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An element e ∈ Q will be the identity element for the operation if

a*e = a = e*a, a ∈ Q

(i) a*b = a − b

If a*e = a,a ≠ 0 ⇒ a −e = a, a ≠ 0 ⇒ e = 0

Also, e*a = a ⇒ e − a = a ⇒ e = 2a

⇒ e = 0 = 2a,a ≠ 0

But the identiry is unique. Hence this operation has no identity.

(ii) a*b = a² + b²

If a*e = a, then a² + e² = a

For a = −2, (−2)² + e² = 4 + e² ≠ −2

Hence, there is no identity element.

(iii) a*b = a + ab

If a*e = a ⇒ a + ae = a ⇒ ae = 0 ⇒ e = 0,a ≠ 0

Also if e*a = a ⇒ e + ea = a ⇒ e = a/(1 - a), a ≠ 1

∴ e = 0 = a/(1 - a), a ≠ 0

But the identity is unique. Hence this operation has no identify.

(iv) a*b = (a − b)²

If a*e = a, then (a − e)² = a. A square is always positive, so for

a = −2,(−2 − e)² ≠ − 2

Hence, there is no identity element.

(v) a*b = ab/4

If a*e = a, then ae /4 = a. 

Hence, e = 4 is the identity element.

∴  a*4 =4 *a = 4a/4 = a

(vi) a*b = ab²

If a*e =a then ae² = a ⇒ e² = 1 ⇒ e = ± 1

But identity is unique. Hence this operation has no identity.

Therefore only part (v) has an identity element.

a * b = (a – b)² = (b – a)²= b * a

hence * is not commutative

(a * b) * c = (a – b)² * c = ((a-b)² – c)²

= (a – b)⁴+ c² – 2c (a – b)² a*(b * c)

= a* (b – c)² = [a -(b – c)]²

= a² + (b – c)² – 2a (b – c)²

hence not associative

The Executive in India has to fulfill following responsibilities:

1. Implementing the laws passed by the Parliament. 

2. Framing policies for governing the nation.

3. Functioning as the members of the Legislature. 

4. Striving to develop the nation through decisions related to the welfare of the people.

(i) On Q, the operation * is defined as a*b = a − b.

It can be observed that for 2,3,4 ∈ Q, we have 2*3 = 2 − 3 = −1 and 3*2 = 3 − 2 = 1 

⇒ 2*3 ≠ 3*2

Thus, the operation is not commutative.

It can also be observed that

(2*3)*4 = (−1)*4 = −1 − 4 = − 5

and 2*(3*4) = 2*(−1) = 2 − (−1) = 3 

2*(3*4) ≠ 2* (3*4)

Thus, the operation * is not associative.

(ii) On Q, the operation * is defined as a*b = a² + b².

For a,b ∈ Q, we have a*b = a² + b² = b² + a² = b*a

Therefore, a*b = b*a

Thus, the operation * is commutative.

It can be observed that

(1* 2)* 3 = (1² + 2²) * 3 (1 + 4) * 4 = 5 * 4 = 5² + 4² = 41

and 1*(2* 3) = 1* (2² + 3²) = 1 * (4 + 9) = 1*13 = 1² + 13² = 170 

⇒ (1*2)* 3 ≠ 1*(2*3) where 1,2, 3 ∈ Q

Thus, the operation * is not associative.

(iii) On Q, the operation is defined as a*b = a + ab 

It can be observed that

1*2 = 1 + 1 × 2 = 1 + 2 = 3, 2*1 = 2 + 2 × 1 = 2 + 2 = 4

⇒ 1*2 ≠ 2*1 where 1, 2 ∈ Q

Thus, the operation * is not commutative.

It can also be observed that

(1*2)*3 = (1 + 1 × 2)*3 = 3*3 = 3 + 3 × 3 = 3 + 9 = 12

and 1*(2* 3) = 1*(2 + 2 × 3) = 1*8 = 1 + 1 × 8 = 9

⇒ (1*2)* 3 ≠ 1*(2*3) where 1,2, 3 ∈ Q

Thus, the operation * is not associative.

