PE 1 - Functions | Algebra - Functions & Graphs
f(x) = 3[x] - 1 and g(x) = {3x – 2} + 4, where [x] is the greatest integer less than or equal to x and {x} is the fractional part of x. Find f(x) + g(x) for all integer values of x.
- (a)
3x + 5
- (b)
3x + 3
- (c)
3x + 1
- (d)
None of these
Answer: Option B
Explanation :
f(x) = 3[x] - 1
As x is an integer,
[x] = x
∴ f(x) = 3[x] - 1 = 3x - 1
Also, g(x) = {3x – 2} + 4
Since x is an integer, (3x – 2) is also an integer.
∴ Fractional part of (3x – 2) = 0
∴ f(x) + g(x) = 3x - 1 + 4 = 3x + 3
Hence, option (b).
Workspace:
Let y = maximum {(2x + 3), (8 - 3x)}. Find the minimum value of y.
- (a)
3
- (b)
5
- (c)
7
- (d)
Cannot be determined
Answer: Option B
Explanation :
In the given function, y will be minimum when
2x + 3 = 8 – 3x
⇒ x = 1
∴ minimum value of y = 2 × 1 + 3 = 5
Hence, option (b).
Workspace:
If y = f(x) = , then which of the following is true.
- (a)
x = f(2y)
- (b)
f(2x) = f(x) - 1
- (c)
x × f(y) = -1
- (d)
None of these
Answer: Option C
Explanation :
We need to check options in this question.
Option (c):
f(y) = = = =
⇒ xf(y) = -1
Hence, option (c).
Workspace:
A function can sometimes reflect on itself, i.e. if y = f(x), then x = f(y). Both of them retain the same structure and form. Which of the following functions has this property?
- (a)
y =
- (b)
y =
- (c)
y =
- (d)
y =
Answer: Option D
Explanation :
In general, any function of the form y = reflects on to itself as we can rearrange it as x = .
Hence, option (d).
Workspace:
If f (x) = 2x + 1 and g(x) = (x - 1)/2, then what is the value of fo(fog)o(gof)(x)?
- (a)
x
- (b)
x2
- (c)
2x + 1
- (d)
(x - 1)/2
Answer: Option C
Explanation :
Here
fog(x) = 2 × + 1 = x …(1)
gof(x) = = x ...(2)
∴ fo(fog)o(gof)(x) = fo(fog)(x) = f(x) = 2x + 1
Hence, option (c).
Workspace:
If f(x) = x2 – 2x and g(x) = 2x + 3, then the minimum value of f(g(x)) - 2x is
- (a)
-1
- (b)
5
- (c)
1/5
- (d)
None of these
Answer: Option B
Explanation :
f(g(x)) = (2x + 3)2 – 2x
⇒ f(g(x)) - 2x = (2x + 3)2 – 2x - 2x
⇒ f(g(x)) - 2x = 4x2 + 8x + 9
The given expression is quadratic. The least value of a quadratic expression ax2 + bx + c occurs at x = - b/2a
∴ Minimum value of f(g(x)) occurs at x = - 8/8 = -1
∴ Minimum value of f(g(x)) is = 4 × (-1)2 + 8 × -1 + 9 = 4 - 8 + 9 = 5
Hence, option (b).
Workspace:
100 chocolates are distributed amongst 11 students such that first child gets x1 chocolates, second child gets x2 chocolates and so on. If x0 = max(x1, x2, …., x11), then the smallest possible value of x0 is
Answer: 10
Explanation :
x0 will be least when maximum out of x1, x2, …, x11 is least.
∴ We need to distribute chocolates such that the person receiving maximum chocolates gets the least number of chocolates. This is possible when everyone gets (almost) equal number of chocolates.
99 chocolates can be distributed equally such that each child gets 9 chocolates.
100 chocolates can be distributed such that 10 children get 9 chocolates each and the 11th child gets 10 chocolates.
∴ x0 = max(x1, x2, …., x11) = max(9, 9, …, 10) = 10
Least possible value of x0 = 10
Hence, 10.
Workspace:
If f(x) = xx-2, then find f(f(f(3)))
- (a)
0
- (b)
1/3
- (c)
2/3
- (d)
None of these
Answer: Option D
Explanation :
f(3) = 33-2 = 3
∴ f(f(3)) = f(3) = 33-2 = 3
∴ f(f(f(3))) = f(3) = 33-2 = 3
Hence, option (d).
Workspace:
Functions f, g and h are defined for x, y and z as follows, where x, y and z are non-zero and unequal numbers.
f(x, y, z) = Minimum of x, y and z
g(x, y, z) =
h(x, y, z) = Minimum of f(x, y, z) and g(x, y, z)
How many of the following statements are true?
- h(a, b, c) = f(a, b, c) where a, b, c > 0
- h(a, b, c) = g(a, b, c) where a, b, c < 0
- h(a, b, c) = g(a, b, c) where |a|, |b|, |c| < 1
- h(|a|, |b|, |c|) = g(|a|, |b|, |c|) where |a|, |b|, |c| < 1
Answer: 1
Explanation :
Statement 1: Let a, b, c > 0 such that a < b < c.
∴ f(a, b, c) = minimum of a, b, c = a
Also, g(a, b, c) = = = c
∴ h(a, b, c) = minimum (a, c) = a
∴ Statement I is correct.
Statement 2: Let a, b, c < 0 such that a < b < c.
∴ f(a, b, c) = minimum of a, b, c = a
Also, g(a, b, c) = = = c
∴ h(a, b, c) = minimum (a, c) = a
∴ Statement II is incorrect.
Statement 3: Let |a|, |b|, |c| < 1 such that a < b < c.
∴ f(a, b, c) = minimum of a, b, c = a
Also, g(a, b, c) =
Here, g(a, b, c) can be any of a or b or c, depending on their signs.
∴ h(a, b, c) = minimum (a, g(a, b, c)) = a
∴ Statement III is incorrect.
Statement 4: Let 0 < |a|, |b|, |c| < 1 such that |a| < |b| < |c|.
∴ f(|a|, |b|, |c|) = minimum of |a|, |b|, |c| = |a|
Also, g(|a|, |b|, |c|) = = = |c|
∴ h(|a|, |b|, |c|) = minimum (|a|, |c|) = |a|
∴ Statement IV is incorrect.
∴ Only 1 statement, i.e., 1st statement is correct.
Hence, 1.
Workspace:
For a function f(x) = x5 – 2x4 + 3x3 – 4x2 – x + a, > 0. What is the range of values of a?
- (a)
a ∈ (3, 9)
- (b)
a ∈ [3, 9)
- (c)
a ∈ (-∞, 3) ∪ (9, ∞)
- (d)
a ∈ [3, 9]
- (e)
Cannot be determined
Answer: Option D
Explanation :
f(x) = x5 – 2x4 + 3x3 – 4x2 – x + a
f(1) = 1 - 2 + 3 - 4 - 1 + a = a – 3
f(-1) = -1 – 2 – 3 – 4 + 1 + a = a – 9
∴ > 0
This is possible when either
Case 1: (a - 3) and (a - 9) both are positive
i.e., a > 9
Case 2: (a - 3) and (a - 9) both are negative
i.e., a < 3.
∴ a ∈ (-∞, 3) ∪ (9, ∞)
Hence, option (d).
Workspace:
Feedback
Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.