COMPLETE SOLUTION-2076 OPT

COMPLETE SOLUTION BY TARA BAHADUR MAGAR

Optional-I (Mathematics)

Time: 3 hrs. F.M.: 100

Attempt all the questions:
Group-A (10  1 = 10)

1. (a) Name trigonometric function which has the following graph.

(b) State remainder theorem.
2. (a) Write down the notational condition for the existence of continuity of a function f (x) at

a point x = a.
(b) Define singular matrix.

3. (a) If  be the angles between the pair of straight lines represented by ax2 + 2hxy + by2 = 0;

a, b  0, what is the value of tan?
(b) Which geometric figure is formed when the intersecting plane is parallel to the generator

of right circular cone?
4. (a) Write down the formula of sin3A.

(b) Express 2sinA.sinB in terms of sum or difference of cosines.

5. (a) If a and b are two vectors and be angle between them then what will be the value of

cos ?

(b) Write down the coordinates of the inversion point P (x, y) of any point P (x, y) with

respect to the circle (x – h) 2 + (y – k) 2 = r2.

Group-B (13 × 2 = 26)

6. (a) If (x + 3) is a factor of x3 – (k – 1) x2 + kx + 54, find the value of k.

(b) Find the values of x, y and z from the given geometric sequence: 1 , x, y, z, 2
8

(b) Find the vertex of the parabola y = x2 – 2x – 3.

7. (a) If A =  4  3  then prove that A – 1 = A.
5  4

(b) According to Cramer’s rule, find the value of D, D1 and x from the linear equations 3x

– y = 7 and x + 2y = 4.

8. (a) For what value of p, line px – 2y + 1 = 0 is perpendicular to the line joining the points (1,

3) and (p, 2)?

(b) Find the equation of a circle with centre (2, 5) and touching x-axis.

9. (a) Prove that: sin 3   cos 3   1 1 sin
22 2

sin   cos 
22

(b) Find the value of cos1300 + cos1100 + cos100

(c) Solve: 4sin25A – 4 sin5A + 1 = 0 (00  A  900)

10. (a) If OA   5  , OB   4  and AOB  90 0 , find the value of m.
m -2

 (b) AD is the median of triangle ABC. If the position vectors of A and D are 3i  5j
 and 3i - 4j respectively, find the position vector of its centroid G.

COMPLETE SOLUTION BY TARA BAHADUR MAGAR

(c) In a continuous series, the lower quartile is 12 and the upper quartile exceeds twice the

lower quartile by 4. Find the quartile deviation and its coefficient.

Group-C (11 × 4 = 44)

11. Given that f (x) = 3x 11 , x  3 & g (x) = x - 3 . If f (x) = g – 1 (x), find the value of x.
x-3 2

12. If the third and eleventh terms of an arithmetic series are 8 and -8 respectively, find the

sum of the first seven terms of the series.

13. For a real valued function f (x) = 4x +1.

(i) What are the values of f (x) at x = 2.9, 2.99 and 2.999?

(ii) What are the values of f (x) at x = 3.1 and 3.01 and 3.001?

(iii) What are the values of lim f (x), lim f (x) and f (2)?
x3 x3

(iv) Is this function continuous at x = 3?

14. Solve by matrix method: x  y  2  y  x  3  6
32

15. Find the single equation of the pair of straight lines passing through the origin and

perpendicular to the lines represented by x2 + xy – 2y2 = 0.

16. Calculate the value of cos2A + cos2 (A + 1200) + cos2 (A-1200)

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

sinA.cosA + sinB.cosB + sinC.cosC = 2sinA.sinB.sinC

18. A vertical pole is divided by any point in the ratio 9:1. If the both of the segment of a

pole subtend equal angles to each other at a distance 20 m away from the foot of the

pole, find the height of pole.

19. The coordinates of vertices of a ABCD are A (1, 1), B (2, 3), C (4, 2) and D (3, 0). Find
the co- ordinates of the vertices of image of a parallelogram ABCD under the rotation

through +900 about origin followed by enlargement E [(0, 0), 3]. Also represents both

object and image on the same graph paper

20. The following table shows the heights of 100 students of class-X.

Calculate the mean deviation and its coefficient of the data from median.