(iv) On Q, the operation * is defined by a*b = (a − b)².

For a,b ∈ Q, we have a*b = (a − b)² and b*a = (b − a)² = [−(a − b)]² = (a − b)²

Therefore, a*b = b*a 

Thus, the operation * is commutative. It can be observed that 

(1* 2) * 3 = (1 – 2)² * 3 = ( 1) * 3 = (1 – 3)² = 4 

and 1 * (2 *3) = 1 * (2 – 3)² = 1 * (1) = (1 – 1)² = 0 

⇒ (1*2)* 3 ≠ 1*(2* 3) where 1,2, 3 ∈ Q 

Thus, the operation * is not associative.

(v) On Q, the operation * is defined as a*b = ab/4

For a,b ∈ Q, we have a * b = ab/4 = ba/4 =b*a 

Therefore, a*b = b*a 

Thus, the operation * is commutative.

For a,b,c ∈ Q, we have a*(b*c) = ab/4 * c = ((ab/4)c)/4 = abc/16 

and a*(b*c) = a* bc/4 = (a(bc/4))/4 = abc/16

Therefore, (a*b)*c = a*(b*c).

Thus, the operation * is associative.

(vi) On Q, the operation is defined as a*b = ab²

It can be observed that for 23 ∈ Q

2 * 3 = 2 x 3² = 18 and 3*2 = 3 x 2² = 12

Hence, 2*3 ≠ 3*2

Thus, the operation is not commutative.

It can also be observed that for 1,2,3 ∈ Q 

(1*2)*3 = (1. 2²)*3 = 4 *3 = 4.3² = 36 

and 1*(2* 3) = 1*(2. 3²) = 1 *18 = 1.18² = 324 

⇒ (1*2)*3 ≠ 1*(2*3)

Thus, the operation * is not associative. 

Hence, the operations defined in parts (ii), (iv), (v) are commutative and the operation defined in part (v) is associative.

a * e = e * a = a 

a – e = a = e – a 

⇒ e = 0 

or 

e = 2a ∀ a ∈ Q 

which is not possible. 

Hence no identity.

1. In the Parliamentary System, the Executive is dependent upon the confidence of the Legislature in them. 

2. If the Legislature feels that the Executive does not work in accordance to its wishes, it can pass a no-confidence motion against the Executive. 

3. If the Legislature passes the no-confidence motion with a majority, then the Executive has to resign. 

4. The Constitution has provided this tool to prevent the Executive from misusing its powers. Hence, no-confidence motion is an effective tool of keeping a check over the Executive.

a ∈ Q, be Q , 

a * b = ab² 

b * a = ba² 

∴ a * b≠ b * a 

hence not commutative. 

(a * b) * c = (ab²) * c = ab²c² 

a * (b * c) = a * (bc²) = ab²c⁴ 

(a * b) * c = a * (b) 

∴ * is not associative.

The functions of the type f(x)/g(x) where f(x) g(x) and g(x) are polynomial functions of x, defined in a domain, where g(x) ≠ 0

(c) 1.5 kg ms^–1.

(c) –(0.9i + 1.2j)

\(\bar F\) = 3\(\hat i\) - 4\(\hat j\)

Magnitude:

\({|\bar F|}\) = \(\sqrt{ 9+ 16}\) = \(\sqrt {25}\) = 5N

F = ma

⇒ m =\(\frac{|\bar F|}{a}\) = \(\frac{5}{10}\) =\(\frac{1}{2}\) = 0.5 kg

Given: f(x) = x + 1,g(x) = 2x - 3

(f + g)(x) = f(x) + g(x) = x + 1 + 2x - 3 = 3x - 2

(f – g)(x) = f(x) – g(x) = x + 1 - 2x + 3 = -x + 4

(f/g)(x) = f(x)/g(x) = (x + 1)/(2x - 3) x ≠ 3/2

A function f : M → R is said to be polynomial function if for each x in IR, 

y = f (x) = a₀ + a_xx + a₂x² + ……………. + a_nxⁿ 

where n is a non-negative integer and a₀,a₁,a₂………………….a_n ∈ R .

