Algebra - Functions & Graphs - Previous Year CAT/MBA Questions
You can practice all previous year OMET questions from the topic Algebra - Functions & Graphs. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.
Consider the function f(x) = (x + 4)(x + 6)(x + 8) ⋯ (x + 98). The number of integers x for which f(x) < 0 is:
- A.
48
- B.
26
- C.
49
- D.
24
- E.
23
Answer: Option D
Explanation :
(x + 4)(x + 6)(x + 8) ⋯ (x + 98) < 0
Critical points are -98, -96, -94, … -8. -6 and -4
For x > -4, the given expression will be positive. Hence, we will reject all values of x > -4.
For -6 < x < -4, the given expression will be negative. The only integral value for x in this range is -5.
For -8 < x < -6, the given expression will be positive.
For -10 < x < -8, the given expression will be negative. The only integral value for x in this range is -9.
The given expression will be negative for x = -5, -9, -13, and so on
Now the only integral value of x between -98 and -96 is -97
∴ f(-97) = -93 × - 91 × -89 × … × -1 × 1
Here we have product of 47 negative terms and 1 positive term, hence the product will be negative.
∴ For -98 < x < -96, the given expression will be negative. The only integral value for x in this range is -97.
Finally, for x < -98, f(x) is product of 48 negative terms, which will give us a positive number. Hence, we will reject all value of x < -98.
The given expression will be negative for x = -5, -9, -13, and so on till -97
Number of possible values of x for which f(x) < 0 = (-5 + 97)/4 + 1 = 24
Hence, option (d).
Workspace:
Given P(x,y) = x2 + xy + y2; Q(x,y) = x2 – xy + y2. Find the value of P(7, Q(9,4))
- A.
4169
- B.
4197
- C.
4089
- D.
4127
Answer: Option B
Explanation :
Q(9, 4) = (9)2 – (9)(4) + (4)2
= 81 – 36 + 16 = 61
∴ P(7,Q(9,4)) = P(7,61)
= (7)2 + (7)(61) + (61)2
= 49 + 427 + 3721 = 4197
Hence, option 2.
Workspace:
If f(x) = ax + b, a and b are positive real numbers and if f(f(x)) = 9x + 8, then the value of a + b is:
- A.
3
- B.
4
- C.
5
- D.
6
- E.
None of these
Answer: Option C
Explanation :
f(x) = ax + b
∴ f(f(x) = a(ax + b) + b
f(f(x) = a2x + ab + b
⇒ a2x + b(a + 1) = 9x + 8
∴ a2 = 9 ⇒ a = 3. and
b(a + 1) = 8
⇒ b(3 + 1) = 8
⇒ b = 2
Thus, a + b = 5
Hence, option (c).
Workspace:
f f(x) = 1/(1 + x), then find the value of f[f{f(x)}] at x = 5.
- A.
7/9
- B.
7/13
- C.
5/13
- D.
5/9
Answer: Option B
Explanation :
f(5) = 1/(1 + 5) = 1/6
f[f(5)] = f(1/6) = 1/[1 + (1/6)] = 6/7
f[f{f(5)}] = f(6/7) = 1/[1 + (6/7)] = 7/13
Hence, option 2.
Workspace:
f is a function for which f(1) = 1 and f(x) = 2x + f(x – 1ؘ) for each natural number x ≥ 2. Find f(31).
- A.
869
- B.
929
- C.
951
- D.
991
- E.
None of the above
Answer: Option D
Explanation :
Given function can be rearranged as,
f(x) – f(x – 1) = 2x
⇒ f(2) – f(1) = 2 × 2
⇒ f(3) – f(2) = 2 × 3
⇒ f(4) – f(3) = 2 × 4
…
⇒ f(31) – f(30) = 2 × 31
Adding all these equations we get
f(31) – f(1) = 2 × (2 + 3 + 4 + … + 31)
f(31) – 1 = 2 × ((31×32)/2-1) = 990
⇒ f(31) = 991
Alternately,
From the given function ;
f(2) = 4 + f(1) = 5 = 22 + 1
f(3) = 6 + f(2) = 6 + 5 = 11 = 32 +2
f(4) = 8 + f(3) = 8 + 11 = 19 = 42 + 3
Hence we can see that f(x) = x2 + (x-1)
Hence f(31) = 312 + 30 = 991.
Hence, option (d).
Workspace:
Find the equation of the graph shown below.
- A.
= 3x – 4
- B.
y = 2x2 – 40
- C.
x = 2y2 – 40
- D.
y = 2x2 + 3x – 19
- E.
x = 2y2 + 3y – 19
Answer: Option E
Explanation :
From the graph, observe that at y = 0, x
= –19
Only x = 2y2 + 3y – 19 satisfies the above criteria.
Hence, option 5.
Workspace:
If f(x2 – 1) = x4 – 7x2 + k1 and f(x3 – 2) = x6 – 9x3 + k2 then the value of (k2 – k1) is
- A.
6
- B.
7
- C.
8
- D.
9
- E.
None of the above
Answer: Option C
Explanation :
Substitute x = 0 in f(x2 – 1) = x4 – 7x2 + k1
⇒ f(–1) = k1 … (i)
Substitute x = 1 in f(x3 – 2) = x6 – 9x3 + k2
⇒ f(–1) = 1 – 9 + k2 … (ii)
From (i) and (ii), k1 = –8 + k2
∴ k2 – k1 = 8
Hence, option 3.
Workspace:
For a positive integer x, define f(x) such that f(x + a) = f(a × x), where a is an integer and f(1) = 4. If the value of f(1003) = k, then the value of ‘k’ will be:
- A.
1003
- B.
1004
- C.
1005
- D.
1006
- E.
None of the above
Answer: Option E
Explanation :
f(1) = 4
f(2) = f(1+1) = f(1 × 1) = f(1) = 4
…
Assume that f(n – 1) = 4
f(n) = f(1 + (n – 1)) = f(1 × (n – 1)) = f(n – 1) = 4
Thus, by induction principle, f(n) = 4, for all n.
∴ f(1003) = k = 4
Hence, option 5.
Workspace:
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