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|3AB|=3³×|A|×|B|

          =27×5×3

           =405

The problem constraints are usually stated in the story problem. The linear inequalities x>=0 and y>=0. These are included because x and y are usually the number of items produced and you cannot produce a negative number of items, the smallest number of items you could produce is zero.

In an optimization problem were the objective function is to be maximized the optimal value is the least upper bound of the objective function values over the entire feasible region. If there is no upper bound, then we say that the optimal value is +∞, while if the feasible region is the empty set, we define the optimal value of a maximization problem to be −∞. Conversely, in an optimization problem were the objective function is to be minimized the optimal value is the greatest lower bound of the objective function values over the entire feasible region. If there is no lower bound, then we say that the optimal value is −∞, while if the feasible region is the empty set, we define the optimal value of a minimization problem to be +∞.

Therefore, every optimization problem has a well-defined optimal value. But not every optimization problem has an optimal solution. For example, consider the optimization problem min {e x : x ∈ R}. this problem has an optimal value of zero, but there is no optimal solution.

(C) (A) and (C) only 

Rb & Cs have nearly same electron gain enthalpy 

electron gain enthalpy = – 46 kj/ml 

Ar & Kr have same ΔH_eq . Value is + 96 kj/ml

Let vector a = vector(i - 2j + 5k) and vector b = vector(-2i + 4j + 2k)

∴ vector(a.b) = vector(i - 2j + 5k), vector(-2i + 4j + 2k) = -2 - 8 + 10 = 0 

Hence the given vectors be perpendicular to each other

Correct option is (B) Both (A) and (R) are true but (R) is not the correct explanation of (A)

Assertion (A): True

Reason (R): True but not correct explanation.

Correct explanation: Expansion of octet not possible for ‘B’.

ans  is.  b.

 since boron can't exhibit+6 O.S. even in the excited state so BF6 can't exist.

also the size of B is small as size decrease along period. but this fact has nothing to do with formation of BF6. hence not correct explanation

(D) Atomic orbital is characterised by the quantum numbers n and l only

Atomic orbital is characterised by n, l, m.

Refers to the below link

∵ x = a cos θ, y = b sin θ

∴ dx/dθ = -a sin θ = dy/dθ = b cos θ 

∴ dy/dx = b cos θ/-a sin θ = - (b/a) cot θ

∵ y = √(x² + ax + 1)

∴ dy/dx = 1/2√(x² + ax + 1)(2x + a)

= ((2x + a)/2√(x² + ax + 1)

Let x,y ∈ Q

then xoy = x + y - xy

= y + x - yx 

= yox

hence o commutative

(A) 26 square unit 

f = {(1, 1), (2, 3), (0, –1), (–1, –3)} and f(x) = ax + b 

(1, 1) ∈ f ⇒

 f(1) = 1 

⇒ a × 1 + b = 1 

⇒ a + b = 1 (0, –1) ∈ f 

⇒ f(0) = –1 ⇒ a × 0 + b = –1

 ⇒ b = –1 

On substituting b = –1 in a + b = 1, 

We obtain a + (–1) = 1 ⇒ a = 1 + 1 = 2. 

Thus, the respective values of a and b are 2 and –1.

R = {(a, b): a, b ∈ N and a = b²} 

(i) It can be seen that 2 ∈ N; however, 2 ≠ 2² = 4. 

Therefore, the statement “(a, a) ∈ R, for all a ∈ N” is not true. 

(ii) It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3². 

Now, 3 ≠ 9² = 81; 

therefore, (3, 9) ∉ N Therefore, the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.

 (iii) It can be seen that (9, 3) ∈ R, (16, 4) ∈ R because 9, 3, 16, 4 ∈ N and 9 = 3² and 16 = 4². 

Now, 9 ≠ 4² = 16; therefore, (9, 4) ∉ N Therefore, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.

