Complex Function Theory Questions and Answers – Continuity

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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)
View Answer

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

Answer: b
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
View Answer

Answer: b
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
View Answer

Answer: b
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
View Answer

Answer: b
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.
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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
View Answer

Answer: a
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
View Answer

Answer: b
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=\(\frac{3\pi}{2}\)
View Answer

Answer: c
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
View Answer

Answer: a
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.
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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
View Answer

Answer: d
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) \(\frac{x^2-13x-90}{x-9}\)
View Answer

Answer: d
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) \(\frac{x-11}{x^3-23x^2+143x-121}\)
View Answer

Answer: c
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.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn