O.-MATH-2076-SOLUTION

By: ENISHCHAL LAMA BAJYU (O. MATH)

ENISHCHAL'S SEE MODEL (b) olb 2 cos A = 3 eP 0 < A < 90° df A sf] dfg
2076 O. MATH WITH slt x'G5 <
GROUPSOALUT1I0O×N1 = 10
If 2 cos A = 3 , what is the value of A for
0 < A < 90° ?

1. (a) 3g3ftLo kmngsf] kl/efiff nV] gx' f];\ .  Here, 2 cosA = 3

Write the definition of a cubic function. or, cos A = 3
2

 Here,a function of the form (x) = ax3 + bx2 or, cos A = cos 30°  A = 30°

+ cx + d; a  0 is called a cubic function. Thus, the required value of A is 30°.
e.g. y = x3, y = 2x3 + 5 etc. are cubic functions.
5. (a) olb p = (a, b) / q = (b, a) eP p . q sf] dfg slt
(b) bO' {cf6] f wgfTds ;ªV\ ofx¿ m / n larsf] u'0ffQ] / xG' 5 < (If p = (a, b) and q = (b, a), what is the
dWodf slt x'G5 <
value of p . q ? )
What is the geometric mean between two
positive numbers m and n?  Here, p . q = (a, b) . (b, a) = ab + ab = 2ab.

 Here, the geometric mean between two (b) lbOPsf] pTjm| d j[Q PsfO jQ[ eP
numbers m and n is mn. OP × OP' sf] gfk slt xf]nf <
Given inversion circle is a unit
2. (a) ;ª\Vof /]vfdf k|fs[lts ;ª\Vofx¿sf] ;dx" / jf:tljs circle. What is the measure of rOP P'
;ª\Vofx¿sf] ;d"x dWo] sg' ;dx" df lg/Gt/tf x'G5 <
OP × OP'?
In which set of numbers, do you find continuity
between set of natural numbers and set of real  Here, the measure of OP × OP' = r2
numbers in number line? i.e. OP × OP' = 1

GROUP B 13 × 2 = 26

 Here, the number line of real numbers has the 6. (a) zi] f ;fWosf] syg n]Vgx' f];\ . pQm ;fWo k|of]u u/L
continuity. x6 – 64 sf] u'0fgv08 (x – 2) xf] , xfO] g olsg ug'{xf;] \ .

(b) dl] 6S« ;  l o  sf] l86/ldgfG6 slt xG' 5 < State remainder theorem. Use it and determine
v e whether (x – 2) is a factor of x6 – 64 or not?

(What is the determinant of the matrix  l o ?)  Here, statement of remainder theorem:
v e If p(x) is a polynomial of degree n and (x – a) is
a divisor of p(x) then p(a) is remainder, where
 Here, determinant of  l o  =  l o  = le – vo the degree of quotient will be (n – 1).
v e v e Given polynomial (P(x)) = x6 – 64
divisor = x – 2 = 0
3. (a) olb em' sfj a / b ePsf b'Oc{ f6] f l;wf
/]vfx¿aLrsf] sf0] f θ eP tan θ slt x'G5 <  x=2
So, Remainder P(2) = 26 – 64 = 64 – 64 = 0
If θ be the angle between two straight lines Thus, the remainder = P(2) = 0 shows that
having slope a and b, what is tan θ? (x – 2) is a factor of given polynomial.

 Here, we have, tan  = ± m1 – m2 (b) kf/faf]nfsf] dfgs ;dLs/0f n]vL y = x2 – 4x + 5 sf]
1+ m1m2 zLif{laGb'sf] lgb{]zfª\s kQf nufpgx' f];\ .

 tan  = ± a–b Write the standard equation of a parabola and
1 + ab find the vertex of y = x2 – 4x + 5.

(b) bL3{ j[Qsf] kl/efiff lbgx' f];\ . (Define ellipse.)  Here, the standard equation of parabola is ;
y = a(x – h)2 + k
 Here, the plane curve formed by intersecting a Now, y = x2 – 4x + 5
cone with a plane surface making an angle of 
with the axis of cone in such a way that the   or, y = x2 – 2.x.2 + 22 + 1
value of  is greater than the semi vertical angle or, y = (x – 2)2 + 1
(α) and less than 90° (α <  < 90°) is called an Comparing it with y = a(x – h)2 + k then,
ellipse. vertex = (h, k) = (2, 1).
In other words, if a plane cuts the cone such
that the angle made by the plane with the axis is (c)  : R → R sf nflu (x) = 3x – 2 eP –1 (4) kQf
greater than the semi-vertical angle, then the nufpgx' f;] \ .
closed curve or the section formed is an ellipse.
For, : R → R, (x) = 3x – 2, find –1 (4).
4. (a) cos 2A nfO{ cos A sf] ¿kdf JoQm ugx'{ f;] \ .
 Here, (x) = 3x – 2
Express cos 2A in terms of cos A.
  let y = (x)
 Here, cos 2A = 2cos2A – 1 is required relation.
or, y = 3x – 2

Interchanging the position of x and y,

x = 3y – 2

or, x + 2 = 3y

 yx +2  i.e.  – (x) = x + 2
3 3

By: ENishchal Lama Bajyu (O.MATH)

Now,  – (4) = 4 + 2 = 6 =2 We know that, equation of circle is;
3 3 (x – h)2 + (y – k)2 = r2

Thus, the value of  – (4) is 2. or, (x – 4)2 + (y – 5)2 = 32
or, x2 – 8x + 16 + y2 – 10y + 25 = 9
7. (a) olb A =  4 –7  eP A sf] ljkl/t d]l6S« ; or, x2 + y2 – 8x – 10y + 41 – 9 = 0
–1 2  x2 + y2 – 8x – 10y + 32 = 0
Thus, the required equation of circle is;
A–1 kQf nufpgx' f];\ . x2 + y2 – 8x – 10y + 32 = 0.
4 –7
If A =  –1 2 , find the inverse matrix A–1 of A. 9. (a) SofNs'n]6/ jf lqsf]0fldtLo tflnsfsf] ko| fu] ljgf
cos 15° sf] dfg kQf nufpg'xf;] \ .
 Here, A =  4 –7 
–1 2 Find the value of cos 15° without using
calculator or trigonometric table.
4 –7
  So, |A| =  –1 2 = 8–7=1  Here,
cos 15° =
Adj (A) =  2 7  1 + cos 2 × 15°
1 4 = 2

We have, A–1 = 1 Adj. (A) = 1  2 7  1 + cos 30°
|A| 1 1 4 2

 2 7  1+ 3 2+ 3 1
1 4 2 2 2
 A–1 = = 2= ×

Thus, the value of A–1 is  2 7 . = 2+ 3 = (2 + 3) × 2
1 4 4 4×2

(b) ;dLs/0fx¿ x + 2y = 7 / 2x – y = 4 df j|mfd/sf] = 4+2 3
D1 D1 D2 D2 8
lgod ko| f]u ubf{ x = D = (– 5) / y= D = (– 5) eP
3+1+2 3
D1 / D2 sf dfgx¿ kQf nufpgx' f;] \ . = 8

In Cramer's rule for solving equations x + 2y = 7 ( 3)2 + 2. 3.1 + 12
D1 D1 D2 D2 22
and 2x – y = 4, x = D = (– 5) and y = D = (– 5) . =

Find the values of D1 and D2. = ( 3 + 1)2
22
 Here, x + 2y = 7 and 2x – y = 4

  So, D1 =  7 2 = – 7 – 8 = – 15 = 3+1
4 –1 22

D2 =  1 7  = 4 – 14 = – 10 Thus, the value of cos 15° is 3 +1 .
2 4 2 2

Thus, the values of D1 and D2 are – 15 and – 10 sin 10x – sin 6x
respectively. cos 10x + cos 6x
(b) kd| fl0ft ug{'xf;] \ (Prove that) : = tan 2x

