 117.6
0-8 7 7 4 8.8 79.2
0.8 12
8 - 16 9 7 + 9 = 16 20.8 12 7.2 79.2
16 - 24 15 16 + 15 = 31 20 15.2 121.6
28
24 - 32 11 31 + 11 = 42 36 ∑f x – Md = 409.8

32 - 40 8 42 + 8 = 50

N = 50

Here,

Median value lies in;
N th
= Size of 2 item

= Size of a 50 th item
2
k

= Size of 25th item

\ 31 is the just greater value than 25 in c.f.
\ Median class = (16 - 24)

Then, N – c.f.
Median (Md) f
=L+ 2 #h

= 16 + 25 – 16 #8
15

= 16 + 4.8

= 20.8

Again,
Mean deviation from median,
/f
M.D. = x – Md
N

= 409.8
50

= 8.196

Coefficient of M.D. = M.D.
Md

= 8.196
20.8

= 0.394

PRIME Opt. Maths Book - X 329

3. Find the mean deviation from mean and it's coefficient of:

Marks 10 - 19 20 - 29 30 - 39 40 - 49 50 - 59

f 4593 2

Solution:
Let us convert the inclusive class into exclusive after finding correction factor.
Correction factor for continuous class interval

=

= 20 – 19
2

= 0.5

Marks f M.V.(x) fx x x–x f x–x
29.34 59.36
9.5 - 19.5 4 14.5 58 14.84 24.20
19.5 - 29.5 5 24.5 122.5 4.84 46.44
29.5 - 39.5 9 34.5 310.5 5.16 45.84
39.5 - 49.5 3 44.5 133.5 15.16 50.32
49.5 - 59.5 2 54.5 109 25.16
∑f x – x = 226.16
N = 25

Here, mean (x) = /fx
N

= 733.5
25

= 29.34

M.D.(X) = /f x–x
N

= 226.16
25

= 9.0464

Coefficient of M.D. = M.D.
x

= 9.0464
29.34

= 0.308

330 PRIME Opt. Maths Book - X

4. Find mean deviation from median and it's coefficient of:

Class less than 12 less than 24 less than36 less than48 less than 60

f3 9 19 22 24

Solution:

Class f c.f. Md M.V. (x) x – Md f x – Md
0 - 12 3 36 64.8
21.6 57.6
12 - 24 9 – 3 = 6 3 + 6 = 9 18 9.6 24
2.4 43.2
24 - 36 19 – 9 = 10 9 + 10 = 19 27.6 30 14.4 52.8
26.4
36 - 48 22 – 19 = 3 19 + 3 = 22 42

48 - 60 24 – 22 = 2 22 + 2 = 24 54

N = 24 ∑f x – Md = 242.4

Here,

Median class lies in
th
= Size of ` N item
2 j

= Size of 12th item

19 is just greater than 12 in c.f.
\ Median class is (24 - 36).

Where, L = 24, f = 10, C.f = 9, h = 12

Then,

Median (Md) =L+ N – c.f. #h
2 f

= 24 + 12 – 9 # 12
10

= 24 + 3.6
= 27.6

Again, = /f x – Md
M.D. (X) N

= 242.4
24

= 10.1

Coefficient of M.D. = M.D.
Md

= 10.1
27.6

= 0.3659

PRIME Opt. Maths Book - X 331

Exercise 8.2

1. Find the mean deviation from mean from the followings. Also find the coefficient of mean
deviation.
i) What is mean deviation? Write down its calculating formula from mean.
ii) If /f x – x = 300, /f = 40 , x = 20 in a continuous class interval data.
iii) If /f x – x = 800, N = 50, Mean = 24 in a continuous class interval frequency
distribution.
iv) Class 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 20 - 24

f 2 3 7 10 2 1

v) Marks 0-8 8 - 16 16 - 24 24 - 32 32 - 40

No. of students 5 9 12 8 6

vi) Age 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100

f 7 11 15 9 8

2. Find the mean deviation from median and it's coefficient from the followings.
i) /f x – Md = 600, ∑f = 25, median = 50 in a continuous class interval data.
ii) If /f x – Md = 800 where median lies in the class 30 - 40 of frequency 5. The
proceeding c.f. value is 18 out of 40 students.

iii) Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50

f 2 4 13 3 3

iv) Class 0 - 5 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30

f 248312

v) Marks 0 - 12 12 - 24 24 - 36 36 - 48 48 - 60

f 5 10 15 12 8

3. PRIME more creative questions:
i) If /f x – x = 600, ∑fx = 1200 out of 60 students, find the coefficient of mean
deviation in a continuous frequency distribution data.
ii) Compute mean deviation from median and it's coefficient of:

Class 10 - 18 20 - 28 30 - 38 40 - 48 50 - 58

f 5 9 12 10 4
iii) Find mean deviation from mean of the following.

