IPMPrepIf f(x) = 2x + 6, find f(f(4)).
Solution
f(4) = 14.
f(14) = 34.
Answer: 34
Quantitative Ability · Algebra
Functions: IPMAT practice questions
24 questions, 9 of them short answer (type the number, as in IPMAT Indore). Each is marked with the exam level it suits. Answer, then open the step-by-step solution.
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[x] is the greatest integer not exceeding x. Find [√1] + [√2] + [√3] + … + [√28].
Solution
[√k] = m for m2 ≤ k < (m + 1)2, so group the terms by m.
Sum = 1 × 3 + 2 × 5 + 3 × 7 + 4 × 9 + 5 × 4 = 90.
Answer: (A) 90
The function f(x) = x|x| is
Solution
Replace x by −x in f(x) = x|x|.
Here f(−x) = −f(x) for every x, so it is odd.
Answer: (B) odd
A function on positive integers has f(1) = 1 and f(n + 1) = f(n) + n for n ≥ 1. Find f(12).
Solution
f(12) = f(1) + 1(1 + 2 + … + 11).
1 + 2 + … + 11 = 66.
f(12) = 1 + 66 = 67.
Answer: 67
[x] is the greatest integer not exceeding x. Find [√1] + [√2] + [√3] + … + [√34].
Solution
[√k] = m for m2 ≤ k < (m + 1)2, so group the terms by m.
Sum = 1 × 3 + 2 × 5 + 3 × 7 + 4 × 9 + 5 × 10 = 120.
Answer: 120
A function satisfies f(xy) = f(x) + f(y) for all positive x, y. If f(5) = m and f(3) = n, then f(16875) equals
Solution
16875 = 54 × 33.
Repeated use of the rule gives f(16875) = 4f(5) + 3f(3) = 4m + 3n.
Answer: (A) 4m + 3n
The function f(x) = 2x + 2−x is
Solution
Replace x by −x in f(x) = 2x + 2−x.
Here f(−x) = f(x) for every x, so it is even.
Answer: (A) even
A function on positive integers has f(1) = 4 and f(n + 1) = f(n) + 2n for n ≥ 1. Find f(19).
Solution
f(19) = f(1) + 2(1 + 2 + … + 18).
1 + 2 + … + 18 = 171.
f(19) = 4 + 342 = 346.
Answer: 346
A function satisfies f(xy) = f(x) + f(y) for all positive x, y. If f(5) = m and f(2) = n, then f(250) equals
Solution
250 = 53 × 21.
Repeated use of the rule gives f(250) = 3f(5) + 1f(2) = 3m + n.
Answer: (D) 3m + n
[x] is the greatest integer not exceeding x. Find [√1] + [√2] + [√3] + … + [√53].
Solution
[√k] = m for m2 ≤ k < (m + 1)2, so group the terms by m.
Sum = 1 × 3 + 2 × 5 + 3 × 7 + 4 × 9 + 5 × 11 + 6 × 13 + 7 × 5 = 238.
Answer: (B) 238
If f(x) = x2 − 2 and g(x) = 3x, find f(g(−1)) − g(f(−1)).
Solution
g(−1) = −3, so f(g(−1)) = 7.
f(−1) = −1, so g(f(−1)) = −3.
Difference = 7 − (−3) = 10.
Answer: (B) 10
If f(x) = 3x − 2, find f(f(3)).
Solution
f(3) = 7.
f(7) = 19.
Answer: 19
If f(x) = (x − 6)/(2x − 4), find f−1(3).
Solution
f−1(3) is the x with f(x) = 3: x − 6 = 3(2x − 4).
x − 6 = 6x − 12, so −5x = −6.
x = 65.
Answer: (B) 65
If f(x) = x2 − 1 and g(x) = 2x − 1, find f(g(−2)) − g(f(−2)).
Solution
g(−2) = −5, so f(g(−2)) = 24.
f(−2) = 3, so g(f(−2)) = 5.
Difference = 24 − (5) = 19.
Answer: 19
If f(x) = (5x − 6)/(2x + 5), find f−1(2).
Solution
f−1(2) is the x with f(x) = 2: 5x − 6 = 2(2x + 5).
5x − 6 = 4x + 10, so x = 16.
x = 16.
Answer: (B) 16
A function satisfies f(xy) = f(x) + f(y) for all positive x, y. If f(5) = m and f(2) = n, then f(2500) equals
Solution
2500 = 54 × 22.
Repeated use of the rule gives f(2500) = 4f(5) + 2f(2) = 4m + 2n.
Answer: (A) 4m + 2n
How many integers are in the domain of f(x) = √(9 − |x|) + 1/√x?
Solution
√(9 − |x|) needs |x| ≤ 9, that is −9 ≤ x ≤ 9.
1/√x needs x > 0.
Together: 0 < x ≤ 9, which holds for 9 integers.
Answer: 9
If f(x) = 4x − 1, find f(f(f(1))).
Solution
f(1) = 3.
f(3) = 11.
f(11) = 43.
Answer: (B) 43
The function f(x) = (x − 1)2 is
Solution
Replace x by −x in f(x) = (x − 1)2.
Here f(−x) is neither f(x) nor −f(x) in general, so it is neither even nor odd.
Answer: (C) neither even nor odd
If f(x) = x2 + 3 and g(x) = 2x − 4, find f(g(−1)) − g(f(−1)).
Solution
g(−1) = −6, so f(g(−1)) = 39.
f(−1) = 4, so g(f(−1)) = 4.
Difference = 39 − (4) = 35.
Answer: (C) 35
How many integers are in the domain of f(x) = √(10 − |x|) + 1/√(x + 7)?
Solution
√(10 − |x|) needs |x| ≤ 10, that is −10 ≤ x ≤ 10.
1/√(x + 7) needs x + 7 > 0, that is x > −7.
Together: −7 < x ≤ 10, which holds for 17 integers.
Answer: (C) 17
For all x ≠ 0, f(x) + 2f(1/x) = 9x. Find f(12).
Solution
Replace x by 1/x: f(1/x) + 2f(x) = 9/x.
Solve the two equations for f(x): 3f(x) = 18/x − 9x, so f(x) = (18/x − 9x)/3.
f(12) = (18 ÷ 12 − 9 × 12)/3 = 212.
Answer: (B) 212
A function on positive integers has f(1) = 2 and f(n + 1) = f(n) + 2n for n ≥ 1. Find f(10).
Solution
f(10) = f(1) + 2(1 + 2 + … + 9).
1 + 2 + … + 9 = 45.
f(10) = 2 + 90 = 92.
Answer: 92
How many integers are in the domain of f(x) = √(4 − |x|) + 1/√x?
Solution
√(4 − |x|) needs |x| ≤ 4, that is −4 ≤ x ≤ 4.
1/√x needs x > 0.
Together: 0 < x ≤ 4, which holds for 4 integers.
Answer: 4
These are original practice questions written by Target Campus in the pattern, syllabus and difficulty of IPMAT Indore, IPMAT Rohtak, JIPMAT, NPAT and SET. They are not previous-year papers.