All Courses
Get the App Signup / Login
Sign in Open App
All Exams  >   Mathematics  >   Mathematics for Competitive Exams  >   All Questions

All questions of Functions of One, Two or Three Variable for Mathematics Exam

Clear all your Mathematics doubts with EduRev

 dx dy is equal to
  • a)
    4 log 8 + e
  • b)
    8 log 8 - 16
  • c)
    8 log 8 - 16 + e
  • d)
     e - 16
Correct answer is option 'C'. Can you explain this answer?

Pavan Kishore answered
First integrate w.r.t x taking y as const
then u get function in terms of pure y and then integrate w.r.t y and asusual substitution

The function f(x) = - 2x3 - 9x2 - 12x + 1 is an increasing function in the interval
  • a)
    -2 < x < -1
  • b)
    -2 < x < 1
  • c)
    - l < x < 2
  • d)
    1 < x < 2
Correct answer is option 'A'. Can you explain this answer?

Vivek Kumar answered
Given that
f(x) = - 2x3 - 9x2 - 12x + 1
- + - 
- 2 - 1
implies f'(x) = - 6x2 - 18x - 12
                   = -6[x2 + 3x + 2]
                   ​= - 6[(x + l ) ( x + 2)]
implies f'(x) > 0, V x ∈ (-2, -1)
Hence, f(x) is increasing in the interval (-2, -1)

The function f(x) = | x + 2 | is not differentiable at a point
  • a)
    x = 2
  • b)
    x = - 2
  • c)
     x = - 1
  • d)
    x = 1
Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered
Given that
Clearly, the rule of the function is changing at x = 0, so we shall test the differentiability of f(x) only at the point x = -2 obviously, being polynomial, at all other points the function is differentiable.


Since Lf '(- 2) ≠ R'f(- 2) , therefore , f(x) is not differentiable at x = - 2 .

Let f(x) = | sin πx |, then
  • a)
    f is continuous nowhere
  • b)
    f is continuous everywhere and differentiable nowhere.
  • c)
    f is continuous everywhere and differentiable every where except at integral values of x.
  • d)
    f is differentiable everywhere.
Correct answer is option 'C'. Can you explain this answer?

Chirag Verma answered
The graph of the function f(x) = | sinπx | is y =f(x)

clearly, the graph have sharp edges at the points 0, ±1, ± 2, ±3, .. therefore function is continuous everywhere but not differentiable everywhere except at integral values of x.

If is α repeated root of the polynomial equation f(x) = 0, then
  • a)
     f(a) = 0 but f'(a) ≠ 0
  • b)
    f(a) ≠ 0 but f'(α) = 0
  • c)
    f'(α) ≠ 0 and f'(α) = 0
  • d)
     f(α) = 0 and f'(α) = 0 
Correct answer is option 'D'. Can you explain this answer?

Arnav Chawla answered


Repeating Root of a Polynomial Equation

An repeated root of a polynomial equation means that the root is a solution to the equation with a multiplicity greater than 1. In this case, if α is a repeated root of the polynomial equation f(x) = 0, it means that when you substitute α into the polynomial equation, it gives you 0.

Understanding the Options

a) f(α) = 0 but f(α) ≠ 0
This option is contradictory because if f(α) = 0, then f(α) cannot be not equal to 0.

b) f(α) ≠ 0 but f(α) = 0
This option is also contradictory because if f(α) = 0, then f(α) cannot be not equal to 0.

c) f(α) ≠ 0 and f(α) = 0
This option is contradictory because it states that f(α) is not equal to 0 and at the same time equal to 0.

Correct Answer: Option 'D'

d) f(α) = 0 and f(α) = 0
This option correctly states that when a root is repeated, it must satisfy the polynomial equation with a value of 0. Therefore, the correct answer is option 'D'.

Let l  = { 1 } ∪ { 2 }  For x ∈ R, let φ(x) = dis (x, l) = ln f{| x —y | : y ∈ l} then
  • a)
    φ is discontinuous some where or R
  • b)
    φ is continuous on R but not differentiable only at x = 1
  • c)
    φ is continuous on R but not differentiable only at x = 1 and 2
  • d)
    φ is continuous on R but not differentiable only at 
    x = 1, 3/2 and 2 
Correct answer is option 'D'. Can you explain this answer?

Chirag Verma answered
Given that,

= ln f{|x -.y| : y ∈ 1}
where l = { l } ∪ { 2 }
= ln f{| x -l| ,| x -2 | } 
the graph of the φ(x) is given by.