Class 0 - 10 0 - 20 0 - 30 0 - 40 0 - 50

f 6 14 25 35 40

332 PRIME Opt. Maths Book - X

iv) Find mean deviation from mean of

Marks more than more than more than more than more than
20 40 60 80 100

f 50 42 31 17 7

v) Find mean deviation from median of the marks secured by 20 students taking a
suitable class interval.

12, 18, 28, 22, 8, 16, 42, 35, 32, 27, 40, 20, 15, 2, 39, 20, 24, 23, 34, 24.

1. i) Show to your teacher. Answer iii) 3.776, 0.325
iv) 16, 0.667 vi) 20, 0.4
ii) 7.5, 0.375 iii) 6.8, 0.272
2. i) 24, 0.48 v) 7.86, 0.38
iv) 4.75, 0.38 iii) 10, 0.4
ii) 20, 0.588
3. i) 0.5 v) 11.92, 0.3725
iv) 20.688
ii) 9.25, 0.272
v) 8

PRIME Opt. Maths Book - X 333

8.3 Standard deviation (Root Mean Square Deviation)

Standard deviation is defined as the square root of the
mean of the square of deviation of the observations from
their arithmetic mean.

Here,
x is the arithmetic mean and deviation d = x – x from mean, then the standard deviation is
formulated as the deviation:

S.D. (σ) = /f (x – x)2
N

Where,
• σ(sigma) is the symbol of standard deviation.

• Coefficient of standard deviation, = S.D = σ
x x

• Coefficient of variance is, C.V. = S.D. # 100% = σ × 100%
x x

• Formula to find the variance is, Variance = (S.D.)2 = (σ)2

• Standard deviation is the best method of the calculation of dispersion because all

the observation are involved and their deviation from the best central tendency

mean.

• It is the best method of the measures of dispersion for the research to establish the

factories, industries etc to analysis the data related to production of items and their

possibility of used in the locality.

• Here we discussed standard deviation for continuous frequency distribution data

only in Grade ‘x' according to National Curriculum Development Centre.

• Standard deviation is the square root of the mean of the square of deviation from

mean. Hence, it is called the root mean square deviation also.

Steps for finding standard deviation and format of the tabulation for different

ways. (Steps to be calculated standard deviation)

i) Using arithmetic mean: /fx
N
• Finding arithmetic mean ^xh = after finding the mid - value (x) of the classes.

• Finding deviation d = x – x and it's square ^x – xh2 and f ^x – xh2 .
• Format of table:

Class f Mid-value (x) fx x x – x (x – x)2 f ^x – xh2

N= ∑fx = Rfx = /f (x – x)2 =
N

Then, S.D. (σ) = /f^x – xh2
N

334 PRIME Opt. Maths Book - X

ii) Direct method : (Without calculating mean)
• Finding mid - value (x), fx, x2, and fx2 in table as,

Class f Mid - Value (x) fx x2 fx2

N = ∑fx = ∑fx2 =
d2 fd2
Here, S.D. = (σ) = /fx2 –a /fx 2
N N
k

iii) Deviation method:
• Finding deviation d = x – A where A is assumed mean.
• Finding fd, d2 and fd2 as the table:

Class f Mid - value (x) d = x – A fd

N= ∑fd = ∑fd2 =

Here, S.D. =

iv) Step - deviation method :

• Finding step - deviation d' = x –A where A is assume mean and ‘h' is height of the
h
class.
• Finding fd, d2 and fd2 as the table.

Class f Mid - Value (x) d' = x–A fd' d'2 fd'2
h

N= ∑fd' = ∑fd'2 =

Here, S.D. = h ×

PRIME Opt. Maths Book - X 335

Worked out Examples

1) Find the standard deviation using arithmetic mean.

Class 0 - 20 20 - 40 40 - 60 60 - 80 80 - 10
10
f 9 11 13 7

Solution:

Let us construct the table to compute S.D.(σ).

