MATHEMATICS PAPER-1A
TIME: 3hrs. Max. Marks.75
Note: This question paper consists of three sections A, B and C.
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Very short answer type questions. 10X2 =20
1. Let ABCDEF be a regular hexagon with center O. Show that
AB AC AD AE AF 3AD 6AO + + + + = = .
2. In ∆ABC, if a, b, c are position vectors of the vertices A, B and C respectively, then
prove that the position vector of the centroid G is 1
and b
prove that
− = , then show that (1 – tan A) (1 + tan B) = 2.
and θ is in the third quadrant, find θ.
6. Prove that { }
1 2 sin log 1 h x x x −
= + +
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, then find AAT
8. If ω is a complex cube root of 1 then show that
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9. If A = {1, 2, 3, 4} and f : A → R is a function defined by
10. If f = {(1, 2), (2, –3), (3, –1)} then find (i) 2 + f, (ii) f .
Short answer type questions.
Answer any five of the following. 5 X 4 = 20
11. Theorem: If A is a non-singular matrix then A is invertible and 1 AdjA A
12. In ∆ABC, if O is the circumcenter and H is the orthocenter, then show that
(i) OA OB OC OH + + =
(ii) HA HB HC 2HO + + =
13. Let a, b, c be mutually orthogonal vectors of equal magnitudes. Prove that the vector
a b c + + is equally inclined to each of a, b, c , the angle of inclination being 1 1
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14. Prove that sin2α + cos2
(α + β) + 2 sin αsinβcos(α + β) is independent of α.
= + and x ∈[0, 2π ] find the values of x .
x
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16. If 1 1 1 Tan x Tan y Tan z π
− − − + + = then prove that x y z xyz + + =
tan tan tan
A B C bc ca ab s + + −
+ + =
2 2 2
Long answer type questions
Answer any five of the following 5 X 7 = 35
18. If f A B : → , g B C : → are two one one onto functions then gof A C : → is also
19. Prove that 2.3+ 3.4 + 4.5+ .... upto n terms
20. If | a | 1,| b | 1,| c | 2 = = = and a (a c) b 0 × × + = , then find the angle between a and c .
21. In a triangle ABC prove that
A B C π π π − − − A B C
cos cos cos 4cos cos cos
2 2 2 4 4 4
+ + =
22. Let an object be placed at some height h cm and let P and Q be two points of
observation which are at a distance 10 cm apart on a line inclined at angle 15° to
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the horizontal. If the angles of elevation of the object from P and Q are 30° and
60° respectively then find h.
b c c a a b a b c
+ + +
c a a b b c 2 b c a
+ + + =
a b b c c a c a b
+ + +
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24. Find the non-trivial solutions, if any, for the following equations.
2x + 5y + 6z = 0, x – 3y – 8z = 0,3x – y – 4z = 0
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Model Paper-2 Solution
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1. Let ABCDEF be a regular hexagon with center O. Show that
AB AC AD AE AF 3AD 6AO + + + + = = .
+ + + + =
( ) ( )
AB AE AD AC AF
+ + + +
( ) ( )
AE ED AD AC CD
= + + + +
( )
AB ED, AF CD
6AO( O is the center and OD AO)
= =
2. In ∆ABC, if a, b, c are position vectors of the vertices A, B and C respectively, then
prove that the positin vector of the centroid G is 1
Let G be the centroid of ∆ABC and AD be the median through the vertex A. (see figure).
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Since the position vector of D is 1
.
and b
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3. For any two vectors a
2 2 2 2 a b a b a b sin ( , ) ( . ) +
2 2 2 2 a b a b a b {1 cos ( , )} ( . ) − +
2 2 2 2 2 2 a b a b a b a b − + cos ( , ) ( . )
2
− = , then show that (1 – tan A) (1 + tan B) = 2.
= ° + ° = − ° = −
tan(90 45) cot 45 1
tan A tan B (1 tan A tan B)
tan A tan B tan A tan B 1
tan B tan A tan A tan B 1 ...(1)
− − =
L.H.S. (1 tan A)(1 tan B)
1 (tan B tan A tan A tan B)
and θ is in the third quadrant, find θ.
