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The relation is defined as

f (x) = { x² , when 0 ≤ x ≤ 3 

            2x, when 3 ≤ x ≤ 5 

It is observed that for 

0 ≤ x ≤ 2, f(x) = x² And 2 ≤ x ≤ 5, f(x) = 2x 

Also, at x = 2, f (x) = 2² = 4 or f(x) = 2 × 2 = 4 

i,e at x = 2, f(x) = 4. 

Therefore, for 0≤ x ≤ 5, the image of f (x) are unique Thus, the given relation ‘f’ is a function. The relation g is defined as g(x) = { x² , when 0 ≤ x ≤ 3 2, when 3 ≤ x ≤ 5 It can be observed that for x = 3 g(x) = 3² = 9 and g (x) = 2 × 3 = 6 Hence, element 3 of the domain of relation ‘g’ corresponds to two different images i.e., 9 and 6. Hence, this relation is not a function.

Given y = 2x – 1 and x ≤ 5 

for x = 1, y = 2 × 1 – 1 = 1 

x = 2, y = 2 × 2 – 1 = 3 

x = 3, y = 2 × 3 – 1 = 5 

x = 4, y = 2 × 4 – 1 = 7 

x = 5, y = 2 × 5 – 1 = 9 

Relation R = {(1, 1), (2, 3), (3, 5), (4, 7), (5, 9)} 

∴ Domain = {1, 2, 3, 4, 5} 

∴ Range = {1, 3, 5, 7, 9}

Liver is the largest solid organ in the body situated in the upper part of the abdomen on the right side. The main function of liver is to filter the blood coming from the digestive tract, before passing it to the rest of the body.

The excess bile secreted by the Liver is stored in the gall bladder.

1. Uvula doses the nasal cavity that opens to the pharynx 

2. Tongue 

3. True

Given n(A x A) = 9 = 3²

⇒n(A) = 3

But (-1, 0) and (0, 1) are in A x A

∴ A= {-1,0,1}.

Remaining elements of A x A: (-1, -1), (-1, 1),

(0,-1), (0,0), (1,-1), (1,0), (1,1).

(1) 7

n(A) = 4

n(A x B) = 28

n(A) x n(B) = 28

4 x n(B) = 28

n(B) = 28/4 = 7

A = {x, y, z} and B = {1, 2}.

Maharshi Parashar was the writer of ‘Vruksha Ayurveda’.

Brahmgupta introduced the type of equations.

Ruminants have four chambers in stomach. They are rumen, reticulum, omasum and abomasum.

Maharashi Patanjali wrote ‘Yogashastra’.

Vastsayana wrote ‘Kamsutra’.

Chakrapanidatta wrote ‘Chikitsa Sangraph’.

Brahmgupta Panchal wrote ‘Prajanan-shastra’.

Given, P = {1,3} and Q = {2, 3, 5} 

∵ n(P) = 2 and n(Q) = 3 

∴ Number of relations = 2^{n(p) x n(Q)}

= 2^{2 × 3}

= \(2^6\)

= 64

    Given, R = {x,y}: x, y ∈ z , x² + y² = 64} 

∵ x ²+ y² = 64 ⇒ y² = 64 – x 

    ∴  R = {(0, 8), (0, - 8), (8, 0), (- 8, 0)

A relation R from a non-empty set A to non empty set B is a subset of the Cartesian product A x B.

Radius of circles are 8cm and 6cm

Area of circle with radius 8cm = π(8)²=64πcm²

Area of circle with radius 6cm = π(6)²=36π cm²

Areas sum = 64π + 36π = 100π cm²

Radius of circle be x cm

Area = πx²

πx² = 100π

x²=100

⇒x=√100=10cm

Given sin^-1 (sin 1550°)

= sin^-1 (sin (1440° + 110°))

= sin^-1 (sin (360° × 4 + 110°))

We know that sin (2nπ + θ) = sin θ

= sin^-1 (sin 110°)

We know that sin^-1 (sin θ) = π – θ, if θ ∈ [π/2, 3π/2]

= 180° - 110°

= 70°

∴ sin^-1 (sin 1550°) = 70°

Given cos^-1 (cos 1540°)

= cos^-1{cos (1440° + 100°)}

= cos ^-1{cos (360° × 4 + 100°)}

We know that cos (2π + θ) = cos θ

= cos -1{cos 100°}

We know that cos^-1 (cos θ) = θ if θ ∈ [0, π]

= 100°

∴ cos^-1 (cos 1540°) = 100°

Given sin^-1 (sin (-600°))

= sin ^-1 (sin (-600 + 360 × 2))

