PE 2 - Functions | Algebra - Functions & Graphs
A function f is defined by f(x) = x + . Consider the following.
- (f(x))2 = f(x2) + 2
- (f(x))3 = f(x3) + 3f(x)
Which of the above is/ are correct?
- A.
1 only
- B.
2 only
- C.
Both 1 and 2
- D.
Neither 1 nor 2
Answer: Option C
Explanation :
f(x2) = x2 +
(f(x))2 = = x2 + + 2 = f(x2) + 2
∴ Statement 1 is true.
(f(x))3 = x3 + + 3 = f(x3) + 3f(x)
∴ Statement 2 is true.
Thus both 1 and 2 are correct
Hence, option (c).
Workspace:
f(x) × f(y) = f(x × y). If f(3) = 5, then what is the value of f(1/3)?
- A.
1
- B.
1/5
- C.
5
- D.
Cannot be determined
Answer: Option B
Explanation :
Given, f(x) × f(y) = f(x × y) and f(3) = 5.
Substitute y = 1, we get
f(x) × f(1) = f(x × 1)
⇒ f(x) × f(1) = f(x)
⇒ f(1) = 1
Now, substitute x = 3 and y = 1/3, we get
f(3) × f(1/3) = f(3 × 1/3)
⇒ 5 × f(1/3) = f(1) = 1
⇒ f(1/3) = 1/5
Hence, option (b).
Workspace:
If f(x) = , x ≠ 1, find f100(2). Here, fn(x) = .
- A.
2
- B.
3
- C.
4
- D.
6
Answer: Option A
Explanation :
f(2) = = 3
f2(2) = f(f(2)) = f(3) = = 2
f3(2) = f(f(f(2))) = f(f(3)) = f(2) = = 3
f4(2) = f(f(f(f(2)))) = f(3) = = 2
f5(2) = f(f(f(f(f(2))))) = f(2) = = 3
∴ f100(2) = f2(2) = 2
Hence, option (a).
Workspace:
If f(2x + 1) = 3x3 + 2x2 + x + 1, where x ∈ R, then f(1 – 2x) = ?
- A.
-3x3 - 2x2 - x - 1
- B.
-3x3 - 2x2 - x + 1
- C.
-3x3 + 2x2 - x + 1
- D.
None of these
Answer: Option C
Explanation :
Given, f(2x + 1) = 3x3 + 2x2 + x + 1
Instead of x if we substitute – x, we get
f(2(-x) + 1) = 3(-x)3 + 2(-x)2 + (-x) + 1
⇒ f(-2x + 1) = -3x3 + 2x2 - x + 1
Hence, option (c).
Workspace:
If f(x) = x – [x], where [x] represents the greatest integer less than or equal to x, then the range of f(x) is given by
- A.
[0, 1]
- B.
(0, ∞)
- C.
[0, 1)
- D.
(0, 1)
Answer: Option C
Explanation :
The function f(x) represents the fractional part of the function as it is the difference between x and the greatest integer function.
For example, if x = 0.499 ⇒ f(x) = 0.499 – 0 = 0.499
If x = 1 ⇒ f(x) = 1 – 1 = 0
∴ The value of this function will always lie between 0 and 1 for all positive and negative values of x.
The range will include 0 as this will be the value when x is an integer but will never include 1.
∴ Range of f(x) = [0, 1)
Hence, option (c).
Workspace:
If f(x − 1) + f(x + 1) = f(x) and f(2) = 7, f(0) = 2, then what is the value of f(1000)?
- A.
7
- B.
9
- C.
-9
- D.
