Activity 2

OBJECTIVE MATERIAL REQUIRED

To verify that the relation R in the set A piece of plywood, some pieces of
L of all lines in a plane, defined by wire (8), plywood, nails, white paper,
R = {( l, m) : l || m} is an equivalence glue.
relation.

METHOD OF CONSTRUCTION
Take a piece of plywood of convenient size and paste a white paper on it. Fix
the wires randomly on the plywood with the help of nails such that some of
them are parallel, some are perpendicular to each other and some are inclined
as shown in Fig. 2.

DEMONSTRATION
1. Let the wires represent the lines l1, l2, ..., l8.
2. l1 is perpendicular to each of the lines l2, l3, l4 (see Fig. 2).

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 3. l6 is perpendicular to l7.
4. l2 is parallel to l3, l3 is parallel to l4 and l5 is parallel to l8.
5. (l2, l3), (l3, l4), (l5, l8), ∈ R

OBSERVATION
1. In Fig. 2, every line is parallel to itself. So the relation R = {( l, m) : l || m}
.... reflexive relation (is/is not)
2. In Fig. 2, observe that l2
l3 . Is l3 ... l2? (|| / || )
So, (l2, l3) ∈ R ⇒ (l3, l2) ... R (∉/∈)
Similarly, l3 || l4. Is l4 ...l3? (|| / || )
So, (l3, l4) ∈ R ⇒ (l4, l3) ... R (∉/∈)
and (l5, l8) ∈ R ⇒ (l8, l5) ... R (∉/∈)

∴ The relation R ... symmetric relation (is/is not)

3. In Fig. 2, observe that l2 || l3 and l3 || l4. Is l2 ... l4 ? (|| / || )
So, (l2, l3) ∈ R and (l3, l4) ∈ R ⇒ (l2, l4) ... R (∈/∉)

Similarly, l3 || l4 and l4 || l2. Is l3 ... l2 ? (|| / || )

So, (l3, l4) ∈ R, (l4, l2) ∈ R ⇒ (l3, l2) ... R (∈,∉)
Thus, the relation R ... transitive relation (is/is not)
Hence, the relation R is reflexive, symmetric and transitive. So, R is an
equivalence relation.

APPLICATION NOTE
This activity is useful in understanding the This activity can be repeated
concept of an equivalence relation. by taking some more wires
in different positions.

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Activity 3
OBJECTIVE MATERIAL REQUIRED
To demonstrate a function which is Cardboard, nails, strings, adhesive
not one-one but is onto. and plastic strips.

METHOD OF CONSTRUCTION
1. Paste a plastic strip on the left hand side of the cardboard and fix three nails
on it as shown in the Fig.3.1. Name the nails on the strip as 1, 2 and 3.
2. Paste another strip on the right hand side of the cardboard and fix two nails in
the plastic strip as shown in Fig.3.2. Name the nails on the strip as a and b.
3. Join nails on the left strip to the nails on the right strip as shown in Fig. 3.3.

DEMONSTRATION
1. Take the set X = {1, 2, 3}
2. Take the set Y = {a, b}
3. Join (correspondence) elements of X to the elements of Y as shown in Fig. 3.3

OBSERVATION
1. The image of the element 1 of X in Y is __________.
The image of the element 2 of X in Y is __________.

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 The image of the element 3 of X in Y is __________.

So, Fig. 3.3 represents a __________ .

2. Every element in X has a _________ image in Y. So, the function is
_________(one-one/not one-one).
3. The pre-image of each element of Y in X _________ (exists/does not exist).
So, the function is ________ (onto/not onto).

APPLICATION NOTE
Demonstrate the same
This activity can be used to demonstrate the
activity by changing the
concept of one-one and onto function.
number of the elements of
the sets X and Y.

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Activity 9
OBJECTIVE MATERIAL REQUIRED
To find analytically the limit of a Paper, pencil, calculator.
function f (x) at x = c and also to check
the continuity of the function at that
point.

METHOD OF CONSTRUCTION

 x 2 – 16 
 , x ≠ 4
1. Consider the function given by f ( x ) =  x – 4 
 10, x = 4 


2. Take some points on the left and some points on the right side of c (= 4)
which are very near to c.
3. Find the corresponding values of f (x) for each of the points considered in
step 2 above.
4. Record the values of points on the left and right side of c as x and the
corresponding values of f (x) in a form of a table.

