Calculus of Variations MCQ Quiz - Objective Question with Answer for Calculus of Variations - Download Free PDF

Last updated on Jul 7, 2025

Latest Calculus of Variations MCQ Objective Questions

Calculus of Variations Question 1:

Define S : = {y ∈ C1[-1, 1] : y(-1) = -1, y(1) = 3}

Let φ be the extremal of the functional J : S → R given by 

Define ||y||:=  for every y ∈ S and let B0(φ, ϵ) := {y ∈ S : ||y - φ||∞ 1(φ, ϵ) := {y ∈ S : ||y - φ|| + ||y' - φ'|| 

  1. φ(x) = 2x + 1 for every x ∈ [-1, 1]
  2. There exists ϵ > 0 such that J[y] ≥ J[φ] for every y ∈ B0(φ, ϵ)
  3. There exists ϵ > 0 such that J[y] ≥ J[φ] for every y ∈ B1(φ, ϵ)
  4. There exists ϵ > 0 such that J[y] ≤ J[φ] for every y ∈ B0(φ, ϵ)

Answer (Detailed Solution Below)

Option :

Calculus of Variations Question 1 Detailed Solution

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Calculus of Variations Question 2:

For any b ∈ ℝ, let S(b) denote the set of all broken extremals with one corner of the variational problem 

minimize J[y] = 

subject to y(0) = 0, y(1) = b

Then which of the following statements are true? 

  1. S(2) has exactly two elements 
  2. S() has exactly one element 
  3. S(2) is empty 
  4. S() has exactly two elements

Answer (Detailed Solution Below)

Option :

Calculus of Variations Question 2 Detailed Solution

Concept:

Broken Extremals in Calculus of Variations:

  • The functional is given by .
  • The Euler-Lagrange equation is based on the derivative of the Lagrangian with respect to .
  • Lagrangian derivative: .
  • In broken extremals, momentum is constant on each smooth segment but can change at the corner point, provided corner conditions hold.
  • The possible constant values of satisfy (constant).

Calculation:

Let .

Solve .

The possible constant solutions for solve .

Consider the roots of

Set derivative .

Critical points are at

.

Also, is a possible slope.

So, possible slopes are , , and 0.

Check for : starting at , ending at .

Possible segments: try to sum contributions of constant slopes on [0, c] and [c, 1].

Each slope is constant, so is linear on each segment.

Consider slopes and in some order.

Example broken extremal:

  • On [0, t₁]: slope , contributes .
  • On [t₁, 1]: slope , contributes .

Total displacement: .

Set equal to 2:

.

This is greater than 1, impossible.

So no broken extremal exists for .

is empty.

Now for :

Repeat same method:

Total displacement:

This is feasible in [0, 1].

Now, the slopes and can be arranged in two ways: positive first or negative first.

Exactly two distinct broken extremals.

Correct statements:

  • Option 1 is false (S(2) is empty).
  • Option 2 is false (S(½) has two elements, not one).
  • Option 3 is true (S(2) is empty).
  • Option 4 is true (S(½) has exactly two elements).

∴ The correct answers are 3 and 4.

Calculus of Variations Question 3:

Let S denote the set of all solutions of the Euler-Lagrange equation of the variational problem: 

minimize J|y| =  

subject to y(0) = 0, y(1) = 0, 

Then the set  is equal to

  1. {-√2, √2}
  2. {-√2, 0, √2}

Answer (Detailed Solution Below)

Option 4 : {-√2, 0, √2}

Calculus of Variations Question 3 Detailed Solution

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Calculus of Variations Question 4:

Let λ ∈ ℝ and K : [0, 1] × [0, 1] → ℝ be a function such that every solution of the boundary value problem  

Satisfies the integral equation

Then

Answer (Detailed Solution Below)

Option 2 :

Calculus of Variations Question 4 Detailed Solution

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Calculus of Variations Question 5:

The extremal of the functional

 y(0) = 0, y(1) = 1, y ∈ C2[0, 1] is

Answer (Detailed Solution Below)

Option :

Calculus of Variations Question 5 Detailed Solution

Concept: 

If  then its extremum is

Explanation:

  y(0) = 0, y(1) = 1,  

According to the question 

 = 0

⇒  = 0

⇒ x + 6y'' = 0

⇒ 

after integrating with respect to x, we get

again integrating with respect to x, we get

 where c and d are constants of integration

now, 

and 

Hence our extremum would be

 

Hence option (2) is correct

Top Calculus of Variations MCQ Objective Questions

Consider the variational problem (P)

J(y(x)) = [(y')2 − y|y| y' + xy] dx, y(0) = 0, y(1) = 0.

Which of the following statements is correct?

  1. (P) has no stationary function (extremal).
  2. y ≡ 0 is the only stationary function (extremal) for (P).
  3. (P) has a unique stationary function (extremal) y not identically equal to 0.
  4. (P) has infinitely many stationary functions (extremal).

Answer (Detailed Solution Below)

Option 3 : (P) has a unique stationary function (extremal) y not identically equal to 0.

