= \(\frac{14}{21},-\frac{1}{7}\)

For x + 2y = 3
a₁ = 1, b₁ = 2 ,c₁ = - 3
For (k-1)x + (k +1)y= (k + 2)
a₂ = (k - 1), b₂= (k + 1), c₂ = - (k + 2)
For no solution, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { c_{ 1 } }{ { c }_{ 2 } } \)
\(\Rightarrow \quad \frac { 1 }{ k-1 } =\frac { 2 }{ k+1 } \neq \frac { 3 }{ k+2 } \)
           I           II          III
From I and II, \(\frac { 1 }{ k-1 } =\frac { 2 }{ k+1 } \)
\(\Rightarrow\)  k + 1 = 2k - 2
\(\therefore\)  k = 3
From II and III, \(\frac { 2 }{ k+1 } \neq \frac { 3 }{ k+2 } \)
\(\Rightarrow\) 2(k + 2) \(\neq\) 3(k + 1)
\(\Rightarrow\) 2k+4\(\neq\)3k+3
\(\therefore\)  k \(\neq\) 1
From I and III, \(\frac { 1 }{ k-1 } \neq \frac { 3 }{ k+2 } \)
\(\Rightarrow \quad k+2\neq 3k-3\)
\(\therefore \quad k\neq -\frac { 5 }{ 2 } \)
Hence, k = 3 but k \(\neq\)1 and \(\\ \\ k\neq -\frac { 5 }{ 2 } \)

Since we will apply the identity involving sec \(\theta\) and tan \(\theta\), let us first convert the LHS (of the identity we need to prove) in terms of sec \(\theta\) and tan \(\theta\) by dividing numerator and denominator by cos \(\theta\).
\(\begin{aligned} \text { LHS } & =\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{(\tan \theta+\sec \theta)-1}{(\tan \theta-\sec \theta)+1}=\frac{\{(\tan \theta+\sec \theta)-1\}(\tan \theta-\sec \theta)}{\{(\tan \theta-\sec \theta)+1\}(\tan \theta-\sec \theta)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{\left(\tan ^2 \theta-\sec ^2 \theta\right)-(\tan \theta-\sec \theta)}{\{\tan \theta-\sec \theta+1\}(\tan \theta-\sec \theta)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{-1-\tan \theta+\sec \theta}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)} \end{aligned}\)
\(=\frac{-1}{\tan \theta-\sec \theta}=\frac{1}{\sec \theta-\tan \theta}\)
which is the RHS of the identity, we are required to prove.

(i)Total number of cards=52
Number of red coloured kings=2
p(a king of red colour)=\({2\over52}={1\over26}\)
(ii)Number of face cards=12
p(a face card)\(={12\over52}={3\over13}\)
(iii)Number of queen of diamonds=1
p(the queen of diamonds)\(={1\over52}\)

AB = \(\sqrt { { (5-1 })^{ 2 }+{ (3-0) }^{ 2 } } =\sqrt { 16+9 } =5\)
DC = \(\sqrt { { (2+2 })^{ 2 }+{ (7-4) }^{ 2 } } =\sqrt { 16+9 } =5\)
BC = \(\sqrt { { (5-2 })^{ 2 }+{ (3-7) }^{ 2 } } =\sqrt { 9+16 } =5\)
AD = \(\sqrt { { (1+2 })^{ 2 }+{ (0-4) }^{ 2 } } =\sqrt { 9+16 } =5\)

Mid-point of AC = \((\frac{1+2}{2}, \frac{0+7}{2}) = (\frac{3}{2},\frac{7}{2})\)
Mid-point of BD = \((\frac{5-2}{2},\frac{3+4}{2}) = (\frac{3}{2},\frac{7}{2})\)
Since AB = DC and BC = AD.
Opposite sides are parallel and diagonals bisect each other.
∴ The given points are the vertices ofa parallelograin. 

