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In ∆ABD and ∆CBE, 

∴ seg AD ≅ seg CE …[Given] 

∠ABD ≅ ∠CBE …[Vertically opposite angles] 

∴ The necessary condition for the two triangles to be congruent by AAS test is 

∠ADB ≅ ∠CEB, or 

∠DAB ≅ ∠ECB

i. In ∆MST and ∆TBM, 

∴ side MS ≅ side TB … [Given] 

m∠MST = m∠TBM = 90° … [Given] 

hypotenuse MT ≅ hypotenuse MT

…[Common side] 

∴ ∆MST ≅ ∆TBM …[by hypotenuse-side test] 

∴ side ST ≅ side BM …[Corresponding sides of congruent triangles] 

∠SMT ≅ ∠BTM …[Corresponding sides of congruent triangles] 

∠STM ≅ ∠BMT …[Corresponding sides of congruent triangles] 

ii. In ∆PRQ and ∆TRS, side PR ≅ side TR … [Given] 

∠PRQ ≅ ∠TRS …[Vertically opposite angles] 

side RQ ≅ side RS … [Given] 

∴ ∆PRQ ≅ ∆TRS …[by SAS test] 

∴ side PQ ≅ side TS …[Corresponding sides of congruent triangles] 

∠RPQ ≅ ∠RTS …[Corresponding sides of congruent triangles]

∠PQR ≅ ∠TSR …[Corresponding sides of congruent triangles] 

iii. In ∆DCH and ∆DCF,

∠DCH ≅ ∠DCF …[Given] 

∠DHC ≅ ∠DFC …[Given] 

side DC ≅ side DC …[Common side] 

∴ ∆DCH ≅ ∆DCF …[by AAS test] 

∴ side HC ≅ side FC …[Corresponding sides of congruent triangles] 

side DH ≅ side DF…[Corresponding sides of congruent triangles] 

∠HDC ≅ ∠FDC ….[Corresponding sides of congruent triangles]

We notice that the two triangles are congruent.

We notice that ∆ABC ≅ ∆LMN.

In cliesgamy, flowers never open at all. Hence foreign pollen will not land on stigma of such flowers. So cliestogamy will only favourself pollination or autogamy.

In cleistogamy, flowers never open at all. Hence, foreign pollen will not land on the stigma of such flowers. So, cleistogamy can favour only self-pollination.

In bisexual flowers the essential organs, the stamens and stigmas, are arranged in such a way that self-pollination becomes impossible. 

For example: in Gloriosa superba, the style is reflexed away from the stamens and in Hibiscus the stigmas project far above the stamens.

x = 2p² + 5 

\(\frac{dx}{dp}\) = 2 x 2p + 0 = 4p 

Elasticity of supply: ηs = \(\frac{p}{x}.\frac{dx}{dp}\)

= \(\frac{p}{2p^2+5}\) x 4p

= \(\frac{4p^2}{2p^2+5}\)

When p = 3, elasticity of supply, ηs = \(\frac{4\times3^2}{2(3)^2+5}\)

= \(\frac{4\times9}{18+5}\)

= \(\frac{36}{23}\)

y = 2x³ – 3x² – 36x + 10 

\(\frac{dy}{dx}\) = 6x² – 6x – 36 = 6(x² – x – 6) 

\(\frac{dy}{dx}\) = 0 gives 6(x² – x – 6) = 0 

6(x – 3) (x + 2) = 0 

x = 3 (or) x = -2 

\(\frac{d^2y}{dx^2}\) = 6(2x – 1)

Case (i): when x = 3, 

\((\frac{d^2y}{dx^2})_{x=3}\) = 6(2 x 3 – 1) 

= 6 x 5 

= 30, positive 

Since \(\frac{d^2y}{dx^2}\) is positive y is minimum when x = 3. 

The local minimum value is obtained by substituting x = 3 in y. 