Given C(t) = 3t. 

To prove that the function is linear 

C(at₁) = 3a(t₁) 

C(bt₂) = 3 b(t₂) 

C(at₁ + bt₂) = 3 [at₁ + bt₂] = 3at₁ + 3bt₂ 

= a(3t₁) + b(3t₂) = a[C(t₁) + b(Ct₂)] 

∴ Superposition principle is satisfied. 

Hence C(t) = 3t is linear function.

Since/is a linear function.

∴ f{x) = ax + b

But (0,-1) ∈ f . f(0) 

∴ a(0) + b ⇒ -1 = b

Similarly, (1, 1) ∈ f 

∴ f(1) = a(1) +a(1) + b

⇒ 1= a + b ∴ a = 2 

∴ f(x) = 2x -1

The function f : R → R defined by f(x) = mx + c, x ∈ R is called linear function, where m and c are constant.

The equation of a circle with (x₁, y₁) and (x₂, y₂) as end points of a diameter is 

(x – x₁)(x – x₂) + (y – y₁)(y – y₂) = 0 

Here the end points of a diameter are (3, 4) and (2, -7) 

So equation of the circle is 

(x – 3 )(x – 2 ) + (y – 4) (y + 7.) = 0 

x² + y² – 5x + 37 – 22 = 0

Let the required circle be 

x² + y² + 2gx + 2fy + c = 0 

The circle passes through 

(1, 0), (-1, 0) and (0, 1) 

(1, 0) ⇒ 1 + 0 + 2g(1) + 2f(0) + c = 0 

2g + c = -1 … (1) 

(-1, 0) ⇒ 1 + 0 + 2g (-1) + 2f(0) + c = 0

-2g + c = -1 … (2) 

(0, 1) ⇒ 0 + 1 + 2g (0) + 2f(1) + c = 0 

2f+ c = -1 ... (3) 

Now solving (1), (2) and (3). 

2g + c = -1 … (1) 

-2g + c = -1 …(2) 

(1) + (2) ⇒ 2c = -2 ⇒ c = -1 

Substituting c = -1 in (1) we get

2g – 1 = -1 

2g = -1 + 1 = 0 ⇒ g = 0 

Substituting c = -1 in (3) we get 

2f – 1 =-1 ⇒ 2f = -1 + 1 = 0 ⇒ f = 0 

So we get g = 0, f= 0 and c = -1 

So the required circle will be 

x² + y² + 2(0) x + 2(0)y – 1 = 0 

(i.e) x² + y² – 1 = 0 ⇒ x² + y² = 1

Area of the circle = 9π 

(i.e) πr² = 9π 

⇒ r² = 9 ⇒ r = 3 

(i.e) radius of the circle = r = 3 

The two diameters are x + y = 5 and x – y = 1 

The point of intersection of the diameter is the centre of the circle = C 

To find C: 

Solving x + y = 5 … (1) 

x – y = 1 …(2) 

(1) + (2) ⇒ 2x = 6 ⇒ x = 3 

Substituting x = 3 in (1) we get 

3 + y = 5 ⇒ y = 5 – 3 = 2 

∴ Centre = (3, 2) and radius = 3

So equation of the circle is 

(x – 3)² + (y – 2)² = 3² 

(i.e) x² + y² – 6x – 4y + 4 =0

The condition for the line y = mx + c to be a tangent to the circle x² + y² = a² is 

c² = a² (1 + m²) 

Here x² + y² = 16 ⇒ a² = 16 

y = 2√2x + c ⇒ m = 2√2 and c = c 

The condition is c² = a² (1 + m²) 