Here, a = -6, d = - 2 +  6 and a_n = 58
a_n = 58
\(\Rightarrow\)    a + ( n - 1 )d = 58  \(\Rightarrow\) - 6 + ( n - 1 )4 = 58
\(\Rightarrow\)    ( n - 1 )4 = 64  \(\Rightarrow\) n - 1 = 16 \(\Rightarrow\) n = 17 (odd)
\(\therefore\)       Middle term = \(\frac{17+1}{2}=\frac{18}{2}\) = 9th term
\(\therefore\)   9th term is the middle term.
Now,    a₉ = a + 8d = - 6 + 8 X 4 = - 6 + 32 = 26

Let usual speed of the train be x km/h Distance = 300 km : Time taken = \(\frac { 300 }{ x } \) hours
Increase speed = (x+5) km/h
Then Time taken = \(\left( \frac { 300 }{ x+5 } \right) \) hours
ATQ \(\frac { 300 }{ x } =\frac { 300 }{ x+5 } =2\) \(\Rightarrow \frac { 300(x+5)-300x }{ x(x+5) } =2\)
\(\Rightarrow 300x+1500-300x=2(x^{ 2 }+5x)\Rightarrow 2x^{ 2 }+10x-1500=0\)
\(\Rightarrow x^{ 2 }+5x-750=0\quad \Rightarrow (x+30)(x-25)=0\)
\(\Rightarrow x=-30\) rejected or x= 25
Usual speed = 25 km/h

OA. OB = OC . OD    (Given)
So, \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OD}}{\mathrm{OB}}\)     (1)
Also, we have  \(\angle\)AOD = \(\angle\)COB   (Vertically opposite angles) (2)
Therefore, from (1) and (2),  \(\Delta\)AOD \(\sim\) \(\Delta\)COB (SAS similarity criterion)
So, \(\angle\)A = \(\angle\)C and \(\angle\)D = \(\angle\)B
(Corresponding angles of similar triangles)

Total number of outcomes = 52
(i) Number of face cards  = 12
\(\therefore\)  Probability of drawing a face card = \(\frac{12}{52}=\frac{3}{13}\)
(ii) Number of cards which are neither king nor red = 24
\(\therefore\)  Probability of drawing a card which is neither a king nor a red card = \(\frac{24}{52}=\frac{6}{13}\)

\(AB=\sqrt { (14-{ 2) }^{ 2 }+(10+2)^{ 2 } } =12\sqrt { 2 } units\)
\(BC=\sqrt { (11-{ 14) }^{ 2 }+(13-10)^{ 2 } } =3\sqrt { 2 } units\)
\(CD=\sqrt { (-1-{ 11) }^{ 2 }+(1-13)^{ 2 } } =12\sqrt { 2 } units\)
\(AD=\sqrt { (-1-{ 2) }^{ 2 }+(1+2)^{ 2 } } =3\sqrt { 2 } units\)
⇒ AB=CD and BC = AD
∴ ABCD is a || gm
Now, \(AC=\sqrt { (11-2)^{ 2 }+({ 13+2) }^{ 2 } } =\sqrt { 306 } \)
⇒ AC²=306 units, AB² = 288 units, BC² =18 units
AB²+BC²=306 units
⇒ AC²=AB²+BC²⇒∠ABC = 90⁰
⇒ ABCD is a rectangle.

(k+4)x²+(k+1) x+1=0
a=k+4,b=k+1,c=1
D=b²-4ac
D=(k+1)²-4(k+4)X1
= k²+2k+1-4k-16
=k²-2k-15
For equal roots D=0
\(\Rightarrow \) k²+2k-15=0
\(\Rightarrow \) (k-5)(k+3)=0
k=5,k=-3

Let Ist term of AP = a and common difference = d.
Now,    a₆ = -10 \(\Rightarrow\)  a +  5d  = -10                 ..(i)
Also,     a₁₀ = -26  \(\Rightarrow\)    a + 9d = -26           ...(ii)
Subtract (i) from (ii),    
a + 9d = - 26
a + 5d = -10
-  -          +
 4d = -16   \(\Rightarrow\) d = - 4
Substituting in (i), we get
a + 5 x ( -4 ) = - 10  \(\Rightarrow\) a = 10
Now,  a₁₅ = a + 14d = 10 + 14 X  - 4 = - 46