Local minimum value = 2(3³) – 3(3²) – 36(3) + 10 

= 2(27) – (27) – 108 + 10 

= 27 – 98 

= -71

Case (ii): when x = -2, 

\((\frac{d^2y}{dx^2})_{x=-2}\) = 6(-2 x 2 – 1) 

= 6 x -5

= -30, negative 

Since \(\frac{d^2y}{dx^2}\) is negative, y is maximum when x = -2. 

Local maximum value = 2(-2)³ – 3(-2)² – 36(-2) + 10 

= 2(-8) – 3(4) + 72 + 10 

= -16 – 12 + 82 

= -28 + 82 

= 54

Demand x = 100 – 2p 

Supply x = 3p – 50 

At equilibrium, demand = supply

100 – 2p = 3p – 50 

-2p – 3p = -100 – 50 

-5p = -150 

p = \(\frac{-150}{-5}\) = 30 

∴ Equilibrium price p_E = 30 

Supply, x = 3p – 50 

Put p = 30, we get 

x = 3(30) – 50 = 90 – 50 = 40 

∴ Equilibrium quantity x_E = 40

AC increases when \(\frac{d}{dx}\)(AC) > 0 

C = 2x – 11 + \(\frac{50}{x}\)

\(\frac{dC}{dx}\) = 2 – 0 + 50(\(\frac{-1}{x^2}\)) = 2 – \(\frac{50}{x^2}\)

\(\frac{d}{dx}\)(AC) > 0 

2 – \(\frac{50}{x^2}\) > 0 

2 > \(\frac{50}{x^2}\)

2x² > 50 

x² > 25 

x > 5

The cost function is C = x³ – 12x² + 48x 

Average cost is minimum, 

When Average Cost (AC) = Marginal Cost (MC) 

Cost function, C = x³ – 12x² + 48x

Average Cost, AC = \(\frac{x^3-12x^2+48x}{x}\) = x² – 12x + 48

Marginal Cost (MC) = \(\frac{dC}{dx}\)

= \(\frac{d}{dx}\)(x³ – 12x² + 48x) 

= 3x² – 24x + 48 

But AC = MC 

x² – 12x + 48 = 3x² – 24x + 48 

x² – 3x² – 12x + 24x = 0 

-2x² + 12x = 0 

Divide by -2 we get, x² – 6x = 0

x(x – 6) = 0 

x = 0 (or) x – 6 = 0 

x = 0 (or) x = 6 

But x > 0 

∴ x = 6 

Output = 6 units

(b) 8

q = 1000 + 8p₁ – p₂

\(\frac{8q}{∂p_1}\) = 8

(b) x² – 7

Profit = Revenue – Cost 

= (2 – x)x – (-2x² + 2x + 7) 

= 2x – x² + 2x² – 2x – 7

= x² – 7

(b) Decreasing function

(c) -1

cot^-1 x + tan^-1 x = π/2 ⇒ u = π/2 so 2u = π 

∴ cos 2u = cos π = -1

R = 14x – x² and C = x(x² – 2) 

C = x³ – 2x 

(i) Average Cost (AC) = \(\frac{Total\,cost}{Output}\) = \(\frac{C(x)}{x}\)

= \(\frac{x^3-2x}{x}\)

= \(\frac{x^3}{x}-\frac{2x}{x}\)

= x² - 2

(ii) Marginal Cost (MC) = \(\frac{dC}{dx}\)

= \(\frac{d}{dx}\)(x³ – 2x) 

= \(\frac{d}{dx}\)(x³) – 2 \(\frac{d}{dx}\)(x) 

= 3x² – 2

(iii) Average Revenue R = 14x – x² 

Average Revenue (AR) = \(\frac{Total\,Revenue}{Output}=\frac{R(x)}{x}\)

= \(\frac{14x-x^2}{x}\)

= \(\frac{14x}{x}-\frac{x^2}{x}\)

= 14 - x

(iv) Marginal Revenue (MR) = \(\frac{dR}{dx}\)

= \(\frac{d}{dx}\)(14x – x²) 