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° + ° = ° =
tan 45 tan11 ( tan 45 1)
6. Prove that { }
= + +
= = = − ⇒ = −
x x e xe e
( ) ( )
y y
e xe e e x x y x x
1 2 sin log 1 e h x x x − ∴ = + +
1 2 sin log 1 h x x x −
sin h x y x y sin h −
e e y y
2 2 1
4 4 2 4 4 2 2 2 1 0 1 log 1
− − = ⇒ = = = + + ⇒ = + +
4 1
ey
x x
{ }
, then find AAT
. Do A and AT
= =
= = −
− −
1 2 1 0 5 2
0 1 2 1 2 1
− − −
1 0 1 2 1 2
2 1 0 1 2 5
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≠ ATA, A and AT
do not commute with respect to multiplication of matrices.
8. If ω is a complex cube root of 1 then show that
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+ ω + ω + ω + ω + ω + ω
2 2 2
1
ω ω
1 [ 1 0]
9. If A = {1, 2, 3, 4} and f : A → R is a function defined by
1 7 13 Range of f is ,1, ,
∴
10. If f = {(1, 2), (2, –3), (3, –1)} then find (i) 2 + f, (ii) f .
2 4 5
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f = {(1, 2), (2, –3), (3, –1)}
i) 2 + f = {(1, 2+2), (2, –3+2), (3, –1+2)}
= {(1, 4), (2, –1), (3, 1)}
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11. Theorem : If A is a non-singular matrix then A is invertible and 1 AdjA A
a b c
1 1 1
a b c
2 2 2
a b c
3 3 3
be a non-singular matrix.
A AdjA a b c B B B
+ + + + + +
a A b B c C a A b B c C a A b B c C
1 1 1 1 1 1 1 2 1 2 1 2 1 3 1 3 1 3
a A b B c C a A b B c C a A b B c C
= + + + + + +
2 1 2 1 2 1 2 2 2 2 2 2 2 3 2 3 2 3
+ + + + + +
a A b B c C a A b B c C a A b B c C
3 1 3 1 3 1 3 2 3 2 3 2 3 3 3 3 3 3
det A 0 0 1 0 0
0 det A 0 det A 0 1 0
= =
0 0 det A 0 0 1
a b c A A A
1 1 1 1 2 3
2 2 2 1 2 3
a b c C C C
3 3 3 1 2 3
Similarly we can prove that AdjA A I
⋅ =
det A
12. In ∆ABC, if O is the circumcenter and H is the orthocenter, then show that
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Sol. Let D be the mid point of BC.
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B C
Let OA a,OB b and OC c = = =
OA 2OD OA AH OH
+ = + =
(Observe that AH 2R cos A,OD R cos A, = =
R is the circum radius of ∆ABC and hence AH 2OD = ).
HA 2HD HA 2(HO OD)
+ = + +
13. Let a, b, c be mutually orthogonal vectors of equal magnitudes. Prove that the
vector a b c + + is equally inclined to each of a, b, c , the angle of inclination
Now, 2 2 2 2 | a b c | a b c 2 a b + + = + + + Σ ⋅
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= λ ⋅ = ⋅ = ⋅ = 3 ( a b b c c a 0) ∵
Let θ be the angle between a and a b c + +
Then a (a b c) a a 1
Similarly, it can be proved that a b c + + inclines at an angle of 1 1
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= α + − α + β + α β α + β
sin 1 sin ( ) 2sin sin cos( )
= + α − α + β + α β α + β
1 [sin sin ( )] 2sin sin cos( )
= + α + α + β α − α − β + α β α + β
1 sin( )sin( ) 2sin sin cos( )
= + α + β −β + α β α + β
1 sin(2 )sin( ) 2sin sin cos( )
= − α + β β + α α + β β
1 sin(2 )sin [2sin cos( )]sin
1 sin(2 )sin [sin( ) sin( )]sin
= − α + β α + α + α + β + α − α −β β
= − α + β α + α + β − β β
1 sin(2 )sin [sin(2 ) sin ]sin
= − α + β α + α + β β − β
1 sin(2 )sin sin(2 )sin sin
Thus the given expression is independent of α.