8. (a) hf8] f ;dLs/0fx¿ mx = ny / nx = – my nfO{   Here, LHS
kl| tlglwTj ug{] Pp6} ;dLs/0f n]Vgx' f;] \ .
= sin 10x – sin 6x
Write the single equation representing the pair of cos 10x + cos 6x
equations mx = ny and nx = – my.
( ) ( )2cos
 Here, mx = ny and nx = – my 10x + 6x sin 10x – 6x
  or, mx – ny = 0 and nx + my = 0 2 2
( ) ( )=
Combining these equations then, 2cos 10x + 6x cos 10x – 6x
(mx – ny) (nx + my) = 0 2 2

or, mx(nx + my) – ny(nx + my) = 0 = sin 2x
or, mnx2 + m2xy – n2xy – mny2 = 0 cos 2x
 mnx2 + (m2 – n2)xy – mny2 = 0
Thus, the single equation of pair of lines is = tan 2x = RHS Proved.
mnx2 + (m2 – n2)xy – mny2 = 0.
(c) xn ug{'xf];\ (Solve) : 2 cos  – 1 = 0 (0   360°)

(b) sG] b| (4, 5) / Jof; 6 PsfO ePsf] jQ[ sf] ;dLs/0f  Here, 2 cos  – 1 = 0
kQf nufpgx' f;] \ .
  or, 2 cos 
Find the equation of circle having centre (4, 5)
and diameter 6 units. or, cos 


 Here, centre = (h, k) = (4, 5) and Since, cos is (+)ve in first and fourth quadrant.

diameter = d = 6 units So, cos cos 60° or cos (360° – 60°)

So, radius (r) = d = 6 = 3 units  or 300°
2 2
Thus,  = 60° or 300° is the solution.

By: ENishchal Lama Bajyu (O.MATH)

10. (a) olb a =  0  / b =  4  eP a / b aLr aGg] GROUP C 11 × 4 = 44
5 4
sf]0f kQf nufpg'xf];\ . 11. bO' { kmngx¿ 2x + 5 / 5g\ . olb
(x) = 8 g(x) = 3x – 4

If a =  0  and b =  4  , find the angle (⸰g)–1(x) Pp6f PsfTds kmng xf] eg] x sf] dfg kQf
5 4 nufpgx' f];\ .

between a and b. Two functions are (x) = 2x + 5 and g(x) = 3x – 4. If
8
0 4
 Here, a =  5  and  =  4  (⸰g)–1(x) is an identity function, find the value of x.
2x + 5
b 8

a.   0 . 4   Here, (x) = and g(x) = 3x – 4
5 4
  So, b = ⸰g(x) = (g(x)) = (3x – 4)

=0×4+5×4 = 2 (3x – 4) + 5
8
 a .  = 20
6x – 8 + 5
b 8

  Again, |a| = 02 + 52 = 5 units =

 = 42 + 42 = 32 = 4 2 units  ⸰g (x) = 6x – 3
8
|b|

If  be the angle between a and  then , For (⸰g)–1, let y = ⸰g(x)

b

cos  = a .b = 20 = 1 or, y = 6x – 3
×4 2 8
|a |  5 2
Interchanging the position of x and y then,
|b|

or, cos  = cos 45° 6y – 3
8
  = 45° x =

Thus, the angle between a and  is 45°. or, 8x = 6y – 3

b

(b) lrqdf BD = DC eP AC = 2AD or, 8x + 3 = 6y

 y = 8x + 3
6
+ BA xG' 5 egL b]vfpgx' f];\ .
A i.e. (⸰g)–1 (x) = 8x + 3
In the figure, BD = DC, prove 6

  By the question; ⸰g–1 is an identity function.
that: AC = 2AD + BA.
 Here, BD = DC BDC So, 8x + 3 = x
6

  (i) BD = BA + AD or, 8x + 3 = 6x

[ ‡ triangle law of vector addition] or, 2x = – 3

  x = – 3
(ii) DC = DA + AC 2

[ ‡ triangle law of vector addition] Thus, the value of x is – 32.

 12. lbOPsf] n]vflrqaf6 p2Z] o Y
(iii) BD = DC [  BD = DC ] C

 kmng z(x, y) = 3x + 2y sf OA
or, BA + AD = DA + AC Y'

   zt{x¿, clwstd / Gog" td
or, BA + AD – DA = AC dfgx¿ kQf nufpg'xf;] \ .

 From the given graph find B
or, BA + AD + AD = AC the constraints, maximum X'
X
 
 AC = 2 AD + BA Proved. and minimum values of the

(c) Pp6f tYofª\ssf] klxnf] rt'yf{z+ 70 / t];f| ] rt'yf+{z objective function
klxnf] rty' f+{zsf] bf]Aa/ eGbf 25 n] w]/} 5 . ;f]
tYofª\ssf] rt'yfz+{ Lo leGgtfsf] u0' ffª\s kQf z(x, y) = 3x + 2y
nufpg'xf];\ .
 Here, co-ordinates of A, B, C and O are ;
The first quartile of a data is 70 and third quartile
is 25 more than the double of the first quartile.  A(1, 0), B(3, 1), C(0, 4) and O(0, 0)
Find the coefficient of the quartile deviation of
the data. Equation of AB; y – y1 = y2 – y1 (x – x1)
x2 – x1

or, y – 0 = 1 – 0 (x – 1)
3 – 1

 Here, first quartile (Q1) = 70 and third quartile or, y = 1 (x – 1)
(Q3) = Q1 × 2 + 25 = 70 × 2 + 25 = 165 2
We know that,
or, 2y = x – 1

Coefficient of QD = Q3 – Q1 = 165 – 70  x – 2y = 1
Q3 + Q1 165 + 70
Taking (0, 0) as the test point of solution set then,

95 19 LHS = x – 2y = 0 – 2 × 0 = 0 < 1
235 47
= = = 0.404 So the corresponding inequality is x – 2y  1.

Thus, the coefficient of QD is 0.404. Equation of BC; B(3, 1) and C(0, 4)

By: ENishchal Lama Bajyu (O. MATH)

Using formula, y – y1 = y2 – y1 (x – x1) 14. lbOPsf hf]8f ;dLs/0fx¿sf] dl] 6S« ; ljlw k|ofu] u/L
x2 – x1 xn ugx'{ f;] \ M 2x – 3y = 7, 4y – 3x + 10 = 0

or, y–1 = 4 – 1 (x – 3) Solve the given pair of equations using matrix
0 – 3 method : 2x – 3y = 7, 4y – 3x + 10 = 0

or, y – 1 = – 3 (x – 3)  Here, given equations are;
3 2x – 3y = 7 and 4y – 3x = – 10
Above equation's can be written as matrix form
or, y – 1 = – x + 3
 2 –3   x   7 
 x+y=4 or, –3 4 y = – 10

Taking (0, 0) as the test point of solution set then, or, AX = B

LHS = 0 + 0 = 0 < 4 or, X = A–1B....(i)

So the corresponding inequality is x + y  4. Where, A =  2 –3 , X =  x  & B =  7 
–3 4 y – 10
The polygonal region lies in first quadrant only. So

other two inequalities are x  0 and y  0. Now, We have, A–1 = |A1 |. adj A

Now, the calculation of maximum and minimum 2 –3 4 3
–3 4 3 2
values: Inverse of A   = 1  
–
Vertices z(x, y) = 3x + 2y Value Remarks 8 9

O (0, 0) 3 × 0 + 2 × 0 0 Minimum =  – 4 – 3 
– 3 – 2
From equation (i),
A (1, 0) 3 × 1 + 2 × 0 3
x – 4 – 3 7
B (3, 1) 3 × 3 + 2 × 1 11 Maximum or,  y  =  – 3 – 2   – 10 

C (0, 4) 3 × 0 + 2 × 4 8 =  – 28 + 30 
– 21 + 20
From the table, the minimum value is 0 and the 2
maximum value is 11. =  –1 
Thus, the required constraints are;
x – 2y  1, x + y  4, x  0, y  0 and the maximum Thus, the required value of x & y are 2 and – 1
and minimum values are 11 and 0 respectively. respectively.