= 14 \(\frac{d}{dx}\)(x) – \(\frac{d}{dx}\)(x²) 

= 14(1) – 2x 

= 14 – 2x

(b) Breakeven point

(b) Average cost = marginal cost

Given, u = x² y³ cos(\(\frac{x}{y}\)) 

i.e., u(tx, ty) = (tx)² (ty)³ cos\((\frac{tx}{ty})\) 

= t² x² t³ y³ cos(\(\frac{x}{y}\)) 

= t⁵ x² y³ cos(\(\frac{x}{y}\)) 

= t⁵ u 

∴ u is a homogeneous function in x and y of degree 5. 

∴ By Euler’s theorem, \(x.\frac{∂u}{∂x}+y.\frac{∂u}{∂y}\) = 5u

Hence Proved.

Given, z = (ax + b)(cy + d) 

Differentiating partially with respect to x we get, 

\(\frac{∂z}{∂x}\) = (cy + d) \(\frac{∂}{∂x}\)(ax + b) [∵ (cy + d) is constant] 

= (cy + d)(a + 0) 

= a(cy + d) 

Differentiating partially with respect to y we get, 

\(\frac{∂z}{∂y}\) = (ax + b) \(\frac{∂}{∂y}\)(cy + d) 

= (ax + b)(c + 0) 

= c(ax + b)

Marginal productivity of labour, \(\frac{∂z}{∂x}\) = 6x – 4y

Marginal productivity of labour when x = 1, y = 2 is

\((\frac{∂z}{∂x})_{(1,2)}\) = 6(1) – 4(1) 

= 6 – 4 

= 2 

Marginal productivity of capital, \(\frac{∂z}{∂y}\) = 0 – 4x(1) + 3(2y) 

= -4x + 6y

Marginal productivity of capital when x = 1, y = 2 is

\((\frac{∂z}{∂y})_{(1,2)}\) = -4(1) + 6(2) 

= -4 + 12 

= 8

(i) Given: Point is (-2, 3, 4) 

To find: the image of the point in yz-plane 

Since we need to find its image in yz-plane, a sign of its x-coordinate will change 

So, Image of point (-2, 3, 4) is (2, 3, 4) 

(ii) Given: Point is (-5, 4, -3) 

To find: image of the point in xz-plane 

Since we need to find its image in xz-plane, sign of its y-coordinate will change 

So, Image of point (-5, 4, -3) is (-5, -4, -3) 

(iii) Given: Point is (5, 2, -7) 

To find: the image of the point in xy-plane 

Since we need to find its image in xy-plane, a sign of its z-coordinate will change 

So, Image of point (5, 2, -7) is (5, 2, 7) 

(iv) Given: Point is (-5, 0, 3) 

To find: image of the point in xz-plane 

Since we need to find its image in xz-plane, sign of its y-coordinate will change 

So, Image of point (-5, 0, 3) is (-5, 0, 3) 

(v) Given: Point is (-4, 0, 0) 

To find: image of the point in xy-plane 

Since we need to find its image in xy-plane, sign of its z-coordinate will change 

So, Image of point (-4, 0, 0) is (-4, 0, 0)

One angle equaling to 60° need not necessarily be a rhombus.

Since, the quadrilateral with opposite angles equal is a parallelogram.

Triplicate ratio of 3 : 5 is 33 : 53 = 27 : 125.

Sub triplicate ratio of 125 : 64 

= 3√125:3√64 = 5:4

In the word “CARROM”, there are 6 letters, out of which 2 R’s are always together can be taken as 1 unit, can be permuted in 5! ways = 120 ways.

The given function is y = a cos x + b sin x 

Differentiating both sides with respect to x, we have

dy/dx = -a sin x + b cos x

Putting values of dy/dx and y in the given differential equation, we have 

L.H.S. = cos x (- a sin x + b cos x) + {a cos x + b sin x) sin x 

= – a sin x cos x + b cos² x + a sin x cos x + b sin² x 

= b (cos² x + sin² x) 

= b × 1 = b = R.H.S 

Thus, y = a cos x + b sin x is a solution of differential equation

cos x(dy/dx) + y sin x = b.

