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 J[y]=∈t11[(y)3+(y)2]dx

Define ||y||:= maxx[1,1]|y(x)| for every y ∈ S and let B0(φ, ϵ) := {y ∈ S : ||y - φ||∞ < ϵ}, B1(φ, ϵ) := {y ∈ S : ||y - φ|| + ||y' - φ'|| < ϵ}. Then which of the following statements are true? 

  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] = 01((y)43(y)2)dx

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(12) has exactly one element 
  3. S(2) is empty 
  4. S(12) 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 J[y]=01((y)43(y)2)dx.
  • The Euler-Lagrange equation is based on the derivative of the Lagrangian L(y) with respect to y.
  • Lagrangian derivative: Ly=4(y)36y.
  • In broken extremals, momentum p(x) is constant on each smooth segment but can change at the corner point, provided corner conditions hold.
  • The possible constant values of y satisfy 4(y)36y=c (constant).

Calculation:

Let p=y.

Solve ddx(4p36p)=0.

The possible constant solutions for p solve 4p36p=k.

Consider the roots of 4p36p:

Set derivative dLdp=12p26.

Critical points are at p2=12

p=±12.

Also, p=0 is a possible slope.

So, possible slopes are 12, 12, and 0.

Check for S(2): starting at y(0)=0, ending at y(1)=2.

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

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

Consider slopes 12 and 12 in some order.

Example broken extremal:

  • On [0, t₁]: slope 12, contributes t12.
  • On [t₁, 1]: slope 12, contributes 1t12.

Total displacement: t121t12=2t112.

Set equal to 2:

2t112=2

2t11=22

2t1=1+22

t1=1+222.

This is greater than 1, impossible.

So no broken extremal exists for b=2.

S(2) is empty.

Now for S(12):

Repeat same method:

Total displacement: 2t112=12

2t11=22

2t1=1+22

t1=12+24

This is feasible in [0, 1].

Now, the slopes 12 and 12 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| = 01(y2+(y)2dx 

subject to y(0) = 0, y(1) = 0, 01y2dx=1

Then the set {ψ(12):ψS} is equal to

  1. {-√2, √2}
  2. {2k:kZ,k0}
  3. {2k:kN}
  4. {-√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  

d2udx2(x)+λu(x)=0,dudx(0)=u(0),dudx(1)=0

Satisfies the integral equation

u(x)+λ01K(x,t)u(t)dt=0

Then

  1. K(x,t)={(1+x)(1t),0xt1 (1+t)(1x),0t<x1
  2. K(x,t)={(1x),0xt1 (1t),0t<x1
  3. K(x,t)={(1x2),0xt1 (1t2),0t<x1
  4. K(x,t)={(1+x)(t1),0xt1 (1+t)(x1),0t<x1

Answer (Detailed Solution Below)

Option 2 : K(x,t)={(1x),0xt1 (1t),0t<x1

Calculus of Variations Question 4 Detailed Solution

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Calculus of Variations Question 5:

The extremal of the functional

J(y)=01[3(y)2xy]dx, y(0) = 0, y(1) = 1, y ∈ C2[0, 1] is

  1. y(x)=x336+3736
  2. y(x)=x336+3736x
  3. y(x)=x3363736x
  4. y(x)=x336+3736

Answer (Detailed Solution Below)

Option :

Calculus of Variations Question 5 Detailed Solution

Concept: 

If J(y)=01[F(x,y,y)]dx, then its extremum is

Fyddx(Fy)=0

Explanation:

J(y)=01[3(y)2xy]dx,  y(0) = 0, y(1) = 1,  yC2[0,1]

According to the question 

F(x,y,y)=3(y)2xy

Fyddx(Fy) = 0

⇒ xddx(6y) = 0

⇒ x + 6y'' = 0

⇒ y=x6

after integrating with respect to x, we get

y=x212+c

again integrating with respect to x, we get

y(x)=x336+cx+d where c and d are constants of integration

now, y(0)=d=0

and y(1)=136+c=1c=3736

Hence our extremum would be

 y(x)=x336+3736x

Hence option (2) is correct

Top Calculus of Variations MCQ Objective Questions

Consider the variational problem (P)

J(y(x)) = 01[(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] = abf(x, y, y')dx satisfies fy - ddx(fy) = 0

Explanation:

J(y(x)) = 01[(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  fy - ddx(fy) = 0 we get