(i.e) c² = 16(1 + 8) = 144 

⇒ c = ± 12

(a) 0, -40/9

x² + y² – 5x – 6y + 9 + λ(4x + 3y -19) = 0 

ie., x² + y² + x(4λ – 5) + y(3λ – 6) + 9 — 19λ = 0 

Since it touches y axis x = 0 

⇒ y² + y(3λ – 6) + 9 – 19λ = 0 

It is a quadratic in y, 

Since the circle touches y axis the roots must be equal 

⇒ b² – 4ac = 0 

ie.,(3λ – 6)² – 4(1) (9 – 19λ) = 0 

Solving we get λ = o or -40/9

x² + y² + 2x – 6y = 0 …(1) 

Here 2g = 2 ⇒ g = 1, 2f = -6 ⇒ f = -3 

Centre of the circle = (-g, -f) = (-1, 3) 

Since the required circle is concentric with (1), its centre is also (-1, 3). 

∴ The equation of the circle whose centre is (-1, 3) and radius 4 is 

(x + 1)² + (y – 3)² = 4² 

⇒ x² + y² + 2x – 6y – 6 = 0

The equation of the tangent to the circle x² + y² + 2 gx + 2fy + c = 0 at (x₁, y₁) is 

xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 9 

So the equation of the tangent to the circle 

x² + y² – 6x + 6y – 8 = 0 at (x₁, y₁) is 

xx₁ + yy₁ - (6(x + x₁)/2) + (6(y + y₁)/2) - 8 = 0

(i.e) xx₁ + yy₁ – 3(x + x₁) + 3(y + y₁) – 8 = 0 

Here (x₁, y₁) = (2, 2) 

So equation of the tangent is 

x(2) + y(2) – 3(x + 2) + 3(y + 2) – 8 = 0 

(.i.e) 2x + 2y – 3x – 6 + 3y + 6 – 8 = 0 

(i.e) -x + 5y – 8 = 0 or x – 5y + 8 = 0 

Normal is a line ⊥r to the tangent 

So equation of normal circle be of the form 5x + y + k = 0 

The normal is drawn at (2, 2) 

⇒ 10 + 2 + k = 0 ⇒ k = -12 

So equation of normal is 5x + y – 12 = 0

(d) -35 < m < 15

x² + y² – 4x – 8y – 5 = 0 

3x – 4y = m Solving (1) and (2) we get 

from (2) ⇒ 3x = 4y + m 

x = (4y + m)/3

Substituting x value in (1) we get

(4y + m/3)² + y² - 4(4y + m/3) - 8y - 5 = 0

It is a quadratic in y and given that the roots are distinct 

⇒ b² – 4ac > 0 

On simplifying we get 

⇒ – 9m² – 18w + 4725 > 0 

⇒ m² + 20m – 525 < 0 

⇒ (m + 35) (m – 15) ≤ 0 

⇒ m lies between -35 and 15 ie„ – 35 < m < 15

(b) 1/3

The given equation is of the form FP² = e²PM² 

i.e., (x – 3)² + (y – 4)² = (1/9)y²

e² = 1/9

e = 1/3

To find the position of a point with regard to a given circle, substitute the point in the equation of the circle if we get a positive value, the point lies outside the circle. 

If we get a -ve value the point lies inside the circle and if we get O then the point lies on the circumference of the circle. 

The given circle is x² + y² – 5x + 2y – 5 = 0 ... (1)

