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cos³A + cos³(120+A) + cos³(240 + A)

= cos³A + (cos 120 cos A - sin 120 sin A)³ + (cos 240 cos A - sin 240 sin A)³

(\(\because\) cos(A+B) = cos A cos B - sin A sin B)³

= cos³A + (-sin 30 cos A - cos 30 sin A)³ + (-sin 30 cos A + cos 30 sin A)³

\(\big(\)\(\because\) cos (90 + \(\theta\)) = -sin\(\theta\), cos (270 - \(\theta\)) = - sin\(\theta\) 

sin(90 + \(\theta\)) = cos\(\theta\), sin(270 - \(\theta\)) = -cos \(\theta\)\(\big)\)

= cos³A - \(\big( \frac{1}{2} cos A + \frac{\sqrt{3}}{2} sin A\big)^3 + (\frac{\sqrt{3}}{2} sin A - \frac{1}{2} cos A\big)^3\)

\(\big(\)\(\because\) sin 30 = \(\frac{1}{2}\) & cos 30 = \(\frac{\sqrt{3}}{2}\)\(\big)\)

= cos³A - (\(\frac{1}{8}\) cos³A + \(\frac{3\sqrt{3}}{8}\) cos²A sin A + \(\frac{9}{8} \) cosAsin²A + \(\frac{3\sqrt{3}}{8} sin^3A\))

+ \(\frac{3\sqrt{3}}{8} \) sin³A - \(\frac{9}{8}\) sin²A cos A + \(\frac{3\sqrt{3}}{8} \) sin A cos² A - \(\frac{1}{8}\) cos³A

= cos³A - \(\frac{1}{4}\) cos³A - \(\frac{9}{4}\) cos A sin²A

= \(\frac{3}{4}\) cos³A - \(\frac{9}{4} \) cos A (1 - cos²A) (\(\because\) sin²A = 1 - cos²A)

= \(\frac{3}{4}\) cos³A - \(\frac{9}{4} \) cos A + \(\frac{9}{4} \) cos³A

 = 3 cos³A - \(\frac{9}{4}\) cos A

= \(\frac{3}{4}\) (4 cos³A - 3 cos A)

= \(\frac{3}{4}\) cos 3A

Correct option is (B) 500 ms^–1 

m x 125 x 200 + m x 2.5 x 10⁴ = 1/2 mv² x 40/100

V = 500 m/s

b. Molluscs

Mussels is an example of – Molluscs.

Correct option: a. A III, B IV, C II, D I

a. Cutting vegetables

Commissary section is used for Cutting vegetables.

x−2<0
x<2
1≤−x+2≤3
1−2≤−x+2−2≤3−2
−1≤−x≤1
1≥x≥−1
x∈[−1,1]
x−2>0
x>2
1≤x−2≤3
3≤x≤5
x∈[3,5]
x∈[−1,1]U[3,5]

Here we want +1 and as such we write cos2A=1−2sin² A and combine the other two terms.
L.H.S. =1−2sin²A+2sin(B+C)sin(C−B)   (Note)
=1−2sin²A−2sinAsin(B−C) [∵sin(−θ)=−sinθ]
=1−2sinA[sinA+sin(B−C)]
=1−2sinA[sin(B+C)+sin(B−C)]
=1−2sinA(2sinBcosC)
=1−4sinAsinBcosC.

4x²+9y²−z²+12xy

=4x²+12xy+9y²−z²

=(2x+3y)²−z²

=(2x+3y+z)(2x+3y−z).

Given:

Ratio of numbers = 4 ∶ 5

Increment in 1^st number = 50%

Decrement in 2^nd number = 27

Concept used:

If X is distributed between A and B in the ratio of a ∶ b, then A = a/(a + b) × X and B = b/(a + b) × X

To convert any ratio to an exact value, we should have to multiply any constant value

Calculation:

Let, numbers be 4x and 5x respectively

Now, 1^st number = 4x + 50% of 4x = 4x + 2x = 6x

2^nd number = 5x - 27

⇒ 6x / (5x - 27) = 3 / 4

⇒ 6x × 4 = (5x - 27) × 3

⇒ 24x = 15x - 81

⇒ 9x = 81

⇒ x = 9

⇒ Numbers are 9 × 4 = 36 and 9 × 5 = 45

The difference of numbers = 45 – 36 = 9

∴ The difference of the numbers is 9.