⋅ + + ⋅ θ = = =
| a || a b c | ( 3) 3
+ + λ λ
(α + β) + 2 sin αsinβcos(α + β) is independent of α.
(α+β) + 2 sinαsinβcos(α+β)
= + and x ∈[0, 2π ] find the values of x
x
cos 1( )suppose tan 0 x >
∴ = + ⇒ = not possible
tan tan 0
cos cos
tan tan 2 tan
∴− = + ⇒ − =
x x x
1 1
x x
1 1
cos cos
x x
− = x x ⇒ = −
x Lies in (iii) or (iv) quadrant
∴ = − x or π π / 6 ( ) 11 / 6
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16. If (i) 1 1 1 Tan x Tan y Tan z π
(ii) 1 1 1
Let 1 1 1 tan tan tan x y z α β γ
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x Tan = α y Tan = β z Tan = γ
Given α β γ π α β π + + = ⇒ + = − x
( ) ( )
= x y z xyz + + =
− − − + + = then prove that x y z xyz + + =
Tan x Tan y Tan z − − − π
+ + = then prove that xy yz zx + + = 1
− − −
= = =
Tan Tan Tan Tan Tan
α β π γ γ
+ = − ⇒ =
Tan Tan Tan Tan Tan Tan Tan Tan Tan α β γ α β γ α β γ + = − + ⇒ + + Tan Tan Tan α β γ
A B C bc ca ab s + + −
tan tan tan
2 2 2
+ + =
(s b s c s c s a s a s b − − − − − − )( ) ( )( ) ( )( )
2 2 2 s cs bs bc s as cs ac s bs as ab − − + + − − + + − − +
3 2 2 2 s as bs cs bc ca ab − − − + + +
bc ca ab s s a b c bc ca ab s s s + + + − + + + + + − 3 2 3 2 2
2 2 2 bc ca ab s s bc ca ab s + + + − + + − 3 4
+ +
∆ ∆ ∆
∆
2 2
( ) ( )
∆ ∆
=
∆ ∆
18. If f A B : → , g B C : → are two one one onto functions then gof A C : → is also one
Sol: i) Let 1 2 x x A , ∈ and 1 2 f x f x ( ) ( ) = .
1 2 1 2 x x A f A B f x f x B , , : ( ), ( ) ∈ → ⇒ ∈
1 2 2 1 2 1 2 f x f x B C f x g f x g f x gof x gof x ( ), ( ) , , ( ) [ ( )] [ ( )] ( )( ) ( )( ) ∈ → ⇒ = ⇒ =
1 2 1 x x A gof x gof A C , ,( )( ) ( ): ∈ = → is one one 1 2 ⇒ x x =
1 2 1 2 1 2 ∴ ∈ = x x A f x f x x x , , ( ) ( )⇒ = .
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∴ → f A B : is one one.
Now ( )( ) ( ) gof x g y z = =
z C g B C ∈ → , : is onto ⇒ ∃ ∈ ∋ = y B f x y ( )
∴ ∈z C x A gof x z ⇒ ∃ ∈ ∋ = ( )( ) .
∴ → gof A C : is onto.