Height (in cm) 95-105 105-115 115-125 125-135 135-145 145-155

No. of students 9 13 25 30 13 10

21. The age of 30 boys of a school is given below.

Age (in years) 0-4 4-8 8-12 12-16 16-20
No. of boys 4 6 10 6 4

(i) What is the average age of the boys?

(ii) What is the standard deviation of the data?

Group-D (4 × 5 = 20)

22. Write one importance of linear programming. Optimize the objective function P = 3x + 2y

subject to the following constraints: x + y ≤ 5, x + 4y ≤ 8, x ≥ 0, y ≥ 0

23. In the figure given alongside, the circles X and Yare interested at

A and B such that the chord AB of the circle Y is the diameter of

the circle X. If the equation of circle Y is x2 + y2 = 1and chord AB

is x + y = 1, find the equation of circle X.

24. In a rhombus BEST, prove by vector method that the diagonals BS and ET bisect to

each other at a right angle.

25. Find the 22 transformation matrix which maps a square PQRS with vertices P (0, 3), Q

(1, 1), R (3, 2) and S (2, 4) into the parallelogram PQRS with vertices P (3, 0), Q (1, 1),

R (2, 3) and S (4, 2). Also, find which transformation is associated to this mapping?

THE END

COMPLETE SOLUTION BY TARA BAHADUR MAGAR

Group- A (10  1 = 10)
1. (a) Solution:

Here, it is the graph of sine function or y = sin.

(b) Solution:
Remainder theorem states that “when a polynomial p (x) of degree > 0 is divided by (x – a)
then the remainder (R) = p (a)”.

2. (a) Solution:

Here,

The notational condition for the existence of continuity of a function f (x) at a point x = a is

lim f(x)  lim f(x)  f(a)
xa  xa 

(b) Solution:
Here, A square matrix whose determinant is zero (0) is called singular matrix.

3. (a) Solution:

Here, the value of tanθ   2 h 2  ab
ab

(b) Solution:
Here,
When the intersecting plane is parallel to the generator of right circular cone then the conic
section so formed is called parabola.

4. (a) Solution:
Here, Sin3A = 3sinA – 4sin3A

(b) Solution:
Here, 2sinA.sinB = cos (A – B) – cos (A + B)

5. (a) Solution:

Here, t cosθ  a.b
a.b

(b) Solution:

Here,
The inversion point P (x, y) of P (x, y) with respect to the circle (x – h) 2 + (y – k) 2 = r2 is

given by

P'(x' y')   (x r 2 (x  h)  h, r 2 (y  k) k) 2  k 
 h)2  (y  k)2 (x  h)2  (y 

Group-B (13 × 2 = 26)

6. (a) Solution:

Here,
Let p(x) = x3 – (k – 1) x2 + kx + 54, D (x) = x + 3
Comparing x + 3 with x –a, we get: a = -3

Since, (x+3) is a factor of p (x).

So, Remainder (R) = p (a) = 0

or, p (-3) = 0
or, (-3)3 – (k – 1) (-3)2 + k(-3) + 54 = 0

or, -27 – (k – 1) 9 – 3k+ 54 = 0

COMPLETE SOLUTION BY TARA BAHADUR MAGAR
or, -27 –9k + 9 - 3k+ 54= 0
or, 36 –12k = 0
or, –12k = -36

k=3

Hence, the required value of k is 3.

(b) Solution:

Here,

In the given GP 1 , x, y, z, 2; first term (a) = 1 , fifth term (t5) = 2
88

Now, fifth term (t5) = 2 or, ar4 = 2 [tn = ar n- 1 ]

or, 1 r4 = 2 or, r4 = 16 r = 2
8

Again, x = second term = ar = 1 2 = 1 ,
84

y = third term = ar2 = 1 22 = 1  4 = 1 and
8 82

z = fourth term ar3 = 1 23 = 1 8 = 1
88

11
Hence, the values of x = , y = and z = 1.

42

(c) Solution:
Here,
Equation of the parabola is y = x2 – 2x – 3.
Comparing it with y = ax2 + bx + c, we get

a = 1, b = - 2 and c = -3

Now,

The vertex (h, k) =  b , 4ac  b2 
2a 4a

  - (-2) , 4 1 (-3)  (-2)2 
21 41

  2 , -12 - 4 
2 4 

 1, -16 
 4

= (1, -4)
Hence, the vertex of the parabola y = x2 – 2x – 3 is (1, -4).

