CRE 5 - Domain & Range | Algebra - Functions & Graphs
test
Answer: 1
Explanation :
Workspace:
test
Answer: 1
Explanation :
Workspace:
How many integral values can x take if y =
Answer: 1
Explanation :
Since y is square root of
⇒ ≥ 0
⇒ ≥ 1
⇒ x2 – 6x + 9 ≤ 0
⇒ (x – 3)2 ≤ 0
This is only possible when x = 3.
Hence, 1.
Workspace:
Find the domain of the definition of the function y = +
- A.
x > 7
- B.
4 ≤ x < 5
- C.
Bothe (a) and (b)
- D.
None of these
Answer: Option C
Explanation :
Find the domain of the definition of the function y = +
For the function to exist, the argument of the logarithmic function should be positive.
∴ > 0
⇒ > 0
⇒ x ∈ (3, 5) ∪ (7, ∞)
Also, (x - 4) ≥ 0 should also be true.
⇒ x ∈ [4, ∞)
∴ x ∈ [4, 5) ∪ (7, ∞)
Hence, option (c).
Workspace:
What is the range of the function f(x) = , x ≠ 0?
- A.
Any real number
- B.
Any integer
- C.
{-1, 1}
- D.
{-1, 0, 1}
Answer: Option C
Explanation :
As we know |x| =
∴ f(x) = =
=
∴ Range = {-1, 1}.
Hence, option (c).
Workspace:
Let f(x) = |x| be a function with domain [-6, 4] and let g(x) = |2x + 3|, Then the domain of (fog)(x) is
- A.
[-6, 4]
- B.
[-7/2, 1/2]
- C.
[-7/2, -1/2]
- D.
None of these
Answer: Option B
Explanation :
(fog)(x) = f[g(x)] = f(|2x + 3|)
Since the domain of f(x) is [-6, 4]
⇒ -6 ≤ |2x + 3| ≤ 4
Since |2x + 3| ≥ 0
⇒ |2x + 3| ≤ 4
⇒ -4 ≤ 2x + 3 ≤ 4
⇒ -7 ≤ 2x ≤ 1
⇒ -3.5 ≤ x ≤ 0.5
∴ Domain of fog(x) is [-7/2, 1/2]
Hence, option (b).
Workspace:
The domain of the function f(x)= is
- A.
x ∈ (-∞, -2) ∪ (5, 7) ∪ (7, ∞)
- B.
x ∈ (-∞, -2] ∪ [5, 7) ∪ (7, ∞)
- C.
x ∈ (-∞, -2) ∪ (5, ∞)
- D.
None of these
Answer: Option B
Explanation :
We are given
∴ (x + 2)(x - 5) ≥ 0
⇒ x ∈ (-∞, -2] ∪ [5, ∞)
Also, the denominator is (x - 7)
∴ x – 7 ≠ 0
⇒ x ≠ 7
∴ Domain is x ∈ (-∞, -2] ∪ [5, 7) ∪ (7, ∞)
Hence, option (b).
Workspace:
If f(x) = log(x2) and g(x) = 2 × log(x), then the domain for which f(x) and g(x) are identical ?
- A.
(-∞, ∞)
- B.
[0, ∞)
- C.
(0, ∞)
- D.
None of these
Answer: Option C
Explanation :
Two functions f(x) and g(x) are identical when,
⇒ Domain of f(x) = Domain of g(x) = X
⇒ f(x) = g(x) for all x ∈ X
Here,
Domain f(x) is R - {0} i.e.,(-∞, 0) ∪ (0, ∞) and
Domain g(x) is R+ i.e.,(0, ∞)
∴ Common domain of f(x) and g(x) is (0, ∞)
Hence, if x∈(0, ∞), then f(x) = g(x)
Hence, option (c).
Workspace:
Which of the following pairs are identical?
- A.
f(x) = , g(x) =
- B.
f(x) = , g(x) =
- C.
Both (a) and (b)
- D.
None of these
Answer: Option B
Explanation :
Option (a):
Domain of f(x) = is R i.e.,x ∈ (-∞,∞)
Domain of g(x) = is R+ ∪ {0} i.e., x ∈ [0, ∞)
∴ ≠
Option (b):
Domain of f(x) = is R - {0} and
Domain of g(x) = is R – {0}
Also, f(x) = g(x) for x ∈ R – {0}
Hence, option (b).
Workspace:
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