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19. Prove that 2.3+ 3.4 + 4.5+ .... upto n terms
Sol: 2, 3, 4..................... n terms 2 ( 1)1 1 n
3, 4, 5........... n terms 3 ( 1)1 2 n
( 6 11) 2.3 3.4 4.5 ......... ( 1)( 2)
For n = 1 L.H.S = 2.3 = 6
1(1 6 11) . . 6
+ + + + + + =
S be the given statement
n n
( 6 11) 2.3 3.4 4.5 ....... ( 1)( 2) ( 2)( 3)
∴ + + + + + + = + + +
k k k k k k k + +
2 2 ( 6 11) 3( 5 6)
k k k k k + + + + +
=
3 2 9 26 18
k k k + + +
=
( 1) 8 18
k k k + + +
=
( 1) ( 1) 6( 1) 11
k k k + + + + +
=
Hence ( ) n S is true for all n N ∈
20. If | a | 1,| b | 1,| c | 2 = = = and a (a c) b 0 × × + = , then find the angle between a and c
Sol. Given that | a | 1,| b | 1,| c | 2 = = =
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Consider a c | a || c | cos ⋅ = θ
2cos ...(1)
Consider a (a c) b 0 × × + =
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(2cos )a (1)c b 0 ...(2)
Squaring on both sides
(4cos )(a) (c) 4cos (a c) b
⇒ θ + − θ ⋅ =
⇒ θ + − θ θ =
4cos (1) (2) 4cos (2cos ) 1
4cos 4 8cos 1
2 2 2 2
2 3 3
4 2
3 5 If cos 150
⇒ = = °
π π θ = − ⇒ θ = π − = = °
2 6 6
5
π
(a, c) 150
6
21. In a triangle ABC prove that
A B C π π π − − − A B C
cos cos cos 4cos cos cos
+ + =
2 2 2 4 4 4
π π π π π π − − − A B C − − A B − C
=
4cos cos cos 2 cos cos 2cos
4 4 4 4 4 4
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= +
π π π π π − + − A B A B − − + −C
cos cos 2cos
{∵ 2cos cos cos cos A B A B A B = + + + ( ) ( )}
= − +
π A B A B + − − π C
cos cos 2cos
2 4 4 4
π π − + − − C A B C A B
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2cos sin 2cos cos
= +
4 4 4 4
π π π π − + + − − − C A B C A B − + − C A B − − + C A B
sin sin cos cos
= − + +
4 4 4 4
= + − −
A B A B A B
2cos sin sin sin
A B A B A B
2cos cos cos cos
π π − + − + + − − − + + − + − C C A B C C A B A B C C A B ∴ − +
sin sin cos
4 4 4
+ + = −
= − + +
= + +
+ + − − + = + +
A B C C A B A B C
π C A B
sin cos cos
2 2 2 2
A B C
cos cos cos
2 2 2
22. Let an object be placed at some height h cm and let P and Q be two points of observation
which are at a distance 10 cm apart on a line inclined at angle 15° to the horizontal. If the
angles of elevation of the object from P and Q are 30° and 60° respectively then find h.
A
h
E
P B
A is the position of the object.
P and Q are points of observation.
∠BPE = 15°, ∠EPA = 30°, ∠EQA = 60°
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P = 30°, Q = 120° and A = 30°
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AP AP 20 20
= ° ⇒ = °
10 3 5 2 3 h 5 2 3 5 6 cm
= = = =
By applying R1⇒R1 + R2 + R3
2(a b c) 2(a b c) 2(a b c)
= + + +
2 c a a b b c
= + + +
By applying R2→R2 – R1 and R3→R3 – R1
2 b c a
= − − −
By applying R1→R1 + R2 + R3
b c c a a b a b c
+ + +
c a a b b c 2 b c a
+ + + =
a b b c c a c a b
+ + +
b c c a a b
+ + +
c a a b b c
+ + +
a b b c c a
+ + +
+ + + + + +
c a a b b c
a b b c c a
+ + +
a b c a b c a b c
+ + + + + +
a b b c c a
+ + +
a b c a b c a b c
+ + + + + +
c a b
− − −
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24. Find the non-trivial solutions, if any, for the following equations.
2x + 5y + 6z = 0, x – 3y – 8z = 0,3x – y – 4z = 0
Sol. The coefficient matrix A =
On interchanging R1 and R2, we get
2 5 6
1 3 8
− −
−
3 1 4
R R 2R , R R 2R we get
→ − → −
2 2 1 3 3 1
Det A = 0 as R2 and R3 are identical.
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Clearly rank (A) = 2, as the sub-matrix
Hence the system has non-trivial solution.
The following system of equations is equivalent to the given system of equations.
On giving an arbitrary value k to z, we obtain the solution set is
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x = 2k, y = –2k, z = k, k ∈ R for k ≠ 0.
