Functions of Several Variables MCQ Quiz - Objective Question with Answer for Functions of Several Variables - Download Free PDF
Last updated on Jun 26, 2025
Latest Functions of Several Variables MCQ Objective Questions
Functions of Several Variables Question 1:
For a positive real number a, √a, denotes the positive square root of a. Consider the function f : ℝ2 → ℝ defined by
Which of the following statements are true?
Answer (Detailed Solution Below)
Functions of Several Variables Question 1 Detailed Solution
Concept:
Differentiability and Continuity of Functions of Two Variables:
- A function
is said to be continuous at a point if . - The partial derivatives of
at are defined as: and . - A function is differentiable at a point if it is continuous there and well-approximated by a linear map locally.
- Differentiability at a point implies the existence of partial derivatives and continuity, but not vice versa.
Calculation:
Given,
Check continuity at
⇒ Along
⇒ Along
⇒ Along any path,
Now, compute partial derivatives:
⇒
⇒ Left and right limits are different ⇒ partial derivative w.r.t x does not exist
⇒
⇒ So,
∴ Correct statements are: 1 and 4
Functions of Several Variables Question 2:
Consider the function f : ℝ2 → ℝ defined by
Which of the following statements are true?
Answer (Detailed Solution Below)
Functions of Several Variables Question 2 Detailed Solution
Concept:
Continuity, Directional Derivatives, and Differentiability of a Function:
- A function
is continuous at a point if . - The directional derivative at
in the direction is defined as . - A function is differentiable at a point if it can be well-approximated by a linear function there; differentiability implies continuity.
- A notable phenomenon is that all directional derivatives may exist at a point even when the function is not continuous and thus not differentiable at that point.
Calculation:
Given,
⇒ Check continuity at
⇒ Along
⇒ As
⇒ Along
⇒ Limits differ
⇒ f is not continuous at (0,0)
⇒ Statement 4 is correct.
⇒ For
⇒ f is differentiable there
⇒ Statement 1 is correct.
⇒ Now, compute directional derivative at
⇒
⇒ Rewrite first term as
⇒ Then, the difference quotient is
⇒ As
⇒ First term tends to
⇒ Second term:
⇒ Thus,
⇒ Statement 2 is correct.
⇒ Since f is not continuous at
f cannot be differentiable there despite all directional derivatives existing
⇒ Statement 3 (differentiability at (0,0)) is false.
∴ Correct statements are: Options 1, 2, and 4.
Functions of Several Variables Question 3:
Define f : R2 → R by f(x, y) = x2 + 2y2 − x for (x, y) ∈ R2. Let D = {(x, y) ∈ R2 ∶ x2 + y2 ≤ 1} and E = {(x, y) ∈ R2 ∶
Dmax = {(a, b) ∈ D ∶ f has absolute maximum on D at (a, b)},
Dmin = {(a, b) ∈ D ∶ f has absolute minimum on D at (a, b)},
Emax = {(c, d) ∈ E ∶ f has absolute maximum on E at (c, d)},
Emin = {(c, d) ∈ E ∶ f has absolute minimum on E at (c, d)}.
Then the total number of elements in the set Dmax ∩ Dmin ∩ Emax ∩ Emin is equal to _________.
Answer (Detailed Solution Below) 0
Functions of Several Variables Question 3 Detailed Solution
Explanation:
Given : f(x, y) = x2 + 2y2 − x for (x, y) ∈ R2
D = {(x, y) ∈ R2 ∶ x2 + y2 ≤ 1} and E = {(x, y) ∈ R2 ∶
f(x, y) = x2 + 2y2 − x
f(x, y) = x2 − x + 2y2
Since
Minimum of
Also maximum of f(x) in D occurs at (0 , ± 1)
And Maximum of f(x) in E occurs at (0 , ± 3 )
Now,
and Dmax ∩ Dmin ∩ Emax ∩ Emin = 0
Then the total number of elements in the set Dmax ∩ Dmin ∩ Emax ∩ Emin = 0
Hence 0 is the answer.
