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Test: Minima & Maxima - 2 - Civil Engineering (CE) MCQ


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10 Questions MCQ Test Engineering Mathematics - Test: Minima & Maxima - 2

Test: Minima & Maxima - 2 for Civil Engineering (CE) 2024 is part of Engineering Mathematics preparation. The Test: Minima & Maxima - 2 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Minima & Maxima - 2 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Minima & Maxima - 2 below.
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Test: Minima & Maxima - 2 - Question 1

The minimum value of (x - 2)(x - 9) is

Detailed Solution for Test: Minima & Maxima - 2 - Question 1

Concept:

Minimum value = (4ac – b2)/4a

Calculation:

(x - 2)(x - 9)

⇒ x2 - 11x + 18

We know ax2 + bx + c = 0

Comparing this with above equation

⇒ a = 1, b = -11 and c = 18 

Minimum value = (4ac – b2)/4a

⇒ (4 × 1 × 18 - (-11)2)/(4 × 1)

⇒ (72 - 121)/4 

⇒ -49/4

∴ Minimum value of (x - 2)(x - 9) = -49/4

Test: Minima & Maxima - 2 - Question 2

What is the maximum value of the expression 1/(2x2 + 5x + 11)? 

Detailed Solution for Test: Minima & Maxima - 2 - Question 2

Given:

Expression: 1/(2x+ 5x + 11)

Concept:

If we get minimum value of expression (2x+ 5x + 11) then we can find maximum value of the expression 1/(2x+ 5x + 11)

Formula Used:

Minimum value of expression (ax2 + bx + c) is = [4ac – b2]/4a

Calculation:

Minimum value of the expression 2x+ 5x + 11

⇒ [4 × 2 × 11 – 52]/[4 × 2]

⇒ [88 – 25]/8

⇒ 63/8

Hence, Maximum value of the expression 1/(2x+ 5x + 11) = 1/(63/8) = 8/63

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Test: Minima & Maxima - 2 - Question 3

Find the maximum value of (x + 8) (7 - x)

Detailed Solution for Test: Minima & Maxima - 2 - Question 3

Given: 

Equation (x + 8) (7 - x)

Concept:  

When, a > 0 gives minimum value, and a < 0 gives maximum value 

For both value be using, 4ac−b2 / 4a

Calculation: 

⇒ (x + 8) (7 - x) = - x2 - x + 56

If we compare it with the equation ax2 + bx + c

then, a = -1, b = -1, c = 56

⇒ 

Hence, the required value is 225/4.

Test: Minima & Maxima - 2 - Question 4

If x2 + y2 = 1, find the maximum value of x2 + 4xy - y2.

Detailed Solution for Test: Minima & Maxima - 2 - Question 4

Given:

x2 + y2 = 1.

x2 + 4xy - y2

Concept Used:

The maximum value of a sin θ + b cos θ is .

Calculation:

Since, x2 + y2 = 1, we can suppose x = cos θ and y = sin θ.

Now, x2 + 4xy - y2

= cos2 θ + 4 sin θ cos θ - sin2 θ

= (cos2 θ - sin2 θ) + 4 sin θ cos θ

= cos 2θ + 2 sin 2θ

Its maximum value is 

= √5.

Test: Minima & Maxima - 2 - Question 5

If x is real, then find the the minimum value of (5x2 - 2x + 7).

Detailed Solution for Test: Minima & Maxima - 2 - Question 5

Given:

x is the real number.

5x2 - 2x + 7

Concept used:

In Quadratic Polynomial ax2 + bx + c

When, a > 0

Minimum value = (4ac - b2)/4a      

Calculation:

5x2 - 2x + 7

Compare the above equation with quadratic polynomial ax2 + bx + c.

a = 5, b = -2 and c = 7

Now,

Minimum value = (4ac - b2)/4a 

⇒ Mininmum value = (4 × 7 × 5 - 4)/20

⇒ Minimum value = 136/20

⇒ Mininmum value = 34/5

∴ 34/5 is the minimum value of 5x2 - 2x + 7. 

Test: Minima & Maxima - 2 - Question 6

If x is real, find the minimum value of 4x2 - 5x + 1 = 0

Detailed Solution for Test: Minima & Maxima - 2 - Question 6

Given:

We have the given expression 4x2 - 5x + 1 = 0

Concept used:

The minimum value of any quadratic expression ax2 + bx + c = 0, (when a > 0) is -D/4a

Where, D = b2 - 4ac

Calculation:

Here, a, b and c is 4, -5 and 1 respectively

So, D = (-5)2 - (4 × 4 × 1)

⇒ 25 - 16

⇒ 9

So, minimum value = (-9)/(4 × 4)

⇒ (-9)/16

∴ The minimum value of given expression is -9/16.

