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

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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)

 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

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. 

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) = | 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.

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 .

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

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?

Chirag Verma 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.

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 

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.

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.

If f(x) is twice differentiable and | f(x) | < α, f'(x)| < β, in the range x > α, then which of the following is correct?
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered
Let x > a and n a positive number then



or 
Now, for maxima or minima of f(h), we have

or 
and 
Hence, the least value of φ(h)

= 2√AB
thus, 

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.

Let

Then, the directional derivative at c = (0, 0) in the direction u = (a1, a2) is
  • a)
  • b)
  • c)
  • d)
    None of these 
Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered
We are given that

Then, the directional derivative at c = (0, 0) in the direction u(a1, a2) is

and 
Hence, option (c) is correct.

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

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?

Veda Institute 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.

Using Rolle’s theorem, the equation. = 0 has atleast one root between 0 and 1, if
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Chirag Verma answered
Consider the function f defined by
Since , f(x) is a polynomial, it is continuous and differentiablex. Consequently f(x) is continuous in the closed interval [0, 1] and differentiable in the open interval [0,1] Also , f(0) = 0 and

i.e. , f(0) = f(1)
Thus, all the three conditions of Rolle’s theorem are satisfied
Hence, there is atleast one value of x in the open interval [0, 1]
where f'(x) = 0
i.e., cos xn + a1.xn-1 + . . . + an-1 + an = 0 
 

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.

Which of the following function is not called the Euler’s integral of the first kind?
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered
Euler’s integral of the first kind is nothing but Beta function. So, here only is not the definition of Beta function.

Consider the function f(x) = , then which one of the following is correct?
  • a)
    f (x) is every where differentiable except x = 0
  • b)
     is empty set.
  • c)
    set of point which f(x) ≥ -1 is countable
  • d)
    None of the above
Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered
 Given that

 
It is clear that f(x) is not continuous at the points x = nπ, n ∈ z, therefore it is not differentiable at the points.
Which f(x) > 1/2, and f (x) ≥ - 1, also there points are not countable. 

Let f(x, y) = then
  • a)
    f(x, y) is continuous at origin
  • b)
    fx exists at origin but not equal to zero.
  • c)
     fy exists at origin but not equal to zero.
  • d)
    None of these.
Correct answer is option 'A'. Can you explain this answer?

Chirag Verma answered
We are given that
f(x,y) = 
Let us take ε > 0 and (x, y) ≠ (0, 0). Consider
|f(x,y) - 0| =
Let x = r cos θ, y = r sin θ
Therefore,
Hence for given ε > 0, there exists δ > 0 such that
Therefore, f(x, y) is continuous at (0, 0). Hence option (a) is correct.
Next,

Hence, option (c) is incorrect.

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