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Test: Partial Derivatives, Gradient- 2 - Civil Engineering (CE) MCQ


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25 Questions MCQ Test Engineering Mathematics - Test: Partial Derivatives, Gradient- 2

Test: Partial Derivatives, Gradient- 2 for Civil Engineering (CE) 2024 is part of Engineering Mathematics preparation. The Test: Partial Derivatives, Gradient- 2 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Partial Derivatives, Gradient- 2 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- 2 below.
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Test: Partial Derivatives, Gradient- 2 - Question 1

 f(x, y) = x2 + xyz + z Find fx at (1,1,1)

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 1

fx = 2x + yz

Put (x,y,z) = (1,1,1)

fx = 2 + 1 = 3.

Test: Partial Derivatives, Gradient- 2 - Question 2

Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally the entire cost of the car, then the share of each of the remaining persons increased by:

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 2

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Test: Partial Derivatives, Gradient- 2 - Question 3

The minimum point of the function f(x) = (x2/3) – x is at 

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 3

Correct Answer :- a

Explanation : f(x) = (x^2/3) - x

f'(x) = 2/3(x-1/2) - 1

f"(x) = -1/3(x-3/2)

For critical points. f′(x)=0

=> 2/3(x-1/2) - 1 = 0 

f has minimum value of x = 1

Test: Partial Derivatives, Gradient- 2 - Question 4

If x=a(θ+ sin θ) and y=a(1-cosθ), then dy/dx will be equal 

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 4

Test: Partial Derivatives, Gradient- 2 - Question 5

The minimum value of function y = x2 in the interval [1, 5] is  

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 5

y =x 2 is strictly increasing function on [1,5]

∴ y= x 2 has minimum value at x = 1 is 1.

Test: Partial Derivatives, Gradient- 2 - Question 6

The function f(x) = 2x3 – 3x2 – 36x + 2 has its maxima at  

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 6

Test: Partial Derivatives, Gradient- 2 - Question 7

What should be the value of λ such that the function defined below is continuous at x = π/22? 

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 7

By the given condition 

Test: Partial Derivatives, Gradient- 2 - Question 8

Consider function f(x) =(x2-4)2 where x is a real number. Then the function has  

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 8

Test: Partial Derivatives, Gradient- 2 - Question 9

If f     where ai (i = 0 to n) are constants, then  

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 9

  - Euler’s theorem for homogeneous function 

Test: Partial Derivatives, Gradient- 2 - Question 10

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 10

Test: Partial Derivatives, Gradient- 2 - Question 11

A point on a curve is said to be an extremum if it is a local minimum or a local maximum. The number of distinct exterma for the curve 3x4 – 16x3 – 24x2 + 37 is 

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 11

Test: Partial Derivatives, Gradient- 2 - Question 12

∇ × ∇ × P, where P is a vector, is equal to  

Test: Partial Derivatives, Gradient- 2 - Question 13

The value of the integral of the function g(x, y) = 4x3 + 10y4 along the straight line segment from the point (0, 0) to the point (1, 2) in the x-y plane is  

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 13

The equation of the line passing through (0,0) and (1,2)  is y = 2x 

Given y x, y ) = 4x3+ 10y4 = 4x3 + 10(2x )4 = 4x3+ 160xy

Test: Partial Derivatives, Gradient- 2 - Question 14

If    is a differentiable vector function and f is a sufficient differentiable scalar function, then curl  

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 14

Test: Partial Derivatives, Gradient- 2 - Question 15

The temperature field in a body varies according to the equation T(x,y) = x3+4xy. The direction of fastest variation in temperature at the point (1,0) is given by 

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 15

Test: Partial Derivatives, Gradient- 2 - Question 16

The divergence of vector  

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 16

Test: Partial Derivatives, Gradient- 2 - Question 17

The divergence of the vector 

Test: Partial Derivatives, Gradient- 2 - Question 18

 Among the following, the pair of vectors orthogonal to each other is 

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 18

Then we say that they are orthogonal.  Choice (c) is correct. 

Test: Partial Derivatives, Gradient- 2 - Question 19

The directional derivative of the scalar function f(x, y, z) = x2 + 2y2 + z at the point P = (1,1, 2) in the direction of the vector 

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 19

Required directional derivatives at P(1,1,-1) 

=2

Test: Partial Derivatives, Gradient- 2 - Question 20

The Gauss divergence theorem relates certain  

Test: Partial Derivatives, Gradient- 2 - Question 21

If P, Q and R are three points having coordinates (3, –2, –1), (1, 3, 4), (2, 1, –2) in XYZ space, then the distance from point P to plane OQR (O being the origin of the coordinate system) is given by  

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 21

The equation of the plane OQR is (O being origin). 

Test: Partial Derivatives, Gradient- 2 - Question 22

Let x and y be two vectors in a 3 dimensional space and <x, y> denote their dot product. 

Then the determinant det 

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 22

Test: Partial Derivatives, Gradient- 2 - Question 23

If a - b = 3 and a2 + b2 = 29, find the value of ab.

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 23

2ab = (a2 + b2) - (a - b)2

   = 29 - 9 = 20

   ab = 10.

Test: Partial Derivatives, Gradient- 2 - Question 24

If a vector R(t)  has a constant magnitude, then  

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 24

On analysing the given (a) option, we find that    will give constant magnitude, so first 
differentiation of the integration will be zero. 

Test: Partial Derivatives, Gradient- 2 - Question 25

For the scalar field    magnitude of the gradient at the point(1,3) is   

Detailed Solution for Test: Partial Derivatives, Gradient- 2 - Question 25

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