The given function is y = 4 sin 3x 

Differentiating both sides with respect to x, we have 

img 177 

Again, differentiating both sides with respect to x, we have 

img 18 

Putting values of d²y/dx² and y in the given differential equation, we have 

L.H.S. = – 36 sin 3x + 9 (4 sin 3x) = – 36 sin 3x + 36 

sin 3x = 0 = R.H.S. 

Thus, y = 4 sin 3x is a solution of differential equation img 20.

The given function is y = ax² + bx + c 

Differentiating both sides with respect to x, we have

img 188 

Again differentiating both sides with respect to x, we have 

img 200 

Which is the given differential equation. 

Thus, y = ax² + bx + c is a solution of differential equation d²y/dx² = 2a.

(i) v = 2x² … (1) 

Differential equation: xy’ = 2y 

Differentiate with respect to ‘x’

y' = 4x 

x = y'/4 

Substitute in (1) 

y = 2x(x)

y = 2x(y'/4)

= xy'/2

On simplifying, 2y = xy’ 

∴ (1) is solution of the given differential equation.

(ii) y = ae^x + be^-x …(1) 

Differential equation: y” – y = 0 

Differentiate with respect to ‘x’ 

y’ = ae^x – be^-x 

Again differentiate with respect to ‘x’ 

y” = ae^x + be^-x 

y” = y ⇒ y” – y = 0 

∴ (1) is the solution of the given differential equation.

Let the cubic polynomial bey = f(x). Since it attains a maximum atx = -1 and a minimum at x = 1.

dy/dx = 0 at x = -1 and 1

dy/dx = k(x + 1)(x + 1) = k(x² - 1)

Separating the variables we have dy = k(x² – 1) dx

∫dy = k∫(x2 - 1) dx 

y = k((x³/3) - x) + c ... (1)

When x = – 1, y = 4 and when x =1,7 = 0 

Substituting the equation (1) we have 

2k + 3c = 12; – 2k + 3c = 0 

On solving we have k = 3 and c = 2. 

Substituting these values in (1) we get the required cubic polynomial y = x³ – 3x + 2.

Answer is (b) (AB)^-1 = B^-1 A^-1

On using the sine rule we know,

a/sin A = b/sin B = c/sin C = k

a = k sin A, b = k sin B, c = k sin C

Let us consider the LHS
a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B)

On substituting the values of a, b, c from sine rule in above equation, we get

a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B) = k sin A (sin B – sin C) + k sin B (sin C – sin A) + k sin C (sin A – sin B)

= k sin A sin B – k sin A sin C + k sin B sin C – k sin B sin A + k sin C sin A – k sin C sin B

Then, upon simplification, we get

= 0

= RHS

Thus proved.