Class f Mid - fx x x–x x – x2 f(x – x )2
Value (x)
–38
0 - 20 9 10 90 –18 1444 12996
324 3564
20 - 40 11 30 330 2
40 - 60 13 50 650 48 22 4 52
42 484
60 - 80 10 70 700 1764 4840
80 - 100 7 90 630 12348
/f (x – x)2 = 33800
N = 50 ∑fx = 240

Here, /fx
N
Mean (x) =

= 2400
50

Then, = 48
S.D. (σ) Rf (x – x)2

=N

= 33800
50

= 676

\ (σ) = 26

2. Find the standard deviation and it's coefficient using direct method.

Class 0 - 12 0 - 24 0 - 36 0 - 48 0 - 60
25
f 2 8 16 21
fx2
Solution: 72
1944
Let us construct the table to compute S.D.(σ). 7200

Class f Mid - Value (x) fx x2
12 36
0 - 12 2 6 408 324
240 900
12 - 24 6 18

24 - 36 8 30

336 PRIME Opt. Maths Book - X

36 - 48 5 42 210 1764 8820
48 - 60 4 54 216 2916 11664
N =25 ∑fx = 786 ∑fx2 = 29700

Then, = /fx2 –a /fx 2
S.D. (σ) N N
k

=

=
= 199.5264
= 14.125

Mean (x) = /fx
N

= 786
25

= 31.44

Coefficient of S.D. = S.D.
x

= 14.125
31.44

= 0.449

3. Find the root mean square deviation by using step - deviation method. Also find C.V.

Class 20 - 29 30 - 39 40 - 49 50 - 59 60 - 69

f 7 9 14 11 9

Solution:

For continuous class interval, (Exclusive class)

Correction factor : 30 – 29 = 0.5
2
Then,

Class f Mid - Value (x) d' = x–A fd' d'2 fd'2
h
19.5 - 29.5 – 14 4 28
29.5 - 39.5 7 24.5 –2 –9 19
39.5 - 49.5 9 34.5 0 00
49.5 - 59.5 14 44.5 –1 11 1 11
59.5 - 69.5 11 54.5 18 4 36
9 64.5 0 ∑fd' = 6
N = 50 ∑fd'2 = 84
1

2

Here, A = 44.5
h = 10

PRIME Opt. Maths Book - X 337

Then.
S.D. = h ×

= 10 × 84 – a 6 2
50 50
k

= 10 × 1.68 – 0.0144

= 10 × 1.6656

= 12.905 /fd'
N
Mean (x) = A + # h = 44.5 + 6 # 10 = 45.7
50

C.V. = S.D. # 100% = 12.905 × 100% = 28.23%
x 45.7

4. If standard deviation of the observation having Σfx2 = 22250 out of 40 students, 12.5, find

the value of arithmetic mean of the continuous frequency distribution. Also find the C.V.

Solution:
In a continuous frequency distribution
∑fx2 = 22250

N = 40

S.D. = 12.5
x =?
We have,

S.D. =

or, 12.5 = 22250 – x 2
40

or, 156.25 = 556.25 – x 2
or, x 2 = 556.25 – 156.25
or, x 2 = 400

\ x = 20.

Again,

C.V. = S.D. # 100 %
x

= 12.5 # 100 %
20

= 62.5 %

338 PRIME Opt. Maths Book - X

Exercise 8.3

1. Compute the followings for the given data of continuous frequency distribution.
i) What is standard deviation? Write down its calculating formula by finding mean.
ii) If /f^x – xh2 = 450, & N = 50, find the standard deviation and coefficient of variance
where, ∑fx = 1000.
iii) If standard deviation of the data is 4 where, /f^x – xh2 = 480 , find the total number
of participants of the data.
iv) If ∑fx = 200, ∑fx2 = 3280 and ∑f = 20, find the value of standard deviation.
v) If N = 25, ∑fx2 = 6100 and standard deviation is 12, find the value of arithmetic
mean.
vi) If x = 12, ∑fx2 = 4880 and standard deviation is 10 of the data, find the total number
of participants.

2. Find the standard deviation of the followings using mean.

i) Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50

f 7 12 15 10 6

ii) Marks 0-8 8 - 16 16 - 24 24 - 32 32 - 40

f 4 8 14 8 6

iii) Age 0 - 12 12 - 24 24 - 36 36 - 48 48 - 60

f 6 8 11 10 5
iv) Marks obtained by the students taking an class interval 20 - 40.

56, 97, 7, 18, 92, 80, 48, 59, 40, 87, 75, 72, 46, 55, 15, 37, 38, 25, 58, 69, 70, 52, 18,
99, 74.
v) 22, 7, 6, 15, 20, 24, 35, 39, 16, 37, 21, 18, 12, 10, 28, 30, 20, 19, 8, 5, 12, 17, 23, 27, 38,
9, 11, 6, 21, 29.