DEMONSTRATION
1. The values of x and f (x) are recorded as follows:

Table 1 : For points on the left of c (= 4).

x 3.9 3.99 3.999 3.9999 3.99999 3.999999 3.9999999

f (x) 7.9 7.99 7.999 7.9999 7.99999 7.999999 7.9999999

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 2. Table 2: For points on the right of c (= 4).

x 4.1 4.01 4.001 4.0001 4.00001 4.000001 4.0000001

f (x) 8.1 8.01 8.001 8.0001 8.00001 8.000001 8.0000001

OBSERVATION
1. The value of f (x) is approaching to ________, as x → 4 from the left.
2. The value of f (x) is approaching to ________, as x → 4 from the right.

3. So, lim f ( x ) = ________ and lim+ f ( x ) = ________.

x→4 x →4

4. Therefore, lim f ( x ) = ________ , f (4) = ________.

x →4

5. Is lim f ( x ) = f (4) ________ ? (Yes/No)

x →4

6. Since f ( c ) ≠ lim f ( x ) , so, the function is ________ at x = 4 (continuous/

x →c
not continuous).

APPLICATION
This activity is useful in understanding the concept of limit and continuity of a
function at a point.

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Activity 13
OBJECTIVE MATERIAL REQUIRED
To understand the concepts of Pieces of wire of different lengths,
decreasing and increasing functions. piece of plywood of suitable size,
white paper, adhesive, geometry
box, trigonometric tables.
METHOD OF CONSTRUCTION
1. Take a piece of plywood of a convenient size and paste a white paper on it.
2. Take two pieces of wires of length say 20 cm each and fix them on the white
paper to represent x-axis and y-axis.
3. Take two more pieces of wire each of suitable length and bend them in the
shape of curves representing two functions and fix them on the paper as
shown in the Fig. 13.

4. Take two straight wires each of suitable length for the purpose of showing
tangents to the curves at different points on them.

DEMONSTRATION
1. Take one straight wire and place it on the curve (on the left) such that it is

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 tangent to the curve at the point say P1 and making an angle α1 with the
positive direction of x-axis.
2. α1 is an obtuse angle, so tanα1 is negative, i.e., the slope of the tangent at P1
(derivative of the function at P1) is negative.
3. Take another two points say P2 and P3 on the same curve, and make tangents,
using the same wire, at P2 and P3 making angles α2 and α3, respectively with
the positive direction of x-axis.
4. Here again α2 and α3 are obtuse angles and therefore slopes of the tangents
tan α2 and tan α3 are both negative, i.e., derivatives of the function at P2 and
P3 are negative.
5. The function given by the curve (on the left) is a decreasing function.
6. On the curve (on the right), take three point Q1, Q2, Q3, and using the other
straight wires, form tangents at each of these points making angles β1, β2,
β3, respectively with the positive direction of x-axis, as shown in the figure.
β1, β2, β3 are all acute angles.
So, the derivatives of the function at these points are positive. Thus, the
function given by this curve (on the right) is an increasing function.

OBSERVATION
1. α1 = _______ , > 90° α2 = _______ > _______, α3 = _______> _______,
tan α1 = _______, (negative) tan α2 = _______, ( _______ ), tan α3 =
_______, ( _______). Thus the function is _______.
2. β1 = _______< 90°, β2 = _______, < _______, β3 = _______ , < _______

tan β1 = _______ , (positive), tan β2 = _______, ( _______ ), tan β3 =

_______( _______ ). Thus, the function is _______.

APPLICATION
This activity may be useful in explaining the concepts of decreasing and
increasing functions.

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Activity 14
OBJECTIVE MATERIAL REQUIRED
To understand the concepts of local A piece of plywood, wires,
maxima, local minima and point of adhesive, white paper.
inflection.

METHOD OF CONSTRUCTION
1. Take a piece of plywood of a convenient size and paste a white paper on it.
2. Take two pieces of wires each of length 40 cm and fix them on the paper on
plywood in the form of x-axis and y-axis.
3. Take another wire of suitable length and bend it in the shape of curve. Fix
this curved wire on the white paper pasted on plywood, as shown in Fig. 14.

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 4. Take five more wires each of length say 2 cm and fix them at the points A, C,
B, P and D as shown in figure.

DEMONSTRATION
1. In the figure, wires at the points A, B, C and D represent tangents to the
curve and are parallel to the axis. The slopes of tangents at these points are
zero, i.e., the value of the first derivative at these points is zero. The tangent
at P intersects the curve.
2. At the points A and B, sign of the first derivative changes from negative to
positive. So, they are the points of local minima.
3. At the point C and D, sign of the first derivative changes from positive to
negative. So, they are the points of local maxima.
4. At the point P, sign of first derivative does not change. So, it is a point of
inflection.

OBSERVATION
1. Sign of the slope of the tangent (first derivative) at a point on the curve to
the immediate left of A is _______.
2. Sign of the slope of the tangent (first derivative) at a point on the curve to
the immediate right of A is_______.
3. Sign of the first derivative at a point on the curve to immediate left
of B is _______.
4. Sign of the first derivative at a point on the curve to immediate right
of B is _______.
5. Sign of the first derivative at a point on the curve to immediate left
of C is _______.
6. Sign of the first derivative at a point on the curve to immediate right
of C is _______.
7. Sign of the first derivative at a point on the curve to immediate left
of D is _______.

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 8. Sign of the first derivative at a point on the curve to immediate right
of D is _______.
9. Sign of the first derivative at a point immediate left of P is _______ and
immediate right of P is_______.
10. A and B are points of local _______.
11. C and D are points of local _______.
12. P is a point of _______.

APPLICATION
1. This activity may help in explaining the concepts of points of local maxima,
local minima and inflection.
2. The concepts of maxima/minima are useful in problems of daily life such
as making of packages of maximum capacity at minimum cost.

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Activity 18
OBJECTIVE MATERIAL REQUIRED
To verify that amongst all the rect- Chart paper, paper cutter, scale,
angles of the same perimeter, the pencil, eraser cardboard, glue.
square has the maximum area.

METHOD OF CONSTRUCTION
1. Take a cardboard of a convenient size and paste a white paper on it.
2. Make rectangles each of perimeter say 48 cm on a chart paper. Rectangles
of different dimensions are as follows:

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 R1 : 16 cm × 8 cm, R2 : 15 cm × 9 cm

R3 : 14 cm × 10 cm, R4 : 13 cm × 11 cm

R5 : 12 cm × 12 cm, R6 : 12.5 cm × 11.5 cm

R7 : 10.5 cm × 13.5 cm
3. Cut out these rectangles and paste them on the white paper on the cardboard
(see Fig. 18 (i) to (vii)).
4. Repeat step 2 for more rectangles of different dimensions each having
perimeter 48 cm.
5. Paste these rectangles on cardboard.

DEMONSTRATION
1. Area of rectangle of R1 = 16 cm × 8 cm = 128 cm2

Area of rectangle R2 = 15 cm × 9 cm = 135 cm2

Area of R3 = 140 cm2

Area of R4 = 143 cm2

Area of R5 = 144 cm2

Area of R6 = 143.75 cm2

Area of R7 = 141.75 cm2

2. Perimeter of each rectangle is same but their area are different. Area of
rectangle R5 is the maximum. It is a square of side 12 cm. This can be verified
using theoretical description given in the note.

OBSERVATION
1. Perimeter of each rectangle R1, R2, R3, R4, R4, R6, R7 is _________.
2. Area of the rectangle R3 ________ than the area of rectangle R5.

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 3. Area of the rectangle R6 _______ than the area of rectangle R5.
4. The rectangle R5 has the diamensions ______ × ______ and hence it is a
________.
5. Of all the rectangles with same perimeter, the ________ has the maximum
area.

APPLICATION
This activity is useful in explaining the idea
of Maximum of a function. The result is also
useful in preparing economical packages.
NOTE

Let the length and breadth of rectangle be x and y.

The perimeter of the rectangle P = 48 cm.
2 (x + y) = 48
or x + y = 24 or y = 24 – x
Let A (x) be the area of rectangle, then
A (x) = xy
= x (24 – x)
= 24x – x2
A′ (x) = 24 – 2x
A′ (x) = ⇒ 24 – 2x = 0 ⇒ x = 12
A′′ (x) = – 2
A′′ (12) = – 2, which is negative
Therefore, area is maximum when x = 12
y = x = 24 – 12 = 12
So, x = y = 12
Hence, amongst all rectangles, the square has the maximum area.

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Activity 19
OBJECTIVE MATERIAL REQUIRED
To evaluate the definite integral Cardboard, white paper, scale,
b pencil, graph paper
∫a (1 − x 2 ) dx as the limit of a sum and
verify it by actual integration.

METHOD OF CONSTRUCTION
1. Take a cardboard of a convenient size and paste a white paper on it.
2. Draw two perpendicular lines to represent coordinate axes XOX′ and YOY′.
3. Draw a quadrant of a circle with O as centre and radius 1 unit (10 cm) as
shown in Fig.19.

The curve in the 1st quadrant represents the graph of the function 1 − x 2 in the
interval [0, 1].