Given as

Sides of a triangle are a = 18, b = 24 and c = 30

On using the formulas,

Cos A = (b² + c² – a²)/2bc

Cos B = (a² + c² – b²)/2ac

Cos C = (a² + b² – c²)/2ab

Therefore now let us substitute the values of a, b and c we get,

Cos A = (b² + c² – a²)/2bc

= (24² + 30² – 18²)/2 × 24 × 30

= 1152/1440

= 4/5

Cos B = (a² + c² – b²)/2ac

= (18² + 30² – 24²)/2 × 18 × 30

= 648/1080

= 3/5

Cos C = (a² + b² – c²)/2ab

= (18² + 24² – 30²)/2 × 18 × 24

= 0/864

= 0

∴ cos A = 4/5, cos B = 3/5, cos C = 0

Given as

Sides of a triangle are a = 4, b = 6 and c = 8

On using the formulas,

Cos A = (b² + c² – a²)/2bc

Cos B = (a² + c² – b²)/2ac

Cos C = (a² + b² – c²)/2ab

Therefore now let us substitute the values of a, b and c we get,

Cos A = (b² + c² – a²)/2bc

= (6² + 8² – 4²)/2 × 6 × 8

= (36 + 64 – 16)/96

= 84/96

= 7/8

Cos B = (a² + c² – b²)/2ac

= (4² + 8² – 6²)/2 × 4 × 8

= (16 + 64 – 36)/64

= 44/64

Cos C = (a² + b² – c²)/2ab

= (4² + 6² – 8²)/2 × 4 × 6

= (16 + 36 – 64)/48

= -12/48

= -1/4

Now considering the LHS

8 cos A + 16 cos B + 4 cos C = 8 × 7/8 + 16 × 44/64 + 4 × (-1/4)

= 7 + 11 – 1

= 17

Thus proved.

Given as 

In a ∆ABC, a = 5, b = 6 and C = 60°

On using the formula,

The area of ∆ABC = 1/2 ab sinθ where, a and b are the lengths of the sides of a triangle and θ is the angle between sides.

Therefore,

Area of ∆ABC = 1/2 ab sinθ

= 1/2 × 5 × 6 × sin 60°

= 30/2 × √3/2

= (15√3)/2 sq. units

Thus proved.

The correct option is:

(A) 189.61 cm³ 

Sum of two number = 95
Let the first number be x, then another number be x + 15 .
According to the question, x+x+15=95
=> 2x + 15 = 95 

=> 2x + 15 - 15 = 95 - 15 [Subtracting 15 from both sides]
=> 2x=80

=> 2x/2=80/2 [Dividing both sides by 2]
=> x = 40
Hence, the first number = 40
And another number = 40 + 15=55

Let one number be x. Therefore, the other number will be x + 15.

According to the question,

x + x + 15 = 95

2x + 15 = 95

On transposing 15 to R.H.S, we obtain

2x = 95 − 15

2x = 80

On dividing both sides by 2, we obtain

2x/2 = 80/2

x = 40

x + 15 = 40 + 15 = 55

Hence, the numbers are 40 and 55.

Perimeter of a sector of circle = 31 cm

Radius = 6.5 cm

Arc length = 31 – (6.5 + 6.5) = 18 cm

Now, Area of sector = 1/2 x Arc length x radius

= 1/2 x 18 x 6.5

= 58.5 cm²

Radius of the circle = 14 cm 

Angle subtend at center = 90° 

By Pythagoras theorem = AB² = OA² + OB² 

= 14² +14² 

AB = \(14\sqrt{2}\)

Area of sector OAB = \(\frac{90}{360}\timesπr^2\)

= \(\frac{1}{4}πr^2\)

= \(\frac{1}{4}\times\frac{22}7\times14\times14\) = 154 cm²

Area of triangle AOB = \(\frac{1}2\times14\times14\) = 98 cm²

So area of minor segment – OACB =area of sector – area of triangle

= 154 – 98 = 56 cm² 

Area of major segment = area of circle - area of minor segment 

= \(\frac{22}7\times14\times14\) - 56

= 44 ×14 – 56 = 560 cm²

Solution: Given, total value of Rs = Rs 300
Total number of coins = 160
Coins of denomination = Re 1, Rs 2 and Rs 5
Number of Rs 2 coins = 3 x number Rs 5 coins

Let the number of coins of Rs 5 = m
Since, the number coins of Rs 2 is 3 times of the number of coins of Rs 5
Therefore, number of coins of Rs 2 = m x 3 = 3m
Now, Number of coins of Re 1 = Total number of coins – (Number of Rs 5 coins + Number of Rs 2 coins)

Therefore,
Number of coins of Re 1 = 160 – (m + 3m) = 160 – 4m
Total Rs = (Re 1 x Number of Re 1 coins) + (Rs 2 x Number of Rs 2 coins) + (Rs 5 x Number of Rs 5 coins)
⇒ 300 = [1 x (160 – 4m)] + (2 x 3m) + (5 x m)
⇒ 300 = (160 – 4m) + 6m + 5m
⇒ 300 = 160 – 4m + 6m + 5m
⇒ 300 = 160 – 4m + 11m
⇒ 300 = 160 + 7m

