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 (x

^{2}-5x=6). When we factorize this, we get,

(x

^{2}-5x+6)=x

^{2}-2x-3x+6

(x

^{2}-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 (x

^{2}-6x+8). When we factorize this, we get,

(x

^{2}-6x+8)=x

^{2}-2x-4x+8

(x

^{2}-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 (x

^{2}-5x). When we factorize this, we get,

(x

^{2}-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 (x

^{2}-6x). When we factorize this, we get,

(x

^{2}-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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