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

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 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 f(x) = (x2 - 1) |x2 - 3x + 2 | + cos (|x|), then set of point of non - differentiability is
  • a)
    {1 ,0 ,2}
  • b)
    {1, 2}
  • c)
    {2}
  • d)
    {0,2}
Correct answer is option 'C'. Can you explain this answer?

Mister Genius answered
Given that

Clearly, the rule of the function is changing of x = 1 and 2. So, we shall test the differentiability of f(x) only at the points x = 1, and 2
Clearly Rf'(1) = Lf'(1) and Rf'(2) ≠ Lf'(2) Hence ,f(x) is not differentiable at x = 2.

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

then at (0, 0),
  • a)
    xfx + yfy = 4f
  • b)
    xfx - yfy = 3f
  • c)
    yfx + xfy = 4f
  • d)
    yfx - xfy = 2f
Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered
For the function f(x,y) to have minimum value at (a,b)
rt – s2>0 and r>0
where, r = ∂2f⁄∂x2, t=∂2f⁄∂y2, s=∂2f⁄∂x∂y, at (x,y) => (a,b)

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.

Given a function f(x) = | x | for all x ∈ R. The function f is
  • a)
    differentiable at zero
  • b)
    continuous at zero
  • c)
    (a) is true (b) is false
  • d)
    both (a) and (b) are true
Correct answer is option 'C'. Can you explain this answer?

Deepanjali Sharma answered
If we plot the graph of given functionthen we see that graph is not breaking at X=0 hence continuous at 0 but we can draw infinitely many tangents at X=0 hence not differentiable .
option (c) is wrong and option (b) is correct

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

What is the domain and range of f?.
Where
  • a)
    (– 1, ∞]
  • b)
    [ ∞, ∞)
  • c)
    (– ∞, ∞)
  • d)
    [– ∞, ∞]
Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered
Here, function is defined in stages, for x ≥ 1: the formula y = x is defined for all values of x greater than or equal to unity. The domain for this part is [1, ∞]. 
For x < 1 is defined for all values less than unity. Then the domain of this part is (– ∞ ,1).
Hence, domain of given function is (– ∞ ,1) ∪ [1, ∞) = (– ∞, ∞).

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