3. Find the root mean square deviation short (direct) method of :

i) Class 0 - 10 10 – 20 20 – 30 30 – 40 40 – 50
10
f 5 10 20 5
24 – 30
ii) Marks 0–6 6 – 12 12 – 18 18 – 24 3

f 2465 26 – 30
3
iii) Age 10 – 14 14 – 18 18 – 22 22 – 26
50 – 59
f 4576 2

iv) Class 10 – 19 20 – 29 30 – 39 40 – 49

f 2376

v) 22, 7, 11, 14, 18, 24, 20, 15, 10, 6, 4, 2, 5, 13, 16, 12, 8, 14, 21, 16

PRIME Opt. Maths Book - X 339

4. PRIME more creative questions:
i) Find the standard deviation and it's coefficient using step - deviation method.

Class 10 - 18 20 - 28 30 - 38 40 - 48 50 - 58

f 7 9 13 11 10
ii) Find the root mean square deviation and coefficient of variation using deviation

method.

Marks less than 10 less than 20 less than 30 less than 40 less than 50

f 5 12 24 34 40

iii) Find the root mean square deviation and C.V. of using step - deviation method.

Marks 0 - 40 8 - 40 16 - 40 24 - 40 32 - 40

f 50 42 32 18 6

iv) Find standard deviation of the given age groups.

Age 1 – 9 11 – 19 21 – 29 31 – 39 41 – 49 51 – 59

f 246414

v) Find C.V. of the data given below.

Marks 0 – 10 0 – 20 0 – 30 0 – 40 0 – 50

f 5 13 25 34 40

5. Project work
i) Collects the marks obtained by the students of grade IX in first terminal examination
by taking class interval 10 and find the coefficient of variation of the marks.
ii) Marks obtained by the class IX students in compulsory maths and optional maths in
first unit test as follows.

Marks 0 – 5 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30

Number of C. Maths 5 16 13 7 5 4
students 2 9 1
O. Maths 7 12 19

a. Find the average mark obtained by the students in such subjects.
b. In which subject, the marks are uniformly distributed?
c. In which subject there is the betterment of mark?

1. i) Show to your teacher Answer iii) 30 iv) 8

ii) 3, 15%

v) 10 vi) 20

2. i) 12.139 ii) 9.43 iii) 14.94 iv) 26.33 v) 9.91
iii) 4.99
3. i) 12.2 ii) 7.168 ii) 12.285, 46.8% iv) 11.079 v) 6.24

4. i) 13.169, 0.37 iii) 9.986, 50.74%

iv) 13.09 v) 47.86%

340 PRIME Opt. Maths Book - X

Statistics

Unit Test Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. Write down the formula of coefficient of standard deviation and coefficient of variation.

2. a) If lower quartile and upper quartile of a continuous frequency distribution are 20
and 60 respectively, find the quartile deviation and its coefficient.

b) If ∑fx = 400 and ∑fx2 = 20000 in a continuous frequency distribution of 40 students,
find the value of standard deviation.

c) If ∑fx = 80 + a, ∑f|x – x| = 100 + 5a out of 20 students where mean deviation of the
continuous class interval is 10, find the value of arithmetic mean.

3. a) Find the quartile deviation of the observations.

Marks 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100
8 4
f 7 9 12
b) Find mean deviation from mean and its coefficient of: 24 – 32 32 – 40
5 3
Class 0–8 8 – 16 16 – 24

f 246

4. Find the coefficient of variation of the observations.

Marks 0 – 12 0 – 24 0 – 36 0 – 48 0 – 60
45 50
f 7 18 33

shared by :- R.P Pandey 341

PRIME Opt. Maths Book - X

Proposed Syllabus with Grid for
First Terminal Examination

S.N. Containts Topics K-1 U-2 A-4 HA-5 TQ TM Periods

1 Algebra i. Function 2 2 2 1 7 18 9
ii. Polynomial

2 Matrices All 22 2 – 6 14 10
3 Co-ordinate 23 2 1 9 21 10
i. Angle betn lines
Geometry ii. Homogeneous eqn 24 3 1 9 28 15
4 Trigonometry
i. Multiple angle
ii. Sub M angle
iii. Transformation

5 Transformation Combination of Tra. 1 1 1 1 4 12 6

6 Statistics Q.D., M.D. 111 – 37 6

Total Questions 10 13 11 4 38

Total Marks 10 26 44 20 100 56

K = Knowledge, U = Understanding, A = Application, HA = Higher ability

Model Question Set for First Terminal Examination

1. a. Group : A [10 × 1 = 10]
b. Define the term inverse function.
State remainder theorem.
2. a.
b. Prove that : Cos2A = 2Cos2A – 1.