Functions of Several Variables Question 4:
Consider the function u: R3 → R given by
Answer (Detailed Solution Below) 41.59 - 41.61
Functions of Several Variables Question 4 Detailed Solution
Explanation:
Given
To Find The partial derivative of u with respect to
Differentiate each term with respect to
For
For
For
Thus
The partial derivative of u with respect to
For
For
For
Thus
Now,
Evaluate at
Substitute
After comparing with
Now,
Hence 41.60 is the correct answer.
Functions of Several Variables Question 5:
Let
Answer (Detailed Solution Below)
Functions of Several Variables Question 5 Detailed Solution
Concept:
Total Derivative :
Explanation:
Formula to Find dz :
Partial derivative with respect to x Treat y as constant and differentiate
Partial derivative with respect to y Treat x as constant and differentiate
Putting the values :
Hence Option(2) is the correct answer.
Top Functions of Several Variables MCQ Objective Questions
Let
Answer (Detailed Solution Below)
Functions of Several Variables Question 6 Detailed Solution
Download Solution PDFConcept:
A function f(x,y) is defined for (x,y) = (a,b) is said to be continuous at (x,y) = (a,b) if:
i) f(a,b) = value of f(x,y) at (x,y) = (a,b) is finite.
ii) The limit of the function f(x,y) as (x,y) → (a,b) exists and equal to the value of f(x,y) at (x,y) = (a,b)
Note:
For a function to be differentiable at a point, it should be continuous at that point too.
Calculation:
Given:
For function f(x,y) to be continuous:
f(a,b) = f(0,0) ⇒ 0 (given)
fx(0, 0) =
fy(0, 0) =
∵ the limit value is defined and function value is 0 at (x,y) = (0,0), ∴ the function f(x,y) is continuous.
Hence, Option 2, 3 & 4 all are correct
Hence, Option 1 is not correct
Hence, The Correct Answer is option 1.
Functions of Several Variables Question 7:
Find the limit of the function f(x, y) = xy2/(x2 + y4) at the point u(√2, √2)?
Answer (Detailed Solution Below)
Functions of Several Variables Question 7 Detailed Solution
Explanation:
=
=
=
= 2√2/6 = √2/3
(2) is correct
Functions of Several Variables Question 8:
Given that there exists a continuously differentiable function g defined by the equation F(x, y) = x3 + y3 - 3xy - 4 = 0 in a neighborhood of x = 2 such that g(2) = 2. find its derivative.
Answer (Detailed Solution Below)
Functions of Several Variables Question 8 Detailed Solution
Solution:
Given function is:
F(x, y) = x3 + y3 – 3xy – 4 = 0
And x = 2 and g(2) = 2
Now,
F(2, 2) = (2)3 + (2)3 – 3(2)(2) – 4
= 8 + 8 – 12 – 4
= 0
So, F(2, 2) = 0
∂F/∂x = ∂/∂x (x3 + y3 – 3xy – 4) = 3x2 – 3y
∂F/∂y = ∂/∂y (x3 + y3 – 3xy – 4) = 3y2 – 3x
Let us calculate the value of ∂F/∂y at (2, 2).
That means, ∂F(2, 2)/∂y = 3(2)2 – 3(2) = 12 – 6 = 6 ≠ 0.
Thus, ∂F/∂y is continuous everywhere.
Hence, by the implicit function theorem, we can say that there exists a unique function g defined in the neighborhood of x = 2 by g(x) = y, where F(x, y) = 0 such that g(2) = 2.
Also, we know that ∂F/∂x is continuous.
Now, by implicit function theorem, we get;
g’(x) = -[∂F(x, y)/∂x]/ [∂F(x, y)/ ∂y]
= -(3x2 – 3y)/(3y2 – 3x)
= -3(x2 – y)/ 3(y2 – x)
= -(x2 – y)/(y2 – x)
Hence, option 3 is correct
Functions of Several Variables Question 9:
Find the limit of sin (y)/x, where (x, y) approaches to (0, 0)?
