# Complex Function Theory Questions and Answers – Continuity

«
»

This set of Complex Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Continuity”.

1. The function f(x, y)=$$\frac{x^2(2xy-3x)+9y}{3x-2y}$$ is which of the following?
a) Discontinuous at origin
b) Discontinuous at (1, 1)
c) Continuous at origin
d) Discontinuous at (2, 2)

Explanation: Origin implies the coordinates (0, 0)
Case 1: $$\lim_{x \to 0 \\ y \to 0}\frac{x^2(2xy-3x)+9y}{3x-2y}$$
$$\lim_{y \to 0}\frac{9y}{-2y}$$
=$$\frac{-9}{2}$$
Case 2: $$\lim_{y \to 0 \\ x \to 0}\frac{x^2(2xy-3x)+9y}{3x-2y}$$
$$\lim_{x \to 0}\frac{-3x^3}{3x}$$
=0
Since, both the limits are not equal, the function is not continuous at origin.

2. The function f(x, y)=$$\frac{2xy-3x}{3x-2y}$$ is continuous at (1, 1).
a) True
b) False

Explanation: Case 1: $$\lim_{x \to 1 \\ y \to 1}\frac{2xy-3x}{3x-2y}$$
$$\lim_{y \to 1}\frac{2y-3}{3-2y}$$=$$\frac{-1}{1}$$=-1
Case 2: $$\lim_{y \to 1 \\ x \to 1}\frac{2xy-3x}{3x-2y}$$
$$\lim_{x \to 1}\frac{2x-3x}{3x-2}$$=$$\frac{-1}{1}$$=-1
Case 3: $$\lim_{y \to mx \\ x \to 0}\frac{2xy-3x}{3x-2y}$$
$$\lim_{x \to 0}\frac{2mx^2-3x}{3x-2mx}$$=$$\frac{-3}{3-2m}$$
Since the cases are not same, f(x, y) is discontinuous at x=1.

3. f(x, y)=$$\frac{3xy+3y+3x}{5x+5y}$$ continuous at origin.
a) True
b) False

Explanation: Case 1: $$\lim_{x \to 0 \\ y \to 0}\frac{3xy+3y+3x}{5x+5y}$$
$$\lim_{y \to 0}\frac{3y}{5y}$$=$$\frac{3}{5}$$
Case 2: $$\lim_{y \to 0 \\ x \to 0}\frac{3xy+3y+3x}{5x+5y}$$
$$\lim_{x \to 0}\frac{3x}{5x}$$=$$\frac{3}{5}$$
Case 3: Along a linear path, $$\lim_{y \to mx \\ x \to 0}\frac{3xy+3y+3x}{5x+5y}$$
$$\lim_{x \to 0}\frac{3mx^2+3mx+3x}{5x}$$=$$\frac{3(m+1)}{5}$$
Since, the cases are not equal, f(x, y) is discontinuous at x=0.

4. If f(x)=$$\frac{x+3}{x^2-5x+6}$$ and g(x)=$$\frac{x-3}{x^2-6x+8}$$, then which of the following is correct?
a) f(x) is continuous at x=2 and g(x) is continuous at x=2
b) f(x) is discontinuous at x=3 and g(x) is discontinuous at x=2
c) f(x) is continuous at x=4 and g(x) is discontinuous at x=4
d) f(x) is discontinuous at x=3 and g(x) is continuous at x=2

Explanation: The denominator of f(x) is (x2-5x=6). When we factorize this, we get,
(x2-5x+6)=x2-2x-3x+6
(x2-5x+6)=(x-2)(x-3)
Since the denominator becomes 0 at x=2 and x=3, f(x) is discontinuous at x=2 and x=3.
The denominator of g(x) is (x2-6x+8). When we factorize this, we get,
(x2-6x+8)=x2-2x-4x+8
(x2-6x+8)=(x-2)(x-4)
Since the denominator becomes 0 at x=2 and x=4, f(x) is discontinuous at x=2 and x=4.

5. If f(x)=$$\frac{x+3}{x^2-5x}$$and g(x)=$$\frac{x-3}{x^2-6x}$$, then which of the following is correct?
a) f(x) is continuous at x=1 and g(x) is continuous at x=0
b) f(x) is discontinuous at x=6 and g(x) is discontinuous at x=5
c) f(x) is continuous at x=5 and g(x) is discontinuous at x=0
d) f(x) is discontinuous at x=5 and g(x) is continuous at x=5

Explanation: The denominator of f(x) is (x2-5x). When we factorize this, we get,
(x2-5x)=x(x-5)
Since the denominator becomes 0 at x=0 and x=5, f(x) is discontinuous at x=0 and x=5.
The denominator of g(x) is (x2-6x). When we factorize this, we get,
(x2-6x+8)=x(x-6)
Since the denominator becomes 0 at x=0 and x=6, f(x) is discontinuous at x=0 and x=6.