-2yy' + x - ddx(2y' - y2) = 0

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

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

If y < 0 then

f(x, y, y') = (y')2 + y2y' + xy

So using  fy - ddx(fy) = 0 we get

2yy' + x - ddx(2y' + y2) = 0

 2yy' + x - 2y'' - 2yy' = 0

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

Hence in both case we get

y'' = x/2

Integrating 

y' = x24+c1

Integrating again

y = x312+c1x+c2

Using  y(0) = 0, y(1) = 0 we get

c2 = 0 and 0 = 112 + c1 ⇒ c1 = - 112

Hence solution is

y = x312x12

Option (3) is correct

The extremal of the functional

J(y)=01[2(y)2+xy]dx, y(0) = 0, y(1) = 1, y ∈ C2[0, 1] is

  1. y=x212+11x12
  2. y=x33+2x23
  3. y=x27+6x7
  4. y=x324+23x24

Answer (Detailed Solution Below)

Option 4 : y=x324+23x24

Calculus of Variations Question 7 Detailed Solution

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Concept: If J(y)=01[F(x,y,y)]dx, then its extremum is

Fyddx(Fy)=0

Explanation:

J(y)=01[2(y)2+xy]dx,  y(0)=0, y(1) = 1,  yC2[0,1]

According to the question 

F(x,y,y)=2(y)2+xy

Fyddx(Fy) = 0

(2(y)2+xy)yddx((2(y)2+xy)y) = 0

⇒ xddx(4y) = 0

⇒ x - 4y'' = 0

y=x4

after integrating  with respect to x, we get

y=x28+c

again integrating with respect to x, we get

y(x)=x324+cx+d

where c and d are constants of integration

now, y(0)=d=0

and y(1)=124+c=1c=2324

Hence our extremum would be y(x)=x324+23x24

Hence option (4) is correct

Calculus of Variations Question 8:

Which of the following is an extremal of the functional I(y)=11(y22xy)dx that satisfies the boundary conditions y(-1) = -1 and y(1) = 1?

  1. x35+6x5
  2. x58+9x8
  3. x36+7x6
  4. x37+8x7

Answer (Detailed Solution Below)

Option 3 : x36+7x6

Calculus of Variations Question 8 Detailed Solution

Concept:

Euler Equation:

Let us examine the functional 

I(y)=x1x(F(x,y,y))dx

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

then Fy - ddxFy) = 0 is called the Euler equation

Calculation:

Given: F( x , y , y' ) = y'2 -2xy

Now by Euler's equation 

Fy - ddx(Fy) = 0

-2x - 2y'' = 0 

y'' = - x 

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

y' = x22 + A 

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

y( x) = x36+ 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) = 16+ A + B 

1 = 16+ A +B 

A + B = 76 . . . . . . (2)

y( -1) =  16 - A + B

-1 = 16 - A + B

- A + B = 76 . . . . . . (3)

Adding equations 2 and 3, we get

B = 0 

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

A = 76

Now from equation (1), 

y(x)  =  x36+76x

Option (3) is correct

Calculus of Variations Question 9:

The cardinality of the set of extremals of 

J[y]=01(y)2dx,

subject to 

y(0) = 1, y(1) = 6, 01ydx=3

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) = abF(x,y,y)dx subject to y(a) = y1, y(b) = y2, J(y) = abG(x,y,y)dx=L then the extrema satisfy the Euler-Lagrange's equation

Hyddx(Hy) = 0

where H = F + λH

Explanation:

J[y]=01(y)2dx, y(0) = 1, y(1) = 6,01ydx=3

Let H = y2+λy

Then using

Hyddx(Hy) = 0

⇒ λ - ddx(2y) = 0

⇒ 2y'' = λ

⇒ y'' = λ/2

Integrating both sides we get

y' = λ2x + a

Again integrating

⇒ y = λx24 + 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 = λx24 + ax + 1

Also, 01ydx=3

⇒ 01(λx24+ax+1)dx = 3

⇒ [λx312+ax22+x]01 = 3

⇒ λ12+a2+1 = 3

⇒ λ12+a2 = 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 I(y)=01y2(dydx)2dx, 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) = 0a F dx where F is function of y and y' only then F satisfy Fy(Fy) = c ...(1)

Explanation:

Here F = y2(dydx)2 = 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 

y22=12x ⇒ 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)] = 0ay10dx, 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)] = x1x2f(x,y,y)dx, y(x1) = y1, y(x2) = yattains 

