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Mechanical Engineering Exam  >  Mechanical Engineering Tests  >  GATE Mechanical (ME) Mock Test Series 2025  >  Test: Partial Derivatives, Gradient- 1 - Mechanical Engineering MCQ

Test: Partial Derivatives, Gradient- 1 - Mechanical Engineering MCQ


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20 Questions MCQ Test GATE Mechanical (ME) Mock Test Series 2025 - Test: Partial Derivatives, Gradient- 1

Test: Partial Derivatives, Gradient- 1 for Mechanical Engineering 2024 is part of GATE Mechanical (ME) Mock Test Series 2025 preparation. The Test: Partial Derivatives, Gradient- 1 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Partial Derivatives, Gradient- 1 MCQs are made for Mechanical Engineering 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 GATE Mechanical (ME) Mock Test Series 2025 for Mechanical Engineering & Test: Partial Derivatives, Gradient- 1 solutions in Hindi for GATE Mechanical (ME) Mock Test Series 2025 course. Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free. Attempt Test: Partial Derivatives, Gradient- 1 | 20 questions in 60 minutes | Mock test for Mechanical Engineering preparation | Free important questions MCQ to study GATE Mechanical (ME) Mock Test Series 2025 for Mechanical Engineering 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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