6. The function f(x)=$$\frac{1}{x^3-3x^2-10x+24}$$is continuous at which of the following?
a) 1
b) 2
c) 3
d) 4

Explanation: When we substitute x=2, x=3 or x=4 in the function, we get the denominator as 0. There by, making the function discontinuous at these points.
At x = 1, we get f(1)=$$\frac{1}{12}$$ which is continuous.

7. The function f(x)=$$\frac{e^{-x}}{x(x-5)(x+2)}$$ is continuous at which of the following?
a) 0
b) 2
c) 5
d) The function is continuous at all points

Explanation: When we substitute x=0, x=5 or x=-2 in the f(x), we get the denominator as 0. This is because these are the points of discontinuity.
At x=2, we get f(2)=$$\frac{-1}{24}$$ which is continuous.

8. Which of the following is true about f(x)=tanx?
a) Continuous at x=$$\frac{\pi}{2}$$
b) Continuous at x=$$-\frac{3\pi}{2}$$
c) Continuous at x=0
d) Continuous at x=$$\frac{3\pi}{2}$$

Explanation: tanx=$$\frac{sinx}{cosx}$$
cosx=0 when x=$$\frac{\pm (n+1)\pi}{2}$$ which are points of discontinuity for the function, f(x)=tanx.

9. Which of the following is true about f(x)=sinx+cosx?
a) Continuous everywhere
b) Continuous at x=$$\frac{\pm n\pi}{2}$$, where n is any integer
c) Continuous at x=$$\frac{\pm (n+1)\pi}{2}$$, where n is any integer
d) Continuous at x=0

Explanation: The values of sinx and cosx never become infinity or non-determinable for any value of x. Therefore, the function f(x)=sinx+cosx is continuous at all points of x.

10. Which of the following is false if both f(x) and g(x) are continuous?
a) f(x)+g(x) is continuous
b) f(x)-g(x)is continuous
c) f(x)*g(x)is continuous
d) $$\frac{f(x)}{g(x)}$$ is continuous

Explanation: For $$\frac{f(x)}{g(x)}$$ to be continuous, the denominator needs to be greater than 0. If not, the function becomes infinity and cannot be determined, thereby becoming discontinuous.

11. Which of the following is discontinuous at x=9?
a) $$\frac{x^2-81}{x-9}$$
b) $$\frac{x^3-20x^2+117x-162}{x-9}$$
c) $$\frac{x^2-11x+18}{x(x-9)}$$
d) $$\frac{x^2-13x-90}{x-9}$$

Explanation: $$\frac{x^2-81}{x-9}$$=$$\frac{(x-9)(x+9)}{x-9}$$=(x-9) which is continuous at x = 9.
$$\frac{x^3-20x^2+117x-162}{x-9}$$=$$\frac{(x-9)^2(x-2)}{(x-9)}$$=(x-9)(x-2) which is continuous at x = 9.
$$\frac{x^2-11x+18}{x(x-9)}$$=$$\frac{(x-2)(x-9)}{x(x-9)}$$=$$\frac{(x-2)}{x}$$ which is continuous at x = 9.
$$\frac{x^2-13x-90}{x-9}$$=$$\frac{(x-18)(x+5)}{x-9}$$ which is discontinuous at x = 9, since the denominator becomes 0.

12. Which of the following functions are continuous at x=11?
a) $$\frac{1}{x^2+x-132}$$
b) $$\frac{1}{x^3-7x^2-40x-44}$$
c) $$\frac{1}{x^2-2x-143}$$
d) $$\frac{x-11}{x^3-23x^2+143x-121}$$

Explanation: $$\frac{1}{x^2+x-132}$$=$$\frac{1}{(x-11)(x+12)}$$ is discontinuous at x=11 since the denominator becomes 0.
$$\frac{1}{x^3-7x^2-40x-44}$$=$$\frac{1}{(x-11)(x+2)^2}$$is discontinuous at x=11 since the denominator becomes 0.
$$\frac{x-11}{x^3-23x^2+143x-121}$$=$$\frac{(x-11)}{(x-11)^2(x-1)}$$=$$\frac{1}{(x-11)(x-1)}$$is discontinuous at x=11 since the denominator becomes 0.
$$\frac{1}{x^2-2x-143}$$is continuous at x=11.

Sanfoundry Global Education & Learning Series – Complex Analysis.

To practice all areas of Complex Analysis, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs! 