Substituting the point (-2, 1) in (1) we get 

4 + 1 – 5(-2) + 2(1) – 5 = 5 + 10 + 2 – 5 = 12 

⇒ (- 2, 1) lies outside the circle 

Substituting the point (0, 0) in (1) we get 

-5 < 0 ⇒ (0, 0) lies inside the circle 

Substituting the point (-4, -3) in (1) we get 

16 + 9 + 20 – 6 – 5 = 34 > 0 

⇒ (- 4, -3) lies outside the circle

2x³ + 13x³ + x – 70 को (x – 2) से भाग करने पर x – 2 = 0 या x = 2 रखने पर

शेषफल = 2(2)³ + 13(2)² + 2 – 70 = 16 + 52 + 2 – 70 = 0

अत: 2x³ + 13x² + x – 70, (x – 2) से पूर्णतया विभाजित होगा।

जब Px³ + 4x² + 3x – 4 को (x – 3) से भाग करने पर x – 3 = 0 या x = 3 रखने पर

शेषफल = P(3)³ + 4(3)² + 3(3) – 4 = 27P + 36 + 9 – 4 = 27P + 41

जब x³ – 4x + P को (x – 3) से भाग करने पर x – 3 = 0 या x = 3 रखने पर

शेषफल = 3³ – 4(3) + P = 27 – 12 + P = 15 + P

∴ 27P + 41 = 15 + P

27P – P = 15 – 41

26P = -26.

P = -26/26 = -1

जब P(x) को (ax – b) से विभाजित किया जाता है तो ax – b = 0 या \(\frac{b}{a}\) रखने पर शेषफल 

= \(\boldsymbol{P}\left(\frac{\boldsymbol{b}}{\boldsymbol{a}}\right)\)

जब x³¹ + 31 को (x + 1) से विभाजित किया जाता है तो x + 1 = 0 या x = 0 – 1 = -1

शेषफल = (-1)³¹ + 31 = -1 + 31 = 30

जब P(x) को (x – a) से विभाजित किया जाता है तो x – a = 0

∴ x = a रखने पर

शेषफल = P(a)

x³ – 2x² + x + 1 को (x – 1) से विभाजित किया जाता है तो x – 1 = 0 या x = 1 रखने पर शेषफल 

= (1)³ – 2(1)² + 1 + 1 = 1 – 2 + 1 + 1 = 1

जब 3x³ – 4x³ + 7x – 5 को (x – 3) से विभाजित किया जाता है तो x – 3 = 0 या x = 3 रखने पर शेषफल

 = 3(3)³ – 4(3)² + 7(3) – 5 

= 3. 27 – 36 + 21 – 5 

= 81 – 36 + 21 – 5 

= 102 – 41 = 61

जब 5x³ – 2x² – 7x + 1 को x से विभाजित किया जाता है तो x = 0 रखने पर

शेषफल = 0 – 0 – 0 + 1 = 1

x³ + 5x² – 2 को (x – 1) से विभाजित किया जाता है तो x – 1 = 0 या x = 1 रखने पर शेषफल 

= (1)³ + 5(1)² – 2 = 1 + 5 – 2 = 4

(25x² – 1) + (1 – 5x)² = 25x² – 1 + 1 + 25x² – 10x = 50x² – 10x = 10x(5x – 1)
x बहुपद का एक गुणनखण्ड है।

यदि f(x) = x⁴ – 2x³ + 3x² – ax + b को (x – 1) से भाग किया जाता है तो x – 1 = 0 या x = 1 रखने पर

शेषफल = (1)⁴ – 2(1)³ + 3(1)² – a(1) + b 

= 1 – 2 + 3 – a + b = 2 – a + b

प्रश्नानुसार, 2 – a + b = 5

-a + b = 5 – 2 = 3 

⇒ – a + b = 3 ………………(1)

प्रश्नानुसार, यदि f(x) को (x + 1) से भाग किया जाता है तो

x + 1 = 0 या x = -1 रखने पर

शेषफल = (-1)⁴ – 2(-1)³ + 3(-1)² – a(-1) + b

 1 + 2 + 3 + a + b

= 6 + a + b

तथा 6 + a + b = 19

a + b = 19 – 6

a + b = 13 ………………(2)

(1) व (2) जोडने पर,

2b = 16

b =16/2 = 8

समी० (2) मे b का मान रखने पर, a + 8 = 13

a = 13 – 8 = 5

अतः a = 5 व b = 8

बहुपद 9x³ + 3x² – 13 को (x – 2) से भाग किया जाता है तो x – 2 = 0 या x = 2 रखने पर,

