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(ABCD) = ar (ABFC) …(i)

ar (∆ABF) = ar (∆AFC)

ar ∆AFC = \(\frac { 1 }{ 2 }\) ar (∆BCD) …(ii)

Also ar (∆ADC) = \(\frac { 1 }{ 2 }\) ar (∆BCD) …(iii)

⇒ ar (∆AFC) = ar (∆ADC) …(iv)

⇒ ar (∆ADF) = ar (∆ADC) + ar (∆ACF) = \(\frac { 1 }{ 2 }\) ar

(∆BCD) + \(\frac { 1 }{ 2 }\) ar (∆BFC) [using (ii) and (iii)]

= \(\frac { 1 }{ 2 }\) ar (∆BFC) + \(\frac { 1 }{ 2 }\) ar (∆BFC)

= ar (∆BFC)

⇒ ar (∆ADF) = ar (∆BFC)

Hence proved.

Given: CD || AE and CY || BE

To prove: ar (∆CBX) = ar (∆AXY)

Proof: Since, ∆ABC and ∆BAY both lie on the same base AB and between the same parallel AB and CY.

ar (∆ABC) = (∆BAY)

⇒ ar (∆ABX) + ar (∆CBX) = ar (∆ABX) + ar (∆AXY)

⇒ ar (∆CBX) = ar (∆AXY)

[Eliminating ar (∆ABX) from both sides]

Hence proved.

2 ∈ Z⁺ , 5 ∈ Z⁺ , 2 – 5 = -3 ∉ Z⁺

hence not a binary operation.

Let P(n): 7ⁿ – 2ⁿ is divisible by 5, for any natural number n.

Now, P(1) = 7¹ - 2¹ = 5, which is divisible by 5.

Hence, P(1) is true.

Let us assume that, P(n) is true for some natural number n = k.

.’. P(k) = 7^k -2^k is divisible by 5

or 7^k – 2^k = 5m, m ∈ N .......(i)

Now, we have to prove that P(k + 1) is true

P(k+ 1): 7^k+1 - 2^k+1

= 7^k.7 - 2^k.2

= (5 + 2)7^k - 2^k.2

= 5.7^k + 2.7^k -2^k.2

= 5.7^k + 2(7^k – 2^k)

= 5 • 7^k + 2(5 m)          (using (i))

= 5(7^k + 2m), which divisible by 5.

Thus, P(k + 1) is true whenever P(k) is true.

So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Given a * b = HCF (a, b), a, b ∈ N

⇒ 22 * 4 = HCF (22, 4) = 2

Required operation table of the operation * is given as,

Any element of set A, say x_i can be connected with the element of set B in p ways. Hence, there are exactly p^q functions.

Here, function f : R₊ →[−5,∞) is given as f(x) = 9x² + 6x − 5. 

Let y be any arbitrary element of [−5,∞). 

Let y = 9x² + 6x − 5 

⇒ y = (3x + 1)² − 1 − 5 = (3x + 1)² − 6 

⇒ (3x + 1)² = y + 6 

⇒ (3x + 1)² = √(y + 6) [as y ≥ − 5 ⇒ y + 6 ≥ 0]

x = (√(y = 6) - 1)/3

Therefore, f is onto, thereby range f = [−5,∞).

Let us define g : [−5,∞)→ R+ as g(y) =  (√(y = 6) - 1)/3

Now, (gof)(x) = g(f(x)) = g(9x² + 6x − 5) = g((3x + 1)² − 6)

= (√((3x + 1)² - 6) + 6 - 1)/3 = (3x + 1 - 1)/3 = x

and (fog)(y) = f(g(y)) = f((√(y + 6) - 1)/3) = [3((√(y + 6) - 1)/3) + 1]² - 6

= (y + 6)2 - 6 = y + 6 - 6 = y

Therefore,gof = I_R+ and fog = I_[−5,∞)

Hence, f is invertible and the inverse of f if given by f^-1(y) = g(y) = (√(y + 6) - 1)/3

fog(x) = f(g(x)) = f(x² + 2) 

= 3(x² + 2) + 1 = 3x² + 6 + 1

⇒ fog(x) = 3x² + 7 

∴ fog(2) = 3 × 2² + 7

= 12 + 7 = 19

(2 * 3) * 4 = (HCF of 2 and 3) * 4 = (1 * 4) = 1

2 * (3 * 4) = 2 * (HCF of 3 and 4) = 2 * 1 = 1

Here, f :R→R is given as f(x) = x³. 