Let the fraction be \(\frac{x}{y}\), where x is the numerator and y is the denominator.
According to the given condition,
y = 2x + 1
So,the fraction = \(\frac{x}{2 x+1}\)
Reciprocal of the fraction \(=\frac{2 x+1}{x}\)
It is given that the sum of the fraction and its reciprocal is 2\(\frac{16}{21}\).
\(\begin{array}{ll} \therefore & \frac{x}{2 x+1}+\frac{2 x+1}{x}=2 \frac{16}{21} \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{x^2+(2 x+1)^2}{x(2 x+1)}=\frac{58}{21} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{x^2+4 x^2+1+4 x}{2 x^2+x}=\frac{58}{21} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{x^2+4 x^2+1+4 x}{2 x^2+x}=\frac{58}{21} \\ \end{array}\)
\(\begin{array}{ll} & \frac{5 x^2+4 x+1}{2 x^2+x}=\frac{58}{21} \end{array}\)
\(\Rightarrow\)   21(5x² + 4x + 1) = 58(2x² + x)
\(\Rightarrow\)   105x² + 84x + 21 = 116x² + 58x
\(\Rightarrow\)    -11x² + 26x + 21 = 0
\(\Rightarrow\)      11x² - 26x - 21 = 0
\(\Rightarrow\)  11x² - 33x + 7x - 21 = 0 
\(\Rightarrow\)  11x(x-3) + 7(x - 3) = 0
\(\Rightarrow\)  (x - 3)(11x + 7) = 0
\(\Rightarrow \quad x=3, x \neq-\frac{7}{11}\)
                                 [x cannot be negative]
\(\Rightarrow\)  x = 3
\(\therefore\)   y = 2(3) + 1 = 6 + 1 = 7
Hence, the required fraction = \(\frac{3}{7}\)

Given, in △ABC and △PQR, AD and PM are their medians, respectively.
\(\therefore \quad \frac{A B}{P Q}=\frac{A C}{P R}=\frac{A D}{P M}\)  ...(i)
To prove △ABC ~ △PQR
Construction Produce AD to E such that AD = DE and produce PM to N such that PM = MN.
Join BE, CE, QN and RN.

In above figures, quadrilaterals ABEC and PQNR are parallelograms because their diagonals bisect each other at D and M, respectively.
\(\therefore\) BE = AC and QN = PR
\(\begin{array}{ll} \Rightarrow & \frac{B E}{A C}=1 \quad \text { and } \frac{Q N}{P R}=1 \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{B E}{A C}=\frac{Q N}{P R} \text { or } \frac{B E}{Q N}=\frac{A C}{P R} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{B E}{Q N}=\frac{A B}{P Q} \\ \end{array}\)      [from Eq. (i)]
\(\begin{array}{ll} \text { or } & \frac{A B}{P Q}=\frac{B E}{Q N} \end{array}\)          ....(ii)
From Eq. (i), we get
\(\frac{A B}{P Q}=\frac{A D}{P M}=\frac{2 A D}{2 P M}=\frac{A E}{P N}\) [since, diagonals bisect each other]
\(\Rightarrow \quad \frac{A B}{P Q}=\frac{A E}{P N}\)    ...(iii)
From Eqs. (ii) and (iii),
\(\frac{A B}{P Q}=\frac{B E}{Q N}=\frac{A E}{P N}\)
\(\Rightarrow \quad \triangle A B E \sim \triangle P Q N \Rightarrow \angle 1=\angle 2\)  ...(iv)
[since, corresponding angles of two similar triangles are equal]
Similarly, we can prove that
\(\begin{aligned} & \triangle A C E \sim \triangle P R N \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \angle 3=\angle 4 \end{aligned}\)   ....(v)
On adding Egs. (iv) and (v), we get
\(\begin{array}{rlrl} \angle 1+\angle 3 & =\angle 2+\angle 4 \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \angle B A C =\angle Q P R \\ \end{array}\)
and \(\begin{array}{rlrl} \frac{A B}{P Q} & =\frac{A C}{P R} \end{array}\)    [from Eq. (i)]
\(\therefore\)   \(\triangle ABC \sim \triangle PQR\) [by SAS similarity criterion]
Hence proved.

