CRE 2 - Even / Odd Functions | Algebra - Functions & Graphs
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Answer the next 5 questions based on the information given below.
Determine which of the following functions are Even or odd or neither.
Type 1: if the function is even.
Type 2: if the function is odd.
Type 3: if the function is neither even nor odd.
f(x) = 2x2.
Answer: 1
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Explanation :
A function is
Condition 1: Even when, f(-x) = f(x)
Condition 2: Odd when, f(-x) = -f(x)
Here, f(x) = 2x2 and
f(-x) = 2(-x)2 = 2x2
∴ f(-x) = f(x)
f(x) satisfies the 1st condition; hence it is an even function.
Hence, 1.
Workspace:
f(x) = 3x3.
Answer: 2
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Explanation :
A function is
Condition 1: Even when, f(x) = f(-x)
Condition 2: Odd when, f(x) = -f(-x)
Here, f(x) = 3x3 and
f(-x) = 3(-x)3 = -3x3
∴ f(-x) = -f(x)
f(x) satisfies the 2nd condition; hence it is an odd function.
Hence, 2.
Workspace:
f(x) = 2x2 + 3x3.
Answer: 3
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Explanation :
A function is
Condition 1: Even when, f(x) = f(-x)
Condition 2: Odd when, f(x) = -f(-x)
Here, f(x) = 2x2 + 3x3 and
f(-x) = 2(-x)2 + 3(-x)3 = 2x2 -3x3
f(x) satisfies neither of the two conditions; hence it is neither an even nor an odd function.
[Note: When an even and an odd function are added / subtracted, the resulting function is neither even nor odd.]
Hence, 3.
Workspace:
f(x) = 2x2 + |x|.
Answer: 1
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Explanation :
A function is
Condition 1: Even when, f(-x) = f(x)
Condition 2: Odd when, f(-x) = -f(x)
Here, f(x) = 2x2 + |x| and
f(-x) = 2(-x)2 + |-x| = 2x2 + |x|
∴ f(-x) = f(x)
f(x) satisfies the 1st condition; hence it is an even function.
[Note: When two even functions are added / subtracted, the resulting function will also be even.]
Hence, 1.
Workspace:
f(x) = 3x3 + x.
Answer: 2
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Explanation :
A function is
Condition 1: Even when, f(x) = f(-x)
Condition 2: Odd when, f(x) = -f(-x)
Here, f(x) = 3x3 + x and
f(-x) = 3(-x)3 + (-x) = -3x3 - x
∴ f(-x) = -f(x)
f(x) satisfies the 2nd condition; hence it is an odd function.
[Note: When two odd functions are added / subtracted, the resulting function will also be odd.]
Hence, 2.
Workspace:
If the function f(x) satisfies the relation f(x + y) + f(x - y) = 2f(x)f(y), such that x and y belong to the set of real numbers, and f(0) ≠ 0, then f(x) is
- (a)
Even
- (b)
Odd
- (c)
Neither even nor odd
- (d)
Both even as well as odd
Answer: Option A
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Explanation :
A function is
Condition 1: Even when, f(x) = f(-x)
Condition 2: Odd when, f(x) = -f(-x)
Given, f(x + y) + f(x - y) = 2f(x)f(y)
Put x = y = 0, we get
f(0) + f(0) = 2f(0)f(0) = 2(f(0))2
⇒ 2f(0) = 2(f(0))2
⇒ f(0) = 1.
Now substituting x = 0, we get
f(y) + f(-y) = 2f(0)f(y)
⇒ f(-y) = f(y)
∴ f(x) satisfies the 1st condition, ; hence it is an even function.
Hence, option (a).
Workspace:
If f(x) is an even function, then the graph y=f(x) will be symmetrical about
- (a)
x - axis
- (b)
y - axis
- (c)
Both the axes
- (d)
None of these
Answer: Option B
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Explanation :
A function is an even function when f(x) = f(-x) i.e., the graph of the function is symmetric around y-axis.
Hence, option (b)
Workspace:
If f(x) = 3xn and g(x) = 5x(n+1), where n is a natural number. Find f(x).g(x) is which kind of function?
- (a)
Even function
- (b)
Odd function
- (c)
Neither Even nor Odd function
- (d)
Cannot be determined
Answer: Option B
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Explanation :
f(x).g(x) = 15x(2n+1),
Now, 2n + 1 is an odd number when n is any natural number.
∴ f(-x).g(-x) = 15(-x)(2n+1) = -15(x)(2n+1)
∴ 15(-x)(2n+1) is an odd function.
Hence, option (b).
Workspace:
Let f(x) = |x - 3| + |x - 4| + |x - 5| and g(x) = [f(x)]2. Then
- (a)
One of f(x) or g(x) is an even function
- (b)
One of f(x) or g(x) is an odd function
- (c)
f(x) and g(x) are both either even or odd
- (d)
None of these
Answer: Option D
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Explanation :
Let us calculate values of f(x) and g(x) for x = 1 and -1.
f(1) = |1 - 3| + |1 - 4| + |1 - 5|
= 2 + 3 + 4 = 9
∴ g(1) = 92 = 81
f(-1) = |-1 - 3| + |-1 - 4| + |-1 - 5|
= 4 + 5 + 6 = 15
∴ g(1) = 152 = 225
Since f(1) ≠ ± f(-1)
⇒ f(x) is neither even nor odd function
Also. Since g(1) ≠ ± g(-1)
⇒ g(x) is neither even nor odd function
Hence, option (d).
Workspace:
Which of the following functions is an odd function?
- (a)
- (b)
- (c)
Both (a) and (b)
- (d)
Neither (a) nor (b)
Answer: Option D
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Explanation :
Consider f(x) = |xx|
f(2) = 22 = 4, and
f(-2) = -2-2 = -1/4
∵ f(2) ≠ f(-2) hence, f(x) is neither even nor odd.
⇒ is neither even nor odd
Now, consider g(x) = |x||x|
g(-x) = |-x||-x| = |x||x|
∴ g(x) is an even function.
⇒ is also an even function
Hence, option (d).
Workspace:
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