7. (a) Solution:

Here, A =  4  43 
5 

Now,

4 3
A

5 4

 16  (15)

 1

COMPLETE SOLUTION BY TARA BAHADUR MAGAR
Again, A1  1  Adjoint of A
A

 1    4 3    4  3  = A
(1)  5 4 5  4

Hence, A – 1 = A.

(b) Solution: … (i)
Given equations are 3x – y = 7 … (ii)

and x + 2y = 4

Coefficient of x Coefficient of y Constant
3 -1 7
4
12

Now,

3 1
D

12

 6 1 7

7 1
and D1  4 2

 14  4

 18

We have, x = D1  18
D7

8. (a) Solution:
Here,

Slope of the line px – 2y + 1 = 0: (m1) = coeff. of x   p  p
coeff. of y  2 2

Slope of line joining the points (1, 3) and (p, 2): (m2) = y2  y1
x2  x1

 23  1
p 1 p 1

For the condition of perpendicularity, m1m2 = - 1

or, p    p 1 1   1
2 

or, -p = -2p + 2 p = 2
Hence, the required value of p is 2.

(b) Solution:

Here,

The centre of circle (h, k) = P (2, 5).

Since, the circle touches X-axis at Q. So, r = k = 5

Now,
The equation of the circle is given by (x – h)2 + (y – k)2 = r2

or, (x – 2)2 + (y – 5)2 = 52
or, x 2 – 4x + 4 + y2 – 10y + 25 = 25
or, x 2+ y2 – 4x – 10y + 4 = 0
Hence, the equation of the circle is x 2+ y2 – 4x – 10y + 4 = 0.

COMPLETE SOLUTION BY TARA BAHADUR MAGAR

9. (a) Solution:

sin 3   cos3 
22
LHS  sin   cos 

22

sin   cos  sin 2   sin  .cos   cos 2  
 2 2  2 22 2

sin   cos  
2 2

 1- 1  2sin  .cos 
2 22

 1  1 sinφ  RHS
2

(b) Solution:

Here,
cos1300 + cos1100 + cos100

 2 cos1300  1100 . cos 130 0  1100   cos 10 0
2 2

 2 cos1200.cos100  cos100

 2  1  cos100  cos100
 2

  cos100  cos100

0

(c) Solution:
Here, 4sin25A – 4 sin5A + 1 = 0

or, (2sin5A) 2– 2sin5A1 + (1)2 = 0
or, (2sin5A – 1) 2= 0
or, 2sin5A– 1 = 0

or, 2sin5A = 1

or, sin5A = 1
2

or, sin5A = sin300, sin(1800 - 300)
or, 5A = 300, 1500

A = 60, 300
Hence, the values of A are 60 and 300.

10. (a) Solution:

Here,

OA   m5  , OB   4  and AOB  90 0
-2

Now, OA.OB  0

or,  5 . 4   0
m 2

or, 20 - 2m  0

or, 20  2m

 m  10

Hence, the required value of m is 2.

(b) Solution:

COMPLETE SOLUTION BY TARA BAHADUR MAGAR
Here,

In ABC, the position vector of A (a) = 3i  5j ,

The position vector of D (d) = 3i - 4 j

The position vector of centroid G (g) =?

Since, the centroid G divides the median AD in the ratio (m1: m2) = 2:1 from vertex A
Now, by using section formula for internal division,

g  m1 b  m2 a  2(3i  4j) 1(3i  5j) = 9i  3j = 3i  j
m1  m2 2 1 3

Hence, the position vector of centroid G (g) is 3i  j .