Clearly, the graph of the function have sharp edges at x = 1,3/2 and 2. Therefore, f(x) is not differentiable at x
= 1, 3/2 and 2.
Hence, φ(x) is continuous on R but not
differentiable at x = 1, 3/2 and 2

The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on x-axis and passing through (2, 1) is
  • a)
    x2 + y2  − x = 0
  • b)
    4x2 + 2y2 − 9y = 0
  • c)
    2x2 + 4y2 − 9x = 0
  • d)
    4x2 + 2y2 − 9x = 0
Correct answer is option 'D'. Can you explain this answer?

Vikram Kapoor answered
∵ Equation of normal at (x, y) is
Y − y = dx/dy (X − x)
Put, y = 0
Then, X = x + y dy/dx
Given, y2 = 2x X
⇒ y2 = 2x ( x + y dy/dx)
⇒ dy/dx = (y2 − 2x2)/(2xy) = ((y/x)2 − 2)/ ( 2y/x)
Put y = vx, we get dx/dy = v + x dv/dx
Then, v + x dv/dx = v2−2/2v
On integrating both sides, we get
ln (2 + v2) + ln|x| = ln c
⇒ ln (|x|(2 + v2)) = ln c
⇒ |x| ( 2 + y2/x2) = c
∵ It passes through (2, 1), then 2 (2 + 1 4 ) = c
⇒ c = 9/2
Then, |x| ( 2 + y2/x2) = 9/2
⇒ 2x2 + y2 = 9/2 |x|
⇒ 4x2 + 2y2 = 9|x|

Let
f(x, y) = 2x2 - xy + 2y2 Then at (1, 2)
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Chirag Verma answered
We are given that
f(x, y) = 2x2 - xy + 2y2
then, 
Hence, = 4 x 1 - 2 = 2
and   = - x + 4y
Hence,  = - l + 4x 2 = 7
Thus,
Hence, option (d) is correct. 

The value of ‘C’ in Rolle’s theorem, where -π/2 < C < π/2 and f(x) = cos x is equal to
  • a)
    π/4
  • b)
    π/3
  • c)
    π
  • d)
    0
Correct answer is option 'D'. Can you explain this answer?

Aisha Gupta answered
An item is subjective and can vary depending on factors such as its condition, rarity, demand, and market trends. Therefore, the value of an item cannot be determined without specific context.

Let 

Then,
  • a)
    f is continuous at (0, 0) and the partial derivatives fx, fy exists at every point o f R2
  • b)
    f is discontinuous at (0,0) and fx, fy does not exists at every point o f R2
  • c)
    f is discontinuous at (0, 0) a n d fx , fy exists at (0,0)
  • d)
    None of the above
Correct answer is option 'C'. Can you explain this answer?

Chirag Verma answered
Let us suppose (x, y) approaches (0, 0) along the line y = mx. Which is a line through the origin. Put y = mx and allows x —> 0, we get

which depends on m, therefore the limit of f(x, y) at (0, 0) does not exists. Hence, f(x, y) is discontinuous at origin.
Now,
since fy exists at origin.

If f(x) is differentiable on interval l and such that |f'(x)| ≤ a on l, then f(x) is
  • a)
     Continuous but not uniformly continuous on l
  • b)
    Uniformly continuous but not continuous on l
  • c)
    Uniformly continuous but not differentiable on l
  • d)
    Continuous, uniformly continuous and differentiable on l
Correct answer is option 'A'. Can you explain this answer?

Chirag Verma answered
For xy ∈ l, by Lagrange’s mean value theorem
where x < c < y
implies f(x)- f(y) = (x - y) f '(c )
implies |f(x) - f(y) | = x - y || f'(c)|
for a given  such that |f(x) - f(y)| < ε, x, y ∈ l. Hence f(x) is uniformly continuous on l. We know that every uniformly continuous function is also continuous.

The derivative of the function f(x) = x2m is
  • a)
    even function
  • b)
    odd function
  • c)
    constant function
  • d)
    None of these 
Correct answer is option 'B'. Can you explain this answer?

Ishika Singhania answered
The derivative of the function f(x) = x^2m is an odd function.

Explanation:

To find the derivative of the function f(x), we can use the power rule of differentiation. According to the power rule, if we have a function f(x) = x^n, the derivative of f(x) is given by:

f'(x) = n * x^(n-1)

In this case, the function f(x) = x^2m can be written as f(x) = (x^2)^m. Applying the power rule, we can find the derivative as follows:

f'(x) = m * (x^2)^(m-1) * 2x

Simplifying further:

f'(x) = 2mx^(2m-1)

Odd Function:

An odd function is a function that satisfies the property f(-x) = -f(x) for all values of x in the domain of the function. In other words, if we replace x with -x in the function and negate the result, it should be equal to the function itself.

Let's substitute -x for x in the derivative we found:

f'(-x) = 2m(-x)^(2m-1)

Now let's negate the result:

-f'(x) = -2m(-x)^(2m-1)

If we simplify this expression, we get:

-f'(x) = 2m(-x)^(2m-1)

We can observe that -f'(x) = f'(-x), which means that the derivative of f(x) satisfies the property of an odd function.

Conclusion:

The derivative of the function f(x) = x^2m is an odd function.

Let f(x + y) = f(x)f(y) for all x and y and f(5) = - 2 and f'(0) = 3. What is the value of f'(5)?
  • a)
    3
  • b)
    1
  • c)
    -6
  • d)
    6
Correct answer is option 'C'. Can you explain this answer?

Chirag Verma answered
f'(0) = 3
implies 
or 
[Since f(5 + 0) = f(5) • f(0)
implies f(5) = f(5) - f(0)
implies f(0) = 1]
Now, we have

[by eq. (1)]
Hence, f'(5) = -6 

Let
  • a)
    f(x, y) is continuous at origin
  • b)
    f(x, y) is differentiable at origin
  • c)
    fx(0,0) ≠ 0
  • d)
     fy(0 ,0) ≠ 0
Correct answer is option 'A'. Can you explain this answer?

Chirag Verma answered
We are given that 
Let us take ε > 0 and x2 + y2 ≠ 0.
Now consider,

We know that, 
or equivalently

Thus,
| f(x ,y )-0 | ≤ | y | = 0- |x| + 1 • |y|
take 
Therefore, for given ε > 0, there exists δ > 0 such that
|f(x, y) - 0 | < ε. whenever 0<|x|<δ,
0< |y |< δ
Hence, 
Thus, f(x,y) is continuous at (0, 0).
Hence, option (a) is correct.
Now,
 f(0 + h ,0 + k) - f(0,0) = f(h , k) - f(*0,0)
= 0 • h + 0 • k + • 
So that A = 0, B = 0 which does not depends on h and k and

Now approaching along h = mk, we get 

which depends on m. Therefore does not exists. Hence, f(x, y) is not differentiable at origin. Hence, option (b) is incorrect.
Next,
Hence, option (c) is incorrect.
And,

Hence, option (d) is incorrect.

For (x, y) ∈ R2, let
  • a)
     fx and fy exists at (0, 0 ) and f is continuous at (0, 0)
  • b)
    fx and fy exists at (0,0) and f is discontinuous at (0,0)
  • c)
    fx and fy does not exist at (0, 0) and f is continuous at (0, 0)
  • d)
    fx and fy does not exists at (0 , 0) and f is discontinuous at (0, 0)
Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered
Let us approaches (0, 0) along the line y = mx which passes through origin. Put y = mx, we get

which depends on m. Hence,  does not exists, Therefore f(x, y) is discontinuous at (0,0).
Now, 
and 

Hence, f(x,y) is discontinuous at (0, 0). But both the partial derivative fx and fy exists at origin.

If  then
  • a)
    f does not exist at (0, 0)
  • b)
    f is continuous at (0, 0)
  • c)
    f is not continuous at (0, 0)
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered
Limit along x → 0 and y → 0:

Limit along y → 0 and x → 0

Limit along the y = mx, then

Since the limit along any path is the same, the limit exists.

If f(x, y) is differentiable at (a, b), then the partial derivatives fx and fy at (a, b)
  • a)
    fx exists but fy does not exists
  • b)
     fx does not exists but fy exists
  • c)
     fx and fy both are exists
  • d)
    fx and fy both are not exists
Correct answer is option 'C'. Can you explain this answer?

Tanvi Sengupta answered
Differentiability of a Function at a Point

To understand why option C is the correct answer, let's first review the concept of differentiability of a function at a point.

A function f(x, y) is said to be differentiable at a point (a, b) if both of the following conditions are satisfied:
1. The partial derivatives fx and fy exist at (a, b).
2. The function is locally linear near the point (a, b), meaning it can be well approximated by a linear function.

