Integral Calculus and Differential Equations Test

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Integral Calculus and Differential Equations Test - 4

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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

A
18

B
10

C
–2.25

D
indete rminate

Solution

∴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.
Question 2
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A
Maximum at x = e

B
minimum at x = e

C
Maximum at x = e-1

D
minimum at x = e-1

Solution
Question 3
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A
2x²

B
√x

C
0

D
1

Solution
Question 4
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The function f(x,y) = 2x² +2xy –y³ has

A
Only one stationary point at (0,0)

B
Two stationary points at (0,0) and (1/6, -1/3)

C
Two stationary points at (0,0) and (1,-1)

D
No stationary point

Solution
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

A
y = 3x -5

B
y = 3x +5

C
3y = x+15

D
3y = x -15

Solution
Question 6
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The distance between the origin and the point nearest to it on the surface z² = 1 + xy is

A
1

B
√3/2

C
√3

D
-2

Solution

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
Question 7
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Given a function
The optimal value of f(x, y)

A
is a minimum equal to 10/3

B
is a maximum equal to 10/3

C
is a minimum equal to 8/3

D
is a maximum equal to 8/3

Solution
Question 8
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For the function f(x) = x² e-x, the maximum occurs when x is equal to

A
2

B
1

C
0

D
-1

Solution
Question 9
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A cubic polynomial with real coefficients

A
Can possibly no extrema and no zero crossings

B
May have up to three extrema and up to 2 zero crossings

C
Cannot have more than two extrema and more than three zero crossings

D
Will always have an equal number of extrema and zero crossings

Solution

So maximum two extrema and three zero crossing
Question 10
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Given y = x² + 2x + 10, the value of

A
0

B
4

C
12

D
13

Solution
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

A
1

B
4

C
3

D
9

Solution
Question 12
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A

B

C

D
Question 13
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The magnitude of the gradient of the function f = xyz³ at (1,0,2) is

A
0

B
3

C
8

D
∞

Solution
Question 14
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The expression curl (grad f), where f is a scalar function, is

A
Equal to ∇² f

B
Equal to div (grad f)

C
A scalar of zero magnitude

D
A vector of zero magnitude

Solution
Question 15
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If the velocity vector in a two –dimensional flow field is given by the vorticity vector, curl

A

B

C

D

Solution
Question 16
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The vector field

A
Divergence free, but not irrotational

B
Irrotational, but not divergence free

C
Divergence free and irrotational

D
Neither divergence free nor irrotational

Solution
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

A
0⁰

B
30⁰

C
45⁰

D
60⁰

Solution
Question 18
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Stokes theorem connects

A
A line integral and a surface integral

B
A surface integral and a volume integral

C
A line integral and a volume integral

D
Gradient of a function and its surface integral
Question 19
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A

B

C

D

Solution

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:
Question 20
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If then is

A
-1/2

B
-1

C
0

D
1

Solution

Concept

Calculation:
Given:

Similarly,

Now,