(i) strong maxima if fyy < 0 for all y'

(ii) weak maxima if fyy < 0 for some y'

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

(iv) weak minima if fyy > 0 for some y'

Explanation:

I[y(x)] = bay10dx, y(0) = 0, y(a) = b

So f = y'10

fy=10y9

So fyy=90y8 > 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) = 23(y2+2y5y+y10)dx, y(0) = e2 , then external is

  1. 1=y3(4x+1e6)
  2. 1=y4(4x2+1e8)
  3. 1=y4(4x+1e8)
  4. 1=y2(4x+1e8)

Answer (Detailed Solution Below)

Option 3 : 1=y4(4x+1e8)

Calculus of Variations Question 12 Detailed Solution

Concept:

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

fyddx(fy) = 0 and fy|x=b = 0

Explanation:

J(y) = 23(y2+2y5y+y10)dx,  y(0) = e2 . Then

fy|x=3 = 0

⇒ 2y' + 2y5 = 0

⇒ y' + y5 = 0

⇒ dyy5=dx

Integrating we get -

1y4=4x4c

Now, y(0) = e2 ⇒ -1/4e8 = c

Hence 1y4=4x+1e8 ⇒ 1=y4(4x+1e8)

(3) is correct

Calculus of Variations Question 13:

The infimum of 01(u(t))2dt of function

{uC1[0,1]} such that u(0) = 0 and {max[0,1]|u|=1} 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 J(u)=t1t2F(t,u,u)dt, u(t1) = u1, u(t2) = u2, will satisfy Euler equation

Fuddt(Fu) = 0

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

Explanation:

J(u) = 01(u(t))2dt, u(0) = 0 and {max[0,1]|u|=1} 

F(t, u, u') = u2 independent of t and u

So u = at + b is the general solution

u(0) = 0 implies b = 0

Hence u(t) = at

{max[0,1]|u|=1} i.e., at x = 1, u = 1

So, 1 = a × 1 ⇒ a = 1

Solution is u(t) = t ⇒ u'(t) = 1

Therefore infimum value

01(u(t))2dt = 0112dt = [t]01 = 1

So infimum value is 1

(3) is correct

Calculus of Variations Question 14:

The cardinality of the set of extremals of 

J[y]=01(y)2dx,

subject to 

y(0) = 2, y(1) = 5, 01ydx=4  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) = abF(x,y,y)dx subject to y(a) = y1, y(b) = y2, J(y) = abG(x,y,y)dx=L then the extrema satisfy the Euler-Lagrange's equation

Hyddx(Hy) = 0

where H = F + λH

Explanation:

J[y]=01(y)2dx, y(0) = 2, y(1) = 5,01ydx=4

Let H = y2+λy

Then using

Hyddx(Hy) = 0

⇒ λ - ddx(2y) = 0

⇒ 2y'' = λ

⇒ y'' = λ/2

Integrating both sides we get

y' = λ2x + a

Again integrating

⇒ y = λx24 + 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 = λx24 + ax + 2

Also, 01ydx=4

⇒ 01(λx24+ax+2)dx = 4

⇒ [λx312+ax22+2x]01 = 4

⇒ λ12+a2+2 = 4

⇒ λ12+a2 = 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

fg(t)=0tf(s)g(ts)ds.

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

  1. 12[exp(t)+sin(t)cos(t)].
  2. 12[exp(t)+sin(t)cos(t)].
  3. 12[exp(t)sin(t)cos(t)].
  4. 12[exp(t)+sin(t)+cos(t)].

Answer (Detailed Solution Below)

Option 1 : 12[exp(t)+sin(t)cos(t)].

Calculus of Variations Question 15 Detailed Solution

Explanation:

fg(t)=0tf(s)g(ts)ds.

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

So, fg(t)=0tessin(ts)ds

Let 

I = fg(t)=0tessin(ts)ds

I = [essin(ts)]0t0t(es)(cos(ts))ds

I = 0 + sin t - 0tescos(ts)ds

I = sin t - {[escos(ts)]0t+0tessin(ts)ds}

I = sin t - e-t - cos t - I 

2I = sin t - e-t - cos t

I = 12[exp(t)+sin(t)cos(t)]

Hence f⋆g(t) = 12[exp(t)+sin(t)cos(t)]

Hence, option (1) is correct

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