Partial Derivatives

The partial derivatives of a function with respect to its variables measure the rate at which the function changes with respect to each variable, while holding the other variables constant.

The partial derivative fx represents the rate of change of f with respect to x, and the partial derivative fy represents the rate of change of f with respect to y.

Explanation of Option C

Option C states that both fx and fy exist at point (a, b). This means that the function f(x, y) has well-defined rates of change with respect to both x and y at this point.

If f(x, y) is differentiable at (a, b), it implies that the function can be well approximated by a linear function near this point. In other words, the function is locally linear.

For a function to be locally linear, it must have well-defined rates of change in all directions. In the case of a function of two variables, this means that both the partial derivatives fx and fy must exist at the point (a, b).

Therefore, option C is the correct answer because if f(x, y) is differentiable at (a, b), it implies that both fx and fy exist at this point.

At what point will be function be continuous ?
  • a)
    At all points of Z
  • b)
    At all points of R ~ Z
  • c)
    At all points of R
  • d)
    Whenever X ∈ Z
Correct answer is option 'D'. Can you explain this answer?

Chirag Verma answered
Let f be the function defined on R by setting f (x) =   when x is not an integer , f(x ) = 0, then f is continuous, when x is an integer.

The value of ‘C’ of Lagrange’s mean value theorem for f( x ) = x (x - 1) in [1 ,2] is given by
  • a)
    5/4
  • b)
    3/2
  • c)
    7/4
  • d)
    11/6
Correct answer is option 'B'. Can you explain this answer?

Chirag Verma answered
Clearly f(x) = x2 - x is continuous in the interval [1, 2] and differentiable in the open interval [1,2]. therefore,
or 2 = 2c - 1
or c = 3/2

Let

then
  • a)
    f(x, y) is continuous at origin
  • b)
    f(x, y) has removable discontinuity of (0,0)
  • c)
    f(x, y) is not differentiable at (0, 0)
  • d)
    None of these 
Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered
ANSWER :- c
Solution :- Consider an approach along the line y=x; then f(x,y)=x^2/(x^2+x^2) = 1/2
for all x≠0. On the other hand, if we approach (0,0) along the line y=2x
f(x,y)=(2x)^2/[x^2+4x^2]
=2/5
So there are two different paths toward the origin, each giving a different limit. Hence, the limit does not exist.
Or more simply, approach along y=0
Therefore , f(x, y) is not differentiable at origin

Let A be a 3 x 4 matrix of rank 2. Then the rank of ATA, where AT denotes the transpose of A. is,
  • a)
    exactly 2
  • b)
    exactly 1
  • c)
    at most 3, and atleast 2
  • d)
    at most 2, but not necessarily 2
Correct answer is option 'A'. Can you explain this answer?

Dipanjan Sharma answered
Rank of a matrix:
The rank of a matrix is the maximum number of linearly independent rows or columns in that matrix. It can be calculated by performing row operations and reducing the matrix to its row echelon form or reduced row echelon form.

Rank of a product of matrices:
For two matrices A and B, the rank of their product AB is at most the minimum of the ranks of A and B. In other words, rank(AB) ≤ min(rank(A), rank(B)).

Rank of a transpose:
The rank of a matrix and its transpose are always the same. So, rank(A) = rank(A^T).

Rank of ATA:
Given a matrix A of rank 2, we need to determine the rank of ATA.

- Since A is a 3 x 4 matrix of rank 2, it means that there are 2 linearly independent rows or columns in A.
- Transposing A will give us a 4 x 3 matrix of rank 2, denoted by A^T.
- Multiplying A and A^T will give us a 3 x 3 matrix, denoted by ATA.
- According to the rank property mentioned earlier, rank(ATA) ≤ min(rank(A), rank(A^T)).
- Since rank(A) = rank(A^T), we can say that rank(ATA) ≤ rank(A).
- But we know that rank(A) = 2, so rank(ATA) ≤ 2.

Conclusion:
From the above analysis, we can conclude that the rank of ATA is at most 2. But since A is a 3 x 4 matrix of rank 2, we have 2 linearly independent rows or columns in A. When we multiply A and A^T to get ATA, we are essentially taking the dot product of rows and columns of A. Since we have 2 linearly independent rows or columns, the dot product will also be linearly independent. Therefore, the rank of ATA is exactly 2.

Hence, the correct answer is option 'A' - exactly 2.

