Test: Partial Derivatives, Gradient- 1 - Civil Engineering (CE) MCQ

# Test: Partial Derivatives, Gradient- 1 - Civil Engineering (CE) MCQ

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## 20 Questions MCQ Test Engineering Mathematics - Test: Partial Derivatives, Gradient- 1

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

### Consider the function f(x) = x2 – x – 2. The maximum value of f(x) in the closed interval [–4, 4] is

Detailed Solution for Test: Partial Derivatives, Gradient- 1 - Question 1

∴ f (x )has minimum at x= 1 / 2 It Shows that a maximum value that will be at x = 4  or x = - 4

At x = 4, f (x )= 10

∴ At x= −4, f (x ) = 18

∴ At x= −4, f (x ) has a maximum.

Test: Partial Derivatives, Gradient- 1 - Question 2
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Test: Partial Derivatives, Gradient- 1 - Question 3
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Test: Partial Derivatives, Gradient- 1 - Question 4

The function f(x) = x3- 6x2+ 9x+25 has

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Test: Partial Derivatives, Gradient- 1 - Question 5

The function f(x,y) = 2x2 +2xy – y3 has

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Test: Partial Derivatives, Gradient- 1 - Question 6

Equation of the line normal to function f(x) = (x-8)2/3+1 at P(0,5) is

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Test: Partial Derivatives, Gradient- 1 - Question 7

The distance between the origin and the point nearest to it on the surface z2 = 1 + xy is

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or pr – q2 = 4 – 1 = 3 > 0 and r = +ve

so f(xy) is minimum at (0,0)

Hence, minimum value of d2 at (0,0)

d2 = x2 + y2 + xy + 1 = (0)2 + (0)2 + (0)(0) + 1 = 1

Then the nearest point is

z2 = 1 + xy = 1+ (0)(0) = 1

or z = 1

Test: Partial Derivatives, Gradient- 1 - Question 8

Given a function

The optimal value of f(x, y)

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Test: Partial Derivatives, Gradient- 1 - Question 9

For the function f(x) = x2e-x, the maximum occurs when x is equal to

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Test: Partial Derivatives, Gradient- 1 - Question 10

A cubic polynomial with real coefficients

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So maximum two extrema and three zero crossing

Test: Partial Derivatives, Gradient- 1 - Question 11

Given y = x2 + 2x + 10, the value of

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Test: Partial Derivatives, Gradient- 1 - Question 12

Consider the function y = x2 – 6x + 9. The maximum value of y obtained when x varies over the interval 2 to 5 is

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Test: Partial Derivatives, Gradient- 1 - Question 13

Consider the function y = x2 – 6x + 9. The maximum value of y obtained when x varies over the interval 2 to 5 is

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Correct Answer :- b

Test: Partial Derivatives, Gradient- 1 - Question 14

Test: Partial Derivatives, Gradient- 1 - Question 15

The magnitude of the gradient of the function f = xyz3 at (1,0,2) is

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Test: Partial Derivatives, Gradient- 1 - Question 16

The expression curl (grad f), where f is a scalar function, is

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Test: Partial Derivatives, Gradient- 1 - Question 17

If the velocity vector in a two – dimensional flow field is given by    the  vorticity vector, curl

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Test: Partial Derivatives, Gradient- 1 - Question 18

The vector field

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Test: Partial Derivatives, Gradient- 1 - Question 19

The angle between two unit-magnitude co-planar vectors P (0.866, 0.500, 0) and Q (0.259, 0.966, 0) will be

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Test: Partial Derivatives, Gradient- 1 - Question 20

Stokes theorem connects

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