Test: Minima & Maxima - 2 - Question 7

What is the maximum value of the expression 1/(3x2 + 4x + 8)? 

Detailed Solution for Test: Minima & Maxima - 2 - Question 7

Given:

The Polynomial = 1/(3x2 + 4x + 8)

Concept used:

If we get minimum value of expression (3x2 + 4x + 8), then we can find maximum value of the expression 1/(3x2 + 4x + 8)

Formula used:

If polynomial ax2 + bx + c, When, a > 0

then Minimum value of polynomial = (4ac – b2)/4a

Calculation:

If we compared (3x2 + 4x + 8) with ax2 + bx + c, then

a = 3, b = 4 and c = 8

Minimum value of the expression 3x2 + 4x + 8

⇒ {(4 × 3 × 8) - (4)2}/(4 × 3)

⇒ (96 - 16)/12

⇒ 80/12

⇒ 20/3

Maximum value of the expression 1/(2x2 + 5x + 11) = 1/(20/3) = 3/20

∴ The maximum value of the expression 1/(3x2 + 4x + 8) is 3/20.

Test: Minima & Maxima - 2 - Question 8

What will be the maximum value of (x – 5)(y – 2)(z – 6) if (x + y + z) = 22?

Detailed Solution for Test: Minima & Maxima - 2 - Question 8

Given:

(x + y + z) = 22

Concept:

Arithmetic Mean > Geometric Mean

Calculation:

(x + y + z) = 22

⇒ (x – 5) + (y – 2) + (z – 6) = 22 – 5 – 2 – 6

⇒ (x – 5) + (y – 2) + (z – 6) = 9

Since, AM > GM

[(x – 5) + (y – 2) + (z – 6)]/3 > 3√[(x – 5)(y – 2)(z – 6)]

⇒ 3√[(x – 5)(y – 2)(z – 6)] < [(x – 5) + (y – 2) + (z – 6)]/3

⇒ 3√[(x – 5)(y – 2)(z – 6)] < 9/3

⇒ 3√[(x – 5)(y – 2)(z – 6)] < 3

⇒ [(x – 5)(y – 2)(z – 6)] < 33

⇒ [(x – 5)(y – 2)(z – 6)] < 27

∴ The maximum value of (x – 5)(y – 2)(z – 6) will be 27.

Test: Minima & Maxima - 2 - Question 9

If x is real, find the maximum value of (-x2 + 3x + 7)

Detailed Solution for Test: Minima & Maxima - 2 - Question 9

Given:

(- x2 + 3x + 7)

Concept used:

dy/dx = 0, the value of x gives the minimum value when d2y/dx2 is greater than 0, and gives the maximum value when d2y/dx2 is less than 0

Calculation:

(- x2 + 3x + 7)      ----(i)

Differentiating (i),

dy/dx = 0

⇒ - 2x + 3 = 0      ----(ii)

⇒ x = 3/2

By double differentiating equation (ii),

d2y/dx2 = - 2 < 0

That means the maximum value of equation (i) is at x = 3/2

Maximum value of equation (i)

⇒ - 9/4 + 9/2 + 7

⇒ (- 9 + 18 + 28)/4

⇒ 37/4

∴ The maximum value is 37/4.

Test: Minima & Maxima - 2 - Question 10

If x is real, then find the maximum value of (-3x2 + 5x + 10).

Detailed Solution for Test: Minima & Maxima - 2 - Question 10

Given:

x is real.

-3x2 + 5x + 10

Concept used:

For quadratic polynomial ax2 + bx + c, When a < 0

Maximum value = (4ac - b2)/4a

Calculation:

-3x2 + 5x + 10

Compare the above equation with quadratic polynomial ax+ bx + c.

a = -3, b = 5, c = 10

Here, a < 0 so

Maximum value = (4ac - b2)/4a

⇒ Maximum value = {4 × (-3) × 10 - 25)}/(-12)

⇒ Maximum value = {-145}/(-12)

⇒ Maximum value = 145/12

∴ 145/12 is the maximum value of -3x2 + 5x + 10.

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