f₁=28, f₂=24

LHS = \(\frac { \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } ++1 }{ (1+\cos { \theta } )\sin { \theta } } \)
\(=\frac { 1+2\cos { \theta } +1 }{ (1+\cos { \theta } )\sin { \theta } } =\frac { 2(\cos { \theta } +1) }{ (1+\cos { \theta } )\sin { \theta } } =2cosec\theta \) 

\(x = {a^{2}\over 2}\) or \(x = {b^{2}\over 2}\)

(a) Since, he pays first instalment of? 1000 and next consecutive months he pay the instalment are 1100, 1200, ..... .
Thus, we get the AP sequence,
1000, 1100, 1200, ...
Here, a = 1000, d = 1100 - 1000 = 100
Now, T₃₀ = a + (30 - 1) d
 = 1000 + 29 \(\times\) 100
= 1000 + 2900 = 3900
Hence, the amount paid by him in 30th instalment is Rs 3900.
(ii) (b) Now, \(S_{30}=\frac{30}{2}[2 a+(30-1) d]\)
= 15 (2 \(\times\) 1000 + 29 \(\times\)100)
= 15 (2000 + 2900)
= 15 \(\times\) 4900 = Rs 73500
(iii) (c) After 30th instalment, he still have to pay = 118000 - 73500= 44500
(iv) (a) The amount in last 40th instalment is
T₄₀ = a + (40 - 1)d
= 1000 + 39 \(\times\) 100
= 1000 + 3900 = Rs 4900
(v) (b) The ratio of Ist instalment to the last instalment is \(\frac{1000}{4900} \text { i.e. } \frac{10}{49}\)

(i) (d):
Total number of participants = 60 + 84 + 108
= 252
(ii) (d):
60 = 22 x 3 x 5
84 = 22 x 3 x 7
108 = 22 x 33
LCM(60, 84, 108) = 22 x 33 x 5 x 7
= 3780
(iii) (a):
60 = 22 x 3 x 5
84 = 22 x 3 x 7
108 = 22 x 33
HCF(60, 84, 108) = 22 x 3
= 12
(iv) (c):
Minimum number of rooms required for all the participants = 252/12
= 21
(v) (d):
Minimum number of rooms required for all = 21 + 1 = 22

(i) (b): The distance between A and C
\(=\sqrt{(8-4)^{2}+(5+3)^{2}}=\sqrt{4^{2}+8^{2}} \)
\(=\sqrt{16+64}=\sqrt{80}=4 \sqrt{5} \text { units }\)
(ii) (a): Let the coordinates of I be (x, y).

Then, by section formula
\(x =\frac{1 \times 8+2 \times 7}{1+2}=\frac{8+14}{3}=\frac{22}{3}\)
\(\text { and } y =\frac{1 \times 5+2 \times 3}{1+2}=\frac{5+6}{3}=\frac{11}{3}\)
Thus, the coordinates of I is \(\left(\frac{22}{3}, \frac{11}{3}\right)\)
(iii) (b): The mid -point of A and C
\(=\left(\frac{8+4}{2}, \frac{5-3}{2}\right)=(6,1)\)
(iv) (b): Let B divides the line segment joining A and C in the ratio k : 1. Then, the coordinates of B will be
\(\left(\frac{8 k+4}{k+1}, \frac{5 k-3}{k+1}\right)\)
\(\text { Thus, we have }\left(\frac{8 k+4}{k+1}, \frac{5 k-3}{k+1}\right)=(7,3)\)
\(\Rightarrow \frac{8 k+4}{k+1}=7 \text { and } \frac{5 k-3}{k+1}=3 \)
\(\text { Consider, } \frac{8 k+4}{k+1}=7 \Rightarrow 8 k+4=7 k+7 \Rightarrow k=3\)
Hence, the required ratio is 3 : 1
(v) (a):\(\because\) B divides AC in the ratio 3 : 4.
\(\therefore\) A, B, C lie on a straight line.