We know that sin (2nπ + θ) = sin θ

= sin (sin 120°)

We know that sin^-1 (sin θ) = π – θ, if θ ∈ [π/2, 3π/2]

= 180° - 120° = 60°

∴ sin -1 (sin (-600°)) = 60°

The given equations are: 

9/x - 4/y  = 8 ……..(i) 

13/x + 7/y = 101 ……..(ii) 

Putting 1/x = u and 1/y = v, we get: 

9u - 4v = 8 …….(iii) 

13u + 7v = 101 ……(iv) 

On multiplying (iii) by 7 and (iv) by 4, we get: 

63u - 28v = 56 ……..(v) 

52u + 28v = 404 ……..(vi) 

On adding (v) from (vi), we get: 

115u = 460 

⇒ u = 4 

⇒ 1/ x = 4 

⇒ x = 1/4 

On substituting x = 1/ 4 in (i), we get: 

9/ 1/ 4 - 4/ y = 8 

⇒36 - 4/ y = 8 

⇒ 4/ y = (36 – 8) = 28 

y = 4/ 28 = 1/7 

Hence, the required solution is x = 1/4 and y = 1/ 7.

The given equations are:

 \(\frac{5}x-\frac{3}y=1\)..........(i)

\(\frac{3}{2x}+\frac{2}{3y}=5\).......(ii)

Putting 1/x = u and 1/y = v, we get: 

5u - 3v = 1 …….(iii) 

⇒ 3/2 u + 2/3 v = 5

⇒ \(\frac{9u+4v}6\) = 5

⇒ 9u + 4v = 30 …….(iv) 

On multiplying (iii) by 4 and (iv) by 3, we get: 

20u - 12v = 4 ……..(v) 

27u + 12v = 90 ……..(vi) 

On adding (iv) and (v), we get: 

47u = 94 ⇒ u = 2 

⇒ 1/x = 2 ⇒ x = 1/2 

On substituting x = 1/2 in (i), we get:

\(\frac{\frac{5}1}2\) - \(\frac{3}y\) = 1

⇒ 10 - 3/y = 1 ⇒ 3/y = (10 – 1) = 9 

y = 3/9 = 1/3 

Hence, the required solution is x = 1/2 and y = 1/3 .

The given equations are:

 \(\frac{9}x-\frac{4}y=8\)........(i)

\(\frac{13}x+\frac{7}y=101 \).......(ii)

Putting 1/x = u and 1/y = v, we get: 

9u - 4v = 8 …….(iii) 

13u + 7v = 101 ……(iv) 

On multiplying (iii) by 7 and (iv) by 4, we get: 

63u - 28v = 56 ……..(v) 

52u + 28v = 404 ……..(vi) 

On adding (v) from (vi), we get: 

115u = 460 ⇒ u = 4 

⇒ 1/x = 4 ⇒ x = 1/4 

On substituting x = 1 4 in (i), we get:

\(\frac{9}{\frac{1}{4}}\) - 4/y = 8 

⇒ 36 - 4/y = 8 

⇒ 4/y = (36 – 8) = 28 

y = 4/28 = 1/7 

Hence, the required solution is x = 1/4 and y = 1/7.

The given equations are

 \(\frac{5}{x+y}-\frac{2}{x-y}=-1\).......(i)

\(\frac{15}{x+y}-\frac{7}{x-y}=10\)......(ii)

Substituting 1/x+y = u and 1/x−y = v in (i) and (ii), we get 

5u – 2v = -1 ……..(iii) 

15u + 7v = 10 …….(iv) 

Multiplying (iii) by 3 and subtracting it from (iv), we get 

7v + 6v = 10 + 3 

⇒ 13v = 13 

⇒ v = 1

⇒x – y = 1 \((∵\frac{1}{x-y}=v)\)......(v)

Now, substituting v = 1 in (iii), we get 

5u – 2 = -1 

⇒5u = 1 

⇒u = 1/5 

x + y = 5 …….(vi) 

Adding (v) and (vi), we get 

2x = 6 ⇒ x = 3 

Substituting x = 3 in (vi), we have 

3 + y = 5 ⇒ y = 5 – 3 = 2 

Hence, x = 3 and y = 2.

