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. 10 X 2 = 20
1. Find the domain of the function
2x 5x 7 f (x)
2. If f(x + y) = f(xy) ∀x, y ∈ R then prove that f is a constant function.
3. ABCDE is a pentagon. If the sum of the vectors AB, AE, BC, DE, ED and AC is
λAC, then find the value of λ.
4. If the position vectors of the points A, B, C are − + − − + + 2 i j k, 4 i 2 j 2k and
6 i 3 j 13k − − respectively and AB AC = λ then find the value of λ.
5. Find the area of the triangle having 3i 4 j + and − + 5 i 7 j as two of its sides.
6. Find the value of 0 0 0 0 sin 330 . cos120 cos 210 . sin 300 +
7. Find the value of 2 2 1 1 cos 52 sin 22
° − °
2 2
3tan tanh tan 3
hx x h x
and E = 0 1
+
2
h x
1 3tan
+
, show that (aI + bE)3
0 0
3
.
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= 0, then find the value of k.
and A2
SECTION B
Short answer type questions.
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Answer any five of the following. 5 X 4 = 20
, show that
θ θ θ φ φ φ
=
θ θ θ φ φ φ
2 2
cos cos sin cos cos sin
cos sin sin cos sin sin
12. If a, b, c are non-coplanar vectors, then test for the collinearity of the following
points whose position vectors are given.
13. Find a unit vector perpendicular to the plane determined by the points P(1, –1, 2),
Q(2, 0, –1) and R(0, 2, 1).
14. Prove that 3
π π + +
1 cos 1 cos
10 10
0 0 sin 10 cos 3 68 x x + = − then find x
15. If x is acute and ( ) ( )
16. If 1 1 1 sin sin sin x y z π
4 4 4 2 2 2 2 2 2 2 2 2
x y z x y z x y y z z x + + + = + + 4 2
17. In a triangle ABC if 1 1 3
− − − + + = then prove that
{ }
+ =
a c b c a b c
+ + + +
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SECTION C
Long answer type questions.
Answer any five of the following. 5 X 7 = 35
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19. If A = (1, –2, –1), B = (4, 0, –3),
C = (1, 2, –1) and D = (2, –4, –5) find the shortest distance between AB and CD.
+ 16n – 1 is divisible by 64 for all positive integers n.
20. If A, B, C are the angles of a triangle then prove that
cos2A+cos2B+cos2C = -4cosAcosBcosC-1
21. In any triangle with usual notation, If r1 + r2 + r3 = r, then show that ∠C = 90°.
22. Solve the equations 2x – y + 8z = 13, 3x + 4y + 5z = 18, 5x – 2y + 7z = 20 by
Matrix inversion method
23. Show that 3
a b 2c a b
c b c 2a b 2(a b c)
c a c a 2b
+ + = + +
24. If f : A→B is a bijection then show that f –1of = IA and fof–1 = IB.
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Modal Paper – 1 Solution
1. Find the domain of the function
2x 5x 7 f (x)
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(x 1)(x 2)(x 3) 0
⇒ − ≠ − ≠ − ≠
(x 1)(x 2)(x 3)
− − −
x 1 0, x 2 0, x 3 0
Domain of f is R {1, 2,3}
∴ −
2. If f(x + y) = f(xy) ∀x, y ∈ R then prove that f is a constant function.
then f(x) = f(x + 0) = f(x⋅ 0) = f(0) = k
∴f is a constant function.
3. ABCDE is a pentagon. If the sum of the vectors AB, AE, BC, DE, ED and AC is λAC,
AB AE BC DC ED AC AC + + + + + = λ
⇒ (AB BC) (AE ED) (DC AC) AC + + + + + = λ
⇒ + + + = λ
AC AD DC AC AC
AC AC AC AC
4. If the position vectors of the points A, B, C are − + − − + + 2i j k, 4i 2 j 2k and
6 i 3 j 13k − − respectively and AB AC = λ then find the value of λ.
