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Civil Engineering (CE) Exam  >  Civil Engineering (CE) Tests  >  Engineering Mathematics  >  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.
Solutions of Test: Partial Derivatives, Gradient- 1 questions in English are available as part of our Engineering Mathematics for Civil Engineering (CE) & Test: Partial Derivatives, Gradient- 1 solutions in Hindi for Engineering Mathematics course. Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free. Attempt Test: Partial Derivatives, Gradient- 1 | 20 questions in 60 minutes | Mock test for Civil Engineering (CE) preparation | Free important questions MCQ to study Engineering Mathematics for Civil Engineering (CE) Exam | Download free PDF with solutions
Test: Partial Derivatives, Gradient- 1 - Question 1
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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
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The function f(x,y) = 2x2 +2xy – y3 has 
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Test: Partial Derivatives, Gradient- 1 - Question 5
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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 6
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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 7
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Given a function 

The optimal value of f(x, y)
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Test: Partial Derivatives, Gradient- 1 - Question 8
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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 9
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A cubic polynomial with real coefficients  
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So maximum two extrema and three zero crossing  

Test: Partial Derivatives, Gradient- 1 - Question 10
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Given y = x2 + 2x + 10, the value of 

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Test: Partial Derivatives, Gradient- 1 - Question 11
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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 12
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Test: Partial Derivatives, Gradient- 1 - Question 13
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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 14
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The expression curl (grad f), where f is a scalar function, is  
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Test: Partial Derivatives, Gradient- 1 - Question 15
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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 16
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The vector field 

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Test: Partial Derivatives, Gradient- 1 - Question 17
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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 18
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Stokes theorem connects  
Test: Partial Derivatives, Gradient- 1 - Question 19
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Detailed Solution for Test: Partial Derivatives, Gradient- 1 - Question 19

Let's solve it using partial diffraction:

Now comparing both sides:

A + B = 1  ----- ---Eq(ii)

-3A + 2B = 0  ----Eq(iiI)

after solving the above equations, we get:

after putting these value in Eq(i), we will get:

Test: Partial Derivatives, Gradient- 1 - Question 20
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If then is
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Concept

Calculation:

Given:

Similarly,

Now,

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