We obtain non-trivial solutions.
TIME: 3hrs. Max. Marks.75
Note: This question paper consists of three sections A, B and C.
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Very short answer type questions. 10X2 =20
1. Let ABCDEF be a regular hexagon with center O. Show that
AB AC AD AE AF 3AD 6AO + + + + = = .
2. In ∆ABC, if a, b, c are position vectors of the vertices A, B and C respectively, then
prove that the position vector of the centroid G is 1
and b
prove that
− = , then show that (1 – tan A) (1 + tan B) = 2.
and θ is in the third quadrant, find θ.
6. Prove that { }
1 2 sin log 1 h x x x −
= + +
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, then find AAT
8. If ω is a complex cube root of 1 then show that
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9. If A = {1, 2, 3, 4} and f : A → R is a function defined by
10. If f = {(1, 2), (2, –3), (3, –1)} then find (i) 2 + f, (ii) f .
Short answer type questions.
Answer any five of the following. 5 X 4 = 20
11. Theorem: If A is a non-singular matrix then A is invertible and 1 AdjA A
12. In ∆ABC, if O is the circumcenter and H is the orthocenter, then show that
(i) OA OB OC OH + + =
(ii) HA HB HC 2HO + + =
13. Let a, b, c be mutually orthogonal vectors of equal magnitudes. Prove that the vector
a b c + + is equally inclined to each of a, b, c , the angle of inclination being 1 1
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14. Prove that sin2α + cos2
(α + β) + 2 sin αsinβcos(α + β) is independent of α.
= + and x ∈[0, 2π ] find the values of x .
x
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16. If 1 1 1 Tan x Tan y Tan z π
− − − + + = then prove that x y z xyz + + =
tan tan tan
A B C bc ca ab s + + −
+ + =
2 2 2
Long answer type questions
Answer any five of the following 5 X 7 = 35
18. If f A B : → , g B C : → are two one one onto functions then gof A C : → is also
19. Prove that 2.3+ 3.4 + 4.5+ .... upto n terms
20. If | a | 1,| b | 1,| c | 2 = = = and a (a c) b 0 × × + = , then find the angle between a and c .
21. In a triangle ABC prove that
A B C π π π − − − A B C
cos cos cos 4cos cos cos
2 2 2 4 4 4
+ + =
22. Let an object be placed at some height h cm and let P and Q be two points of
observation which are at a distance 10 cm apart on a line inclined at angle 15° to
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the horizontal. If the angles of elevation of the object from P and Q are 30° and
60° respectively then find h.
b c c a a b a b c
+ + +
c a a b b c 2 b c a
+ + + =
a b b c c a c a b
+ + +
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24. Find the non-trivial solutions, if any, for the following equations.
2x + 5y + 6z = 0, x – 3y – 8z = 0,3x – y – 4z = 0
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Model Paper-2 Solution
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1. Let ABCDEF be a regular hexagon with center O. Show that
AB AC AD AE AF 3AD 6AO + + + + = = .
+ + + + =
( ) ( )
AB AE AD AC AF
+ + + +
( ) ( )
AE ED AD AC CD
= + + + +
( )
AB ED, AF CD
6AO( O is the center and OD AO)
= =
2. In ∆ABC, if a, b, c are position vectors of the vertices A, B and C respectively, then
prove that the positin vector of the centroid G is 1
Let G be the centroid of ∆ABC and AD be the median through the vertex A. (see figure).
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Since the position vector of D is 1
.
and b
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3. For any two vectors a
2 2 2 2 a b a b a b sin ( , ) ( . ) +
2 2 2 2 a b a b a b {1 cos ( , )} ( . ) − +
2 2 2 2 2 2 a b a b a b a b − + cos ( , ) ( . )
2
− = , then show that (1 – tan A) (1 + tan B) = 2.
= ° + ° = − ° = −
tan(90 45) cot 45 1
tan A tan B (1 tan A tan B)
tan A tan B tan A tan B 1
tan B tan A tan A tan B 1 ...(1)
− − =
L.H.S. (1 tan A)(1 tan B)
1 (tan B tan A tan A tan B)
and θ is in the third quadrant, find θ.
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° + ° = ° =
tan 45 tan11 ( tan 45 1)
6. Prove that { }
= + +
= = = − ⇒ = −
x x e xe e
( ) ( )
y y
e xe e e x x y x x
1 2 sin log 1 e h x x x − ∴ = + +
1 2 sin log 1 h x x x −
sin h x y x y sin h −
e e y y
2 2 1
4 4 2 4 4 2 2 2 1 0 1 log 1
− − = ⇒ = = = + + ⇒ = + +
4 1
ey
x x
{ }
, then find AAT
. Do A and AT
= =
= = −
− −
1 2 1 0 5 2
0 1 2 1 2 1
− − −
1 0 1 2 1 2
2 1 0 1 2 5
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≠ ATA, A and AT
do not commute with respect to multiplication of matrices.
8. If ω is a complex cube root of 1 then show that
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+ ω + ω + ω + ω + ω + ω
2 2 2
1
ω ω
1 [ 1 0]
9. If A = {1, 2, 3, 4} and f : A → R is a function defined by
1 7 13 Range of f is ,1, ,
∴
10. If f = {(1, 2), (2, –3), (3, –1)} then find (i) 2 + f, (ii) f .
2 4 5
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f = {(1, 2), (2, –3), (3, –1)}
i) 2 + f = {(1, 2+2), (2, –3+2), (3, –1+2)}
= {(1, 4), (2, –1), (3, 1)}
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11. Theorem : If A is a non-singular matrix then A is invertible and 1 AdjA A
a b c
1 1 1
a b c
2 2 2
a b c
3 3 3
be a non-singular matrix.
A AdjA a b c B B B
+ + + + + +
a A b B c C a A b B c C a A b B c C
1 1 1 1 1 1 1 2 1 2 1 2 1 3 1 3 1 3
a A b B c C a A b B c C a A b B c C
= + + + + + +
2 1 2 1 2 1 2 2 2 2 2 2 2 3 2 3 2 3
+ + + + + +
a A b B c C a A b B c C a A b B c C
3 1 3 1 3 1 3 2 3 2 3 2 3 3 3 3 3 3
det A 0 0 1 0 0
0 det A 0 det A 0 1 0
= =
0 0 det A 0 0 1
a b c A A A
1 1 1 1 2 3
2 2 2 1 2 3
a b c C C C
3 3 3 1 2 3
Similarly we can prove that AdjA A I
⋅ =
det A
12. In ∆ABC, if O is the circumcenter and H is the orthocenter, then show that
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Sol. Let D be the mid point of BC.
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B C
Let OA a,OB b and OC c = = =
OA 2OD OA AH OH
+ = + =
(Observe that AH 2R cos A,OD R cos A, = =
R is the circum radius of ∆ABC and hence AH 2OD = ).
HA 2HD HA 2(HO OD)
+ = + +
13. Let a, b, c be mutually orthogonal vectors of equal magnitudes. Prove that the
vector a b c + + is equally inclined to each of a, b, c , the angle of inclination
Now, 2 2 2 2 | a b c | a b c 2 a b + + = + + + Σ ⋅
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= λ ⋅ = ⋅ = ⋅ = 3 ( a b b c c a 0) ∵
Let θ be the angle between a and a b c + +
Then a (a b c) a a 1
Similarly, it can be proved that a b c + + inclines at an angle of 1 1
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= α + − α + β + α β α + β
sin 1 sin ( ) 2sin sin cos( )
= + α − α + β + α β α + β
1 [sin sin ( )] 2sin sin cos( )
= + α + α + β α − α − β + α β α + β
1 sin( )sin( ) 2sin sin cos( )
= + α + β −β + α β α + β
1 sin(2 )sin( ) 2sin sin cos( )
= − α + β β + α α + β β
1 sin(2 )sin [2sin cos( )]sin
1 sin(2 )sin [sin( ) sin( )]sin
= − α + β α + α + α + β + α − α −β β
= − α + β α + α + β − β β
1 sin(2 )sin [sin(2 ) sin ]sin
= − α + β α + α + β β − β
1 sin(2 )sin sin(2 )sin sin
Thus the given expression is independent of α.
⋅ + + ⋅ θ = = =
| a || a b c | ( 3) 3
+ + λ λ
(α + β) + 2 sin αsinβcos(α + β) is independent of α.
(α+β) + 2 sinαsinβcos(α+β)
= + and x ∈[0, 2π ] find the values of x
x
cos 1( )suppose tan 0 x >
∴ = + ⇒ = not possible
tan tan 0
cos cos
tan tan 2 tan
∴− = + ⇒ − =
x x x
1 1
x x
1 1
cos cos
x x
− = x x ⇒ = −
x Lies in (iii) or (iv) quadrant
∴ = − x or π π / 6 ( ) 11 / 6
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16. If (i) 1 1 1 Tan x Tan y Tan z π
(ii) 1 1 1
Let 1 1 1 tan tan tan x y z α β γ
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x Tan = α y Tan = β z Tan = γ
Given α β γ π α β π + + = ⇒ + = − x
( ) ( )
= x y z xyz + + =
− − − + + = then prove that x y z xyz + + =
Tan x Tan y Tan z − − − π
+ + = then prove that xy yz zx + + = 1
− − −
= = =
Tan Tan Tan Tan Tan
α β π γ γ
+ = − ⇒ =
Tan Tan Tan Tan Tan Tan Tan Tan Tan α β γ α β γ α β γ + = − + ⇒ + + Tan Tan Tan α β γ
A B C bc ca ab s + + −
tan tan tan
2 2 2
+ + =
(s b s c s c s a s a s b − − − − − − )( ) ( )( ) ( )( )
2 2 2 s cs bs bc s as cs ac s bs as ab − − + + − − + + − − +
3 2 2 2 s as bs cs bc ca ab − − − + + +
bc ca ab s s a b c bc ca ab s s s + + + − + + + + + − 3 2 3 2 2
2 2 2 bc ca ab s s bc ca ab s + + + − + + − 3 4
+ +
∆ ∆ ∆
∆
2 2
( ) ( )
∆ ∆
=
∆ ∆
18. If f A B : → , g B C : → are two one one onto functions then gof A C : → is also one
Sol: i) Let 1 2 x x A , ∈ and 1 2 f x f x ( ) ( ) = .
1 2 1 2 x x A f A B f x f x B , , : ( ), ( ) ∈ → ⇒ ∈
1 2 2 1 2 1 2 f x f x B C f x g f x g f x gof x gof x ( ), ( ) , , ( ) [ ( )] [ ( )] ( )( ) ( )( ) ∈ → ⇒ = ⇒ =
1 2 1 x x A gof x gof A C , ,( )( ) ( ): ∈ = → is one one 1 2 ⇒ x x =
1 2 1 2 1 2 ∴ ∈ = x x A f x f x x x , , ( ) ( )⇒ = .
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∴ → f A B : is one one.
Now ( )( ) ( ) gof x g y z = =
z C g B C ∈ → , : is onto ⇒ ∃ ∈ ∋ = y B f x y ( )
∴ ∈z C x A gof x z ⇒ ∃ ∈ ∋ = ( )( ) .
∴ → gof A C : is onto.