(c) Solution:
Here,
Lower quartile (Q1) = 12,
Upper quartile (Q3) = 212 + 4 = 28
Quartile deviation (Q.D.) =?
Coefficient of quartile deviation =?
Now, Q.D. = Q3  Q1 = 28 12 = 16 = 8
2 22
Again,
Coefficient of quartile deviation = Q3  Q1 = 28 12 = 16 = 0.4
Q3  Q1 28 12 40

Hence, the required Q.D.is 8 and its coefficient is 0.4

Group-C (11 × 4 = 44)

11. Solution:

Here,

3x 11 , x  3 and g (x) = x-3

The given functions are f (x) = 2 .

x-3

Now, let g (x) = y then y = x - 3
2

Interchanging the role of x and y, we get

x  y - 3 or, y  3  2x or, y  2x  3 g-1 (x)  2x  3
2

According to question, f (x) = g – 1(x)

3x  11  2x  3
x-3

or, (x  3)(2x  3)  3x  11

or,2x 2  3x  6x  9  3x  11

or,2x 2  6x  20  0

or,2(x2  3x 10)  0

or, x 2  (5  2)x 10  0

or, x 2  5x  2x 10  0

or, x(x  5)  2(x  5)  0

or,(x  5)(x  2)  0

COMPLETE SOLUTION BY TARA BAHADUR MAGAR
Either, x – 5 = 0 x = 5 OR, x + 2 = 0 x = - 2

Hence, the required value of x is 5 or – 2.

12. Solution:

Let a and d denote the first term and common difference of an arithmetic series.

From fist condition:

Third term (t3) = 8 [tn = a + (n – 1)d] a = 8 – 2d … (i)
or, a + 2d = 8 … (ii)

From second condition:

Eleventh term (t11) = -8  a + 10d = - 8

Substituting the value of a in equation (ii), we get

8 – 2d + 10d = - 8 or, 8d = -16 d = -2

Also, putting the value of d in equation (i), we get
a = 8 – 2(-2) = 12

Again, sum of first seven terms (S7) =?
We have,

Sn  n [2a  (n  1)d]
2

S7  7 [2 12  (7  1)(2)]
2

 7 [24  6(2)]
2

 7 (24  12)
2

 7 12
2

 42

Hence, the sum of first seven terms of the series is 42.

13. Solution:

Here,

The given real valued function is f (x) = 4x + 1

Now,

(i)

X 2.9 2.99 2.999 x3-

f (x) = 4x + 1 12.6 12.96 12.996 f (x)  13
(ii)

x 3.1 3.01 3.001 x3+

f (x) = 4x + 1 13.4 13.04 13.004 f (x)  13

(iii) From (i), lim f(x)  13and from (ii), lim f(x)  13
x 3 x 3

Functional value of f (x) at x = 3, f (3) = 43 + 1 = 13

(iv) Since, lim f(x)  lim f(x)  f(3) . Hence, the function f (x) is continuous at x = 3.
x 3 x3

14. Solution:
Here,
The given equations are:

COMPLETE SOLUTION BY TARA BAHADUR MAGAR

x  y  2  6 or, 3x  y - 2  6 or,3x  y  2  18 3x  y  20 ...(i)
33

y  x  3  6 or, 2y  x - 3  6 or,x  2y  3  12 x  2y  15 ...(ii)
22

Expressing equations (i) and (ii) in matrix form. We get

 3 21 x    20 
1 y 15

or, AX = B where A =  3 21 , B =  20  and X =  x 
1 15 y

X =A–1B … (iii)

Also, determinant of A = 3 1
 61 5
12

Since, A  0 So, A -1 exists and the given system has a unique solution.

Again, A – 1 = 1  Ad joint of A
A

= 1  2 -31  1  2 -31
5 -1 5 -1

Putting the value of A -1 in equation (iii), we get

 x   1  2 3- 1 20 
y 5 -1 15

 1  - 40 -15 
5 20  45

 1  25 
5 25

  55

Equating the corresponding elements, we get x = 5 and y = 5

Hence, x = y = 5.