The relation f is defined as f = {(ab, a + b): a, b ∈ Z} 

We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set B. Since 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ f i.e., (12, 8), (12, –8) ∈ f It can be seen that the same first element i.e., 12 corresponds to two different images i.e., 8 and –8. Thus, relation f is not a function. 

In the figure ΔAED, ΔBFC

AD = BC. DE = CF

AE = BF = 2 unit

∴ x coordinate of C

= 7 + BF = 7 + AE = 7 + 2 = 9

y coordinate of C = 3 + FC = 3 + ED = 3 + 4 = 7

C has the coordinates (9, 7)

The line 2x - y + 1 = 0

x = 2 then 4 - y + 1 = 0

y = 5; (2,5)

The line 3x - 2y + 3 = 0

x = 3 then 9 - 2y + 3 = 0

2y = 12;

y = 6; (3, 6)

4x - 2y + 2 = 0

3x + 2y + 3 = 0

(1) + (2), x - 1 = 0

x = 1

x = 1 (put value in (1))

4x 1 - 2y + 2 = 0

y = 3 intersecting points = (1, 3)

3.28 

= 328/100 

= 164/50 

= 82/25 Ans

An integer is a whole number (not a fractional number) that can be positive, negative, or zero. Examples of integers are: -5, 1, 5, 8, 97, and 3,043. Examples of numbers that are not integers are: -1.43, 1 3/4, 3.14, . 09, and 5,643.1.

2x² - 5x + 3 = 0

2x²/2 - 5x/2 + 3/2 = 0 (dividing both sides by 2)  

x² - 5x/2 + 3/2 = 0

[x² - 5x/2 + (5/4)²] - (5/4)² + 3/2 = 0

[x - 5/4]² - 25/16 + 3/2 = 0

[x - 5/4]² - (25-24)/16 = 0

[x - 5/4]² - 1/16 = 0

[x - 5/4]² = 1/16

x - 5/4 = ± 1/4   (square rooting on both sides)  

x = 1/4 + 5/4  or x = -1/4 + 5/4

x = (1+5)/4   or x = (-1+5)/4

x = 6/4   or x = 4/4

x = 3/2  or  x = 1

Solution:
i. Since the given number has a terminating decimal expansion, it is a rational number of the form p/q and q is of the form 2^m x 5ⁿ.
The prime factors of p and q will be either 2 or 5 or both.

ii. Since the decimal expansion is neither terminating nor recurring, the given number is an irrational number.
iii. Since the decimal expansion is nonterminating repeating, the given number is a
rational number of the form p/q and q is not of the form 2^m x 5ⁿ.
The prime factors of q will also have a factor other than 2 or 5.

It is (B) an irrational number 

The product of a non zero rational and an irrational number is always irrational 

Correct answer is (D) always a non-zero number

Option : (A) \(\frac{\vec i-3\vec j +5\vec k}{\sqrt{35}}\)

Option : (D) 5√2

Option : (A) \(\frac{6}{5\sqrt 2}\)

Option : (A) cos^-1\(\frac{1}{6}\)

Option : (D) \(\frac{\pi}{12}\) 

Option : (A) 3

Option : (C) \(\frac{1}{\sqrt {14}}\),\(\frac{2}{\sqrt {14}}\),\(\frac{3}{\sqrt {14}}\)

Option : (C) 18

Option : (C) 1

Option : (A) 0

Option : (A) \(\frac{6^{5x}}{25}\)(5x - 1) + K

Option : (B) 28 

\(4\hat{i}.(7\hat{i} - 8\hat{j} + 3\hat{k})\)

\(= 28\,\hat{i}.\hat{i} - 32\,\hat{i}.\hat{j} + 12\,\hat{i}.\hat{k}\)

\(= 28(\because \hat{i}.\hat{i} = 1\,\, \&\,\, \hat{i}.\hat{j} = \,\hat{i}.\hat{k} = 0)\)

Option : (B) 28