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Given:

Number between 100 to 300

Formula used:

If a number divide by p and q so the total number

Divisible by p + divisible by q – Divisible by L.C.M. of p and q

Calculation:

Divisible by 11

⇒ [(297 – 110)/11] + 1

⇒ (187/11) + 1

⇒ 17 + 1

⇒ 18

Divisible by 10

⇒ [(300 – 100)/10] + 1

⇒ (200/10) + 1

⇒ 20 + 1

⇒ 21

LCM of 10 and 11 is 110

Divisible by 110

⇒ [(220 – 110)/110] + 1

⇒ (110/110) + 1

⇒ 1 + 1

⇒ 2

Divisible by 10 and 11

⇒ 18 + 21 - 2

⇒ 37

Let the two consecutive integers are x and (x + 1)

According to question

(x)² + (x + 1)² = 3445

⇒ x² + x² + 1 + 2x = 3445

⇒ 2x² + 2x + 1 = 3445

⇒ 2x² + 2x – 3444 = 0

⇒ x² + x – 1722 = 0

Using D = b² - 4ac

= (1)² - 4(1)(-1722)

= 6889

x = (- b + √D)/2a, or (-b - √D)/2a

⇒ (- 1 + √6889)/2(1), or (- 1 - √6889)/2(1)

⇒ (- 1 + 83)/2, or -1 – 83/2

⇒ 82/2, or -84/2

⇒ 41, or -42

x is not negative, therefore the positive numbers are 41 and 42 i.e. (41 + 1)

Given:

a² + b² + c² = 2(6a – 8b + 12c) – 244

Formula used:

(x + y)² = x² + y² + 2xy

(x – y)² = x² + y² – 2xy

Calculation:

a² + b² + c² = 2(6a – 8b + 12c) – 244

⇒ a² + b² + c² = 12a – 16b + 24c – 244

⇒ a² + b² + c² – 12a + 16b – 24c + 244 = 0

⇒ a² – 12a +  b² +16b +  c²  – 24c + 244 = 0

⇒ a² – (2 × 6)a +  b² + (2 × 8)b +  c²  – (2 × 12) c + 244 = 0

For perfect square add and subtract y²  

⇒ a² – (2 × 6)a + 6² – 6² + b² + (2 × 8)b + 8² – 64 + c²  – (2 × 12) c + 12² – 12² + 244 = 0

⇒ (a – 6)² + (b + 8)² + (c – 12)² – 244 + 244 = 0

⇒ (a – 6)² + (b + 8)² + (c – 12)² = 0

Here,

a – 6 = 0

⇒ a = 6

b + 8 = 0

⇒ b = - 8

c – 12 = 0

⇒ c = 12

So, a + b + c

⇒ 6 – 8 + 12

⇒ 10

∴ The value of a + b + c is 10

Given:

The polynomial is -2x² + 3x + 9

Formula Used:

If polynomial ax² + bx + c, then

Min/Max value of polynomial = c – b²/4a

Calculation:

For polynomial -2x² + 3x + 9

a = -2, b = 3, and c = 9

Maximum value of the polynomial = c – b²/4a

⇒ 9 – (3)²/4 × (-2)

⇒ (9 × 8 + 9)/8

⇒ (72 + 9)/8

⇒ 81/8

∴ The maximum value of -2x² + 3x + 9 is 81/8.

Given:

a + b + c = 0

Formula Used:

If a + b + c = 0 then

a³ +b³ + c³ = 3abc

Calculation:

Substitute the value a3 +b3 + c3 = 3abc in the given expression

\(\Rightarrow \frac{{{a^3}\; + \; {b^3}\; + \;{c^3}}}{{abc}} = \frac{{3abc}}{abc}\)

\(\Rightarrow \frac{{{a^3} + {b^3} + {c^3}}}{{abc}} = 3\)

∴ The correct answer is 3

Calculation:

x² + 8x + 15 = 0

⇒ x² + 5x + 3x + 15 = 0

⇒ x(x + 5) + 3(x + 5) = 0

⇒ (x + 5) (x + 3) = 0

∴ The required answer is (x + 5) (x + 3)

Given:

Our quadratic equation is x2 – 8x + k = 0

And α – β = 2      ----(1)

Formula used:

General form of quadratic equation in terms of roots is x2 – (α + β)x + αβ  = 0

Calculation:

Our quadratic equation is x2 – 8x + k = 0

By comparing with the general form of quadratic equation,

We get α + β = 8      ----(2)

From eq (1) and eq(2) we get,

α = 5 and β = 3

Now, k = αβ = 15

∴ The required value of k is 15

Given: 

x² + 1/x² = 7

Concept: 

(a + b)² = a² + b² + 2ab

(a + b)³ = a³ + b³ + 3ab(a + b)

If b = 1/a

(a + 1/a)² = a² + 1/a² + 2

(a + 1/a)³ = a³ + 1/a³ + 3(a + 1/a)

Calculation:  

⇒ (x + 1/x)² = x² + 1/x² + 2

⇒ (x + 1/x)² = 7 + 2

⇒ x + 1/x = √9

⇒ x + 1/x = 3

Now, 

⇒ (x + 1/x)³ = x³ + 1/x³ + 3(x + 1/x)

⇒ 3³ = x³ + 1/x³ + 3 × 3

⇒ x³ + 1/x³ = 27 - 9

⇒ x³ + 1/x³ = 18

∴ The required value is 18.