13. lrqdf Pp6f cfg'kflts kmngsf] Y 15. ;dLs/0f 3x2 + xy – 10y2 = 0 n] k|ltlglwTj ug{]
/v] fx¿;u“ ;dfgfGt/ x'g] / laGb' (1, 0) eP/ hfg] Ps
nv] flrq lbOPsf] 5 . pQm 3 hf]8L /v] fx¿sf] ;dLs/0f kQf nufpg'xf];\ .

n]vflrq sg' } Pp6f sDklgsf] 1 Find the equation of a pair of lines passing through
the point (1, 0) and parallel to the lines represented by
gfkmf÷ gfS] ;fg;“u ;DalGwt 5 . X' –4 –2 O 2 4 6 X the equation 3x2 + xy – 10y2 = 0.

pQm jj|m s'g–s'g laGb'df ljlR5Gg Y'  Here, given equation of pair of lines;
5 / lsg < sf/0f nV] g'xf;] \ . 3x2 + xy – 10y2 = 0

In the figure, the graph of a rational function is or, 3x2 + 6xy – 5xy – 10y2 = 0
or, 3x (x + 2y) – 5y(x + 2y) = 0
shown. The graph represent the profit/loss of a or, (x + 2y) (3x – 5y) = 0
Either, x + 2y = 0 ......(i) or, 3x – 5y = 0 .........(ii)
company. At which points does the function is Equation of line parallel to line (i) is x + 2y = k
It passes through (1, 0).
discontinuous and why ? Give reason. So, 1 + 2 × 0 = k
 k=1
 Here, So, the equation of line parallel to (i) is;

Points of Reason x + 2y = 1
 x + 2y – 1 = 0...............(iii)
discontinuities Again, equation of line parallel to the line (ii) is;

(i) x = – 2 (i) There is a jump at x = – 2. 3x – 5y = k
So the curve is It passes through (1, 0).
discontinuous at x = – 2. So, 3 × 1 – 5 × 0 = k
 k=3
(ii) x = 0 (ii) There is a hole at x = 0. So, the equation of line parallel to the line (ii) is
So the curve is
discontinuous at x = 0. 3x – 5y = 3
 3x – 5y – 3 = 0....................(iv)
(iii) x = 4 (iii)There is a gap at x = 4. Combining (iii) and (iv) then,

So the curve is discontinuous (x + 2y – 1) (3x – 5y – 3) = 0
at x = 4. or, x(3x – 5y – 3) + 2y(3x – 5y – 3) – 1(3x – 5y – 3) = 0
or, 3x2 – 5xy – 3x + 6xy – 10y2 – 6y – 3x + 5y + 3 = 0
(iv) x = 6 (iv) There is a jump at x = 6.  3x2 + xy – 10y2 – 6x – y + 3 = 0
So the curve is Thus, the required equation of pair of lines is;
discontinuous at x = 6. 3x2 + xy – 10y2 – 6x – y + 3 = 0.

By: ENishchal Lama Bajyu (O. MATH)

16. k|dfl0ft ug{x' f;] \ (Prove that): We have, FAC = ACE = 45° and

( )2sin2 c – A = (1 – tan A)2 FAD = ADB = 60°
4 1 + tan2 A
From right angled AEC, tan  = AE
EC
 Here, LHS
or, tan 45° = AE
( )= c EC
2sin2 4 – A
or, 1 = AE  AE = EC
 = 2 sin2 (45° – A) EC

= 1 – cos 2 (45° – A) [‡ 1 – cos 2A = 2 sin2A] From right angled ABD, tan  = AB
BD
= 1 – cos (90° – 2A)
or, tan 60° = 24
= 1 – sin 2A [‡ cos (90° – ) = sin ] BD
2 tan A
= 1– 1 + tan2A or, 3 = 24
BD

= 1 + tan2A – 2 tan A or, BD = 24 = 8 3
1 + tan2A 3

= (1 – tan A)2 = RHS Proved. Now,
1 + tan2 A
CD = EB = AB – AE = 24 – EC = 24 – BD = 24 – 8 3

17. olb A + B + C = 180° eP kd| fl0ft ug'x{ f];\ M  CD = 10.14 m
If A + B + C = 180°, prove that: Thus, the height of the pole is 10.14 m.

2 sin A cos A + sin B – sin C = 4 sin A sin B cos C 19. olb d]l6«S; M1 n] X- cIfdf xg' ] k/fjtg{ / dl] 6S« ; M2 n]
2 2 2 2 2 pbu\ d laGb'sf] jl/kl/ 180° df xg' ] kl/j|md0fnfO{
hgfp5“ g\ eg] M1M2 n] PsfO ju{nfO{ :yfgfGt/0f ubf{
 Here, A + B + C = 180° aGg] rt'e'{hsf] zLif{laGb'x¿sf] lgb{]zfª\s kQf nufpg'xf];\ .

  or, A + B = 180° – C If M1 is the matrix represented by reflection on X-axis
and M2 is the matrix represented by rotation through
 A+B = 90°– C 180° about origin then transform unit square by M1M2
2 2 and write the co-ordinates of vertices of image

Taking sin and cos on both the sides then,

( ) ( )sinA+B C C quadrilateral.
2 = sin 90 – 2 = cos 2
0 1 1 0
( ) ( )cosA+B 90 – C = C Here, object = unit square =  0 0 1 1
2 = cos 2 sin 2 

So, LHS  We have, (x, y)  (x, – y) when reflected in x- axis.

A A So, x = 1.x + 0.y
2 2
= 2 sin cos + sin B – sin C – y = 0.x + (–1)y

= sin A + sin B – sin C  M1 =  1 0 
0 –1
( ) ( )= 2 sin A+B A–B
2 cos 2 – sin C Again, we know that (x, y)  (– x, – y) when

( )= C A–B C C rotated through 180° about the origin.
2 2 2 2
2 cos cos – 2 sin cos So, – x = (– 1).x + 0.y

[ ( ) ]= – y = 0.x + (– 1).y

2 cos C cos A–B – sin C  M2 =  –1 0 
2 2 2 0 –1

[ ( ) ( )]=C cos A–B A+B So, transformation matrix is;
2 cos 2 2 – cos 2
 1 0   –1 0 
C A B M1M2 = 0 –1 0 –1
2 2 2
= 2 cos 2 sin sin  –1+0 0 – 0   –1 0 
0–0 0 + 1 0 1
= 4 sin A sin B cos C = RHS Proved. = =
2 2 2
 Now, Image = TM × object

18. 24 ld6/ cUnf] Pp6f :tDesf] 6'Kkf]af6 Pp6f vDafsf] = M1M2  0 1 1 0 
0 0 1 1
6K' kf] / kmb] nfO{ cjnfs] g ubf{ cjglt sf]0fx¿ jm| dzM 45°
/ 60° kfOof] eg] vDafsf] prfO kQf nufpg'xf];\ .
From the top of a tower 24 meter high, the angles of =  –1 0   0 1 1 0 
0 1 0 0 1 1
depression of the top and the foot of a pole are
0 + 0 –1+0 –1+0 0 + 0
observed to be 45° and 60° respectively. Find the =  0 + 0 0+0 0+1 0 + 1 

height of the pole. FA

Here, let AB = 24 m be the  0 –1 –1 0 
0 0 1 1
height of a tower and CD be =

the height of a pole. 0 –1 –1 0
0 0 1 1
Let FAC = 45° and FAD = 45° E   Image =  
60° be the angles of depression. C

Then, ? 60° Thus, the co-ordinates of vertices of image
Height of the pole (CD) = ? D
B quadrilateral are (0, 0), (– 1, 0), (– 1, 1) and (0, 1).

By: ENishchal Lama Bajyu (O. MATH)

20. tn lbOPsf] tYofªs\ sf] dlWosfaf6 dWos leGgtf kQf GROUP D 4 × 5 = 20

nufpgx' f;] \ . 22. Pp6f dflg;n] ?= 3900 C0f dfl;s ls:tfdf ltg]{ ljrf/
u/]5 . k|To]s ls:tf 7Ls cl3Nnf] ls:tf eGbf ?= 20 n]
Find the mean deviation from median of the data sd 5 . klxnf] ls:tf jfktsf] /sd ?= 400 5 . hDdf
/sd slt cf]6f ls:tfdf ltg{ ;lsG5 < sf/0f lbgx' f];\ .
given below: 
A person pays a loan of Rs 3900 in monthly
Class interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 installments, each installment being less than the
former by Rs 20. The amount of first installment is Rs
Frequency 5 8 15 16 6 400. In how many installments will the entire amount
be paid ? Give reason.
 Here,Calculation of mean deviation from median

Class x  c.f *D* = *D*
Interval *x – Md|

0 - 10 5 5 5 23 115  Here, Sum of money (Sn) = Rs 3900
  Common difference (d) = – 20
10 - 20 15 8 13 13 104

20 - 30 25 15 28 3 45 First installment (a) = Rs 400

30 - 40 35 16 44 7 112 No. of terms (n) = ?

40 - 50 45 6 50 17 102 We know that, Sn = n [2a + (n – 1) d]
2
N 3*D*
= 50 = 478 or, 3900 = n [2 × 400 + (n – 1) (– 20)]
2

For median, or, 7800 = n (800 – 20n + 20)

th th or, 7800 = n(820 – 20n)

item item = 25th item
( ) ( )N = 50 or, 7800 = 20n(41 – n)
2 2 or, 390 = 41n – n2

= 20 - 30 class interval or, n2 – 41n + 390 = 0
or, n2 – 26n – 15n + 390 = 0
We have,

( )Median = L + i N – c = 20 + 10 (25 – 13) or, n(n – 26) – 15(n – 26) = 0
f 2 15
or, (n – 26) (n – 15) = 0

= 20 + 2 × 12 Either, n = 26 or, n = 15
3
When n = 26 then,

ˆ Median = 28 t26 = a + 25d
Now, mean deviation from median: = 400 + 25 × (– 20)

MD = 3*D* = 478 = 9.56 = 400 – 500
N 50
= – 100 (Impossible)

Thus, the mean deviation from median is 9.56. When n = 15 then

21. lbOPsf] tYofªs\ sf] :t/Lo leGgtf kQf nufpg'xf];\ t15 = a + 14d
= 400 + 14 × (– 20)
Find the standard deviation of the given data:
= 400 – 280

Class Interval 10-20 20-30 30-40 40-50 50-60 = 120

Frequency 4 10 12 8 6  t15 = 120
Thus, in 15 installments, the entire amount will be

Here, from the given data : paid.

CI m  .x d = m – ¯x d2 d2 23. lbOPsf] lrqdf a/fa/ jQ[ x¿ A
/ B sf sG] bl| aGb'x¿ j|mdzM X / x2 + y2 – 2x + 6y + 1 = 0
10-20 15 4 60 – 20.5 420.25 1681

20-30 25 10 250 – 10.5 110.25 1102.5 Y 5g\ . olb X sf] lgb]{zfª\s A B

30-40 35 12 420 – 0.5 0.25 3 (2, 3) / jQ[ B sf] ;dLs/0f X(2, 3) Y
x2 + y2 – 2x + 6y + 1 = 0 5 eg]
40-50 45 8 360 9.5 90.25 722
j[Q A sf] ;dLs/0f lgsfNg'xf;] \ .
50-60 55 6 330 19.5 380.25 2281.5
;fy} /v] f XY sf] ;dLs/0f klg
Total  = 40 1420 5790
kQf nufpg'xf]; .

Here, N =  = 40, m = 1420  In the given figure, the centres of two equal circles A

Mean (x– ) m 1420 and B are X and Y respectively. If the coordinates of X
N 40
  = = = 35.5 are (2, 3) and the equation of circle B is

Again, d2 = 5790 x2 + y2 – 2x + 6y + 1 = 0 then find the equation of the

circle A. Also, find the equation of line XY.

So, standard deviation () = d2  Here, given equation of circle B is;
N x2 + y2 – 2x + 6y + 1 = 0

= 5790 = 144.75 = 12.03 or, x2 – 2x + y2 + 6y + 1 = 0
40 or, x2 – 2 . x . 1 + 12 + y2 + 2 . 3y + 32 = 12 + 32 – 1
or, (x – 1)2 + (y + 3)2 = 32
Thus, standard deviation is 12.03.

By: ENISHCHAL LAMA BAJYU (O. MATH)

Comparing it with (x – h)2 + (y – k)2 = r2, we get, 25. olb E1 = [(1, 2); 4] / E2 = [(1, 2); 1 ] lj:tf/x¿ eP
2
radius (r) = 3 units and centre Y(1, – 3)

which is also the radius of circle A. E1BE2 kQf nufpgx' f;] \ . s'g} Pp6f ;dsf]0fL lqeh' sf]
lgbz{] fª\sx¿ nv] flrqsf] dfWodaf6 nv] L E1BE2 åf/f
So, in a circle A, radius (r) = 3 units and pQm lqeh' nfO{ :yfgfGt/0f ug'{xf;] \ / kf| Kt k|ltlaDanfO{

centre (h, k) = (2, 3)

We have the equation of circle is; klg ;fx] L nv] flrqdf b]vfpg'xf];\ .
(x – h)2 + (y – k)2 = r2
If E1 = [(1, 2); 4] and E2 = [(1, 2); 1 ] are enlargements,
or, (x – 2)2 + (y – 3)2 = 32 2
or, x2 – 4x + 4 + y2 – 6y + 9 = 9

 x2 + y2 – 4x – 6y + 4 = 0

Again, equation of line XY is given by, find E1BE2. Write the co-ordinates of a right-angled
triangle in a graph. Transform the triangle by E1BE2
y – y1 = y2 – y1 (x – x1) and represent the image in same graph.
x2 – x1

or, y – 3 = –6 (x – 2)  Here, from the graph the co-ordinates of right
–1

or, y – 3 = 6x – 12 angled ABC are A(0, 1), B(1, 0) and C(1, 1).

 6x – y = 9 1
2
Thus, the required equation of circle A is;  We have given E1 = [(1, 2);4] and E2 = [(1, 2); ]
x2 + y2 – 4x – 6y + 4 = 0 and equation of lines

XY is 6x – y = 9. [ ]So, E1 E2 = (1, 2); 4 × 1 = [(1, 2); 2]
2
24. lqe'hdf u?' TjsG] bs| f] kl/efiff lbg'xf;] \ . PQR df

dWo/v] f QS / u?' Tjs]Gb| G 5g\ . olb O pb\ud laGb' xf] We know that,

( )eg,] Object Image
kd| fl0ft ugx{' f;] \ M OG = 1 OP + OQ + OR P(x, y) E[(a, b), k] P'(k(x – a) + a, k(y – b) + b)
3 E[(1, 2), 2]
P(x, y) P'(2(x – 1) + 1, 2(y – 2) + 2)
Define centroid of a triangle. In PQR, QS is a median
= P'(2x – 2 + 1, 2y – 4 + 2)
and G is the centroid. If O is the origin then prove

( )that:
OG = 1  = P'(2x – 1, 2y – 2)
3 OP + OQ + OR A'(2 × 0 – 1, 2 × 1 – 2)
= A'(– 1, 0)
 Here, the centroid is a point lying in G A(0, 1) E[(1, 2), 2] B'(2 × 1 – 1, 2 × 0 – 2)
median of a triangle which divides B(1, 0) E[(1, 2), 2]
the median in the ratio of 2 : 1.

 let O be the origin and G be the

centroid of PQR. = B'(1, – 2)
In PQR; QS is a median. C(1, 1) E[(1, 2), 2] C'(2 × 1 – 1, 2 × 1 – 2)
So, S is the mid point of PR.

(i) OS = 1 (OP + OR ) = C'(1, 0)
2
Now the graphical representation is shown below:
Since, centroid (G) divides median (QS) in the Thus, the co-ordinates of final image are as follows:
ratio of 2 : 1 = m : n A'(– 1, 0), B'(1, – 2) and C'(1, 0).

So, using section formula

(ii) OG = m OS + n OQ Y
m + n

2 
OS + 1 OQ P
= 2+1 C
C'
2 × 1 (OP + OR ) + OQ A B
2 3
= [from (i)]

= OP + OR + OQ X' A' X
3 O

ˆ OG = 1 (  +  
3 OP OQ + OR )

Thus, the position vector of centroid

i.e. OG = 1 (  +   Proved. Y' B'
3 OP OQ + OR )
–0–

ENISHCHAL'S SEE MODEL 5. (a) Pp6} cf/lDes laGb' ePsf a / b aLrsf] :sn] f/
2076 O. MATH WITH u0' fgkmnnfO{ kl/efifLt ug{x' f];\ .
GROUP A SOLU[ 5T×I(O1 +N1 ) = 10 ]
Define the scalar product between a and b
having same initial point.

( )1. b 1 'n' n] hxf“  Here, if a and b are two vectors having same
(a) ;q" r = a s] hgfpb“ 5 initial point then the scalar product of a and b
n + 1 df is defined by a. b = |a |.|b| cos 

;ª\st] x¿sf cfg} k|rlnt cy{ 5g\ <
What does 'n' represent in the formula
1 (b) olb ;“us} f] pTjm| d jQ[ sf] I
( )r = b cw{Jof; 8 cm eP OP × OP'
a n + 1 where the symbols have their own kQf nufpgx' f];\ . r
OP
usual meanings? If the radius of the adjoining P'
  Here, a, b, r represent the first term, last term, inversion circle is 8 cm then
find OP × OP'.
common ratio of a GP. Now, n represents
number of geometric means between a and b.

(b) ax'kbLosf] z]if ;fWo pNnv] ug{x' f;] \ .  Here, the radius of the given inversion circle is
r = 8 cm
State the remainder theorem of polynomial.) Now, OP × OP' = r2 = (8 cm)2 = 64 cm2.
 Here, if (x) is a polynomial of degree n and
GROUP B [ 13 × 2 = 26 ]
(x – a) is a divisor of (x) then (a) is remainder.

2. (a) afofa“ f6 lnOg] ;LdfGt dfgnfO{ ;fª\s]lts k:| tt' Ldf
nV] gx' f;] \ . 6. (a) olb (x + 2) = 2x + 3 eP – 1(4) kQf nufpg'xf];\ .
Write the symbolic notation of left hand limit.
If (x + 2) = 2x + 3 then find – 1(4).
 Here, the notation of left hand limit at the point
 Here, (x + 2) = 2x + 3
x = a is lim (x).
x  a– When x  x – 2 then,

 m i sf] or, (x – 2 + 2) = 2(x – 2) + 3
n a
(b) l86/ldGofG6 kQf nufpg'xf;] \ . or, (x) = 2x – 4 + 3

m i or, (x) = 2x – 1 = y
n a
Find the determinant of  . Now, interchanging the role of x and y then,

2y – 1 = x

 Here,  m i  = m × a – n × i = am – in. or, 2y = x + 1
n a
or, y = x + 1
2
3. (a) Pp6f ;dsf]0fL ;fn] Lsf] cfwf/;u“ k|ltR5]bg ug{]
;dtn ;tx ;dfgfGt/ eP o;af6 s'g zfªl\ ss or,  – 1 (x) = x + 1
efu aGb5 < 2

If the intersection plane is parallel to the base of   – 1(4) = 4 + 1 = 5
cone, what conic does it form? 2 2
 Here, if the intersection plane is parallel to the
Thus the value of – 1 (4) is 52.
base of cone then the conic represents a circle.

(b) ax2 + 2hxy + by2 = 0 n] hgfpg] /]vfx¿ aLrsf] (b) x3 – 6x2 + 11x – 8 df slt hf8] \bf Pp6f u0' fgv08
sf]0f kQf nufpg] ;q" s] xG' 5 < (x – 3) ePsf] ax'kbLo agfpg ;lsG5 <

What is the formula of angle between the lines What must be added to x3 – 6x2 + 11x – 8 to make
represented by ax2 + 2hxy + by2 = 0? it a polynomial having a factor (x – 3)?
 Here, if  be the angle between the lines
represented by ax2 + 2hxy + by2 = 0 then  Here, let k must be added to (x) to make it a
polynomial having a factor (x – 3).
tan  =  2 h2 – ab Now, (x) = x3 – 6x2 + 11x – 8 + k has a factor
+b
a (x – 3) i.e. remainder (3) = 0.
or, 33 – 6 × 32 + 11 × 3 – 8 + k = 0
( ) or, 27 – 54 + 33 – 8 + k = 0
 = tan – 1  2 h2 – ab or, k + 60 – 62 = 0
a+b or, k – 2 = 0

4. (a) sin 2A nfO{ tan A sf] ¿kdf JoQm ug{x' f;] \ .  k=2
Express sin 2A in terms of tan A. Thus, 2 must be added to the given polynomial.
2 tan A
  Here, sin 2A = 1 + tan2A

(b)  sf] sg' Go"gsf0] fLo dfgn] lqsf]0fldtLo ;dLs/0f (c) Pp6f ju{ ;dLs/0fsf] kf/fafn] f Y - cIf;“u
tan  – 3 = 0 nfO{ ;Gtn' g ub{5 < ;dldlt ePsf] 5 . olb ;f] kf/fafn] f (0, 0) / (2, 4)
eP/ hfG5 eg] ;dLs/0f kQf nufpgx' f];\ .
What acute value of  satisfies the trigonometric
The parabola of a quadratic equation is
equation tan  – 3 = 0? symmetric with Y-axis. If its passes through (0, 0)
and (2, 4), find the equation.
 Here, tan  = 3 = tan 60°
  = 60°

By: ENishchal Lama Bajyu (O.MATH)

 Here, the equation of the parabola symmetric And, y = 2
with y-axis and passing through (0, 0) is; x
y = ax2 ...............(i)
It passes through (2, 4). or, 2x = y
So, 4 = a × 22
or, 4 = 4a or, 2x – y = 0

 a=1 Now, (x – 2y) (2x – y) = 0
Putting value of a = 1 in (i) then, or, 2x2 – xy – 4xy + 2y2 = 0
or, 2x2 – 5xy + 2y2 = 0
y = 1.x2
 y = x2 Thus, the required homogeneous equation is
Thus, the required equation of the parabola is y = x2. 2x2 – 5xy + 2y2 = 0.

7. (a) olb A =  p 3  / B =  1 3  eP p sf] dfg (b) jQ[ (x – 2)2 + (y – 3)2 = 49 sf] s]Gb|laGb' / cwJ{ of;
2 1 2 4 kQf nufpg'xf];\ .

slt x“'bf AB Pp6f Psn dl] 6S« ; x'G5 < Find the centre and radius of circle
p 3 1 3 (x – 2)2 + (y – 3)2 = 49.
If A =  2 1  and B =  2 4 , for what value   Here, given equation is (x – 2)2 + (y – 3)2 = 49
or, (x – 2)2 + (y – 3)2 = 72 ........(i)
of p, AB is a singular matrix? Equating (i) with (x – h)2 + (y – k)2 = r2
Then we get, h = 2, k = 3, r = 7
 Here, A =  p 3  and B =  1 3 
2 1 2 4  centre (h, k) = (2, 3)
Thus, the center of the circle is (2, 3) and radius
of the circle is 7 units.

 p 3   1 3  9. (a) tflnsf jf SofNsn' ]6/sf] ko| fu] gu/L (sin 75° +
2 1 2 4 sin 15°) sf] dfg kQf nufpgx' f;] \ .
 AB = . Find the value of (sin 75° + sin 15°) without using

=  p + 6 3p + 12  table or calculator.
2 + 2 6 + 4
 Here, sin 75° + sin 15°

If AB is a singular matrix then its determinant = 2 sin 75° + 15° . cos 75° – 15°
2 2
will be zero.
90° 60°
or, |AB| = 0 = 2 sin 2 . cos 2

or,  p + 6 3p + 12  = 0 = 2 sin 45° . cos 30°
4 10
1 3 3 3
or, 10(p + 6) – 4(3p + 12) = 0 = 2× 2 × 2 = 2 = 2

or, 10p + 60 – 12p – 48 = 0 3
2
or, – 2p + 12 = 0 Thus, the required value is .

or, – 2p = – 12

 p=6 (b) kd| fl0ft ugx{' f;] \ (Prove that):

Thus, the value of p is 6. 1 + cos 2 + sin 2
1 – cos 2 + sin 2
(b) ;dLs/0fx¿ 2x + 3y = 12 / x + y = 5 df D sf] dfg = cot 
– 1 5 . jm| fd/sf] lgod ko| f]u u/L x sf] dfg kQf
nufpgx' f];\ .   Here, LHS = 1 + cos 2 + sin 2
1 – cos 2 + sin 2
In the equations 2x + 3y = 12 and x + y = 5, the
value of D = – 1. Find the value of x using = (1 + cos 2) + sin 2
Cramer's rule. (1 – cos 2) + sin 2

  Here, 2x + 3y = 12, x + y = 5 and D = – 1 = 2 cos2  + 2 sin  cos 
2 sin2  + 2 sin  cos 
12 3
 D1 =  5 1   By Cramer's rule = 2 cos  (cos  + sin )
2 sin  (sin  + cos )

= 12 × 1 – 5 × 3 = 12 – 15 = – 3 = cos  = cot  = RHS
sin 
x = D1 = – 3 =3
D – 1
(c) Gog" sf]0fsf] nflu xn ugx{' f];\ M
Thus, the value of x is 3. Solve for acute angle:

8. (a) hf]8f /]vfx¿ x = 2 / y = 2 af6 b'O{ l8u|Lsf] ju{ 2 sin2  – 3 sin  + 1 = 0
y x
 Here, 2 sin2  – 3 sin  + 1 = 0
;d3ftLo ;dLs/0f kQf nufpg'xf];\ . or, 2 sin2  – 2sin – sin  + 1 = 0
Find the homogeneous equation of second
or, 2 sin  (sin  – 1) – 1 (sin  – 1) = 0
x y
degree from the pair of lines y = 2 and x = 2. or, (sin  – 1) ( 2 sin  – 1) = 0

Either, sin  – 1 = 0 .....(i)

  Here, given equations are: or, 2 sin  – 1 = 0 .......(ii)

x =2 From (i), sin = 1 = sin 90°
y
From (ii), sin  = 1 = sin 30°
or, x = 2y 2

or, x – 2y = 0 Thus, the required acute angle is 30°.

By: ENishchal Lama Bajyu (O.MATH)

10. (a) olb a + b + c = 0, |a| = 6, |b| = 10 / |c| = 14 eP GROUP C [ 11 × 4 = 44 ]
a / b aLrsf] sf0] f kQf nufpgx' f;] \ .
11. b'O{cf]6f kmngx¿ (x) = x + 5 / ⸰g (x) = 3x – 3 5g\ .
If a + b + c = 0, |a| = 6, |b| = 10 and |c| = 14, find 4 8 4 8
the angle between a and b.
olb (g⸰)–1(x) Pp6f PsfTds kmng eP x sf] dfg kQf
  Here, if  be the angle between a and b then nufpg'xf;] \ .

a .b = |a | .| b | cos  Two functions are (x) = x + 5 and ⸰g (x) = 3x – 38.
4 8 4

  Now, a + b = – c If (g⸰)– 1 (x) is an identity function, find the value of x.

or, (a + b )2 = (–c )2  Here, (x) = x + 5 = 2x + 5
4 8 8

or, (a )2 + 2a .b + (b )2 = (–c )2 ⸰g(x) = 3x – 3 = 6x – 3
or, a2 + 2ab cos  + b2 = c2 4 8 8

or, 36 + 2 × 6 × 10 × cos  + 100 = 196 Now, (g(x)) = 6x – 3
8
or, 120 cos  = 196 – 136 = 60
2 g(x) + 5 6x – 3
or, cos  = 60 = 1 = cos 60° or, 8 = 8
120 2

Thus, the required angle is  = 60°. or, 2 g(x) + 5 = 6x – 3

(b) laGb'x¿ A / B sf l:ylt e]S6/x¿ jm| dzM (8i + 6j ) or, 2 g(x) = 6x – 3 – 5
/ (3 i + j ) 5g\ . AB nfO{ 2 : 3 sf] cgk' ftdf or, 2 g(x) = 6x – 8
leqk6\6Laf6 ljefhg ug]{ laGb' C sf] l:ylt eS] 6/
kQf nufpgx' f;] \ . or, 2 g(x) = 2(3x – 4)

The position vectors of the points A and B are or, g(x) = 2(3x – 4)
2
(8i + 6j ) and (3i + j ) respectively. Find
the position vector of the point C, which divides or, g(x) = 3x – 4
AB in the ratio of 2 : 3 internally.
Again, (g⸰) (x) = g((x))

( )=
g 2x + 5
8

( )=
  Here, if C divides AB in the ratio of 2 : 3 then, 3 2x + 5 –4
8

OC = m. OB + n. OA = 6x + 15 – 4
m+n 8 1

= 2.( 3i + j ) + 3. (8i + 6j ) = 6x + 15 – 32
2+3 8

6i + 2j + 24i + 18j = 6x – 17
5 8
=
6x – 17
30i + 20j  (g⸰) (x) = 8
5 = 6i + 4j
= We have, (g⸰) (x) = 6x – 17 =y
8
Thus, the position vector of the point C is 6i + 4j .
Interchanging the role of x and y then,
(c) Pp6f lg/Gt/ >]0fLdf klxnf] rty' fz+{ sf] dfg
rty' f{+zLo leGgtfsf] bO' { u0' ff 5 . rt'yf{+zLo 6y – 17 =x
leGgtfsf] u0' ffªs\ kQf nufpg'xf];\ . 8

In a continuous data, the value of the first or, 6y – 17 = 8x
quartile is two times of the value of quartile
deviation. Find the coefficient of the quartile or, 6y = 8x + 17
deviation.
or, y = 8x + 17
6

  Here, in a continuous data, put quartile  (g⸰)– 1(x) = 8x + 17
6
deviation (QD) = x then first quartile (Q1) = 2x
Q3 – Q1 But, according to question (g⸰)– 1(x) is an identity
2
Now, QD = function.

Q3 – 2x So, 8x + 17 =x
2 6
or, x =

or, 2x = Q3 – 2x or, 8x + 17 = 6x

 Q3 = 4x or, 8x – 6x = – 17

Again, coefficient of QD = Q3 – Q1 or, 2x = – 17
Q3 + Q1
 x = – 17
4x – 2x 2x 2 1 2
= 4x + 2x = 6x = 6 = 3 = 0.33
Thus, the value of x is – 127.
Thus, the coefficient of QD is 0.33.

By: ENishchal Lama Bajyu (O.MATH)

12. Pp6f AP sf tLgcf]6f kbx¿sf] of]ukmn 21 5 . olb Writing equations (i) and (ii) in matrix form,
bf];|f] ;ªV\ of 1 n] 36fOof] / t];f| ] ;ª\Vof 1 n] a9fOof]
eg] Pp6f GP aGb5 . tL ;ª\Vofx¿ kQf nufpg'xf;] \ .  3 5   x  =  24 
5 –2 y 9
The sum of three numbers in an AP is 21. If the
second number is reduced by 1 and the third term is Put A =  3 5 , X =  x , B =  24 
increased by 1, we obtain the GP. Find the numbers. 5 –2 y 9

 Let, a – d, a, a + d are the three numbers in AP. Now, |A| =  3 5  = 3 × (– 2) – 5 × 5
In 1st case, a – d + a + a + d = 21 5 –2

or, 3a = 21 = – 6 – 25 = – 31  0

or, a = 21  It has unique solution.
3
We know A– 1 = Adj (A) = 1  – 2 –5 
 a=7 |A| – 31 – 5 3
In 2nd case, a – d, a – 1, a + d + 1 are in GP,
Now, A.X = B
t2 t3 or, X = A– 1B
 common ratio = t1 = t2
x – 2 –5 24
or, a–1 = a + d + 1 or,  y  = – 1  – 5 3   9 
a–d a – 1 31

or, 7–1 = 7 + d + 1 = – 1  – 2 × 24 + (– 5) 9 
7–d 7 – 1 31 – 5 × 24 + 3 ×9

or, 6 = 8 + d 1  – 48 – 45 
7–d 6 31 – 120 + 27
= –

or, 36 = (8 + d) (7 – d) = – 1  – 93 
or, 36 = 56 + 7d – 8d – d2 31 – 93
or, 36 = 56 – d – d2
or, d2 + d – 56 + 36 = 0   x  =  3 
or, d2 + d – 20 = 0 y 3
or, d2 + 5d – 4d – 20 = 0
Thus, x = 3 and y = 3 is the solution.
or, d(d + 5) – 4(d + 5) = 0
or, (d + 5) (d – 4) = 0 15. P(3, 6), Q(10, 7) / R(7, 3) Pp6f j[Qsf] kl/lwsf laGb'x¿
x'g\ . olb PRQ = 90° eP j[Qsf] ;dLs/0f / cw{Jof;
Either, d + 5 = 0  d = – 5 RC sf] ;dLs/0f kQf nufpgx' f];\ .

or, d – 4 = 0  d = 4 P(3, 6), Q(10, 7) and R(7, 3) are the points at the

If d = – 5 then, AP  7 + 5, 7, 7 – 5  12, 7, 2 circumference of a circle. If PRQ = 90° then find the
equation of circle and the equation of radius (RC).
If d = 4 then, AP  7 – 4, 7, 7 + 4  3, 7, 11
Thus, the required numbers are: 12, 7, 2 or 3, 7, 11

13. kmng h(x) = 4x2 – 25 sf] lg/Gt/tf jf ljlR5Ggtf  Here, let, (3, 6) = (x1, y1) and (10, 7) = (x2, y2)
2x – 5 R(7, 3)

x = 2.5 df klxrfg ug'{xf];\ . 4x2 – 25
2x – 5
Identify whether the function h(x) = has P(3, 6) Q(10, 7)

continuity or discontinuity at x = 2.5.

 Here, (x) = 4x2 – 25 We have, Equation of circle in diameter form:
2x – 5
(x – x1) (x – x2) + (y – y1) (y – y2) = 0
The functional value = (2.5) = 4 (2.5)2 – 25
2 × 2.5 – 5 or, (x – 3) (x – 10) + (y – 6) (y – 7) = 0
or, x2 – 10x – 3x + 30 + y2 – 7y – 6y + 42 = 0
= 25 – 25 = 0
5 – 5 0  x2 + y2 – 13x – 13y + 72 = 0

The functional value is in indeterminate form. So Which is the required equation of circle.

the function is discontinuous at x = 2.5. For equation of RC, C is the midpoint of PQ.

14. dl] 6«S; ljlwaf6 xn ugx{' f;] \ M ( )So, midpoint of PQ = x1 + x2‚ y1 + y2
2 2
Solve by the matrix method:
( )= 3 +210‚ 6 + 7
3 2 2
3x + 5y = 5x – 2y =  4 3 
24 9
( ) C 13 ‚ 13 = (x1, y1) and
3 2 2 2
3x + 5y 5x – 2y  4 3 
 Here, 24 = 9 = R = (7, 3) = (x2, y2)
We have,
3x + 5y 5x – 2y
or, 24 = 9 =9–8 Equation of CR is; y – y1 = y2 – y1 (x – x1)
x2 – x1
3x + 5y 5x – 2y
or, 24 = 9 =1 3 – 13
( )or, 13 7 – 2 13
 3x + 5y = 24 .....(i) y – 2 = x – 2
13
 5x – 2y = 9 .....(ii) 2

By: ENishchal Lama Bajyu (O.MATH)

6 – 13 17. olb A + B + C = sin c eP kd| fl0ft ug{x' f;] \ M

or, 2y – 13 = 2 (2x – 13) If A + B + C = sin c then prove that:
2 – 2
14 13 cos A + cos B + cos C = 4 cos A cos B cos C – 1
2 2 2
2

or, 2y – 13 = 6 – 13 × 2 × 2x – 13  Here, A + B + C = sin c
2 2 1 2
or, A + B + C = sin 180° = 0
or, 2y – 13 = –7 (2x – 13)
2 2
or, A + B + C =0
or, 2y – 13 = – 14x + 91 2 2 2
or, 14x + 2y = 104
Thus, the equation of the radius is 14x + 2y = 104. or, A + B = – C ..........(i)
2 2 2

16. k|dfl0ft ug'{xf];\ (Prove that): LHS

sin6 c + sin6 3c + sin6 5c + sin6 7c = 5 = cos A + cos B + cos C
8 8 8 8 4
= 2 cos A + B cos A – B + cos C
2 2
 Here, LHS
c 3c 5c 7c
= sin6 8 + sin6 8 + sin6 8 + sin6 8 ( ) ( )= A + B . cos A – B + cos C
2 cos 2 2 2 2

( ) ( )= 8c
sin68c + sin638c + sin6 8 – 3c + sin6 8c – c ( ) ( )= C A B cos2 C
8 8 8 2 cos – 2 . cos 2 – 2 + 2 2 –1

( ) ( )= c
sin68c + sin638c + sin6 c – 3c + sin6 c – 8 ( )= 2 cos C . cos A – B + 2 cos2 C –1
8 2 2 2 2

( ) ( )=sin6 c sin6 3c 3c 6 c 6 [ ( ) ]=
8 8 8 8
+ + sin + sin 2 cos C cos A – B + cos C –1
2 2 2 2
c 3c 3c c
= sin6 8 + sin6 8 + sin6 8 + sin6 8 [ ( ) ( )]= C A B A + B
2 2 2 2
sin6 c sin6 3c 2 cos cos – + cos – –1
8 8
= 2 + 2 [ ( ) ( ) ]=

[ ]= 2 sin6 c + sin6 3c 2 cos C cos A – B + cos A+B –1
8 8 2 2 2 2

[ ( )]= c 48c– c [ ( ) ( )]=2cosC cos A – B A + B –1
2 sin6 8 + sin6 8 2 2 2 + cos 2 2

[ ( )]= c 2c– c [ ( ) ( )]= C A + B A – B –1
2 sin6 8 + sin6 8 2 cos 2 cos 2 2 + cos 2 2

( )= 2 sin6 c + cos c 6 = 2 cos C × 2 cos A . cos B –1
8 8 2 2 2

[ ]= 2 sin6 c + cos6 c = 4 cos A cos B cos C – 1 = RHS
8 8 2 2 2

( ) ( ) = 2 sin2 c 3 cos2 c 3 18. Pp6f w/x/fsf] prfO o;dfly /xs] f] Whb08sf]
8 8
+ prfOsf] cfwf 5 . olb hldgsf] s'g} laGba' f6 w/x/fsf]

( ) ( ) = 2 sin28c + cos28c 3 – 3.sin28c.cos28c sin28c + cos28c 6'Kkfsf] pGgtfz+ sf]0f 30° eP Wjhb08sf] 6K' kfsf]

pGgtfz+ sf0] f kQf nufpgx' f;] \ .

[ ]= 2 (1)3 – 3 sin2 c cos2 c × 1 The height of a tower is half the height of the flagstaff
8 8
at its top. The angle of elevation of the top of the

( )= 2 1 – 3 × 2 sin c cos c 2 tower as seen from any point on the ground is 30°.
4 8 8
Find the angle of elevation of the top of the flagstaff

( )= 2c 2 from the same point. C
2 × 1 – 2 × 34. 8
sin  Here, let, height of the

( )= 3 c 2 tower AB = x m and height 2x m
2 4
2 – . sin of the flagstaff BC = 2x m

= 2 – 3 (sin 45°)2 Angle of elevation of the B
2 top of the tower from any
point 'O' on the ground is xm
( )= 2 – 3 12 AOB = 30° 
2 2 Angle of elevation of the A
O 30°
3 1
= 2 – 2 × 2 top of the flagstaff from the

= 2 – 3 = 8–3 = 5 = RHS Proved. same point AOC = °
4 4 4
(suppose).

By: ENishchal Lama Bajyu (O.MATH)

Now, in right angled OAB, 20. tn lbOPsf] tYofª\ssf] dlWosfaf6 dWos leGgtf kQf
nufpg'xf];\ .
So, tan 30° = AB or, 1 = x Find the mean deviation from median of the data
OA 3 OA
given below:
or, x 3 = OA  OA = 3x ......(i)
Class interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
And, in right angled AOC,
Frequency 5 8 15 16 6
AC AB + BC x + 2x
tan  = OA = OA = 3x  Here, calculation of MD from median,

3x 3. 3 CI f cf m |D| = |m – Q2| f |D|
3x 3 115
= = = 3 0 - 10 5 5 5 23

or, tan  = tan 60° 10 - 20 8 13 15 13 104

or,  = 60° 20 - 30 15 28 25 3 45
Thus, required angle of elevation is 60°.
30 - 40 16 44 35 7 112

19. Pp6f 2 × 2 dl] 6«S; kQf nufpgx' f;] \ h;n] zLif{laGb'x¿ 40 - 50 6 50 45 17 102
A(2, 3), B(4, 3), C(4, 5) / D(2, 5) ePsf] Pp6f ju{ ABCD
nfO{ zLif{laGb'x¿ A'(3, 2), B'(3, 4), C'(5, 4) / D'(5, 2) Total 50 478
ePsf] ju{ A'B'C'D' df :yfgfGt/0f ub5{ .
( ) N th
Find the 2 × 2 transformation matrix which transforms Median (Q2) lies in 2
a square ABCD with vertices A(2, 3), B(4, 3), C(4, 5) term
and D(2, 5) into a square A'B'C'D' with vertices
A'(3, 2), B'(3, 4), C'(5, 4) and D'(5, 2). ( )= 50 th
2
= 25th term

 Median class is (20 - 30)

a b ( ) (Q2) = i N – cf
 Here, let,  c d  be the 2 × 2 transformation Median  + f 2

matrix which transforms a square ABCD in to a = 20 + 10 (25 – 13)
15
square A'B'C'D'.
2
Now,  a b   2 4 4 2  =  3 3 5 5  = 20 + 3 ×12 = 20 + 8 = 28
c d 3 3 5 5 2 4 4 2
Now, mean deviation MD = |D| = 478 = 9.56
 2a + 3b 4a + 3b 4a + 5b 2a + 5b  N 50
or, 2c + 3d 4c + 3d 4c + 5d 2c + 5d
Thus, required MD is 9.56.