3. a. Prove that : Cos40° + Cos20° = 3 Cos10°.
b. Write down the conditions of the two straight lines being perpendicular and parallel.
Write down the formula of calculating angle between the line pairs of homogeneous
4. a. equation px2 + 2hxy + qy2 = 0
In what condition inverse of a matrix can be defined? Also write down the formula
b. of A-1.
5. a. Write down the formula to find the variables x and y using Cramer's rule.
Find the single transformation ro2roirg1inw. here r1 is reflection about y-axis and r2
b. rotation about +90° with Centre is

Find 2 × 2 matrix represented by reflection about x - axis.

Group : B [13 × 2 = 26]

6. a. If f-1(x) = 3x – 2 , find f(x) and f(2).
b. 2

c. Prove that (x + 2) and (x – 5) are the factors of the polynomial x3 – 19x – 30.

7. a. 3 3 +1
b. If Cos330° = , prove that cos165° = –
2 2 2 P(4, 3)
8. a.
b. Prove that : = 2Cosq

Prove that : Cos20° – Cos40° – Cos80° = 0

If Sin A = 1 (a + 1 ) , find SinA. A(3, –2) Q B(–1, 4)
3 2 a

Prove that PQ is perpendicular to AB in the given diagram.

342 PRIME Opt. Maths Book - X

9. a. If the straight line joining the points (m, –3) and (7, 3) is parallel to the line 3x – 2y
b. + 10 = 0, find the value of 'm'.
c. Find the angle between the line pairs of 3 x2 – 4xy + 3 y2 = 0

10. a. If a matrix A = –3 7 is inverse of the matrix –5 7 , find the value of p and q.
b. < F < F
–2
c. 5 p q

If a matrix A = 2 –1 , find the determinant of A2 – 3A + 5I where I is 2 × 2 unit
< F
matrix. 3
–2

If A'(3, 7) is the image of a point A under a combined transformation of translation

about T = 1 and rotation about +270° with center origin, find the co-ordinate of A.
<F
2 –2 0
< F
What transformation is represented by the matrix ? Also find the image of
a point A[3, –2] under the matrix. 0 –2

Group : C [11 × 4 = 44]

11. If f(x) = 2x – 3 , g(x) = 3x + 7 and gof(x) = f-1(x), find the value of 'x'.
2

12. Solve the polynomial : x3 + 4x2 – 11x – 30 = 0
13. Solve the equations by matrix method of 3x – y = 7 and 5x + 2y = 19.

4 3 x 17 , find the value of x and y by using Cramer's rule.
14. If < F <F = <F
3 y
–1 3

15. Prove that q = Tan-1 8± m1 – m2 B as the angle between any two straight lines.
1 + m1 m2

16. Find the equation of straight line passes through a point (2, –3) which is perpendicular
to the straight line of equation 4x – 6y + 7 = 0
17. If the line pairs of 4x2 – 12xy + (k + 3)y2 = 0 are coincident to each other, find the value of
'k'. Also find the separate equations.

18. Prove that : Sin2q – Cos2qcos2b = sin2b – Cos2b.Cos2q

19. Prove that : Sin20°.Sin40°.Sin60°.Sin80° = 3
16

20. If 2Tana = 3Tanb, prove that : Tan(a – b) = Sin2b
5 – Cos2b

21. Find the 2 × 2 transformation matrix which transform the unit square to a parallelogram

0132
having vertices < F
0 3 7 4

Group : D [4 × 5 = 20]
22. If x3 – 8x2 + (2p + 1)x + 20 has a factor (x – 4), find the value of p. Also find the other

factors of it.

23. Prove that : Tanq + 2Tan2q + 4Tan4q + 8Cot8q = Cotq
24. Find the equation of straight line passes through a point (1, –2) which makes 135° angle

with the line 3x – 4y + 5 = 0?
25. Transform a triangle having vertices A(1, 3), B(–1, 1) and C(3, –4) under reflection about

x + 2 = 0 followed by rotation about –90° with centre origin. Also plot the object and
image in graph.