Suppose, f(x) = f(y),where x,y ∈ R ⇒ x³ = y³ …(i) 

Now, we need to show that x = y 

Suppose, x ≠ y, their cubes will also not be equal. x³ ≠ y³

However, this will be a contradiction to Eq. i). 

Therefore, x = y. Hence, f is injective.

Number of functions from X to Y = 3² = 9.

To prove f is invertible, it is sufficient to prove f is one - one onto.

Here,

f(x) = 5x² + 6x - 9

One - one : 

Let x₁,x₂ ∈ R,. then,

f(x₁) = f(x₂)

⇒ 5x₁² + 6x₁ - 9 = 5x₂² + 6x₂ - 9

⇒ 5x₁² + 6x₁  - 5x₂² - 6x₂ = 0

⇒ 5(x₁² - x₂²) + 6(x₁ - x₂) = 0 

⇒ 5(x₁ - x₂)(x₁ + x₂) +6(x₁ - x₂) = 0

⇒ (x₁ - x₂) (5x₁ + 5x₂ + 6) = 0

⇒ x₁ - x₂ = 0 [ ∵ 5x¹ + 5x² + 6 ≠ 0]

⇒ x₁ = x₂

i.e., f is one - one function.

Onto : 

Let f(x) = y

∴ y = 5x² + 6x -9 

⇒ 5x² + 6x - (9 + y) = 0 

\(⇒x = {-6 \pm \sqrt{36\,+\,4\,\times\,5(9\,+\,y)} \over 10}\)

\(⇒x = {-6 \pm \sqrt{216\,+\,20y} \over 10}\) 

\(⇒x = {\pm \sqrt{54\,+\,5y\,-\,3} \over 5}\) 

\(⇒x = \frac{\sqrt{54+5y}\,-\,3}{5}\)  [ ∵ x ∈ R₊]

Obviously, 

∀ y ∈ [-9, ∞] the value of x ∈ R₊

⇒ f is onto function.

Hence, f is one - one onto function, i.e., invertible.

Also, f is invertible with

\(f^{-1}(y) = \frac{\sqrt{54+5y}\,-\,3}{5}\)

Obviously ‘gof’ function is defined as 

gof :  A → C such that 

gof(1) = g(f(1)) = g(4) = 5

gof(2) = g(f(2)) = g(5) = 6

gof(3) = g(f(3)) = g(4) = 5

Hence, gof : A → C is given by gof = {(1, 5), (2, 6), (3, 5)}

As g (x) = g (-x) = |x| for all x ∈ Z , 

∴ g is not one-one 

i.e. y is not injective 

As f : N → Z and g : Z → Z gof : N → Z 

let x₁, x₂, ∈ N such that gof (x₁) = gof (x₂) 

⇒ g (x₁) = g (x₂) 

⇒ | x₁ | = | x₂ | 

⇒ x₁= X₂ (both x₁ ,x₂ >0) 

Hence g o f is injective.

The number of binary operations on the set consisting n elements is \(n^{n^2}\) . 

Here n = 3. 

Therefore, total number of binary operation S = \((3)^{3^2}\) = 3⁹.