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DEMONSTRATION
1. Let origin O be denoted by P0 and the points where the curve meets the
x-axis and y-axis be denoted by P10 and Q, respectively.
2. Divide P0P10 into 10 equal parts with points of division as, P1, P2, P3, ..., P9.
3. From each of the points, Pi , i = 1, 2, ..., 9 draw perpendiculars on the x-axis
to meet the curve at the points, Q1, Q2, Q3 ,..., Q9. Measure the lengths of
P0Q0, P1 Q1, ..., P9Q9 and call them as y0, y1 , ..., y9 whereas width of each part,
P0P1, P1P2, ..., is 0.1 units.
4. y0 = P0Q0 = 1 units
y1 = P1Q1 = 0.99 units
y2 = P2Q2 = 0.97 units
y3 = P3Q3 = 0.95 units
y4 = P4Q4 = 0.92 units
y5 = P5Q5 = 0.87 units
y6 = P6Q6 = 0.8 units
y7 = P7Q7 = 0.71 units
y8 = P8Q8 = 0.6 units
y9 = P9Q9 = 0.43 units
y10 = P10Q10 = which is very small near to 0.

5. Area of the quadrant of the circle (area bounded by the curve and the two
axis) = sum of the areas of trapeziums.

(1 + 0.99 ) + ( 0.99 + 0.97 ) + ( 0.97 + 0.95 ) + ( 0.95 + 0.92 ) 

1  
= × 0.1  + (0.92 + 0.87) + (0.87 + 0.8) + (0.8 + 0.71) + (0.71+ 0.6) 
2  + (0.6 + 0.43) + (0.43) 
 
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 = 0.1 [0.5 + 0.99 + 0.97 + 0.95 + 0.92 + 0.87 + 0.80 + 0.71 + 0.60 + 0.43]

= 0.1 × 7.74 = 0.774 sq. units.(approx.)

1
6. Definite integral = ∫0 1 – x 2 dx

1
 x 1 – x 2 1 −1  1 π 3.14
= + sin x  = × = = 0.785sq.units
 2 2  0 2 2 4

Thus, the area of the quadrant as a limit of a sum is nearly the same as area
obtained by actual integration.

OBSERVATION
1. Function representing the arc of the quadrant of the circle is y = ______.
1

2. Area of the quadrant of a circle with radius 1 unit = ∫ 1– x 2 dx = ________.

0

sq. units
3. Area of the quadrant as a limit of a sum = _______ sq. units.
4. The two areas are nearly _________.

APPLICATION
This activity can be used to demonstrate the
concept of area bounded by a curve. This
activity can also be applied to find the
approximate value of π.
NOTE
Demonstrate the same activity
by drawing the circle x2 + y2 = 9
and find the area between x = 1
and x = 2.

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Activity 20
OBJECTIVE MATERIAL REQUIRED
To verify geometrically that Geometry box, cardboard, white
paper, cutter, sketch pen, cellotape.
( )








c× a + b = c× a + c×b

METHOD OF CONSTRUCTION
1. Fix a white paper on the cardboard.
2. Draw a line segment OA (= 6 cm, say) and let it represent c .

3. Draw another line segment OB (= 4 cm, say) at an angle (say 60°) with OA.



Let OB = a

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4. Draw BC (= 3 cm, say) making an angle (say 30°) with OA . Let BC = b
5. Draw perpendiculars BM, CL and BN.
6. Complete parallelograms OAPC, OAQB and BQPC.

DEMONSTRATION






1. OC = OB +BC = a + b , and let ∠COA = α .

( )







2. c × a + b = c a + b sin α = area of parallelogram OAPC.



3. c × a = area of parallelogram OAQB.



4. c × b = area of parallelogram BQPC.

5. Area of parallelogram OAPC = (OA) (CL)

= (OA) (LN + NC) = (OA) (BM + NC)

= (OA) (BM) + (OA) (NC)

= Area of parallelogram OAQB + Area of parallelogram BQPC




= c+a + c × b

So, c × ( a + b ) = c × b + c × b

Direction of each of these vectors c × (a + b ), c × a and c × b is perpendicular
to the same plane.








So, c × (a + b ) = c × a + c × b.

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OBSERVATION



c = OA = OA = _______



a + b = OC = OC = ______

CL = ______




c × ( a + b) = Area of parallelogram OAPC

= (OA) (CL) = _____________ sq. units (i)

c × a = Area of parallelogram OAQB

= (OA) (BM) = _____ × _____ = ______ (ii)

c × b = Area of parallelogram BQPC

= (OA) (CN) = _____ × _____ = ______ (iii)

From (i), (ii) and (iii),

Area of parallelogram OAPC = Area of parallelgram OAQB + Area of

Parallelgram ________.





Thus c × (a + b | = c × a + c × b

( )








c × a, c × b and c × a + b are all in the direction of _______ to the plane
of paper.

