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
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
View Answer
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
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
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
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
View Answer
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
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
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
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
View Answer
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
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
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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