Given f(x) = (x² + 3x + 5)/(x² - 5x + 4)

f(x) is defined for all real numbers except x² – 5x + 4 = 0

But x² -5x + 4 = (x - 4) (x - 1)

Domain of f= R-{1,4}

Let f : X → Y is invertible 

⇒ f is one and onto and 

f^-1 : Y ⇒ X is defined as f^-1(y) = x 

y : f (x) ∀ x ∈ X and y ∈ Y 

let y₁,y₂∈ y f^-1 (y₁) = f^-2(y₂) 

fof¹ (y₁) = fof¹ (y2) 

I_y (y₁) = I_y(y₂) 

⇒ y₁ = y₂∴ f₁ is one-one 

∀ x ∈ X, ∋ y ∈ Y such that 

f¹(y) = x, hence f¹ is onto 

hence invertible. 

let g = (f¹)^-1 

gof^-1 = I_y and f^-1og = lx 

∀ x ∈ X, I_x (x) = x 

fof^-1(x) = f^-1 [g(x)] = x 

fof^-1 [g (x)] = f (x) 

(fof^-1) (g (x)) = f (x) 

g(x) = f (x) 

g = f 

(f^-1)^-1 = f

Given, 

R = {(x, y) : x ∈ N, x < 5, y = 3}

⇒ R = {(1, 3), (2, 3), (3, 3), (4, 3)}

Hence, required inverse relation is 

R^–1 = {(3, 1), (3, 2), (3, 3), (3, 4)} 

∴ Domain of R^–1 = {3}

And Range of R^–1 = {1, 2, 3, 4}

f(x) = x+ lxl + 1 ≥ 1 + 1 ∀ x ∈ N 

(∵ v ≥ 1 x ∈ N) 

⇒ f(x) ≥ 2 ∀ x ∈ N . R_f ^ N as 1 t R_f 

Hence f is not onto 

gof : → ∼ → N such that 

gof (x) = g (f (x)) = g (x + 1) = (x+1) -1 

[ ∵ ∀ x ∈ N,x+ 1 > 1] 

⇒ gof (x) = x V x ∈ N 

∴ Range of gof = N as gof is the identity f 

Hence gof is onto.

Total number of all onto functions from the set {1, 2, 3, …, n) to itself is n!

Given, A = {2, 3, 4}, B = {2, 5, 6, 7}

(i) Let f: A → B denote a mapping

f = {(x, y): y = x + 3} or

f = {(2, 5), (3, 6), (4, 7)}, which is an injective mapping.

(ii) Let g: A → B denote a mapping such that g = {(2, 2), (3, 2), (4, 5)}, which is not an injective mapping.

(iii) Let h: B → A denote a mapping such that h = {(2, 2), (5, 3), (6, 4), (7, 4)}, which is one of the mapping from B to A.

    f is a rational function 

∴ Domain of f = R – {x : x² – 5x + 6 = 0} 

i. e., for function to be defined x² – 5x + 6 ≠ 0 

⇒ x² - 3x - 2x + 6 ≠ 0 ⇒ x (x – 3) – 2 (x – 3) ≠ 0 

⇒ (x – 3) (x – 2) ≠ 0 

∴ Domain = R – {2, 3}

Given,

f: A → B and g: B → A satisfy g o f = I_A

It’s clearly seen that function ‘g’ is inverse of ‘f’.

So, ‘f’ has to be one-one and onto.

Hence, ‘g’ is also one-one and onto.

Given ∀ a, b ∈ Z, aRb if and only if a – b is divisible by n.

Now, for

aRa ⇒ (a – a) is divisible by n, which is true for any integer a as ‘0’ is divisible by n.

Thus, R is reflective.

Now, aRb

So, (a – b) is divisible by n.

⇒ – (b – a) is divisible by n.

⇒ (b – a) is divisible by n

⇒ bRa

Thus, R is symmetric.

Let aRb and bRc

Then, (a – b) is divisible by n and (b – c) is divisible by n.

So, (a – b) + (b – c) is divisible by n.

⇒ (a – c) is divisible by n.

⇒ aRc

Thus, R is transitive.

So, R is an equivalence relation.

(i) Let f: N → N, be a mapping defined by f (x) = x²

For f (x₁) = f (x₂)

Then, x₁² = x₂²

x₁ = x₂ (Since x₁ + x₂ = 0 is not possible)

Further ‘f’ is not onto, as for 1 ∈ N, there does not exist any x in N such that f (x) = 2x + 1.

(ii) Let f: R → [0, ∞), be a mapping defined by f(x) = |x|

Then, it’s clearly seen that f (x) is not one-one as f (2) = f (-2).