Concept:

Base Rule

If b raised to the xth power is equal to b raised to the yth power, that implies that x = y.” 

\(\rm b^x = b^y \) ⇒ x = y

Calculations:

Given equation is 4x - 3(2x + 3) + 128 = 0

⇒ \(\rm (2^2)^x - 3 (2^x.2^3) + 128 = 0\)

⇒ \(\rm (2^x)^2 - 24 (2^x) + 128 = 0\)

⇒ \(\rm (2^x)^2 - 16 (2^x) - 8(2^x)+ 128 = 0\)

⇒ \(\rm (2^x - 16)(2^x - 8) = 0\)

⇒ \(\rm 2^x = 16 \;\;\text{or}\;\; 2^x = 8\)

⇒ \(\rm 2^x = 2^4 \;\;\text{or}\;\; 2^x = 2^3\)

⇒ x = 4 or x = 3

The roots of the equation 4x - 3(2x + 3) + 128 = 0 are 4 and 3

Its Sum = 4 + 3 = 7

Given f(x) = (3x - 4)/5 

Let y = (3x - 4)/5 

⇒ 3x - 4 = 5y 

⇒ x = (5y + 4)/3 

⇒ f^-1(x) = (5x + 4)/3

The density of mercury at 25°C is found from Appendix Table to be 13.534 kg/m³
By the formula
Patm=ρgH₀=13.534 x 9.806 x 0.750/1000
=99.54 kPa

\(P_c=I^2R=(4.8)^2(40)\)

\(=(23.04)(40)=921.6W\)

Read The above math tricks!

A = {3,5,7,a}, B = {4,6,25,27,54,100}

given, a \(\in\) A, b \(\in\) B and a is factor of b such that a < b.

Possible pair If a = 2 then b = 4, 6, 54, 100

If a = 4 then b = 100

If a = 6 then b = 54

If a = 25 then b = 100

If a = 27 then b = 54

If a = 9 then b = 27, 54

If a = 10 then b = 100

If a = 18 then b = 54

If a = 20 then b = 100

If a = 50 then b = 100

Therefore, the set of ordered pair (a,b) are {(2,4), (2,6),(2,54),(2,100),(4,100),(6,54),(9,27),(9,54),(10,100),(18,54),(20,100),(25,100),(27,54),(50,100),(100,100)}

Given

Inner dimensions of wooden box are 2 m by 1.2 m by 0.75 m

Thickness of the wood = 2.5 cm = 2.5/100 m = 0.025 m

Now

External dimensions of wooden box are

= (2 + 2 x 0.025) by (1.2 + 2 x 0.025) by (0.75 + 2 x 0.025)

= (2+0.05) by (1.2+0.05) by (0.75+0.5)

= 2.05 by 1.25 by 0.80

Thus

Volume of solid = External volume of box - Internal volume of box

= 2.05 x 1.25 x 0.80 m³ − 2 x 1.2 x 0.75 m³

= 2.05 − 1.80 = 0.25 m³

Given, the cost of wood = Rs 5400 for 1 m³

Hence, Total cost = Rs 5400 x 0.25 = Rs 5400 x 25/100 = Rs 54 x 25

= Rs 1350

It is given that A × B = {(a, x), (a, y), (b, x), (b, y)} 

We know that the Cartesian product of two non-empty sets P and Q is defined as P × Q = {(p, q): p ∈ P, q ∈ Q} 

∴ A is the set of all first elements and B is the set of all second elements. 

Thus, A = {a, b} and B = {x, y} 

(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C) We have B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ 

∴ L.H.S. = A × (B ∩ C) = A × Φ = Φ

 A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}

 A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}  

∴ R.H.S. = (A × B) ∩ (A × C) = Φ 

∴ L.H.S. = R.H.S Hence, A × (B ∩ C) = (A × B) ∩ (A × C)

 (ii) To verify: A × C is a subset of B × D

 A × C = {(1, 5), (1, 6), (2, 5), (2, 6)} 

A × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)} 

We can observe that all the elements of set A × C are the elements of set B × D. Therefore, A × C is a subset of B × D. 