Given:

The given polynomial is –x² – 8x + 7

Formula used:

For a given polynomial ax² + bx + c

Min/Max value of a polynomial = c – b²/4a or (4ac – b²)/4a

Calculation:

For polynomial –x² – 8x + 7,

Value of a = –1, b = –8, and c = 7

Maximum value of the polynomial = c – b²/4a

⇒ Maximum value of the polynomial = 7 – (–8)²/(4 × (–1)

⇒ Maximum value of the polynomial = 7 + 64/4

⇒ Maximum value of the polynomial = 7 + 16

∴ Maximum value of –x² – 8x + 7 is 23

GIVEN:

\(x - \frac 3 x = 6,\; x \ne 0,\)

FORMULA USED:

(a - b)² = (a² + b² - 2ab), (a³ - b³) = (a - b) (a² + b² + ab)

CALCULATION:

(a - b)2 = (a2 + b2 - 2ab)

⇒ \(x - \frac 3 x = 6,\; x \ne 0,\)

Squaring both sides

⇒ (x - 3/x)² = (6)²

⇒ x² + 9/x² - 6 = 36

⇒ x2 + 9/x² = 42

\(\frac {x^4 - \frac {27}{x^2}}{x^2 - 3x - 3}\)

 Divide by x into numerator and denominator

\(⇒ \frac{{{x^3} - \frac{{27}}{{{x^3}}}}}{{x - 3 - \frac{3}{x}}}\)

\( ⇒ \frac{{\left( {x - \frac{3}{x}} \right)\left( {{x^2} + \frac{9}{{{x^2}}} + 3} \right)}}{{\left( {x - \frac{3}{x}} \right) - 3}}\)

\(⇒ \;\frac{{6 \times \left( {42 + 3} \right)}}{{6 - 3}}\)

\( ⇒ \;\frac{{6 \times 45}}{3}\)

⇒ 90

∴ \( \frac{{{x^3} - \frac{{27}}{{{x^3}}}}}{{x - 3 - \frac{3}{x}}}\) = 90

Given:

The sum of digits is 14 and the number is 21 less when the digits are reversed

Calculation:

Let the once place digit is y and tens place digit is x

∴ The number is = 10x + y

Now, the sum of digits is 14

∴ x + y = 14     --- (1)

And the number is 21 less when the digits are reversed

∴ 10x + y = x + 10y + 18

⇒ x - y = 2      --- (2)

By equation (1) and (2)

∴ x = 8 and y = 6

Hence number = 86.

Given:

The given expression is \({64^{{{128}^{256}}}}\)

Concept Used:

Concept of unit digit

Calculation:

The given expression can be written as 64^{128 × 128 × 128………… × 256 times}

Again the expression can be simplified as 64^{128 × (128 × 128 × 128………… × 255 times)}

64^{128 × (128 × 128 × 128………… × 255 times)} = 64^{4 × 32 × (128 × 128 × 128………… × 255 times)}

Now, let 32 × (128 × 128 × 128………… × 255 times) = n

64^{4 × 32 × (128 × 128 × 128………… × 255 times)} = 64^{4 × n}

Now, unit digit of 64^{4 × n}

64^{4 × n} = (4)^{4 × n}

For any value of n 4^{4 × n} always gives the unit digit as 6

∴ The unit digit is 6.

Given:

X = 256

Y = -113

Z = -135

Formula used:

⇒ x³ + y³ + z³ – 3xyz = (1/2) × (x + y + z)[(x – y)² + (y – z)² + (z – x)²]

Calculation:

∵ x³ + y³ + z³ – 3xyz = (1/2) × (x + y + z)[(x – y)² + (y – z)² + (z – x)²]

⇒ (x³ + y³ + z³ – 3xyz)/[(x – y)² + (y – z)² + (z – x)²] = (1/2) × (x + y + z)

⇒ (x³ + y³ + z³ – 3xyz)/[(x – y)² + (y – z)² + (z – x)²] = (1/2) × (256 – 113 – 135)

= (1/2) × (256 – 248)

= (1/2) × 8

= 4