15. Solution:

Here,

The given equation is x2 + xy – 2y2 = 0

or, x2 + (2 – 1)xy – 2y2 = 0

or, x2 + 2xy – xy – 2y2 = 0

or, x (x + 2y) – y(x + 2y) = 0

or, (x + 2y)(x – y) = 0

Either, x + 2y = 0 … (i)

Or, x – y = 0 … (ii)

The equation of straight line  to line (i) is 2x – y + c = 0

Since, it passes through origin i.e., (0, 0). So, 20 – 0 +c = 0 c = 0
Pitting the value of c in equation (iii), we get. 2x – y = 0 … (A)

The equation of straight line  to line (ii) is x + y + k = 0 … (iv)

Since, it passes through origin i.e., (0, 0). So, 0 + 0 + k = 0 k = 0
Putting the value of k in equation (iv), we get. x + y = 0 … (B)

Again, combining equations (A) and (B), we get
(2x – y)(x + y) = 0

or, 2x2 + 2xy –xy – y2 = 0
or, 2x2 + xy – y2 = 0 which is required equation.

COMPLETE SOLUTION BY TARA BAHADUR MAGAR

16. Solution:
Here, cos2A + cos2 (A + 1200) + cos2 (A-1200)

 1 [2cos 2 A  2cos 2 (A 1200 )  2cos 2 (A 1200 )]
2

 1 [1 cos2A 1 cos2(A 1200 ) 1 cos2(A 1200 )]
2

 1 [3  cos2A  cos(2A  2400 )  cos(2A  2400 )]
2

 1   cos2A  2cos (2A  240 0 )  (2A  240 0 ) cos (2A  240 0 )  (2A  240 0 ) 
2 3 2 2


  1 3  cos2A  2cos2A.cos2400
2

 1   cos2A  2cos2A. - 1 
2 3  2

 1 3  cos2A  cos2A.

2

3
2

Hence, the value of cos2A + cos2 (A + 1200) + cos2 (A-1200) is 3 .
2

17. Solution:

Here,

A + B + C = 1800 A + B = 1800 - C

Taking sin and cos on both side successively, we get
sin (A + B) = sin (1800 – C) = sinC
& cos (A + B) = cos (1800 – C) = -cosC

Now,

LHS = sinA.cosA  sinB.cosB  sinC.cosC

 1 (2sinA.cosA  2sinB.cosB  2sinC.cosC)
2

 1 (sin2A  sin2B  sin2C)
2

 1 2sin 2A  2B .cos 2A  2B   
2 2  2  sin2C

 1 2sin(A  B)cos(A - B)  sin2C

2

 1 2sin 2A  2B .cos 2A  2B   
2 2  2  sin2C

 1 2sin(A  B)cos(A - B)  sin2C

2

 1 2sinC.cos(A - B)  2sinC.cosC

2

 1  2sinCcos(A - B)  cosC

2

 sinCcos(A - B) - cos(A  B)

 sinCcosA.coB  sinA.sinB - cosA.cosB  sinA.sinB

 sinC.2sinA.sinB

 2sinA.sinB.sinC  RHS

COMPLETE SOLUTION BY TARA BAHADUR MAGAR

18. Solution:

Let AB be the height of the vertical pole, C be a point on it which

divides it in the ratio of 9:1, D be a point on the ground, , ADC and

CDB be the angles subtended by the segments AC and BD

respectively.

Then, AC = 9x m., BC = x m , ADC = BDC =  (say)AB = 10x

m and ADB = 2, BD = 20 m

Now, from right angled BCD, tan = BC  x ... (i)
CD 20

Also, from right angled ADB, tan2 = AB
CD

or, 2tanθ  10x
1  tan 2θ 20

2 x  [From (i)]
or,  20   x

1   x 2 2
 20 

x
or, 10  x

400  x 2 2

400
or, 40  1

400  x 2 2
or,80  400  x 2

or, x 2  320

 x  8 5  17.8885
Hence the height of the pole is 10x = 10  17.8885 = 178.885 m.

19. Solution:
Here,
The coordinates of vertices of ROSE are R (1, 1), O (2, 3), S (4, 2) and
E (3, 0)
Now,
Rotating ABC through positive 900 about centre at origin, we get

P (x, y) R [+900, (0, 0)] P (-x, -y)

 R (1, 1) R (-1, 1)
O (2, 3) O (-3, 2)
S (4, 2) S (-2, 4)
E (3, 0) E (0, 3)

Thus, R (-1, 1), O (-3, 2), S (-2, 4) and E (0, 3) are the vertices of image ROSE.
Again,
Enlarging A’B’C’ by enlargement E [(0, 0), 3], we get

We have,

P (x, y) E [(0, 0); k] P’ (kx, ky)
 R (-1, 1) E [(0, 0); 3] R (-3, 3)
O (-3, 2) O (-9, 6)

COMPLETE SOLUTION BY TARA BAHADUR MAGAR

S (-2, 4) S (-6, 12)

E (0, 3) E (0, 9)

Thus, R (-3, 3), O (-9, 6), S (-6, 12) and E (0, 9) are the vertices of image ROSE.

