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4. (b) Define even and odd functions with examples. Decide whether the equation x2
+ y2 = 1 is a function? Explain.
There are different types of functions in mathematics that we study. We can determine
whether a function is even or odd algebraically or graphically. Even and Odd functions
can be checked by plugging in the negative inputs (-x) in place of x into the function f(x)
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5. and considering the corresponding output value. Even and odd functions are classified on
the basis of their symmetry relations. Even and odd functions are named based on the fact
that the power function f(x) = xn is an even function, if n is even, and f(x) is an odd
function if n is odd.
Generally, we consider a real-valued function to be even or odd. To identify if a function
is even or odd, we plug in -x in place of x into the function f(x), that is, we check the
output value of f(-x) to determine the type of the function. Even and odd functions are
symmetrical. Let us first understand their definitions.
Even and Odd Functions Definition
Even Function - For a real-valued function f(x), when the output value of f(-x) is
the same as f(x), for all values of x in the domain of f, the function is said to be
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6. an even function. An even function should hold the following equation: f(-x) = f(x),
for all values of x in D(f), where D(f) denotes the domain of the function f. In other
words, we can say that the equation f(-x) - f(x) = 0 holds for an even function, for
all x. Let us consider an example, f(x) = x2.
f(-x) = (-x)2 = x2 for all values of x, as the square of a negative number is the same
as the square of the positive value of the number. This implies f(-x) = f(x), for all x.
Hence, f(x) = x2 is an even function. Similarly, functions like x4, x6, x8, etc. are even
functions.
Odd Function - For a real-valued function f(x), when the output value of f(-x) is
the same as the negative of f(x), for all values of x in the domain of f, the function
is said to be an odd function. An odd function should hold the following equation:
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7. f(-x) = -f(x), for all values of x in D(f), where D(f) denotes the domain of the
function f. In other words, we can say that the equation f(-x) + f(x) = 0 holds for an
odd function, for all x. Let us consider an example, f(x) = x3.
f(-x) = (-x)3 = -(x3) for all values of x, as the cube of a negative number is the same
as the negative of the cube of the positive value of the number. This implies f(-x) =
-f(x), for all x. Hence, f(x) = x3 is an odd function. Similarly, functions like x5, x7,
x9 etc. are odd functions.
Both Even and Odd Functions - A real-valued function f(x) is said to be both
even and odd if it satisifies f(-x) = f(x) and f(-x) = -f(x) for all values of x in
the domain of the function f(x). There is only one function which is both even and
odd and that is the zero function, f(x) = 0 for all x. We know that for zero function,
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8. f(-x) = -f(x) = f(x) = 0, for all values of x. Hence, f(x) = 0 is an even and odd
function.
Neither Even Nor Odd Function - A real-valued function f(x) is said to be neither
even nor odd if it does not satisfy f(-x) = f(x) and f(-x) = -f(x) for atleast one value
of x in the domain of the function f(x). Let us consider an example to understand
the definition better. Consider f(x) = 2x5 + 3x2 + 1, f(-x) = 2(-x)5 + 3(-x)2 + 1 = -
2x5 + 3x2 + 1 which is neither equal to f(x) nor -f(x). Hence, f(x) = 2x5 + 3x2 + 1 is
neither even nor odd function.
In the first quadrant (where x and y coordinates are all positive), all six trigonometric
ratios have positive values. In the second quadrant, only sine and cosecant are positive. In
the third quadrant, only tangent and cotangent are positive. In the fourth quadrant, only
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9. cosine and secant are positive. Based on these signs, we will categorize them as even and
odd functions.
If a trigonometric ratio is even or odd can be checked through a unit circle. An angle
measured in anticlockwise direction is a positive angle whereas the angle measured in the
clockwise direction is a negative angle.
The graph of an even function is symmetric with respect to the y-axis. In other words,
the graph of an even function remains the same after reflection about the y-axis. For any
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10. two opposite input values of x, the function value will remain the same all along the
curve.
Whereas the graph of an odd function is symmetric with respect to the origin. In other
words, the graph of an odd function is at the same distance from the origin but in opposite
directions. For any two opposite input values of x, the function has opposite y values.
Here are a few examples of even and odd functions.
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11. The graph of an even function is symmetric with respect to the y-axis. In other words,
the graph of an even function remains the same after reflection about the y-axis. For any
two opposite input values of x, the function value will remain the same all along the
curve.
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12. Whereas the graph of an odd function is symmetric with respect to the origin. In other
words, the graph of an odd function is at the same distance from the origin but in opposite
directions. For any two opposite input values of x, the function has opposite y values.
Here are a few examples of even and odd functions.
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13. x 3
f (x)
x
.
(c) Find the domain of the function
Interval Notation:
(−∞,3)∪(3,∞)(-∞,3)∪(3,∞)
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14. Set-Builder Notation:
{x|x≠3}
Q. 2 (a) If A is a square matrix, then show that A +At is symmetric.
Let B=A+AT.
Now,
BT
=(A+AT)T
=AT+(AT)T
=AT+A [ Since (AT)T=A]
=B.
So, A+AT is a symmetric matrix.
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15. (b) Determine the solution of the system of equations by
(i) Matrix method
(ii) Cramer’s rule: 10x1
+ 4x2 = 46
–5x1 + 6x2 = 9
Let
x1 = x
x2 = y
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26. Q. 4 (a) Find two numbers whose sum and product are 14 and 45 respectively.
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28. (b) Find the market equilibrium point for the following supply and demand
functions:
Demand: P = –3q + 26
Supply: P = 4q – 9
Demand = Supply
–3q + 26 = 4q – 9
-3q – 4q = -26 – 9
-7q = -33
q = 33 / 7 = 4.71
p = -3(33/7) + 26
p = -99/7 + 26
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29. p = -14.14 + 26 = 12
p = 11.9
(c)Find the x and y intercept if:
y = x2 – 2x – 8
Let x = 0
Y = -8
y-intercept = (0, -8)
Let y = 0
0 = x2 – 2x – 8
x2 – 2x – 8 = 0
x2 – 4x + 2x – 8 = 0
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30. x (x-4) + 2(x-4) = 0
(x-4) (x+2) = 0
X = 4, -2
X intercept = (4,0) and (-2, 0)
List the intersections.
x-intercept(s): (4,0) ,(-2,0)
y-intercept(s): (0, −8)
Q. 5 (a) Simplify the following in binary system:
i) {[10001101)2 x (2335)10} – (278)10
= {[10001101)2 x (100100011111)2} - (100010110)2
First First we will solve [10001101)2 x (100100011111)2}
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