Sol. Let O be the origin and OA 2 i j k = − + − , OB 4 i 2 j 2k = − + + , OC 6 i 3 j 13k = − −
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− = λ −
4 i 2 j 2k 2i j k
− + + + − + =
λ − − + − +
6 i 3 j 13k 2 i j k
− + + = λ − −
2 i j 3k 8i j 12k
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Comparing i coefficient on both sides
2 1
λ = − ⇒ λ =
8 4
−
5. Find the area of the triangle having 3i 4 j + and − + 5 i 7 j as two of its sides.
Given AB 3i 4 j, BC 5 i 7 j = + − +
CA AB BC 3i 4 j 5i 7 j
= − − = − − + −
∴ Area of ∆ABC = 1
| AB AC |
×
2
1 1 3 4 0 k( 33 8)
= = − −
6. Find the value of 0 0 0 0 sin 330 . cos120 cos210 . sin 300 +
sin 330 sin 360 30 sin 30
( ) 0 0 0 0 0 1 3 3 cos120 cos 210 ; sin 300 sin 360 60
( ) 0 0 0 0 1
= − = − = −
= − = − = − = −
2 2 2
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0 0 0 0 1 1 3 3 sin 330 cos120 cos 210 . sin 300 1
7. Find the value of 2 2 1 1 cos 52 sin 22
+ = − × − + − − =
° − °
2 2
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2 2 [ cos A sin B cos(A B)cos(A B)]
= ° + ° ° − °
− = + −
1 1 1 1 cos 52 22 cos 52 22
2 2 2 2
− −
3 3 1 3 3
2 2 2 4 2
3tan tanh tan 3
hx x h x
tan tan 2 tanh 3 tanh(2 )
+
h x
1 3tan
+
hx h x
+
1 tan tan 2
hx h x
3
2
hx x hx
3 3
tanh 3tan tanh 3tan
x hx x hx
2 2 2
1 tanh 2 tan 1 3tan
+ + +
x hx hx
+ +
x h x h x
0 1
, show that
0 0
1 0 0 1 a b
+ =
0 1 0 0 0 a
a b a b a 2ab
=
0 a 0 a 0 a
2
2
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=
2 3 2
a 2ab a b a 3a b
0 a 0 a 0 a
2 3
a 0 0 3a b
= +
= +
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− + = − − − +
4 4 8 4k 0 0
2 k 4 k 0 0
⇒ 8 + 4k = 0 ⇒ 4k = –8 ⇒ k = –2
1 0 0 1
0 1 0 0
= 0, then find the value of k.
2 4 2 4 0 0
= − −
1 k 1 k 0 0
θ θ θ φ φ φ =
2 2
cos cos sin cos cos sin
θ θ θ φ φ φ
cos sin sin cos sin sin
π π θ − φ = ⇒ θ = + φ
θ = + φ = − φ
cos cos sin
θ = + φ = φ
θ θ θ
∴
θ θ θ
=
θ θ θ
∴
θ θ θ
φ φ φ
φ φ φ
φ − φ φ
sin sin cos
sin cos cos
− φ φ φ
2
cos cos sin
cos sin sin
2
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φ − φ φ
=
− φ φ φ
sin sin cos
φ φ − φ φ φ φ − φ φ
2 2 2 2 3 3
sin cos sin cos sin cos sin cos
=
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− φ φ + φ φ − φ φ + φ φ
sin cos sin cos sin cos sin cos
φ φ φ
φ φ φ
cos sin sin
3 3 2 2 2 2
cos cos sin
2
12. If a, b, c are non-coplanar vectors, then test for the collinearity of the following points
whose position vectors are given.
i) Show that a 2b 3c,2a 3b 4c − + + − , − + 7b 10c are collinear.
Sol. Let OA a 2b 3c, OB 2a 3b 4c = − + = + −
AB OB OA a 5b 7c
= − = + −
AC OC OA a 5b 7c
= − = − − +
AC a 5b 7c [a 5b 7c]
= − − + = − + −
∴ Given vectors are collinear.
13. Find a unit vector perpendicular to the plane determined by the points P(1, –
1, 2), Q(2, 0, –1) and R(0, 2, 1).
Sol. Let O be the origin and
OP i j 2k,OQ 2 i k,OR 2 j k = − + = − = +
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P R OR OP i 3 j k
= − = − + −
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= + − − − + −
i (0 6) j( 1 2) k(3 0)
| PQ PR | 3 4 1 1 3 6
∴ The unit vector perpendicular to the plane passing through
P, Q and R is PQ PR
+ + + + = ± = ±
14. Prove that 3
×
| PQ PR |
×
3(2i j k) 2 i j k
3 6 6
π π + +
1 cos 1 cos
π π + + = .
