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The given expression is 

⇒ 1113 × 1111

We can write this expression as

⇒ (1100 + 13) × (1100 + 11)

⇒ (1100 × 1100) + (1100 × 11) + (13 × 1100) + (13 × 11)

⇒ 1210000 + 12100 + 14300 + 143

⇒ 1236543

Hence, the correct answer is 1236543.

 Multiplication rule of '11'

To multiply any digit by 11, simply add the digits of the number together and then put this sum between the original two digits.

For e.g. Multiply 13 by 11.

⇒ 13 × 11

Add 1 and 3 = 1 + 3 = 4

Now, put '4' between 1 and 3, wet get the multiplication of 13 and 11.

⇒ 13 × 11 = 143

The order of magnitude is the power of 10, when a number is written in its standard form. 

• 500,000 in its standard form is:
• 5.0×10⁵
• Hence, the order of magnitude is 5!

Just to clarify, the standard form of any number is that number written as a single digit followed by a decimal point and decimal places, which is multiplied with a power of 10.

Here are a few examples:

10=1.0×10¹

1489,000=1.489×10⁶

Hence, the order of magnitude is 6.

The gamma function then is defined as the analytic continuation of this integral function to a meromorphic function that is holomorphic in the whole complex plane except zero and the negative integers, where the function has simple poles.

Correct option:

(D) a = 1/|λ|

Correct option:

(C) 8/7

Given : n = 9; Mean = n = 9; Mode = (n – 2) = 9 – 2 = 7.

1. The total frequency N should be large. 

2. The expected frequencies (E) should be 5 or more. If any E is less than 5, it should be pooled with adjacent frequency. 

3. If any parameter is estimated, one degrees of freedom should be lessened.

1. The procedure of calculation is difficult. 

2. If one of the value is added or removed,’ whole procedure has to be repeated.

‘Interpolation’ is the technique of estimating the value of dependent variable (y) for any intermediate value of the independent variable (x).

‘Extrapolation’ is the technique of estimating the value of dependent variable (y) for any value of independent variable (x) which is outside the given series.

Chronological arrangement of statistical data is called time series.

If q = 0.4; 

∴ p = 1 – q = 1 – 0.4 = 0.6, 

Mean = p = 0.6 

Variance = pq = 0.6 × 0.4 = 0.24

answer is not equal

The correct answer is -15 

-15 should be the answer 

f(x) = (x-1)² + 2(x-2)² + 3(x-3)² + ......+n(x-n)²

∴f'(x) = 2(x-1) + 4(x-2)+6(x-3)+ ....+ 2n (x-n)

= 2x(1 + 2 + 3 + ....+ n) -2 (1² + 2² + 3² + .... + n²)

= \(\frac{2x\times n(n+1)}{2}-2\times n\frac{n(n+1)(2n+1)}{6}\)

= n(n + 1)x - \(\frac{n(n+1)(2n+1)}{3}\) 

= \(\frac{n(n+1)}{3}\) (3x -(2n + 1)), n>1

∴ f''(x) = n(n+1)>0 

∴ point where f'(x) = 0 is point of minima.

f'(x) = 0 gives x = \(\frac{2n+1}{3}\) 

Hence, x = \(\frac{2n+1}{3}\) is point of minima of given function f(x),

Hence, option (2) is correct.

Correct answer is (c) Fan

2.) (-5)^x+1 x (-5)⁵ = 1

⇒ (-5)^x+1+5 = 1 (∵ a^maⁿ = a^m+n)

⇒ (-5)^x+6 = 1 = (-5)⁰ (∵ a⁰ = 1)

⇒ x+6 = 0

⇒ x = -6

3.) \(\frac{(-2)^3\times(-2)^7}{3\times4^6}=\frac{(-1)^{3+7}2^3\times2^7}{3\times(2^2)^6}\) (∵ (ab)^m = a^mb^m)

\(=\frac{(-1)^{10}(2)^{10}}{3\times2^{12}}\) (∵ a^maⁿ = a^m+n and (a^m)ⁿ = a^mn)

\(=\frac{1}{3\times2^{12-10}}\) (∵ (-1)¹⁰ = 1 and \(\frac{a^m}{a^n}=\frac{1}{a^{n-m}})\)

\(=\frac{1}{3\times2^2}\)

\(=\frac{1}{3\times4}=\frac{1}{12}.\)

Given:

Sum of the three prime numbers = 84

Concept used:

Prime number = It is the number that has only two factors, one and itself

If the sum of three prime numbers is even then one prime number must be 2

Calculation:

Let the one prime number be x, then another prime number is x + 24

Sum of the three prime number = 84

⇒ 2 + x + x + 24 = 84

⇒ 2x = 58

⇒ x = 29

Another prime number = x + 24 = 29 + 24 = 53

∴ The greatest prime number is 53 

Given :

Total sum of money = 8200

A had Rs. 500 more than B and C had Rs. 300 more than A.

Concept used:

Using linear equation. 

Calculation :

Let the money of B be 'x'

So, money of A = x + 500

Now, money of C = x + 500 + 300 = x + 800

According to the Question ,

x + x + 500 + x + 800 = 8200

⇒ 3x + 1300 = 8200

⇒ 3x = 8200 – 1300

⇒ 3x = 6900

⇒ x = 2300

So, C's share = 2300 + 800 = Rs. 3100

∴ C's share is Rs. 3100.

Given quadratic equation is px (x – 2) + 6 = 0 

⇒ px² − 2px + 6 = 0 … (1)

By comparing equation (1)with ax² + bx + c = 0, 

we get a = p, b = −2p and c = 6. 

Since given that the given quadratic equation has equal roots. 