शेषफल = 9(2)³ + 3(2)² – 13 = 72 + 12 – 13 = 71

बहुपद 2x³ – 5x + a को (x – 2) से भाग किया जाता है तो x – 2 = 0 या x = 2 रखने पर,

शेषफल = 2(2)³ – 5(2) + a = 16 – 10 + a या 6 + a

प्रश्नानुसार, 6 + a = 71

a = 71 – 6 = 65

जब Px³ + 9x³ + 4x – 10 को (x + 3) से भाग किया जाता है तो x + 3 = 0 या x = -3 रखने पर

शेषफल = P(-3)³ + 9(-3)² + 4(-3) – 10 = -22

-27P + 81 – 12 – 10 = -22

-27P = -22 – 81 + 22

– 27P = -81

P = -81/27 = 3

P(x) = 3x² – 4x + √17, यदि x = 3 

शेषफल = 3(3)² – 4 × 3 +√17

= 27 – 12 + √17 = 15 + √17

• Nowadays, the computer can be used to watch television programs on the Internet. People can also watch movies, listen to music and play games on the computer.
• Multimedia is used to develop special effects in movies and animations.

• Computers are used to control machines and the continuous running of machinery.
• A robot is an automatic programmable machine and can be used in hundreds of applications like assembling and spray painting of cars etc.

• Using Computer Aided Learning (CAL) & Computer Based Training
• Students can readily learn and get new skills at their own convenient speed. 
• Students can get timely and high-quality training on demand.
• Students can solve different kinds of problems quickly and efficiently, by using computers.

Hybrid computers are used in Process control (industries) and Robotics.

Mainframe computers are used in business process applications and supercomputers are used in scientific applications and calculations.

Intel core i9 5 generation and AMD’s FX processors.

t_n = a_n = n³ - 2

(i) a₁ = 1³ – 2 = 1 – 2 – 1 

a₂ = 2³ – 2 = 8 – 2 = 6 

a₃ = 3³ – 2 = 27 – 2 = 25 

a₄ = 4³ – 2 = 64 – 2 = 62 

∴ The first four terms are -1, 6, 25, 62, …

(ii) a_n = (-1)^{n + 1} n(n + 1) 

a₁ = (-1)^{1 + 1} (1) (1 +1) 

= (-1)² (1) (2) = 2 

a₂ = (-1)^{2 + 1} (2) (2 + 1) 

= (-1)³ (2) (3) = -6 

a₃ = (-1)^{3 + 1} (3) (3 + 1) 

= (-1)⁴ (3) (4) = 12 

a₄ = (-1)^{4 + 1} (4) (4 + 1) 

= (-1)⁵ (4) (5) = -20 

∴ The first four terms are 2, -6, 12, -20,…

(iii) a_n = 2n² – 6 

a₁ = 2(1)² – 6 = 2 – 6 = -4 

a₂ = 2(2)² – 6 = 8 – 6 = 2 

a₃ = 2(3)² – 6 = 18 – 6 = 12

a₄ = 2(4)² – 6 = 32 – 6 = 26 

∴ The first four terms are -4, 2, 12, 26, …

(i) a_n = 5n/(n + 2)

a₆ = 5(6)/(6 + 2) = 30/8 = 15/4

a₁₃ = 5(13)/(13 + 2) = (5 x 13)/15 = 13/3 

a₆ = 15/4, a₁₃ = 13/3

(ii) a_n = -(n² – 4) 

a₄ = -(4² – 4) 

= – (16 – 4) 

= -12

a₁₁ = -(11² – 4) 

= – (121 – 4) 

= – 117 

a₄ = -12 and a₁₁ = -117

The main difference is microcomputers are single-user computers and mini-computers are multi-user computers.

(i) (1² + 1);(2² + 1),(3² + 1),(4² + 1) … 

n^th term is n² + 1 

a_n = n² + 1

(ii) (1 - 1/1), (2 - 1/2), (3 - 1/3) … 

n^th term is (n - 1)/n

a_n = (n - 1)/n

(iii) [5(1) -2], [5(2) – 2], [5(3) – 2], [5(4) – 2] …

The n^th term is 5n – 2 

a_n = 5n – 2