Calculus of Variations Question 6 Detailed Solution

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Concept:

Euler-Lagrange Equation: The extremal of the functional J[y] = f(x, y, y')dx satisfies  - () = 0

Explanation:

J(y(x)) = [(y')2 − y|y| y' + xy] dx, y(0) = 0, y(1) = 0

If y > 0 then

f(x, y, y') = (y')2 − y2y' + xy

So using   - () = 0 we get

-2yy' + x - (2y' - y2) = 0

- 2yy' + x - 2y'' +2yy' = 0

y'' = x/2....(i)

If y

f(x, y, y') = (y')2 + y2y' + xy

So using   - () = 0 we get

2yy' + x - (2y' + y2) = 0

 2yy' + x - 2y'' - 2yy' = 0

y'' = x/2....(ii)

Hence in both case we get

y'' = x/2

Integrating 

y' = 

Integrating again

y = 

Using  y(0) = 0, y(1) = 0 we get

c2 = 0 and 0 =  + c1 ⇒ c1 = - 

Hence solution is

y = 

Option (3) is correct

The extremal of the functional

 y(0) = 0, y(1) = 1, y ∈ C2[0, 1] is

Answer (Detailed Solution Below)

Option 4 :

Calculus of Variations Question 7 Detailed Solution

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Concept: If  then its extremum is

Explanation:

  y(0)=0, y(1) = 1,  

According to the question 

 = 0

 = 0

⇒  = 0

⇒ x - 4y'' = 0

after integrating  with respect to x, we get

again integrating with respect to x, we get

where c and d are constants of integration

now, 

and 

Hence our extremum would be 

Hence option (4) is correct

Calculus of Variations Question 8:

Which of the following is an extremal of the functional  that satisfies the boundary conditions y(-1) = -1 and y(1) = 1?

Answer (Detailed Solution Below)

Option 3 :

Calculus of Variations Question 8 Detailed Solution

Concept:

Euler Equation:

Let us examine the functional 

For an extreme value, the boundary points of the admissible curves being fixed  y( x1 ) = y1 and y( x2 ) = y2 

then  - ) = 0 is called the Euler equation

Calculation:

Given: F( x , y , y' ) = y'2 -2xy

Now by Euler's equation 

 - () = 0

-2x - 2y'' = 0 

y'' = - x 

Now integrating on both sides with respect to x, we get 

y' =  + A 

Now again, integrating on both sides with respect to x, we get 

y( x) = + Ax + B . . . . . . (1)

where A and B are constant.

Using boundary conditions y(1) = 1 and y( -1) = -1

Now from equation 1, we get 

y(1) = + A + B 

1 = + A +B 

A + B =  . . . . . . (2)

y( -1) =   - A + B

-1 =  - A + B

- A + B =  . . . . . . (3)

Adding equations 2 and 3, we get

B = 0 

Putting the value of B in equation (2), we get 

A = 

Now from equation (1), 

y(x)  =  +x

Option (3) is correct

Calculus of Variations Question 9:

The cardinality of the set of extremals of 

subject to 

y(0) = 1, y(1) = 6, 

is

  1. 0
  2. 1
  3. 2
  4. countably infinite

Answer (Detailed Solution Below)

Option 2 : 1

Calculus of Variations Question 9 Detailed Solution

Concept:

A typical isoperimetric problem is to find an extremum of 

I(y) =  subject to y(a) = y1, y(b) = y2, J(y) =  then the extrema satisfy the Euler-Lagrange's equation

 = 0

where H = F + λH

Explanation:

, y(0) = 1, y(1) = 6,

Let H = 

Then using

 = 0

⇒ λ -  = 0

⇒ 2y'' = λ

⇒ y'' = λ/2

Integrating both sides we get

y' = x + a

Again integrating

⇒ y =  + ax + b

y(0) = 1, y(1) = 6

⇒ b = 1

and

6 = λ/4 + a + 1

⇒ 5 = λ/4 + a

⇒ λ + 4a = 20.....(i)

So, we get y =  + ax + 1

Also, 

⇒  = 3

⇒  = 3

⇒  = 3

⇒  = 2

⇒ λ + 6a = 24....(ii)

Subtracting (i) from (ii) we get

2a = 4 ⇒ a = 2

Putting a = 2 in (i) we get

λ + 8 = 20 ⇒ λ = 12

Hence extremal is

y = 3x2 + 2x + 1

Hence the functional has only one extremal.