( )






Therefore c × a + b = c × a + ________.

Mathematics 155

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APPLICATION
Through the activity, distributive property of vector multiplication over addition
can be explained.

NOTE
This activity can also be per-
formed by taking rectangles
instead of parallelograms.

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Activity 21
OBJECTIVE MATERIAL REQUIRED
To verify that angle in a semi-circle is Cardboard, white paper, adhesive,
a right angle, using vector method. pens, geometry box, eraser, wires,
paper arrow heads.
METHOD OF CONSTRUCTION
1. Take a thick cardboard of size 30 cm × 30 cm.
2. On the cardboard, paste a white paper of the same size using an adhesive.
3. On this paper draw a circle, with centre O and radius 10 cm.

4. Fix nails at the points O, A, B, P and Q. Join OP, OA, OB, AP, AQ, BQ, OQ
and BP using wires.
5. Put arrows on OA, OB, OP, AP, BP, OQ, AQ and BQ to show them as vectors,
using paper arrow heads, as shown in the figure.

DEMONSTRATION







1. Using a protractor, measure the angle between the vectors AP and BP , i.e.,
∠ APB = 90°.

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2. Similarly, the angle between the vectors AQ and BQ , i.e., ∠ AQB = 90°.
3. Repeat the above process by taking some more points R, S, T, ... on the
semi-circles, forming vectors AR, BR; AS, BS; AT, BT; ..., etc., i.e., angle
formed between two vectors in a semi-circle is a right angle.

OBSERVATION
By actual measurement.













OP = OA = OB = OQ = r = a = p = _______ ,









AP = _______ , BP = _______, AB = ______







AQ = _______ , BQ = _______

 2


 2


 2


 2

AP + BP = ________, AQ + BQ = ________







So, ∠APB = ________ and AP.BP ________ ∠AQB = ________ and







AQ.BP = ________
Similarly, for points R, S, T, ________
∠ARB = ________, ∠ASB = ________, ∠ATB = ________, ________
i.e., angle in a semi-circle is a right angle.
APPLICATION
This activity can be used to explain the
concepts of

(i) opposite vectors

(ii) vectors of equal magnitude

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(iii) perpendicular vectors
(iv) Dot product of two vectors.
NOTE

Let OA = OB = a = OP = p



 


 


 
OA = – a , OB = a , OP = p









  


  
AP = – OA + OP = a + p ., BP = p – a .






      2  2
AP. BP =( p + a ) .( p – a ) = p – a = 0
.

( 2 
since p = a )




So, the angle APB between the vectors AP and




BP is a right angle.








Similarly, AQ. BQ = 0 , so, ∠AQB = 90° and so on.

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Activity 27
OBJECTIVE MATERIAL REQUIRED
To explain the computation of A piece of plywood, white paper
conditional probability of a given pen/pencil, scale, a pair of dice.
event A, when event B has already
occurred, through an example of
throwing a pair of dice.

METHOD OF CONSTRUCTION
1. Paste a white paper on a piece of plywood of a convenient size.
2. Make a square and divide it into 36 unit squares of size 1cm each
(see Fig. 27).
3. Write pair of numbers as shown in the figure.

Fig. 27

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DEMONSTRATION
1. Fig. 27 gives all possible outcomes of the given experiment. Hence, it
represents the sample space of the experiment.
2. Suppose we have to find the conditional probability of an event A if an event
B has already occurred, where A is the event “a number 4 appears on both
the dice” and B is the event "4 has appeared on at least one of the dice”i.e,
we have to find P(A | B).
3. From Fig. 27 number of outcomes favourable to A = 1
Number of outcomes favourable to B = 11
Number of outcomes favourable to A ∩ B = 1.
NOTE
11
4. (i) P (B) = , 1. You may repeat this activity by
36
taking more events such as the
probability of getting a sum 10 when
1 a doublet has already occurred.
(ii) P (A ∩ Β) =
36 2. Conditional probability
P (A | B) can also be found by first
P(A ∩ B) 1 taking the sample space of event B
(iii) P (A | B) = = . out of the sample space of the
P(B) 11 experiment, and then finding the
probability A from it.
OBSERVATION
1. Outcome(s) favourable to A : _________, n (A) = _________.
2. Outcomes favourable to B : _________, n (B) = _________.
3. Outcomes favourable to A ∩ B : _________, n (A ∩ B) = _________.
4. P (A ∩ B) = _________.
5. P (A | B) = _________ = _________.

APPLICATION
This activity is helpful in understanding the concept of conditional probability,
which is further used in Bayes’ theorem.

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