7 9 1 1 cos 1 cos
10 10
π π π π + + + +
3 7 9 1 cos 1 cos 1 cos 1 cos
10 10 10 10
π π π π
3 3 1 cos 1 cos 1 cos 1 cos
= + + + π − + π −
10 10 10 10
π π π π
3 3 1 cos 1 cos 1 cos 1 cos
= + + − −
10 10 10 10
π π
= − −
3 3 sin sin sin sin
= =
− + − +
5 1 5 1 ( 5 1) ( 5 1)
= = ×
4 4 16 16
[( 5 1)( 5 1)] (5 1) 4 16 1
= = = = =
15. If x is acute and ( ) ( )
Sol. Given ( ) ( )
10 10
π π π π
10 10 10 10
2 2 2 2
3
2 2
− + −
16 16 16 16 16 16 16 16 16
× × × ×
2 2 2
0 0 sin 10 cos 3 68 x x + = − then find x
0 0 sin 10 cos 3 68 x x + = −
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⇒ + = + −
sin( 10) sin 90 3 68
x x
( )
10 1 22 3 n
( )
0
x
( ) ( )
0
∴ + = + − +
x n x π
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x = is not acute
( ) ( )
10 2 1 22 3 , 2 1
x k x n k
+ = + − + = +
16. If 1 1 1 sin sin sin x y z π
4 4 4 2 2 2 2 2 2 2 2 2
x y z x y z x y y z z x + + + = + + 4 2
Let 1 1 1 sin sin sin x y z α β γ
sin sin sin α β γ = = = x y z
α β γ π α β π γ + + = ⇒ cos cos ( + = − ) ( )
2 2 2 1 1 1 − − − = − − x y xy z
2 2 2 1 1 1 − − = − − x y xy z
( ) ( ) 2 2 2 2 2 2
1 1 1 2 1 − − = + − − − x y x y z xy z
1 1 2 1 − − + = + − − − x y x y x y z xy z
Squaring on both sides we have
2 2 2 2 z x y xy z − − = − − 2 1
( )2 4 4 4 2 2 2 2 2 2 2 2
z x y x z x y y z x y z + + − + − = − 2 2 2 4 1
4 4 4 2 2 2 2 2 2 2 2 2
x y z x y z x y y z x z + + − = + + 4 2 2 2
− − − + + = then prove that
{ }
− − −
= = =
2 2 2 2 2 2 2 2
Let 1 1 cos cos
− − + = then prove that
x y
− −
a b
θ φ
and
= =
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θ φ θ = = =
cos cos cos cos sin sin (θ φ α θ φ θ φ + =) ⇒ −
2 2 1 1 cos xy x y
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2 2 cos 1 1 xy x y
ab a b
Squaring on both sides we have
2 2 2 2
x y x y
2 2 2 2
a b a b
− − − = α
2 2
− = − − α
cos 1 1
+ = − −
2 2 2 2 2 2
x y x y x y
2 2 2 2 2 2
a b a b a y
+ = − − +
cos 1 cos x xy y
cos 1 sin x xy y
∴ − + = − α α
2
− + = − α α
2
17. In a triangle ABC 1 1 3
a c b c a b c
+ + + +
Given 1 1 3
( ) ( )
a c b a b c a
1 1 b a b bc a ac
a c b c ab ac bc c
+ + + + +
2 2 2 2 2 2 a b c ab ab C ab a b c abc C + − = ⇒ 2 cos = + − = ∵ 2 cos
+ =
a c b c a b c
+ + + +
+ =
+ + + +
+ = ⇒
a c b c
+ = ⇒ =
3
+ b a b c
3
− + + b c +
a
2 2
+ + +
{ }
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cos 60
ab C C
= = ⇒ = cos
1 0
ab C =
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Sol. Let S(n) be the statement.
+ 16n – 1 is divisible by 64.
+ 16n – 1 is divisible by 64 for all positive integers n.
+ 16.1 – 1 = 64 is divisible by 64.
Assume that the statement S(n) is true for n = k
+ 16n – 1 is divisible by 64
+ 16k – 1 = 64 M ...(1)
(∵ M is an integer)
We show that the statement S(n) is true for n = k + 1
i.e. we show that 49k+1 + 16(k+1) – 1 is divisible by 64.
× 49 = (64 M – 16k + 1) × 49
49k+1 + 16(k + 1) – 1 = (64M – 16k + 1)49 + 16(k + 1) – 1
= 64× 49 M – 49 × 16k + 49 + 16k + 16 – 1
= 64 × 49 M – 48 × 16k + 64
= 64 × 49 M – 64 × 12k + 64
= 64 N [∵ N is an integer]
∴S(n) is true for n = k + 1
∴ By the principle of mathematical induction, S(n) is true for all n ∈N.
19. If A = (1, –2, –1), B = (4, 0, –3),
C = (1, 2, –1) and D = (2, –4, –5) find the shortest distance between AB and CD.
Sol. Let O be the origin
Let OA i 2 j k,OB 4 i 3k = − − = −
OC i 2 j k,OD 2 i 4 j 5k = + − = − −
The vector equation of a line passing through A, B is
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= − − + − − + +
= − − + + −
where a i 2 j k, b 3i 2 j 2k
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The vector equation of a line passing through C, D is
i 2 j k t(4 i 3k i 2 j k)
i 2 j k t(3i 2 j 2k)
= − − = + −
i 2 i k s[2 i 4 j 5k i 2 j k]
= + − + − − − − +
i 2 j k s[ i 6 j 4k]
where c i 2 j k, d i 6 j 4k
= + − = − −
i[ 8 12] j[ 12 2] k[ 18 2]
= − − − − + + − −
20 i 10 j 20k 10[ 2 i j 2k]
= − + − = − + −
| b d | 10 4 1 4 10 3 30
× = + + = ⋅ =
a c i 2 j k i 2 j k 4 j
− = − − − − + = −
[a c b d] (a c) (b d)
− ⋅ − ⋅ ×
| b d | | b d |
4 j 10[ 2 i j 2k] 10[4] 40 4
= = = =
∴ The shortest distance between the lines = 4/3.
20. . If A, B, C are the angles of a triangle then prove that
cos2A+cos2B+cos2C = -4cosAcosBcosC-1
SOL. cos2A+cos2B+cos2C =
30 30 30 3
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2 2 2 2 2cos cos cos 2
A B A B C
= +
2cos cos 2cos 1
A B A B C
= + − + −
2cos cos 2cos 1
= − − + −
C A B C
2cos cos 2cos 1
= − − + −
( ) ( )
2cos cos cos 1
C A B C
= − − + −
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( ) ( ) ( ) ( )
2cos cos cos 1
C A B A B
= − − + − + −
( ) ( ) ( )
C A B A B
2cos cos cos 1
= − − − + −
c A B C
2
2
2
π
21. In any triangle with usual notation , If r1 + r2 + r3 = r, then show that ∠C = 90°.
r r r r 1
A B C r r 4R sin cos cos
...(1)
C A B A B 4R cos sin cos cos sin
= +
2 2 2
A B C 4R cos sin cos
2 2 2 2 2
2 2 2
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A B C r r 4R sin sin sin
− = −
C A B A B 4R sin sin sin cos cos
= −
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C A B 4R sin cos
= −
2 2 2
A B C 4R cos cos sin
2 2 2
2 2 2 2 2
+
r r 2 C
3 2
r r 2 C
tan
tan tan 45 From(1)
22. solve the equations 2x – y + 8z = 13
5x – 2y + 7z = 20 by Matrix inversion method
Sol. Matrix inversion method :
x
, X =
and D =
y
z
A 28 10 38
B (21 25) 4
= − = − − =
C 6 20 26
= = − − = −
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A ( 7 16) 9
= − = − − + = −
B (14 40) 26
= = − = −
C ( 4 5) 1
= − = − − + = −
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A 5 32 37
= = − − = −
B (10 24) 14
= − = − − =
AdjA B B B 4 26 14
Det A a A b B c C 1 1 1 1 1 1
− −
A A A 38 9 37
= = −
−
C C C 26 1 11
2 38 ( 1)4 8( 26)
= ⋅ + − + −
76 4 208 136
AdjA 1 A 4 26 14
= = − −
X A D 4 26 14 18
= = − −
− −
38 9 37
−
26 1 11
− −
38 9 37 13
−
26 1 11 20
= − − +
− −
494 162 740
52 468 280
− − +
338 18 220
Solution is x = 3, y = 1, z = 1.
23. Show that 3
Sol. L.H.S. C1→C1 + (C2 + C3)
a b 2c a b
c b c 2a b 2(a b c)
c a c a 2b
+ + = + +
+ +
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2(a b c) a b
2(a b c) b c 2a b
2(a b c) a c a 2b
2(a b c) 1 b c 2a b
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(R2→R2 – R1, R3→R3 – R1)
2(a b c) 0 a b c 0
+ + + +
1 a b
1 a c a 2b
1 a b
0 0 a b c
+ +
+ +
24. If f : A→B is a bijection then show that f –1of = IA
Since f : A→B is bijection we have its inverse f –1 : B→A is also a bijection.
f : A→B and f –1 : B→A then f –1of : A→A
Let a ∈A, since f–1 : B→A is onto, there exists b∈B such that f–1(b) = a⇒f(a) = b.
(f–1of)(a) = f–1(f(a)) = f–1(b) = a = IA
f–1: B→ A and f : A B fof–1 : B→B.
Let b∈B, since f : A→B is onto, there exists a∈A such that f(a) = bf–1(b) = a.
(fof–1)(b) = f(f–1(b)) = f(a) = b = IB
have same domain A.
(a)
→ ⇒
have same domain B.
(b).
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. 10 X 2 = 20
1. Find the domain of the function
2x 5x 7 f (x)
2. If f(x + y) = f(xy) ∀x, y ∈ R then prove that f is a constant function.
3. ABCDE is a pentagon. If the sum of the vectors AB, AE, BC, DE, ED and AC is
λAC, then find the value of λ.
4. If the position vectors of the points A, B, C are − + − − + + 2 i j k, 4 i 2 j 2k and
6 i 3 j 13k − − respectively and AB AC = λ then find the value of λ.
5. Find the area of the triangle having 3i 4 j + and − + 5 i 7 j as two of its sides.
6. Find the value of 0 0 0 0 sin 330 . cos120 cos 210 . sin 300 +
7. Find the value of 2 2 1 1 cos 52 sin 22
° − °
2 2
3tan tanh tan 3
hx x h x
and E = 0 1
+
2
h x
1 3tan
+
, show that (aI + bE)3
0 0
3
.
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= 0, then find the value of k.
and A2
SECTION B
Short answer type questions.
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Answer any five of the following. 5 X 4 = 20
, show that
θ θ θ φ φ φ
=
θ θ θ φ φ φ
2 2
cos cos sin cos cos sin
cos sin sin cos sin sin
12. If a, b, c are non-coplanar vectors, then test for the collinearity of the following
points whose position vectors are given.
13. Find a unit vector perpendicular to the plane determined by the points P(1, –1, 2),
Q(2, 0, –1) and R(0, 2, 1).
14. Prove that 3
π π + +
1 cos 1 cos
10 10
0 0 sin 10 cos 3 68 x x + = − then find x
15. If x is acute and ( ) ( )
16. If 1 1 1 sin sin sin x y z π
4 4 4 2 2 2 2 2 2 2 2 2
x y z x y z x y y z z x + + + = + + 4 2
17. In a triangle ABC if 1 1 3
− − − + + = then prove that
{ }
+ =
a c b c a b c
+ + + +
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SECTION C
Long answer type questions.
Answer any five of the following. 5 X 7 = 35
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19. If A = (1, –2, –1), B = (4, 0, –3),
C = (1, 2, –1) and D = (2, –4, –5) find the shortest distance between AB and CD.
+ 16n – 1 is divisible by 64 for all positive integers n.
20. If A, B, C are the angles of a triangle then prove that
cos2A+cos2B+cos2C = -4cosAcosBcosC-1
21. In any triangle with usual notation, If r1 + r2 + r3 = r, then show that ∠C = 90°.
22. Solve the equations 2x – y + 8z = 13, 3x + 4y + 5z = 18, 5x – 2y + 7z = 20 by
Matrix inversion method
23. Show that 3
a b 2c a b
c b c 2a b 2(a b c)
c a c a 2b
+ + = + +
24. If f : A→B is a bijection then show that f –1of = IA and fof–1 = IB.
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Modal Paper – 1 Solution
1. Find the domain of the function
2x 5x 7 f (x)
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(x 1)(x 2)(x 3) 0
⇒ − ≠ − ≠ − ≠
(x 1)(x 2)(x 3)
− − −
x 1 0, x 2 0, x 3 0
Domain of f is R {1, 2,3}
∴ −
2. If f(x + y) = f(xy) ∀x, y ∈ R then prove that f is a constant function.
then f(x) = f(x + 0) = f(x⋅ 0) = f(0) = k
∴f is a constant function.
3. ABCDE is a pentagon. If the sum of the vectors AB, AE, BC, DE, ED and AC is λAC,
AB AE BC DC ED AC AC + + + + + = λ
⇒ (AB BC) (AE ED) (DC AC) AC + + + + + = λ
⇒ + + + = λ
AC AD DC AC AC
AC AC AC AC
4. If the position vectors of the points A, B, C are − + − − + + 2i j k, 4i 2 j 2k and
6 i 3 j 13k − − respectively and AB AC = λ then find the value of λ.