Answer (Detailed Solution Below)
Functions of Several Variables Question 9 Detailed Solution
Given:
f(x, y) =
Concept Used:
Putting y = mx in the function and checking whether the function is free from m then limit will exist if not then limit will not exist.
Solution:
We have,
f(x, y) = \(\frac{siny}{x}\) (x, y) → (0, 0)
Put y = mx
So,
lim (x, y) → (0, 0) \(\frac{siny}{x}\)
⇒ lim x → 0
We cannot eliminate m from the above function.
Hence limit does not exist.
Functions of Several Variables Question 10:
Let
Answer (Detailed Solution Below)
Functions of Several Variables Question 10 Detailed Solution
Concept:
A function f(x,y) is defined for (x,y) = (a,b) is said to be continuous at (x,y) = (a,b) if:
i) f(a,b) = value of f(x,y) at (x,y) = (a,b) is finite.
ii) The limit of the function f(x,y) as (x,y) → (a,b) exists and equal to the value of f(x,y) at (x,y) = (a,b)
Note:
For a function to be differentiable at a point, it should be continuous at that point too.
Calculation:
Given:
For function f(x,y) to be continuous:
f(a,b) = f(0,0) ⇒ 0 (given)
fx(0, 0) =
fy(0, 0) =
∵ the limit value is defined and function value is 0 at (x,y) = (0,0), ∴ the function f(x,y) is continuous.
Hence, Option 2, 3 & 4 all are correct
Hence, Option 1 is not correct
Hence, The Correct Answer is option 1.
Functions of Several Variables Question 11:
Let f : R2 → R be defined by
Which of the following statements holds regarding the continuity and the existence of partial derivatives of f at (0, 0)?
Answer (Detailed Solution Below)
Functions of Several Variables Question 11 Detailed Solution
Concept:
If f(x,y) is a function, where f partially depends on x and y and if we differentiate f with respect to x and y then the derivatives are called the partial derivative of f. The formula for partial derivative of f with respect to x taking y as a constant is given by;
⇒
Given:
Calculation:
By the defination of partial derivatives
⇒ fx(0, 0) =
⇒ fy(0, 0)
If we moving along the curve v = mx2
⇒
⇒ f(x, y) is not continuous.
Both partial derivatives of f exist at (0, 0) and f is not continuous at (0,0)
Functions of Several Variables Question 12:
Let if possible,
Answer (Detailed Solution Below)
Functions of Several Variables Question 12 Detailed Solution
Explanation:
Put x2 + y2 = t, ∴ (x, y) → (0, 0) ⇒ t → 0
= 1
∴ α exists.
Now,
Let us approaches along y = mx, which is (0, 0) Hence,
Which depends on m.
Hence β does not exists.
∴ α exists but β does not exists.
Functions of Several Variables Question 13:
if f(x, y) = xy then
Answer (Detailed Solution Below)
Functions of Several Variables Question 13 Detailed Solution
Concept :
If f is a function of n variables, then its partial derivatives are the functions
Calculations :
Here f is a function of two variables, so partial derivative of f with respect to variable y is defined as
Using L'Hospital rule , we get
Thus, option 2 is correct.
Functions of Several Variables Question 14:
f(x, y) =
Answer (Detailed Solution Below)
Functions of Several Variables Question 14 Detailed Solution
Concept :
The directional derivative of the function f in the direction
Calculations :
=
Now substitute x = y = 1 ,
Thus option 1 is correct.
Functions of Several Variables Question 15:
Answer (Detailed Solution Below)
Functions of Several Variables Question 15 Detailed Solution
Concept:
Polar Coordinates: For functions with radial symmetry, converting to polar coordinates and evaluating the limit as r approaches 0 can simplify the calculation.
Explanation:
If x = rcosθ , y = rsinθ
⇒
⇒
⇒ limit doesn't exist