∴ D = b²– 4ac = 0 (∵ for equal roots D = 0) 

⇒(−2p)² − 4 × p × 6 = 0 

⇒ 4p² − 24p = 0 

⇒ 4p (p – 6) = 0 

⇒ p = 0 or p – 6 = 0 

⇒ p = 0 or p = 6 but p ≠ 0 

(∵ If p = 0 then given equation is not quadratic equation) 

Hence the value of p is 6, for which the given quadratic equation have equal roots.

Since x > 0

∴ \(\frac{x}5\) , \(\frac{x}4\) , \(\frac{x}2\) , x and \(\frac{x}3\) all are positive.

Now, arranging numbers into ascending order which is  \(\frac{x}5\) , \(\frac{x}4\) , \(\frac{x}3\), \(\frac{x}2\) and x.

Total number are n = 5 which is odd.

∴ The median is \(\frac{n+1}2\)th term = \(\frac{5+1}2\)th term = 3^rd term.

Hence, the 3^rd number in sequence of ascending order of numbers is the median of given numbers. 

∴ The median of numbers is \(\frac{x}3\).

Given that the median of number is 8.

∴ \(\frac{x}3\) = 8 ⇒ x = 3 × 8 = 24. 

Hence, the value of x is 24.

Given quadratic equation is x² + 6x + 6 = 0 

By comparing given quadratic equation with ax² + bx + c = 0, 

we get a =1, b = 6 and c = 6. 

The discriminant of given quadratic equation is 

D = b² − 4ac = 6² − 4 × 1 × 6 = 36 – 24 = 12 > 0 (real and distinct roots) 

Hence, the roots of given quadratic equation are real and distinct.

And the roots are \(\frac{-b±\sqrt{D}}{2a}\) = \(\frac{-6±\sqrt{12}}{2\times1}\) = \(\frac{-6±2\sqrt{3}}{2}\) = -3 ± \(\sqrt3\).

Hence, the roots (or solution) of given quadratic equation are x = -3 +√3 and x = -3 − √3.

Let the required numbers are a – 3d, a – d, a + d and a + 3d which are in AP with common difference 2d. 

Given that sum of these required numbers is 20. 

∴ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20

⇒ 4a = 20 ⇒ a = \(\frac{20}4\) = 5. 

Hence, a = 5. 

Also given that the sum of squares of required numbers is 120. 

∴ (a − 3d)² + (a − d)² + (a + d)² +(a + 3d)² = 120 

⇒ a² – 6ad + 9d² + a² – 2ad +d ² +a² + 2ad + d² + a² + 6ad + 9d² =120 

⇒ 4a² + 20d² = 120 

⇒ 4 × 5² + 20d² = 120 ⇒ 20d² = 120 – 4 × 25 = 120 – 100 = 20 

⇒ d² = \(\frac{20}{20}\) = 1 

⇒ d = ±1. 

Case I :- If d = 1 and a = 5. 

Then the number are a – 3d, a – d, a + d, a + 3d 

= 5 – 3, 5 – 1, 5 + 1, 5 + 3 = 2, 4, 6, 8 . 

Case II:- If d = –1 and a = 5. 

Then the numbers are a – 3d, a – d, a + d, a + 3d 

= 5 – 3 ×–1, 5 – (–1), 5 – 1, 5 + 3 (–1) = 5 + 3, 5 + 1, 5 – 1, 5 –3 = 8, 6, 4, 2. 

Hence the required numbers are 2, 4, 6 and 8.

Given that the sum of first nth term of AP is S_n = \(\frac{1}2\) (3n² + 7n). 

∴ The n^th term of AP is a_n = S_n –S_n–1 

⇒ a_n = \(\frac{1}2\) (3n² + 7n) –\(\frac{1}2\) [3(n − 1)² + 7(n − 1)] 

= \(\frac{1}2\) [3n² + 7n – 3(n − 1)² − 7(n − 1)] 

= \(\frac{1}2\) [3n² + 7n – 3(n² − 2n + 1) − 7(n − 1)] 

= \(\frac{1}2\) [3n² + 7n – 3n² + 6n – 3 – 7n + 7] 

= \(\frac{1}2\) [6n + 4] = 3n + 2. 

Hence, the nth terms of given AP is an = 3n + 2. 

Now, put n = 20, 20^th term = 3n + 2 = 3 × 20 + 2 = 60 + 2 = 62. 

Hence, the 20^th term of AP is a₂₀ = 62.

GIVEN:

\(\frac b a = 0.7\)

CALCULATION:

\(\frac {a-b}{a+b} + \frac {11}{34}\)

 \(⇒ \frac {1-b/a}{1 + b/a} + \frac {11}{34}\)

⇒ (1 - .7)/(1 + .7) + 11/34

⇒ .3/1.7 + 11/34

⇒  .17 + .33

⇒ .5

∴ \(\frac {a-b}{a+b} + \frac {11}{34}\) = .5

7n²+ 9x - 1

n = \(\frac{-9\pm\sqrt{81+28}}{14}=\frac{-9\pm{109}}{14}\)

Zeros are \(\frac{-9+\sqrt{109}}{14}\) & \(\frac{-9-\sqrt{109}}{14}\)

Given:

A person walks a distance from point A to B at 15 km/h, and from point B to A at 30 km/h.

He takes 3 hours to complete the journey, 

Concept used:

Distance = velocity × time

Velocity = distance/time

Time = distance/velocity

Calculation:

Let the distance between A to B be x km

According to the question,

x/15 + x/30 = 3

⇒ 3x/30 = 3

⇒ x = 30 km

∴ The distance between A to B is 30 km.

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