-7
Answer: Option C
Explanation :
Given, f(x − 1) + f(x + 1) = f(x)
We can rewrite the equation as,
f(x + 1) = f(x) – f(x – 1)
Substituting x = 1, we get
f(2) = f(1) - f(0)
⇒ 7 = f(1) – 2
⇒ f(1) = 9
∴ f(0) = 2, f(1) = 9 and f(2) = 7
Now, substituting x = 2, we get
f(3) = f(2) - f(1) = 7 – 9 = -2
Similarly,
f(4) = f(3) - f(2) = -2 – 7 = -9
f(5) = f(4) – f(3) = -9 – (-2) = -7
f(6) = f(5) – f(4) = -7 – (-9) = 2 = f(0)
f(7) = f(6) – f(5) = 2 – (-7) = 9 = f(1)
f(8) = (7) – f(6) = 9 – 2 = 7 = f(2)
Hence, value of f(x) repeats after every 6 integral value of x.
∴ f(6x) = f(0) = 2
f(6x + 1) = f(1) = 9,
f(6x + 2) = f(2) = 7,
f(6x + 3) = f(3) = -2,
f(6x + 4) = f(4) = -9,
f(6x + 5) = f(5) = -7
Now, f(1000) = f(6 × 166 + 4) = f(4) = -9
Hence, option (c).
Workspace:
Let f(x) be a function such that f(x + y) = f(x) + f(y), for all real x and y. If it is given that f(11) = 55, what is f(1) + f(2) + f(3) + … + f(10)?
Answer: 275
Explanation :
Given, f(x + y) = f(x) + f(y)
Substitute y = 0
⇒ f(x + 0) = f(x) + f(0)
⇒ f(0) = 0
Now, substitute x = y = 1, we get
f(1 + 1) = f(1) + f(1)
⇒ f(2) = 2f(1)
Now, substitute x = 2 and y = 1, we get
f(2 + 1) = f(2) + f(1) = 2f(1) + f(1) = 3f(1)
∴ f(3) = 3f(1)
Now, substitute x = 3 and y = 1, we get
f(3 + 1) = f(3) + f(1) = 3f(1) + f(1) = 4f(1)
∴ f(4) = 4f(1)
By observing the pattern, we can say that
f(x) = x × f(1)
∴ f(11) = 11 × f(1) = 55
⇒ f(1) = 5
Now, we have to calculate the value of f(1) + f(2) + f(3) + … + f(10)
= f(1) + 2f(1) + 3f(1) + … + 11f(1)
= f(1) × [1 + 2 + 3 + … + 11]
= 5 × 55
= 275
Hence, 275.
Workspace:
Given f(x) = ax + 1 and g(x) = 2x + 3. If fog = gof, find a.
- A.
4/3
- B.
4
- C.
5
- D.
None of these
Answer: Option A
Explanation :
fog = f(2x + 3) = a(2x + 3) + 1 = 2ax + 3a + 1
gof = g(ax + 1) = 2(ax + 1) + 3 = 2ax + 5
⇒ 2ax + 3a + 1 = 2ax + 5
⇒ a = 4/3
Hence, option (a).
Workspace:
If f(x + 2) = 3x − 4, then what is f(x − 2)?
- A.
3x - 4
- B.
3x - 8
- C.
3x - 16
- D.
None of these
Answer: Option C
Explanation :
Given, f(x + 2) = 3x – 4
Now substitute x + 2 with x – 2 i.e., substitute x with x – 4
f(x – 4 + 2) = 3(x - 4) – 4
⇒ f(x – 2) = 3x – 16
Hence, option (c).
Workspace:
If f(x/y) = f(x) − f(y) (for y ≠ 0), then which of these is equal in value to f(2/72)?
- A.
f(6)
- B.
2f(6)
- C.
-2f(6)
- D.
-f(6)
- E.
None of these
Answer: Option C
Explanation :
f(x/y) = f(x) − f(y)
Putting x = y, we get,
f(1) = f(x) − f(x) = 0
Putting x = 1 and y = 1/x, we get,
f(x) = f(1) − f(1/x) = 0 − f(1/x)
∴ f(1/x) = −f(x)
Also, putting x = x and y = 1/x, we get,
f(x2) = f(x) − f(1/x) = f(x) − (−f(x))
∴ f(x2) = 2f(x)
Now, f(2/72) = f(1/36) = - f(36) = -f(62) = -2f(6)
Hence, option (c).
Workspace:
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