At last, representing the ROSE and its images on the same graph paper

20. Solution:
Here,
Computation of the mean deviation from the median:
Height (in cm) m No. of students (f) c. f. m – Md  fm – Md 

95-105 100 9 9 26 234

105-115 110 13 22 16 208

115-125 120 25 47 6 150

125-135 130 30 77 4 120

135-145 140 13 90 14 182

145-155 150 10 100 24 240

N = 100 fm – Md = 1134

Now,
Position of median =  N  thi item = 100 thiitem = = 50thi item

2  2 

From the c.f. column, the c.f . just greater than or equal to 50 is 77and its corresponding class

is (125-135) Median class = (125-125) where L = 125, c.f = 47, f = 30 and i= 10.

Also,

 N  c.f.  = 125   50  47  10 = 126
Median = L   2 i  30 
 f 
 

Again,

COMPLETE SOLUTION BY TARA BAHADUR MAGAR

M.D. from median = f m  Md = 1134 = 11.34

N 100

Hence, the mean deviation from median is 11.34

Lastly, coefficient of M.D. = M.D.frommedian = 11.34 = 0.09
median 126

Hence, the coefficient mean deviation from median is 0.09.

21. Solution:
Here,
Computation of the standard deviation from the median:

Ages (in years) No. of boys (f) M fm m2 fm2
4 16
0-4 4 2 8 36 216
100 1000
4-8 6 6 36 196 1176
324 1296
8-12 10 10 100  fm2 = 3704

12-16 6 14 84

16-20 4 18 72

N = 30 fm = 300

Now,
(i)

X   fx
N
 300
30
 10
Hence, average age is 10 years.

(ii) σ  fm2    fm 2

N N

 3704   300 2
30  30 

 123.467  100

 23.467
= 4.84
Hence, the standard deviation of the data is 3.683

Group-D (4 × 5 = 20)

22. Solution:

Here,

First part:

Importance of linear programming: Linear programming is a mathematical technique of

optimization of an objective function under the given constraints. It is used in business to plan

COMPLETE SOLUTION BY TARA BAHADUR MAGAR

the maximum production, maximum profit and minimum investment with the use of limited

resources available.

Next part:
The equation of boundary line corresponding to x + y ≤ 5 is x + y = 5 y = 5 – x

x 1 23

y 4 32

The boundary line passes through the points (1, 4), (2, 3) and (3, 2). So, plotting these points

and joining them on a graph paper.

Taking (0, 0) as the testing point, we get 0+ 0 < 6 or, 0 < 6 (True)
Hence the half- plane of x + y ≤ 6 contains the origin.

Also,
The equation of boundary line corresponding to x +4y ≤ 8 is x +4y = 8

y = 8  x x0 48
4 y2 10

The boundary line passes through the points (0, 2), (4, 1) and (8, 0). So, plotting these points

and joining them on the same graph paper.

Taking (0, 0) as the testing point, we get 0 + 40 < 8 or, 0 < 8 (True)
Hence the half- plane of x +4y ≤ 8 contains the origin.
Again, x  0  x = {0, 1, 2, …} with boundary line x = 0 (y-axis) and y  0
 y = {0, 1, 2, …} with boundary line y = 0 (x-axis). Thus, x  0 and y  0 represent the first

quadrant.

From the above graph, the shaded region OABC is the feasible region with vertices O (0, 0),

A (5, 0), B (4, 1) and C (0, 2).At last, tabulating the values of given objective function

P = 3x + 2y at each vertex of feasible region.