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19. Prove that 2.3+ 3.4 + 4.5+ .... upto n terms
Sol: 2, 3, 4..................... n terms 2 ( 1)1 1 n
3, 4, 5........... n terms 3 ( 1)1 2 n
( 6 11) 2.3 3.4 4.5 ......... ( 1)( 2)
For n = 1 L.H.S = 2.3 = 6
1(1 6 11) . . 6
+ + + + + + =
S be the given statement
n n
( 6 11) 2.3 3.4 4.5 ....... ( 1)( 2) ( 2)( 3)
∴ + + + + + + = + + +
k k k k k k k + +
2 2 ( 6 11) 3( 5 6)
k k k k k + + + + +
=
3 2 9 26 18
k k k + + +
=
( 1) 8 18
k k k + + +
=
( 1) ( 1) 6( 1) 11
k k k + + + + +
=
Hence ( ) n S is true for all n N ∈
20. If | a | 1,| b | 1,| c | 2 = = = and a (a c) b 0 × × + = , then find the angle between a and c
Sol. Given that | a | 1,| b | 1,| c | 2 = = =
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Consider a c | a || c | cos ⋅ = θ
2cos ...(1)
Consider a (a c) b 0 × × + =
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(2cos )a (1)c b 0 ...(2)
Squaring on both sides
(4cos )(a) (c) 4cos (a c) b
⇒ θ + − θ ⋅ =
⇒ θ + − θ θ =
4cos (1) (2) 4cos (2cos ) 1
4cos 4 8cos 1
2 2 2 2
2 3 3
4 2
3 5 If cos 150
⇒ = = °
π π θ = − ⇒ θ = π − = = °
2 6 6
5
π
(a, c) 150
6
21. In a triangle ABC prove that
A B C π π π − − − A B C
cos cos cos 4cos cos cos
+ + =
2 2 2 4 4 4
π π π π π π − − − A B C − − A B − C
=
4cos cos cos 2 cos cos 2cos
4 4 4 4 4 4
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= +
π π π π π − + − A B A B − − + −C
cos cos 2cos
{∵ 2cos cos cos cos A B A B A B = + + + ( ) ( )}
= − +
π A B A B + − − π C
cos cos 2cos
2 4 4 4
π π − + − − C A B C A B
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2cos sin 2cos cos
= +
4 4 4 4
π π π π − + + − − − C A B C A B − + − C A B − − + C A B
sin sin cos cos
= − + +
4 4 4 4
= + − −
A B A B A B
2cos sin sin sin
A B A B A B
2cos cos cos cos
π π − + − + + − − − + + − + − C C A B C C A B A B C C A B ∴ − +
sin sin cos
4 4 4
+ + = −
= − + +
= + +
+ + − − + = + +
A B C C A B A B C
π C A B
sin cos cos
2 2 2 2
A B C
cos cos cos
2 2 2
22. Let an object be placed at some height h cm and let P and Q be two points of observation
which are at a distance 10 cm apart on a line inclined at angle 15° to the horizontal. If the
angles of elevation of the object from P and Q are 30° and 60° respectively then find h.
A
h
E
P B
A is the position of the object.
P and Q are points of observation.
∠BPE = 15°, ∠EPA = 30°, ∠EQA = 60°
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P = 30°, Q = 120° and A = 30°
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AP AP 20 20
= ° ⇒ = °
10 3 5 2 3 h 5 2 3 5 6 cm
= = = =
By applying R1⇒R1 + R2 + R3
2(a b c) 2(a b c) 2(a b c)
= + + +
2 c a a b b c
= + + +
By applying R2→R2 – R1 and R3→R3 – R1
2 b c a
= − − −
By applying R1→R1 + R2 + R3
b c c a a b a b c
+ + +
c a a b b c 2 b c a
+ + + =
a b b c c a c a b
+ + +
b c c a a b
+ + +
c a a b b c
+ + +
a b b c c a
+ + +
+ + + + + +
c a a b b c
a b b c c a
+ + +
a b c a b c a b c
+ + + + + +
a b b c c a
+ + +
a b c a b c a b c
+ + + + + +
c a b
− − −
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24. Find the non-trivial solutions, if any, for the following equations.
2x + 5y + 6z = 0, x – 3y – 8z = 0,3x – y – 4z = 0
Sol. The coefficient matrix A =
On interchanging R1 and R2, we get
2 5 6
1 3 8
− −
−
3 1 4
R R 2R , R R 2R we get
→ − → −
2 2 1 3 3 1
Det A = 0 as R2 and R3 are identical.
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Clearly rank (A) = 2, as the sub-matrix
Hence the system has non-trivial solution.
The following system of equations is equivalent to the given system of equations.
On giving an arbitrary value k to z, we obtain the solution set is
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x = 2k, y = –2k, z = k, k ∈ R for k ≠ 0.
We obtain non-trivial solutions.
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