Therefore the cardinality of the set of extremals is 1

(2) is correct

Calculus of Variations Question 10:

Let y = ϕ(x) be the extremizing function for the functional  subject to y(0) = 0, y(1) = 1. Then ϕ(1/4) is equal to 

  1. 1/2
  2. 1/4
  3. 1/8
  4. 1/12

Answer (Detailed Solution Below)

Option 1 : 1/2

Calculus of Variations Question 10 Detailed Solution

Concept:

We know that if I(y) =  F dx where F is function of y and y' only then F satisfy  = c ...(1)

Explanation:

Here F =  = y2y'2

So using (1)​we get

y2y'- y'2y2y' = c

 - y2y'2= c 

⇒ yy' = c'

⇒ y dy = c' dx

Integrating both side

y2 / 2 = c'x + d 

Substituting y(0) = 0 we get

d = 0

and substituting y(1) = 1 we get

1 / 2 = c'

Hence we get 

 ⇒ y2 = x ⇒ y = √(x)

Therefore we get ϕ(x) = √(x). Hence ϕ(1/4) = √(1/4) ⇒ ϕ(1/4) = 1/2

Option (1) is correct   

Calculus of Variations Question 11:

The extremum of the functional I[y(x)] = , y' > 0 & a > 0 attains 

  1. Strong maxima
  2. Weak maxima
  3. Strong minima
  4. Weak minima

Answer (Detailed Solution Below)

Option 4 : Weak minima

Calculus of Variations Question 11 Detailed Solution

Concept:

The extremum of functional I[y(x)] = , y(x1) = y1, y(x2) = yattains 

(i) strong maxima if  

(ii) weak maxima if  

(iii) strong minima if  > 0 for all y'

(iv) weak minima if  > 0 for some y'

Explanation:

I[y(x)] = , y(0) = 0, y(a) = b

So f = y'10

So  > 0 if y' > 0

Hence the extremum of the functional attains weak minima.

Option (4) is correct

Calculus of Variations Question 12:

If J(y) = , y(0) = e2 , then external is

Answer (Detailed Solution Below)

Option 3 :

Calculus of Variations Question 12 Detailed Solution

Concept:

The extremal of the functional J(y) = , y(a) = y1, y(b) = arbitrary then satisfy Lagrange equation

 = 0 and  = 0

Explanation:

J(y) = ,  y(0) = e2 . Then

 = 0

⇒ 2y' + 2y5 = 0

⇒ y' + y5 = 0

⇒ 

Integrating we get -

Now, y(0) = e2 ⇒ -1/4e8 = c

Hence  ⇒ 

(3) is correct

Calculus of Variations Question 13:

The infimum of  of function

 such that u(0) = 0 and  is equal to

  1. 0
  2. 1/2
  3. 1
  4. 2

Answer (Detailed Solution Below)

Option 3 : 1

Calculus of Variations Question 13 Detailed Solution

Concept:

The extremal of the functional , u(t1) = u1, u(t2) = u2, will satisfy Euler equation

 = 0

If F is independent of t and u then u = at + b is the extremize function.

Explanation:

J(u) = , u(0) = 0 and  

F(t, u, u') =  independent of t and u

So u = at + b is the general solution

u(0) = 0 implies b = 0

Hence u(t) = at

 i.e., at x = 1, u = 1

So, 1 = a × 1 ⇒ a = 1

Solution is u(t) = t ⇒ u'(t) = 1

Therefore infimum value

 =  = [t]01 = 1

So infimum value is 1

(3) is correct

Calculus of Variations Question 14:

The cardinality of the set of extremals of 

subject to 

y(0) = 2, y(1) = 5,   is

  1. 0
  2. 1
  3. 2
  4. countably infinite

Answer (Detailed Solution Below)

Option 2 : 1

Calculus of Variations Question 14 Detailed Solution

Concept:

A typical isoperimetric problem is to find an extremum of 

I(y) =  subject to y(a) = y1, y(b) = y2, J(y) =  then the extrema satisfy the Euler-Lagrange's equation

 = 0

where H = F + λH

Explanation:

, y(0) = 2, y(1) = 5,

Let H = 

Then using

 = 0

⇒ λ -  = 0

⇒ 2y'' = λ

⇒ y'' = λ/2

Integrating both sides we get

y' = x + a

Again integrating

⇒ y =  + ax + b

y(0) = 2, y(1) = 5

⇒ b = 2

and

5 = λ/4 + a + 2

⇒ 3 = λ/4 + a

⇒ λ + 4a = 12.....(i)

So, we get y =  + ax + 2

Also, 

⇒  = 4

⇒  = 4

⇒  = 4

⇒  = 2

⇒ λ + 6a = 24....(ii)

Subtracting (i) from (ii) we get

2a = 12 ⇒ a = 6

Putting a = 6 in (i) we get

λ + 24 = 12 ⇒ λ = -12

Hence extremal is

y = -3x2 + 6x + 1

Hence the functional has only one extremal.

Therefore the cardinality of the set of extremals is 1

(2) is correct

Calculus of Variations Question 15:

For any two continuous functions f, g : ℝ → ℝ, define

Which of the following is the value of f ⋆ g(t) when f(t) = exp(-t) and g(t) = sin(t)?  

  1. .
  2. .
  3. .
  4. .

Answer (Detailed Solution Below)

Option 1 : .

Calculus of Variations Question 15 Detailed Solution

Explanation:

f(t) = exp(-t) = e-t and g(t) = sin(t)

So, 

Let 

I = 

I = 

I = 0 + sin t - 

I = sin t - 

I = sin t - e-t - cos t - I 

2I = sin t - e-t - cos t

I = 

Hence f⋆g(t) = 

Hence, option (1) is correct

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