The Cartesian product R × R represents the set R × R={(x, y) : x, y Î R}
which represents the coordinates of all the points in two dimensional space and the
cartesian product R × R × R represents the set R × R × R ={(x, y, z) : x, y, z Î R}
which represents the coordinates of all the points in three-dimensional space.

It is given that 

n(A) =3 and n(B) =2; and (x, 1), (y, 2), (z, 1) are in A×B. 

We know that

 A = Set of first elements of the ordered pair elements of A × B 

B = Set of second elements of the ordered pair elements of A × B. 

∴ x, y, and z are the elements of A; and 1 and 2 are the elements of B. 

Since n(A) = 3 and n(B) = 2, it is clear that A = {x, y, z} and B = {1, 2}. 

The relation R from A to A is given as R = {(x, y): 3x – y = 0, where x, y ∈ A} 

i.e., R = {(x, y): 3x = y, where x, y ∈ A}

 ∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)} The domain of R is the set of all first elements of the ordered pairs in the relation.

 ∴ Domain of R = {1, 2, 3, 4} The whole set A is the codomain of the relation R. 

∴ Codomain of R = A = {1, 2, 3… 14} The range of R is the set of all second elements of the ordered pairs in the relation. ∴Range of R = {3, 6, 9, 12} 

We know that if n(A) = p and n(B) = q, then n(A × B) = pq. 

∴ n(A × A) = n(A) × n(A) It is given that n(A × A) = 9 

∴ n(A) × n(A) = 9 ⇒ n(A) = 3 The ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A×A. 

We know that   A × A = {(a, a): a ∈ A}. Therefore, –1, 0, and 1 are elements of A. 

Since   n(A) = 3, it is clear that A = {–1, 0, 1}. The remaining elements of set A × A are (–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1). 

Vol of cuboid=l*b*h=1*9*81=729cm³
Vol. of cube=vol. of cuboid....
a³=l*b*h
729=a³
a=9cm....
Now SA=6*a²
=6*9*9=81*6=486cm²

Given set of lines px + qy + r = 0

given condition 3p + 2q + 4r = 0

⇒3/4p+1/2q+r=0

⇒ All lines pass through a fixed point (3/4,1/2)

Consider,

n[(A x B)∩(A x C)]=8

n[A x (B∩C)] = 8

But n(A x B) = n(A) x n(B)

=> n(A) x n(B∩C) = 8

=> n(A) x 2=8

=> n(A) = 8/2

=> n(A) = 4

\(n((A \cup B) \cup C) = n(A\cup B) + n(C) - n((A \cup B)\cap C)\)

\((\because n (A \cup B) = n(A + n(B) - n(A \cap B))\)

\(= n(A) + n(B) - n(A \cap B) + n(C) - n((A \cap C) \cup(B \cap C))\)

\(= n(A) + n(B) + n(C) - n(A \cap B) - (n(A\cup C) + n(B \cap C) - n(A \cap C) \cap (B \cap C))\)

\(= n(A) + n(B) + n(C) - n(A \cap B) - n(A\cap C) - n(B \cap C) + n(A \cap B\cap C) \)

Hence Proved.

We have relation R on set A = {1, 2, 3, 4} which is defined as R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (3, 2)}.

Since, (1, 1), (2, 2), (3, 3), (4, 4) ∈ R.

Therefore, (a, a)  R, ∀a ∈  A = {1, 2, 3, 4}.

Hence, R is a reflexive relation on set A.

Now, (1, 2) ∈  R but (2, 1) ∉ R.

Therefore, relation R is not symmetric.

Now, (1, 3) ∈ R, (3, 2) ∈ R and (1, 2) ∈ R.

Therefore, if there is any (a, b) ∈ R and (b, c) ∈ R then (a, c) ∈ R for any a, b and c ∈ A.

Therefore, relation R is transitive relation.

Hence, relation R is reflexive and transitive but not symmetric.

Types of immunity are 2. 

Factors influencing child’s learning are mental, physiological and environmental.   

Audio-visual Aid is cinema.