=  3 3 5 5  21. tnsf] tYofªs\ af6 :t/Lo leGgtf lgsfNgx' f;] \ M
2 4 4 2 Calculate the standard deviation from the following data:

Equating the corresponding elements then, x 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50

2a + 3b = 3 .......(i) ƒ5 4 4 6 1

2a + 5b = 5 ........(ii)  Here, the calculation of SD,
x  m d = m – A d2 d d2
Now, solving (i) and (ii),

2a + 5b = 5 0 - 10 5 5 – 20 400 – 100 2000

2a + 3b = 3 10 - 20 4 15 – 10 100 – 40 400

–– – 20 - 30 4 25 0 00 0

2b = 2  b=1 30 - 40 6 35 10 100 60 600

From (i), 40 - 50 1 45 20 400 20 400

2a + 3 × 1 = 3 N = 20 – 60 3400

or, 2a + 3 = 3 Let, assumed mean (A) = 25

or, 2a = 3 – 3 ( )Now, SD = d2 d 2
N N
or, a = 0 –
2

 a=0 ( )= 3400 – – 60 2
20 20
2c + 3d = 2 .......(iii)
= 170 – (– 3)2 = 170 – 9 = 161
2c + 5d = 2 ........(iv)  SD = 12.6885 = 12.69
Thus, required S.D is 12.69.
And, solving (iii) and (iv)

2c + 5d = 2

2c + 3d = 2 GROUP D [ 4 × 5 = 20 ]

––– 22. tnsf zt{x¿sf] cfwf/df z = 2x – 3y sf] Gog" td dfg
kQf nufpg'xf;] \ .
2d = 0  d=0
Find the minimum value of z = 2x – 3y under the
From (iii), 2c + 3d = 2 following constraints.
x + y x – y  0, x  – 1 and y  2
or, 2c + 3 × 0 = 2

or, 2c + 0 = 2

or, 2c = 2

  c=1  Here, the corresponding equation of x + y  0 is x + y = 0

Thus, required 2 × 2 matrix  a b  =  0 1  x01
c d 1 0 y 0 –1

By: ENishchal Lama Bajyu (O.MATH)

Let (0, 2) be the test point then, 24. ;dafx' rte' h'{ sf] ljs0f{x¿ ;dsf0] f xg' ] u/L kl| tR5b] g
x + y = 0 + 2 = 2 > 0 (True) xG' 5g\ egL eS] 6/ ljlwaf6 kd| fl0ft ug'x{ f];\ .
So, towards the test point is shaded. Prove by vector method that, the diagonals of
The corresponding equation of x – y ≤ 0 is x – y = 0
rhombus intersect to each other at a right angle.
x02
 Here, Given: OABC is a rhombus C B
y02 in which OA = AB = CB = OC. If

Let, (0, 2) be the test point then, OA = a then CB = a and if
x – y = 0 – 2 = – 2 < 0 (True)
So, towards the test point is shaded. OC = c then AB = c . OA
x ≥ – 1 represents the half plane right from x = – 1
y ≤ 2 represents the half plane downward from y = 2. To Prove: The diagonals OB and AC intersect to

Y each other at a right angle.

Proof:

(i) |a | = |c | = OA = AB = BC = OC [Given]

B y=2 (ii) AC = OC – OA = c – a [Subtraction law]
A
C (iii) OB = OA + AB = a + c [Addition law]
X' O X
(iv) OB.AC = (a + c ).(c – a ) = (c )2 – (a )2
= c2 – a2 = a2 – a2 = 0
[From 1, 2, 3]

Y' (v) OB  AC [Since, dot product is zero.]

From the graph, the vertices of polygonal region i.e. The diagonals of rhombus intersect to each
are; O(0, 0), A(2, 2), B(– 1, 2), C(– 1, 1).
Calculation of minimum value: other at a right angle.

Vertices z = 2x – 3y Value Remarks 25. lqeh' CAT sf zLif{laGb'x¿ C(2, 5), A(– 1, 3) / T(4, 1)
5g\ . lqeh' CAT nfO{ pbu\ d laGbs' f] jl/kl/ 90° n]
wgfTds lbzfdf kl/jm| d0f u/L kml] / E[(0, 0), 2] df
O(0, 0) 2 × 0 – 3 × 0 0
lj:tf/Ls/0f ubf{ aGg] k|ltlaDax¿sf] lgbz]{ fª\sx¿ kQf
A(2, 2) 2 × 2 – 3 × 2 – 2 nufpgx' f];\ . ;fy} j:t' / kl| tlaDax¿nfO{ pxL
nv] flrqdf k:| tt' ugx'{ f];\ .
B(– 1, 2) 2 × (– 1) – 3 × 2 – 8 Minimum
C(2, 5), A(– 1, 3) and T(4, 1) are the vertices of a triangle
C(– 1, 1) 2 × (– 1) – 3 × 1 – 5 CAT. Find the coordinates of the vertices of the image of
 Thus, the minimum value is – 8 at (– 1, 2). ΔCAT under the rotation of positive 90° about origin
followed by enlargement E[(0, 0), 2]. Represent the
23. /]vf 2x – y + 7 = 0 ;“u 45° sf] sf]0f agfpg] / laGb' object and images on the same graph paper.
(4, 5) eP/ hfg] /]vfx¿sf] ;dLs/0f kQf nufpgx' f;] \ .
 Here, C(2, 5), A(– 1, 3) and T(4, 1) are the vertices
Find the equation of the straight lines passing
through the point (4, 5) and making an angle of 45° of CAT. At first rotating CAT under the rotation
with the line 2x – y + 7 = 0. of positive 90° about origin,

 Here, given equation is 2x – y + 7 = 0 ......(i) 2 Object Image
coeff. of x – 1 P(x, y) R[(0, 0), + 90°] P'(– y, x)
Slope of equation (i) is m1 = – coeff. of y = – = 2

Equation of the line passing through (4, 5) is C(2, 5) R[(0, 0), + 90°] C'(– 5, 2)
y – y1 = m(x – x1) or, y – 5 = m(x – 4) .....(ii)
A(– 1, 3) R[(0, 0), + 90°] A'(– 3, – 1)
Angle between both lines is  = 45°

( )Now, tan  = 
m1 – m2 T(4, 1) R[(0, 0), + 90°] T'(– 1, 4)
1 + m1m2
Again, enlarging C'A'T' by E[(0, 0), 2] then,
( )or, tan 45° = 
2–m Object E[(0, 0), 2] Image
1 + 2.m P(x, y) P'(2x, 2y)

or, 1.(1 + 2m) =  (2 – m)
Taking + ve sign
1 + 2m = 2 – m Taking –ve sign C'(– 5, 2) E[(0, 0), 2] C"(– 10, 4)
or, 2m + m = 2 – 1 A'(– 3, – 1) E[(0, 0), 2] A"(– 6, – 2)
or, 3m = 1 1 + 2m = – 2 + m
or, 2m – m = – 2 – 1

or, m = 1 or, 1.m = – 3 T'(– 1, 4) E[(0, 0), 2] T"(– 2, 8)
3 or, m = – 3
Y
From (ii), From (ii),
y – 5 = m(x – 4)
1 y – 5 = m(x – 4) T''
3 or, y – 5 = – 3(x – 4)
or, y – 5 = (x – 4) or, y – 5 = – 3x + 12 C
or, y – 5 + 3x – 12 = 0 C'' T'
or, 3y – 15 = x – 4
or, – 15 + 4 = x – 3y
or, x – 3y + 11 = 0 or, 3x + y – 17 = 0 C' A
T
Thus, required lines are: X' O X

x – 3y + 11 = 0 and 3x + y – 17 = 0. A'' A'

Y'