PRIME Opt. Maths Book - X 343

Proposed Syllabus with Grid for
Second Terminal Examination

S.N. Containts Topics K-1 U-2 A-4 HA-5 TQ TM Periods

1 Algebra i. Sequence & Series 23 2 1 8 21 18
ii. Graph and L.P.

2 Limit and All 1– 1 – 25 8
Continuity

3 Matrices All 121 – 49 6

4 Co-ordinate i. Conic Section 22 1 1 6 15 10
Geometry – 8 20 18
ii. Circle
5 Trigonometry
i. Conditional 23 3

ii. Tri. Equation
iii. Heights & distance

6 Transformation Transformation- 1 – 1 1 3 10 6
Matrix

7 Vector All 1 2 – 1 4 10 6

8 Statistics Standard Deviation – 1 2 – 3 10 8

9 First Term Review 6

Total Questions 10 13 11 4 38

Total Marks 10 26 44 20 100 86

K = Knowledge, U = Understanding, A = Application, HA = Higher ability

Model Question Set for Second Terminal Examination

1. a. Group : A [10 × 1 = 10]
b. If pre-image of f(x) = x + 1 is 5, find the element of image.
What do you mean by zero of the polynomial?
2. a. Write down the sequence 0.9, 0.99, 0.999, 0.9999, ............... in the limit notation.
b. What is singular matrix?
What is the slope of a straight line which is parallel to the another straight line of
3. a. 2
slope –3 ?
b.
4. a. Write down the formula of equation of circle having ends of diameter (x1, y1) and
(x2, y2). 2SinA.
b. Prove that : 1 – Cos2A =
5. a. SinA

b. If 2SinA – 1 = 0, find A where 0 ≤ A ≤ 90°.
Write down the conditions of any two vectors a and b being perpendicular and

parallel.

If translations T1 = <1F and T2 = <–3F , what will be the combined translation T2oT1?
2 1

344 PRIME Opt. Maths Book - X

2x + 1 Group : B [13 × 2 = 26]
2
6. a. If f(x) = , prove that fof-1 is an identity function.
b.
c. Is 1536 a term of the sequence 3, 6, 12, 24, ...........................?

7. a. Find the vertex of parabola of the quadratic equation x2 – 4x – 12 = 0.

b. 3 –1 AA-1
8. a. If a matrix A = < F , prove that is an unit matrix.
–4
b. 1

9. a. 2 m and = 25, find the value of m.
b. If A = < F A2
c. –1
1
10. a. Find the separate equation of x2 – 2xyCot2a – y2 = 0

b. Find the equation of circle having radius 4 units which O
c. touches x-axis at point (3, 0)

Prove that : Cot 22 1c – Tan 22 1c = 2 a ?c
2 2 B

If CosA + CosB = 1 , SinA + SinB = 1 , find Tan A+B .
3 4 2

Solve (0 ≤ q ≤ p) : Sin2q + Sinq = 0 A P
If a + b + c = 0 where a, b and c are the unit vectors, find the
angle between the vectors b and c .

Find OP in terms of a and b from the adjoining diagram where 3AP = 2PB .

If coefficient of quartile deviation of the continuous class interval is 0.5 where upper
quartile is 30, find the lower quartile.

Group : C [11 × 4 = 44]
11. Solve the polynomial y = x3 – 19x – 20 and y = 10.

12. How many AMs are there between 124 and 76 where the first mean and last mean are in
the ration 3:2?

13. Examine the function f(x) = 2 for x ≤ 0 for continuous or discontinuous by showing
in graph. for x ≥ 0
2

14. Solve the equations 3x – 2y = 8 and 5x – 3y = 14 by using Cramer's rule.

15. Find the equation of straight line passes through the centroid of DABC having vertices
A(2, 5), B(–2, –4) and C(4, 4) which is perpendicular to the side BC.

16. Prove that : Cos3qCos3q + Sin3q.Sin3q = Cos32q.

17. If A, B and C are the angles of a triangle,

Prove that : Sin2 A + Sin2 B + Sin2 C = 1 – 2Sin A Sin B Sin C
2 2 2 2 2 2

18. The angle of elevations of the top of two poles from mid-way between their bases which
are 120m apart are found to be complementary where one pole is double than the other.
Find the height of the taller pole.

19. Find 2 × 2 transformation matrix which transform the points A(1, –2) → A'(4, 7) and B(3,
1) → B'(5, 7).

20. Find mean deviation from median and its coefficient of the observations given below.

Marks 0-12 12-24 24-36 36-48 48-60

f 5 7 12 9 7

PRIME Opt. Maths Book - X 345