But |x| ≥ 0, so range is [0, ∞].

Therefore, f (x) is onto.

(iii) Let f: R → R, be a mapping defined by f (x) = x²

Then clearly f (x) is not one-one as f (1) = f (-1). Also range of f (x) is [0, ∞).

Therefore, f (x) is neither one-one nor onto.

Given, A = [–1, 1]

(i) f: [-1, 1] → [-1, 1], f (x) = x/2

Let f (x₁) = f(x₂)

x₁/ 2 = x₂

So, f (x) is one-one.

Also x ∈ [-1, 1]

x/2 = f (x) = [-1/2, 1/2]

Hence, the range is a subset of co-domain ‘A’

So, f (x) is not onto.

Therefore, f (x) is not bijective.

(ii) g (x) = |x|

Let g (x₁) = g (x₂)

|x₁| = |x₂|

x₁ = ± x₂

So, g (x) is not one-one

Also g (x) = |x| ≥ 0, for all real x

Hence, the range is [0, 1], which is subset of co-domain ‘A’

So, f (x) is not onto.

Therefore, f (x) is not bijective.

(iii) h (x) = x|x|

Let h (x₁) = h (x₂)

x₁|x₁| = x₂|x₂|

If x₁, x₂ > 0

x₁² = x₂²

x₁² – x₂² = 0

(x₁ – x₂)(x₁ + x₂) = 0

x₁ = x₂ (as x₁ + x₂ ≠ 0)

Similarly for x₁, x₂ < 0, we have x₁ = x₂

It’s clearly seen that for x₁ and x₂ of opposite sign, x₁ ≠ x₂.

Hence, f (x) is one-one.

For x ∈ [0, 1], f (x) = x² ∈ [0, 1]

For x < 0, f (x) = -x² ∈ [-1, 0)

Hence, the range is [-1, 1].

So, h (x) is onto.

Therefore, h (x) is bijective.

(iv) k (x) = x²

Let k (x₁) = k (x₂)

x₁² = x₂²

x₁ = ± x₂

Therefore, k (x) is not one-one.

Given,

f(x) = 1/(2 – cos x) ∀ x ∈ R

Let y = 1/(2 – cos x)

2y – ycos x = 1

cos x = (2y – 1)/ y

cos x = 2 – 1/y

Now, we know that -1 ≤ cos x ≤ 1

So,

-1 ≤ 2 – 1/y ≤ 1

-3 ≤ – 1/y ≤ -1

1 ≤ – 1/y ≤ 3

1/3 ≤ y ≤ 1

Thus, the range of the given function is [1/3, 1].

Answer is (D)

We have, n(A) = m and n(B) = n

n(A x B) = n(A). n(B) = mn

Total number of relation from A to B = Number of subsets of A x B = 2^mn

So, total number of non-empty relations = 2^mn – 1

Two parallel lines l and m be cut by a transversal lines s and t, forming angles.

From the figure, we have,

∠1 = ∠3 = 50° … [∵Corresponding angles]

And,

∠1 + x = 180° … [∵linear pair]

= x° =180° – 50

= x° = 130°

Now,

∠2 = ∠4 = 65° … [∵Corresponding angles]

And,

∠2 + y = 180° … [∵linear pair]

= 65 + y = 180°

= y = 180 – 65

= y = 115°

∴The value of x =130° and y = 115°

We know that,

∠ACD = ∠ACE + ∠ECD

∠ACD = 35° + 22°

∠ACD = 57° = ∠BAC

Thus, lines BA and CD are intersected by the line AC such that, ∠ACD = ∠BAC

So, the alternate angles are equal

Therefore, AB ∥ CD ……1

Now,

∠ECD + ∠CEF = 35° + 45° = 180°

This, shows that sum of the angles of the interior angles on the same side of the transversal CE is 180°

So, they are supplementary angles

Therefore, EF ∥ CD …….2

From equation 1 and 2

We conclude that, AB ∥ EF

It is given that,

AB || CD

EF ⊥ CD

∠GED = 126º

∠GEF + ∠FED = 126º

∠GEF + 90º = 126º

∠GEF = 36º

∠AGE and ∠GED are alternate interior angles.