S.N. Vertex P = 3x + 2y Remarks

1. O (0, 0) 3.0 + 2.0 = 0 Minimum

2. A (5, 0) 3.5 + 2.0 = 15 Maximum

3. B (4, 1) 3.4 + 2.1 = 14

4. C (0, 2) 3.0+2.2 = 4

Hence, the maximum value of F (x, y) = 3x + 2y is 15 at A (5, 0) and the minimum value is 0 at O (0, 0).

23. Solution: … (i)
… (ii)
Here,
The equation of circle-Y is x2 + y2 = 1

The equation of chord AB of circle-Y is x+ y = 1 y = 1 - x

Now,

Substituting the value of y in equation (i), we get
x2 + (1 – x)2 = 1
or, x2 + 1 – 2x + x2 = 1

COMPLETE SOLUTION BY TARA BAHADUR MAGAR
or, 2x2 – 2x = 0

or, 2x ( x – 1) = 0

Either, 2x = 0 x = 0

OR, x -1 = 0 x = 1
When, x = 0, from equation (ii); y = 1 – 0 = 1
When, x = 1, from equation (ii); y = 1 – 1 = 2

Hence, the ends of the chord AB are (0, 1) and (1, 0).

Since, chord AB of circle-Y is the diameter of the circle-X.

So, the ends of diameter AB for the circle-X are (0, 1) and (1, 0)

Again,

Equation of the circle-X is given by (x – x1) (x – x2) + (y – y1) (y – y2) = 0
or, (x – 0) (x – 1) + (y –1) (y – 0) = 0
or, x2 – x + y2 – y = 0
or, x2 + y2 – x – y = 0
Hence, the equation of the circle-X is x2+ y2 – x – y = 0.

24. Solution:

Given: In rhombus BEST; BE= ES = ST = TB and BS and ET are the

diagonals.

To prove: (i) Diagonals BS and ET bisect each other
(ii) BS and ET are perpendicular to each other i.e. ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ = 0

Assumption: M is the mid-point of diagonal BS and N is the mid-point of diagonal ET.

Proof:

(i) ⃗⃗⃗⃗⃗⃗ 1 ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ [By mid-point theorem]
2

(ii) ⃗⃗⃗⃗⃗ 1 ⃗⃗⃗⃗⃗ [Being N the mid-point of diagonal ET]
2

= 1 ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ [By parallelogram law of vector addition]
2

From (i) and (ii), we get ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ i.e., mid-point M of diagonal BS and N of diagonal ET

coincide. Therefore, the diagonals of rhombus bisect each other.

Also, (i) In BES; ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ [By  law of vector addition]

(ii) In EST; ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗⃗⃗ [By  law of vector addition]

Again, ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗⃗⃗

(⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ) ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ [⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ]

(⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ) ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗

[

Since, ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ 0 So, BS  ET
Hence, the diagonals of rhombus bisect each other at right angle. QED

25. Solution:

Here,
Writing the vertices of square ABCD and image parallelogram A’B’C’D’ in matrix form as:

Object = Square PQRS =  0 1 3 2 
3 1 2 4

Image = Parallelogram PQRS=  3 1 2 4 
0 1 3 2

COMPLETE SOLUTION BY TARA BAHADUR MAGAR

Let 2  2 transformation matrix (T.M.) =  a b 
c d

Now, we have,

Image = T.M.  Object

or,  3 1 2 4  =  a b   0 1 3 2 
0 1 3 2 c d 3 1 2 4

or,  3 1 2 4  =  0  3b ab 3a  2b 2a  4b 
0 1 3 2 0  3d cd 3c  2d 2c  4d

Equating corresponding elements, we get

3b = 3 … (i)

a+b=1 … (ii)

3d = 0 … (iii)

c+d=1 … (iv)

From (i), 3b = 3 b = 1

From (ii), a + b = 1 or, a + 1 = 1a = 0

From (i), 3d = 0 d = 0

From (ii), c + d = 1 or, c + 0 = c = 1

Hence, required 2  2 transformation matrix =  a b  =  0 1 
c d 1 0

Again, let P (x, y) be an object point.

Then image = T.M. Object point = 10 1  xy   0  y0   xy
0 x 

P (x, y) P (y, x)

Hence, the mapping is associated to the reflection about the line y = x.

THE END