After transposing 160 to LHS, we get
300 – 160 = 7m
⇒ 140 = 7 m

After dividing both sides by 7, we get

140/7 = 7m/7

=> 140/7 = m

=> m = 20

Thus, number of coins of Rs 5 = 20
Now, since, number of coins of Re 1 = 160 – 4m
Thus, by substituting the value of m, we get
Number of coins of Re 1 = 160 – (4 x 20) = 160 – 80 = 80
Now, number coins of Rs 2 = 3m

Thus, by substituting the value of m, we get
Number of coins of Rs 2 = 3m = 3 x 20 = 60
Therefore,
Number of coins of Re 1 = 80
Number of coins of Rs 2 = 60
Number of coins of Rs 5 = 20

Let the number of winners be x. Therefore, the number of participants who did not win will be 63 − x.

Amount given to the winners = Rs (100 × x) = Rs 100x

Amount given to the participants who did not win = Rs [25(63 − x)] = Rs (1575 − 25x)

According to the given question,

100x + 1575 − 25x = 3000

On transposing 1575 to R.H.S, we obtain

75x = 3000 − 1575

75x = 1425

On dividing both sides by 75, we obtain

75x/75 = 1425/75

x = 19

Hence, number of winners = 19

Area of the sector = 17.6 cm² (given)

We know, Area of the sector = θ/360 (πr²) square units

This implies,

θ/360 (πr²) = 17.6

56/360 x 22/7 x r² = 17.6

r² = 36

or r = 6

Radius of the circle is 6 cm.

Circumference of a circle = 8 cm

Central angle = 72°

Now, Circumference of a circle = 2πr

8 = 2πr

r = 14/11 cm

Area of sector = θ/360 x (πr²)

= 72/360 x 22/7 x 14/11 x 14/11

= 1.02 cm²

Let the number of Rs 5 coins be x. 

Number of Rs 2 coins = 3 × Number of Rs 5 coins = 3x

Number of Re 1 coins = 160 − (Number of coins of Rs 5 and of Rs 2)

= 160 − (3x + x) = 160 − 4x

Amount of Re 1 coins = Rs [1 × (160 − 4x)] = Rs (160 − 4x)

Amount of Rs 2 coins = Rs (2 × 3x)= Rs 6x

Amount of Rs 5 coins = Rs (5 × x) = Rs 5x

It is given that the total amount is Rs 300.

∴ 160 − 4x + 6x + 5x = 300 160 + 7x = 300

On transposing 160 to R.H.S, we obtain

7x = 300 - 160

7x = 140

On dividing both sides by 7, we obtain

7x/7 = 140/7

x = 20

Number of Re 1 coins = 160 − 4x 

= 160 − 4 × 20 = 160 − 80 = 80 

Number of Rs 2 coins = 3x = 3 × 20  = 60 

Number of Rs 5 coins = x = 20

For a circle having radius = r cm 

● Circumference or perimeter of a circle = 2πr cm

● Area of circle = πr²  cm² = \(\frac{1}{2}\) × Circumference × Radius

Radius of circle = r = 30 cm

Area of minor segment = Area of sector – Area of triangle …(1)

Area of major segment = Area of circle – Area of minor segment …(2)

Area of sector = θ/360 x πr²

= 60/360 x 3.14 x 30 x 30

= 471 cm²

Area of triangle = √3/4 (side)² (Since it form a equilateral triangle)

= √3/4 x 30 x 30

= 389.7 cm²

(1) ⇨

Area of minor segment = 471 – 389.7 = 81.3 cm²

(2) ⇨

Area of major segment = π(30²) – 81.3 = 2744.7 cm²

Answer:

Area of major segment is 2744.7 cm² and of minor segment is 81.3 cm².