The given equations are

 \(\frac{44}{x+y}+\frac{30}{x-y}=10\)............(i)

\(\frac{55}{x+y}-\frac{40}{x-y}=13\)..........(ii)

Putting 1/x+y = u and 1/x−y = v, we get: 

44u + 30v = 10 ……..(iii) 

55u + 40v = 13 …….(iv) 

On multiplying (iii) by 4 and (iv) by 3, we get: 

176u + 120v = 40 …..(v) 

165u + 120v = 39 …..(vi) 

On subtracting (vi) and (v), we get: 

11u = 1 

⇒u = 1/11 

⇒ 1/x+y = 1/11 ⇒ x + y = 11 …….(vii) 

On substituting u = 1/11 in (iii), we get: 

4 + 30v = 10 

⇒30v = 6 

⇒v = 6/30 = 1/5 

⇒ 1/x−y = 1/5 ⇒ x – y = 5 …….(viii) 

On adding (vii) and (viii), we get 

2x = 16 

⇒ x = 8 

On substituting x =8 in (vii), we get: 

8 + y = 11 

⇒ y = 11 – 8 = 3 

Hence, the required solution is x = 8 and y = 3.

The given equations are

 \(\frac{3}{x+y}+\frac{2}{x-y}=2\)....(i)

\(\frac{9}{x+y}+\frac{4}{x-y}=1\)......(ii)

Substituting 1/x+y = u and 1/x−y = v, we get: 

3u + 2v = 2 ……..(iii) 

9u - 4v = 1 …….(iv) 

On multiplying (iii) by 2, we get: 

6u + 4v = 4 …..(v) 

On adding (iv) and (v), we get: 

15u = 5 

⇒u = 5/15 = 1/3 

⇒ 1/x+y = 1/3 

⇒ x + y = 3 …….(vi) 

On substituting u = 1/3 in (iii), we get 

1 + 2v = 2 

⇒2v = 1 

⇒v = 1/2 

⇒ 1/x−y = 1/2 ⇒ x – y = 2 …….(vii) 

On adding (vi) and (vii), we get 

2x = 5

⇒ x = 5/2 

On substituting x = 5/2 in (vi), we have 

5/2 + y = 3

⇒ y = \((3-\frac{5}2)\)= 1/2 

Hence, the required solution is x = 5/2 and y = 1/2 .

The given equations can be written as

 \(\frac{5}x+\frac{2}y=6\)........(i)
\(\frac{-5}x+\frac{4}y=-3\).......(ii)

Adding (i) and (ii), we get 

6/y = 3 ⇒ y = 2 

Substituting y = 2 in (i), we have 

5/x + 2/2 = 6 ⇒ x = 1 

Hence, x = 1 and y = 2.

The given equations are

 \(\frac{10}{x+y}+\frac{2}{x-y}=4\).............(i)

\(\frac{15}{x+y}-\frac{9}{x-y}=-2\).......(ii)

Substituting 1/x+y = u and 1/x−y = v in (i) and (ii), we get: 

10u + 2v = 4 ……..(iii) 

15u - 9v = -2 …….(iv) 

Multiplying (iii) by 9 and (iv) by 2 and adding, we get: 

90u + 30u = 36 – 4

⇒120u = 32 

⇒u = 32/120 = 4/15 

⇒x + y = 15/4 \((∵\frac{1}{x+y}=u)\).........(v)

On substituting u = 4/15 in (iii), we get: 

10 × 4/15 + 2v = 4 

8/3 + 2v = 4 

⇒2v = 4 - 8/3 = 4/3 

⇒v = 2/3 

⇒ x – y = 3/2 \((∵\frac{1}{x-y}=v)\)........(vi)

Adding (v) and (vi), we get 

2x = 15/4 + 3/2 ⇒2x = 21/4 ⇒x = 21/8 

Substituting x = 21/8 in (v), we have 

21/8 + y = 15/4 ⇒y = 15/4 - 21/8 = 9/8 

Hence, x = 21/8 and y = 9/8

The given equations are

 \(\frac{5}{x+1}+\frac{2}{y-1}\) = \(\frac{1}2\)..........(i)

 \(\frac{10}{x+1}-\frac{2}{y-1}\)= \(\frac{5}2\).........(ii)

 Substituting 1/x+1 = u and 1/y−1 = v, we get: 

5u - 2v = 1/2 ……..(iii) 

10u + 2v = 5/2 …….(iv) 

On adding (iii) and (iv), we get: 

15u = 3 

⇒u = 3/15 = 1/5 

⇒ 1/x+1 = 1/5 ⇒ x + 1 = 5 ⇒ x = 4 

On substituting u = 1/5 in (iii), we get 

5 × 1/5 - 2v = 1/2 ⇒ 1 – 2v = 1/2 

⇒2v = 1/2 ⇒v = 1/4 

⇒ 1/y−1 = 1/4 ⇒ y – 1 = 4 ⇒ y = 5 

Hence, the required solution is x = 4 and y = 5.