Sol. Let O be the origin and OA 2 i j k = − + − , OB 4 i 2 j 2k = − + + , OC 6 i 3 j 13k = − −
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− = λ −
4 i 2 j 2k 2i j k
− + + + − + =
λ − − + − +
6 i 3 j 13k 2 i j k
− + + = λ − −
2 i j 3k 8i j 12k
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Comparing i coefficient on both sides
2 1
λ = − ⇒ λ =
8 4
−
5. Find the area of the triangle having 3i 4 j + and − + 5 i 7 j as two of its sides.
Given AB 3i 4 j, BC 5 i 7 j = + − +
CA AB BC 3i 4 j 5i 7 j
= − − = − − + −
∴ Area of ∆ABC = 1
| AB AC |
×
2
1 1 3 4 0 k( 33 8)
= = − −
6. Find the value of 0 0 0 0 sin 330 . cos120 cos210 . sin 300 +
sin 330 sin 360 30 sin 30
( ) 0 0 0 0 0 1 3 3 cos120 cos 210 ; sin 300 sin 360 60
( ) 0 0 0 0 1
= − = − = −
= − = − = − = −
2 2 2
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0 0 0 0 1 1 3 3 sin 330 cos120 cos 210 . sin 300 1
7. Find the value of 2 2 1 1 cos 52 sin 22
+ = − × − + − − =
° − °
2 2
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2 2 [ cos A sin B cos(A B)cos(A B)]
= ° + ° ° − °
− = + −
1 1 1 1 cos 52 22 cos 52 22
2 2 2 2
− −
3 3 1 3 3
2 2 2 4 2
3tan tanh tan 3
hx x h x
tan tan 2 tanh 3 tanh(2 )
+
h x
1 3tan
+
hx h x
+
1 tan tan 2
hx h x
3
2
hx x hx
3 3
tanh 3tan tanh 3tan
x hx x hx
2 2 2
1 tanh 2 tan 1 3tan
+ + +
x hx hx
+ +
x h x h x
0 1
, show that
0 0
1 0 0 1 a b
+ =
0 1 0 0 0 a
a b a b a 2ab
=
0 a 0 a 0 a
2
2
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=
2 3 2
a 2ab a b a 3a b
0 a 0 a 0 a
2 3
a 0 0 3a b
= +
= +
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− + = − − − +
4 4 8 4k 0 0
2 k 4 k 0 0
⇒ 8 + 4k = 0 ⇒ 4k = –8 ⇒ k = –2
1 0 0 1
0 1 0 0
= 0, then find the value of k.
2 4 2 4 0 0
= − −
1 k 1 k 0 0
θ θ θ φ φ φ =
2 2
cos cos sin cos cos sin
θ θ θ φ φ φ
cos sin sin cos sin sin
π π θ − φ = ⇒ θ = + φ
θ = + φ = − φ
cos cos sin
θ = + φ = φ
θ θ θ
∴
θ θ θ
=
θ θ θ
∴
θ θ θ
φ φ φ
φ φ φ
φ − φ φ
sin sin cos
sin cos cos
− φ φ φ
2
cos cos sin
cos sin sin
2
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φ − φ φ
=
− φ φ φ
sin sin cos
φ φ − φ φ φ φ − φ φ
2 2 2 2 3 3
sin cos sin cos sin cos sin cos
=
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− φ φ + φ φ − φ φ + φ φ
sin cos sin cos sin cos sin cos
φ φ φ
φ φ φ
cos sin sin
3 3 2 2 2 2
cos cos sin
2
12. If a, b, c are non-coplanar vectors, then test for the collinearity of the following points
whose position vectors are given.
i) Show that a 2b 3c,2a 3b 4c − + + − , − + 7b 10c are collinear.
Sol. Let OA a 2b 3c, OB 2a 3b 4c = − + = + −
AB OB OA a 5b 7c
= − = + −
AC OC OA a 5b 7c
= − = − − +
AC a 5b 7c [a 5b 7c]
= − − + = − + −
∴ Given vectors are collinear.
13. Find a unit vector perpendicular to the plane determined by the points P(1, –
1, 2), Q(2, 0, –1) and R(0, 2, 1).
Sol. Let O be the origin and
OP i j 2k,OQ 2 i k,OR 2 j k = − + = − = +
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P R OR OP i 3 j k
= − = − + −
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= + − − − + −
i (0 6) j( 1 2) k(3 0)
| PQ PR | 3 4 1 1 3 6
∴ The unit vector perpendicular to the plane passing through
P, Q and R is PQ PR
+ + + + = ± = ±
14. Prove that 3
×
| PQ PR |
×
3(2i j k) 2 i j k
3 6 6
π π + +
1 cos 1 cos
π π + + = .
7 9 1 1 cos 1 cos
10 10
π π π π + + + +
3 7 9 1 cos 1 cos 1 cos 1 cos
10 10 10 10
π π π π
3 3 1 cos 1 cos 1 cos 1 cos
= + + + π − + π −
10 10 10 10
π π π π
3 3 1 cos 1 cos 1 cos 1 cos
= + + − −
10 10 10 10
π π
= − −
3 3 sin sin sin sin
= =
− + − +
5 1 5 1 ( 5 1) ( 5 1)
= = ×
4 4 16 16
[( 5 1)( 5 1)] (5 1) 4 16 1
= = = = =
15. If x is acute and ( ) ( )
Sol. Given ( ) ( )
10 10
π π π π
10 10 10 10
2 2 2 2
3
2 2
− + −
16 16 16 16 16 16 16 16 16
× × × ×
2 2 2
0 0 sin 10 cos 3 68 x x + = − then find x
0 0 sin 10 cos 3 68 x x + = −
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⇒ + = + −
sin( 10) sin 90 3 68
x x
( )
10 1 22 3 n
( )
0
x
( ) ( )
0
∴ + = + − +
x n x π
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x = is not acute
( ) ( )
10 2 1 22 3 , 2 1
x k x n k
+ = + − + = +
16. If 1 1 1 sin sin sin x y z π
4 4 4 2 2 2 2 2 2 2 2 2
x y z x y z x y y z z x + + + = + + 4 2
Let 1 1 1 sin sin sin x y z α β γ
sin sin sin α β γ = = = x y z
α β γ π α β π γ + + = ⇒ cos cos ( + = − ) ( )
2 2 2 1 1 1 − − − = − − x y xy z
2 2 2 1 1 1 − − = − − x y xy z
( ) ( ) 2 2 2 2 2 2
1 1 1 2 1 − − = + − − − x y x y z xy z
1 1 2 1 − − + = + − − − x y x y x y z xy z
Squaring on both sides we have
2 2 2 2 z x y xy z − − = − − 2 1
( )2 4 4 4 2 2 2 2 2 2 2 2
z x y x z x y y z x y z + + − + − = − 2 2 2 4 1
4 4 4 2 2 2 2 2 2 2 2 2
x y z x y z x y y z x z + + − = + + 4 2 2 2
− − − + + = then prove that
{ }
− − −
= = =
2 2 2 2 2 2 2 2
Let 1 1 cos cos
− − + = then prove that
x y
− −
a b
θ φ
and
= =
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θ φ θ = = =
cos cos cos cos sin sin (θ φ α θ φ θ φ + =) ⇒ −
2 2 1 1 cos xy x y
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2 2 cos 1 1 xy x y
ab a b
Squaring on both sides we have
2 2 2 2
x y x y
2 2 2 2
a b a b
− − − = α
2 2
− = − − α
cos 1 1
+ = − −
2 2 2 2 2 2
x y x y x y
2 2 2 2 2 2
a b a b a y
+ = − − +
cos 1 cos x xy y
cos 1 sin x xy y
∴ − + = − α α
2
− + = − α α
2
17. In a triangle ABC 1 1 3
a c b c a b c
+ + + +
Given 1 1 3
( ) ( )
a c b a b c a
1 1 b a b bc a ac
a c b c ab ac bc c
+ + + + +
2 2 2 2 2 2 a b c ab ab C ab a b c abc C + − = ⇒ 2 cos = + − = ∵ 2 cos
+ =
a c b c a b c
+ + + +
+ =
+ + + +
+ = ⇒
a c b c
+ = ⇒ =
3
+ b a b c
3
− + + b c +
a
2 2
+ + +
{ }
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cos 60
ab C C
= = ⇒ = cos
1 0
ab C =
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Sol. Let S(n) be the statement.
+ 16n – 1 is divisible by 64.
+ 16n – 1 is divisible by 64 for all positive integers n.
+ 16.1 – 1 = 64 is divisible by 64.
Assume that the statement S(n) is true for n = k
+ 16n – 1 is divisible by 64
+ 16k – 1 = 64 M ...(1)
(∵ M is an integer)
We show that the statement S(n) is true for n = k + 1
i.e. we show that 49k+1 + 16(k+1) – 1 is divisible by 64.
× 49 = (64 M – 16k + 1) × 49
49k+1 + 16(k + 1) – 1 = (64M – 16k + 1)49 + 16(k + 1) – 1
= 64× 49 M – 49 × 16k + 49 + 16k + 16 – 1
= 64 × 49 M – 48 × 16k + 64
= 64 × 49 M – 64 × 12k + 64
= 64 N [∵ N is an integer]
∴S(n) is true for n = k + 1
∴ By the principle of mathematical induction, S(n) is true for all n ∈N.
19. If A = (1, –2, –1), B = (4, 0, –3),
C = (1, 2, –1) and D = (2, –4, –5) find the shortest distance between AB and CD.
Sol. Let O be the origin
Let OA i 2 j k,OB 4 i 3k = − − = −
OC i 2 j k,OD 2 i 4 j 5k = + − = − −
The vector equation of a line passing through A, B is
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= − − + − − + +
= − − + + −
where a i 2 j k, b 3i 2 j 2k
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The vector equation of a line passing through C, D is
i 2 j k t(4 i 3k i 2 j k)
i 2 j k t(3i 2 j 2k)
= − − = + −
i 2 i k s[2 i 4 j 5k i 2 j k]
= + − + − − − − +
i 2 j k s[ i 6 j 4k]
where c i 2 j k, d i 6 j 4k
= + − = − −
i[ 8 12] j[ 12 2] k[ 18 2]
= − − − − + + − −
20 i 10 j 20k 10[ 2 i j 2k]
= − + − = − + −
| b d | 10 4 1 4 10 3 30
× = + + = ⋅ =
a c i 2 j k i 2 j k 4 j
− = − − − − + = −
[a c b d] (a c) (b d)
− ⋅ − ⋅ ×
| b d | | b d |
4 j 10[ 2 i j 2k] 10[4] 40 4
= = = =
∴ The shortest distance between the lines = 4/3.
20. . If A, B, C are the angles of a triangle then prove that
cos2A+cos2B+cos2C = -4cosAcosBcosC-1
SOL. cos2A+cos2B+cos2C =
30 30 30 3
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2 2 2 2 2cos cos cos 2
A B A B C
= +
2cos cos 2cos 1
A B A B C
= + − + −
2cos cos 2cos 1
= − − + −
C A B C
2cos cos 2cos 1
= − − + −
( ) ( )
2cos cos cos 1
C A B C
= − − + −
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( ) ( ) ( ) ( )
2cos cos cos 1
C A B A B
= − − + − + −
( ) ( ) ( )
C A B A B
2cos cos cos 1
= − − − + −
c A B C
2
2
2
π
21. In any triangle with usual notation , If r1 + r2 + r3 = r, then show that ∠C = 90°.
r r r r 1
A B C r r 4R sin cos cos
...(1)
C A B A B 4R cos sin cos cos sin
= +
2 2 2
A B C 4R cos sin cos
2 2 2 2 2
2 2 2
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A B C r r 4R sin sin sin
− = −
C A B A B 4R sin sin sin cos cos
= −
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C A B 4R sin cos
= −
2 2 2
A B C 4R cos cos sin
2 2 2
2 2 2 2 2
+
r r 2 C
3 2
r r 2 C
tan
tan tan 45 From(1)
22. solve the equations 2x – y + 8z = 13
5x – 2y + 7z = 20 by Matrix inversion method
Sol. Matrix inversion method :
x
, X =
and D =
y
z
A 28 10 38
B (21 25) 4
= − = − − =
C 6 20 26
= = − − = −
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A ( 7 16) 9
= − = − − + = −
B (14 40) 26
= = − = −
C ( 4 5) 1
= − = − − + = −
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A 5 32 37
= = − − = −
B (10 24) 14
= − = − − =
AdjA B B B 4 26 14
Det A a A b B c C 1 1 1 1 1 1
− −
A A A 38 9 37
= = −
−
C C C 26 1 11
2 38 ( 1)4 8( 26)
= ⋅ + − + −
76 4 208 136
AdjA 1 A 4 26 14
= = − −
X A D 4 26 14 18
= = − −
− −
38 9 37
−
26 1 11
− −
38 9 37 13
−
26 1 11 20
= − − +
− −
494 162 740
52 468 280
− − +
338 18 220
Solution is x = 3, y = 1, z = 1.
23. Show that 3
Sol. L.H.S. C1→C1 + (C2 + C3)
a b 2c a b
c b c 2a b 2(a b c)
c a c a 2b
+ + = + +
+ +
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2(a b c) a b
2(a b c) b c 2a b
2(a b c) a c a 2b
2(a b c) 1 b c 2a b
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(R2→R2 – R1, R3→R3 – R1)
2(a b c) 0 a b c 0
+ + + +
1 a b
1 a c a 2b
1 a b
0 0 a b c
+ +
+ +
24. If f : A→B is a bijection then show that f –1of = IA
Since f : A→B is bijection we have its inverse f –1 : B→A is also a bijection.
f : A→B and f –1 : B→A then f –1of : A→A
Let a ∈A, since f–1 : B→A is onto, there exists b∈B such that f–1(b) = a⇒f(a) = b.
(f–1of)(a) = f–1(f(a)) = f–1(b) = a = IA
f–1: B→ A and f : A B fof–1 : B→B.
Let b∈B, since f : A→B is onto, there exists a∈A such that f(a) = bf–1(b) = a.
(fof–1)(b) = f(f–1(b)) = f(a) = b = IB
have same domain A.
(a)
→ ⇒
have same domain B.
(b).
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