Let
f(x,y) = x3 + y3- 63 (x + y) + 12xy,
then
  • a)
    the function has three stationary points
  • b)
    the function is minimum at (-7, -7)
  • c)
    the function is maximum at (3, 3)
  • d)
    the function has neither minimum nor a maximum at (5,-1)
Correct answer is option 'D'. Can you explain this answer?

Rudra Sethi answered
Given Function:
The given function is:
f(x,y) = x^3y^3 - 63xy + 12xy

Stationary Points:
To find the stationary points of the function, we need to find the values of x and y where the partial derivatives of the function with respect to x and y are equal to zero.

Taking the partial derivative of f(x,y) with respect to x:
∂f/∂x = 3x^2y^3 - 63y + 12y

Taking the partial derivative of f(x,y) with respect to y:
∂f/∂y = 3x^3y^2 - 63x + 12x

Setting both partial derivatives equal to zero and solving for x and y, we get:
3x^2y^3 - 63y + 12y = 0
3x^3y^2 - 63x + 12x = 0

The Function has Three Stationary Points:
To determine the number of stationary points, we need to solve the system of equations formed by setting the partial derivatives equal to zero.

By solving the system of equations, we can find the values of x and y where the function has stationary points. If the system has three distinct solutions, then the function has three stationary points.

The Function is Minimum at (-7, -7):
To determine whether the function is a minimum or maximum at a specific point, we need to consider the second-order partial derivatives of the function.

Calculating the second-order partial derivatives of f(x,y):
∂^2f/∂x^2 = 6xy^3
∂^2f/∂y^2 = 6x^3y
∂^2f/∂x∂y = 9x^2y^2 - 63
∂^2f/∂y∂x = 9x^2y^2 - 63

To determine the nature of the critical point, we can use the second-order partial derivatives test. However, in this case, we can see that the second-order partial derivatives are not needed to determine the nature of the point.

The Function is Maximum at (3, 3):
Similarly, to determine whether the function is a minimum or maximum at a specific point, we need to consider the second-order partial derivatives of the function.

Calculating the second-order partial derivatives of f(x,y):
∂^2f/∂x^2 = 6xy^3
∂^2f/∂y^2 = 6x^3y
∂^2f/∂x∂y = 9x^2y^2 - 63
∂^2f/∂y∂x = 9x^2y^2 - 63

To determine the nature of the critical point, we can use the second-order partial derivatives test. However, in this case, we can see that the second-order partial derivatives are not needed to determine the nature of the point.

The Function has Neither Minimum nor Maximum at (5, -1):
To determine whether the function has a minimum or maximum at a specific point, we need to consider the

Write the roaster form for A = {x : x2 – 5x + 6 = 0}
 
  • a)
    {2}
  • b)
    {2,3}
  • c)
    {3}
  • d)
    Φ
Correct answer is option 'B'. Can you explain this answer?

Diya Srinivasan answered
The roaster form for A = {x : x^2 < 10}="" />

A = {-√10, -√9, -√8, -√7, -√6, -√5, -√4, -√3, -√2, -√1, √1, √2, √3, √4, √5, √6, √7, √8, √9}

If any four numbers are selected and they are multiplied, then the probability that the last digit will be 1, 3, 5 or 7 is
  • a)
    1/5
  • b)
    7/625
  • c)
    16/625
  • d)
    20/625
Correct answer is option 'C'. Can you explain this answer?

Vikram Kapoor answered
The total number of digits in any number at the units place is 10.
Therefore, n (S) = 10
If the last digit is 1, 3, 5 or 7, then it is necessary that the last digit in each number must be 1, 3, 5 or 7.
Therefore, n (A) = 4
P (A) = 4 / 10 = 2 / 5
Hence, the required probability is (2 / 5)4 = 16 / 625.

Chapter doubts & questions for Functions of One, Two or Three Variable - Mathematics for Competitive Exams 2025 is part of Mathematics exam preparation. The chapters have been prepared according to the Mathematics exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Mathematics 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

Chapter doubts & questions of Functions of One, Two or Three Variable - Mathematics for Competitive Exams in English & Hindi are available as part of Mathematics exam. Download more important topics, notes, lectures and mock test series for Mathematics Exam by signing up for free.

Mathematics for Competitive Exams

98 videos|28 docs|30 tests

Signup to see your scores go up within 7 days!

Study with 1000+ FREE Docs, Videos & Tests
10M+ students study on EduRev
© EduRev
Education Revolution
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily