Complex Analysis Multiple Choice Questions 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)
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.
2. The function f(x, y)=(frac{2xy-3x}{3x-2y}) is continuous at (1, 1).
a) True
b) False
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
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
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
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.
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
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