The given equations are

 \(\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}4\).........(i)

\(\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=-\frac{1}8\)

\(\frac{1}{3x+y}-\frac{1}{3x-y}=-\frac{1}4\) (Multiplying by 2) ……(ii) 

Substituting 1/3x+y = u and 1/3x−y = v in (i) and (ii), we get: 

u + v = 3/4 ……..(iii) 

u – v = − 1/4 …….(iv) 

Adding (iii) and (iv), we get: 

2u = 1/2

⇒ u = 1/4 

⇒3x + y = 4 \((∵\frac1{3x+y}=u)\)........(v)

Now, substituting u = 1/4 in (iii), we get: 

1/4 + v = 3/4 

v = 3/4 - 1/4 

⇒v = 1/2

 ⇒ 3x – y = 2 \((∵\frac1{3x-y}=v)\)....(vi)

Adding (v) and (vi), we get 

6x = 6 ⇒x = 1 

Substituting x = 1 in (v), we have 

3 + y = 4 ⇒y = 1 

Hence, x = 1 and y = 1.

The given equations are 

3/x+y + 2/x-y = 2 ……(i) 

9/x+y - 4/x-y = 1 ……(ii) 

Substituting 1/x+y= u and 1/x-y = v, we get: 

3u + 2v = 2 ……..(iii) 

9u - 4v = 1 …….(iv) 

On multiplying (iii) by 2, we get: 

6u + 4v = 4 …..(v) 

On adding (iv) and (v), we get: 

15u = 5 

⇒u = 5/15 = 1/3 

⇒ 1/x+y = 1/3 

⇒ x + y = 3 …….(vi) 

On substituting u = 1/3 in (iii), we get 

1 + 2v = 2 

⇒2v = 1 

⇒v = 1/2 

⇒ 1/x-y = 1/2 

⇒ x – y = 2 …….(vii) 

On adding (vi) and (vii), we get 

2x = 5

⇒ x = 5/2 

On substituting x = 5/2 in (vi), we have 

5/2 + y = 3 

⇒ y = (3 – 5/2 ) = 1/2 

Hence, the required solution is x = 5/2 and y = 1/2 .

The given equations are

 \(\frac{1}{2(x+2y)}+\frac{5}{3(3x-2y)}=-\frac{3}2\)......(i)

\(\frac{1}{4(x+2y)}-\frac{3}{5(3x-2y)}=\frac{61}{60}\).......(ii)

 Putting 1/x+2y = u and 1/3x−2y = v, we get: 

1/2 u + 5/3 v = - 3/2 ……..(iii) 

5/4 u – 3/5 v = 61/60 …….(iv) 

On multiplying (iii) by 6 and (iv) by 20, we get: 

3u + 10v = -9 …..(v) 

25u – 12v = 61/3 …..(vi) 

On multiplying (v) by 6 and (vi) by 5, we get 

18u + 60v = -54 …….(vii) 

125u – 60v = 305/3 ……..(viii) 

On adding(vii) and (viii), we get: 

143u = 305/3 – 54 = 305−162/3 = 143/3 

⇒u = 1/3 = 1/x+2y

⇒x + 2y = 3 …….(ix) 

On substituting u = 1/3 in (v), we get: 

1 + 10v = -9 

⇒10v = -10 

⇒v = -1 

⇒ 1/3x−2y = -1 ⇒3x – 2y = -1 …….(x) 

On adding (ix) and (x), we get: 

4x = 2 

⇒x = 1/2 

On substituting x = 1/2 in (x), we get: 

3/2 – 2y = -1 

2y = (3/2 + 1) = 5/2 

y = 5/4 

Hence, the required solution is x = 1/2 and y = 5/4.

The given equations are 

5/ x+y - 2/ x-y = -1 ……(i) 

15/ x+y - 7/x-y= 10 ……(ii) 

Substituting 1/x+y = u and 1/x-y = v in (i) and (ii), we get 

5u – 2v = -1 ……..(iii) 

15u + 7v = 10 …….(iv) 

Multiplying (iii) by 3 and subtracting it from (iv), we get 

7v + 6v = 10 + 3 

⇒13v = 13 

⇒v = 1

⇒x – y = 1 (∵ 1/ x-y = ) …..(v) 

Now, substituting v = 1 in (iii), we get 

5u – 2 = -1 

⇒5u = 1 

⇒u = 1/ 5 

x + y = 5 …….(vi) 

Adding (v) and (vi), we get 

2x = 6 

⇒ x = 3 

Substituting x = 3 in (vi), we have 

3 + y = 5 ⇒ y = 5 – 3 = 2 

Hence, x = 3 and y = 2. 

The given equations are

 \(\frac{2}{3x+2y}+\frac{3}{3x-2y}=\frac{17}{5}\)...........(i)

\(\frac{5}{3x+2y}+\frac{1}{3x-2y}=2\).........(ii)

Substituting 1/3x+2y = u and 1/3x−2y = v, in (i) and (ii), we get: 

2u + 3v = 17/5 ……..(iii) 

5u + v = 2 …….(iv) 

Multiplying (iv) by 3 and subtracting from (iii), we get: 

2u – 15u = 17/5 – 6 

⇒ –13u = −13/5 ⇒u = 1/5

⇒3x + 2y = 5 (∵ 1/3x+2y = ) …..(v) 

Now, substituting u = 1/5 in (iv), we get 

1 + v = 2 ⇒ v = 1 

⇒ 3x – 2y = 1 (∵ 1/3x−2y = ) …….(vi)

Adding(v) and (vi), we get: 

⇒ 6x = 6 ⇒ x = 1 

Substituting x = 1 in (v), we get: 

3 + 2y = 5 ⇒ y = 1 

Hence, x = 1 and y =1.

Ratio : It is a tool to compare two or more quantities (or numbers) having same units by division. 

First term is called antecedent and second term is called consequent.

Mixture 60

Milk 7x=35

Water 5x=25

Milk is not added and kept as it is so 

5x=35   x=7 there by 7x=7*7=49  

Already there is 25 litres of water in the mixture 

so additionally we need to add 24 litres of water.

Given cos^-1 (cos 680°)

= cos^-1 (cos (720° - 40°)) 

= cos^-1 (cos (2 × 360° - 40°)) 

= cos^-1 (cos 40°) 

= 40° 

∴ cos^-1 (cos 680°) = 40°

Given sin^-1 (sin 3π/5) 

= sin^-1 [sin (π – 2π/5)] 

= sin^-1 (sin 2π/5) 

= 2π/5 

∴ sin^-1 (sin 3π/5) = 2π/5

Given |x| ≤ 1

⇒ ± x ≤ 1

⇒ x ≤ 1 or –x ≤ 1

⇒ x ≤ 1 or x ≥ -1

⇒ x ∈ [-1, 1]

Now also given cos (sin^-1 x + cos^-1 x)

We know that sin^-1 x + cos^-1 x = π/2

∴ cos (sin^-1 x + cos^-1 x) = cos (π/2) = 0

Given |x| ≤ 1 

⇒ ± x ≤ 1 

⇒ x ≤ 1 or –x ≤ 1 

⇒ x ≤ 1 or x ≥ -1 

⇒ x ∈ [-1, 1] 

Now also given cos (sin^-1 x + cos^-1 x) 

We know that sin^-1 x + cos^-1 x = π/2 

∴ cos (sin^-1 x + cos^-1 x) = cos (π/2) = 0

We know that the value of sec^-1 (1/2) is undefined as it is outside the range i.e. R – (-1, 1).

The set of values of sec^-1 (1/2) is ɸ.

Domain of sec^-1 x is R – (-1,1).

As, -1/2 is outside domain of sec^-1 x.

Which means there is no set of value of sec^-1 1/2.

So, the solution set of sec^-1 1/2 is null set or ɸ.

Given cos^-1 (cos 14π/3) 

= cos^-1 [cos (4π + 2π/3)] 

= cos^-1 (cos 2π/3) 

= 2π/3 

∴ cos -1 (cos 14π/3) = 2π/3

(b) 2/5

Given,

cos (sin^-1 2/5 + cos^-1 x) = 0

So, this can be rewritten as

sin^-1 2/5 + cos^-1 x = cos^-1 0

sin^-1 2/5 + cos^-1 x = π/2

cos^-1 x = π/2 – sin^-1 2/5

cos^-1 x = cos^-1 2/5 [Since, cos^-1 x + sin^-1 x = π/2]

Hence,

x = 2/5

(a) [1, 2]

We know that, sin^-1 x is defined for x ∈ [-1, 1]

So, f(x) = sin^-1 √(x – 1) is defined if

0 ≤ √(x – 1) ≤ 1

0 ≤ x – 1 ≤ 1

1 ≤ x ≤ 2

Hence,

x ∈ [1, 2]

Because the winds that bring rainfall blow from various directions.

Climate is an average weather condition that exists in an area over a long period of time. Conditions related to climate or weather are called climatic conditions.

• India possesses a very large area of forests.
• India has evergreen, deciduous, thorny and tidal forests.
• India ranks 10th in terms of variety in vegetation.
• These forests provide livelihood to numerous people.
• We get several medicinal herbs, paper, timber, sealing wax, rubber, etc. from forests.
• Thus, forests are invaluable treasure of our country.