∠AGE = ∠GED = 126º

However, ∠AGE + ∠FGE = 180º (Linear pair)

126º + ∠FGE = 180º

∠FGE = 180º − 126º = 54º

∠AGE = 126º, ∠GEF = 36º, ∠FGE = 54º

From the question,

CE∥BA, ∠BAC = 80°, ∠ECD = 35°.

Now,

(i) ∠BAC = ∠ACE = 80° … [∵ Alternate angles]

(ii) ∠ACB,

= ∠ACB + ∠ACD = 180° … [∵ Linear pair]

= ∠ACB + ∠ACE + ∠ECD = 180°

= ∠ACB + 80 + 35 = 180

= ∠ACB + 125 = 180

= ∠ACB = 180 – 115

= ∠ACB = 65°

(iii) ∠ABC

Let us consider Δ ABC,

∠ABC + ∠ ACB + ∠BAC = 180°

= ∠ABC + 65 + 80 = 180

= ∠ABC + 145 = 180

= ∠ABC = 180 – 145

= ∠ABC = 35°

Given that, AB ∥ CD and CD ∥ EF

Sum of the interior angles,

∠CEF + ∠ECD = 180°

130° + ∠ECD = 180°

∠ECD = 180° – 130°

∠ECD = 50°

We know that alternate angles are equal

∠BAC = ∠ACD

∠BAC = ∠ECD + ∠ACE

∠ACE = 70° – 50°

∠ACE = 20°

Therefore, ∠ACE = 20°

Given that EF ⊥ CD; ∠GED = 126° 

i. e., ∠FED = 90° 

and ∠GEF = ∠GED – ∠FED ∠GEF = 126° – 90° = 36° 

In ∆GFE ∠GEF + ∠FGE + ∠EFG = 180° 

36 + ∠FGE + 90° = 180

∠FGE = 180° – 126° = 54° 

∠AGE = ∠GFE + ∠GEF 

(exterior angle in ∆GFE) 

= 90°+ 36°= 126°

We know, If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel 

From figure: 

∠EDC = ∠DCA = 100° 

Lines DE and AC are intersected by a transversal DC such that the pair of alternate angles are equal. 

So, DE || AC.

Given: CD || EF 

∠FEC + ∠ECD = 180° [Sum of co-interior angles is supplementary to each other] 

∠ECD = 180° – 130° = 50° 

Also, BA || CD 

∠BAC = ∠ACD = 70° [Alternative angles of parallel lines are equal] 

But, ∠ACE + ∠ECD = 70°

∠ACE = 70° - 50° = 20°

(b) Two-dimensional series

Artificial base line is a method of magnifying the small fluctuations in a time series. There are certain instances where the dependent variables have got very small differences in values and the smallest value is very far distant from the origin. Under these circumstances, if we start our scale from zero on the Y-axis, the plotted values will show a straight line. In order to present such values, we use the technique of ‘artificial base line’.

These are the two-dimensional diagrams. Only one extension (height/length) is considered in one-dimensional diagrams, while two-dimensional diagrams are constructed considering two extensions: height and width. The areas of two dimensional diagrams are in proportion to item- values, hence these are also called surface or area diagrams.

(d) Double Bar Diagram

The following are the merits of Graphical Presentation:

1. Diagrammatic presentation implies presenting data-items in simple and easy manner. 

2. The statistical numbers shown by diagrammatic presentation are kept in mind for a longer period of time. 

3. It is a very attractive and effective means. 

4. Various facts can be compared using diagrams. 

5. Pictures, besides providing information about data items, also entertain us. 

6. This helps in calculating the averages like median, mode and quartiles.

7. The relationship of correlation between two variables is also gained with its help. 

8. Special knowledge or training is not required to understand it.

Correct Answer is: (a) Bar diagram

Following are three ways of presenting data : 

1. Subjective or Descriptive Presentation 

2. Tabulation 

3. Diagrammatic Representation of Data items

Correct option is (a) 10