CRE 1 - Basics (Functions) | Algebra - Functions & Graphs
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If f(x,y) = x2 - y3, then f(3, -2) =
Answer: 17
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Explanation :
Given, f(x,y) = x2 - y3
Putting x = 3 and y = -2, we get
f(3, -2) = (3)2 – (-2)3
⇒ f(3, -2) = 9 + 8 = 17.
Hence, 17.
Workspace:
A function f is defined such that it satisfies f(x) = f(x – 1) – f(x – 2). If x is a natural number and f(1) = 0, f(0) = 1, then the value of f(5) is equal to
- (a)
0
- (b)
1
- (c)
-1
- (d)
2
Answer: Option B
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Explanation :
Given, f(x) = f(x – 1) – f(x – 2)
f(0) = 1 and f(1) = 0
Putting x = 2, f(2) = f(1) – f(0) = 0 – 1 = -1
Putting x = 3, f(3) = -1 – 0 = -1
Putting x = 4, f(4) = -1 – (-1) = 0
Putting x = 5, f(5) = 0 – (-1) = 1
Hence, option (b).
Workspace:
If g is a function such that g(0) = 2, g(1) = 3 and g(x + 2) = 2g(x) – g(x + 1) then g(5) is
- (a)
13
- (b)
-3
- (c)
-1
- (d)
-2
Answer: Option A
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Explanation :
Given, g(x + 2) = 2g(x) – g(x + 1)
g(0) = 2 and g(1) = 3
Putting x = 0, g(2) = 2g(0) – g(1) = 4 – 3 = 1
Putting x = 1, g(3) = 6 – 1 = 5
Putting x = 2, g(4) = 2 – 5 = -3
Putting x = 3, g(5) = 10 – (-3) = 13
Hence, option (a).
Workspace:
A function f (x) is defined, satisfying the relation f(x + y) = f(x) × f (y) for every real value of x and y ; Given f (1) = 2 and f(1) + f(2) + f(3) + ……f(n) = 510. The value of n is
- (a)
5
- (b)
6
- (c)
7
- (d)
8
Answer: Option D
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Explanation :
Given, f(x + y) = f(x) × f (y) and f(1) = 2.
Putting x = y = 1, f(2) = f(1) × f(1) = 22
Putting x = 1 and y = 2, f(3) = f(1) × f(2) = 23
… and so on.
We can see that f(x) = 2x.
Now, we have, f(1) + f(2) + f(3) + ……f(n) = 510
⇒ 21 + 22 + 23 + … + 2n = 510
⇒ 2n+1 – 2 = 510
⇒ 2n+1 = 512 = 29
⇒ n + 1 = 9
⇒ n = 8.
Hence, option (d).
Workspace:
Given, f(1) = 1 and n × f(x) = f(nx), then f(1) + f(2) + f(3) + … + f(100) =
- (a)
5100
- (b)
4950
- (c)
5050
- (d)
None of these
Answer: Option C
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Explanation :
Given, f(1) = 1 and f(nx) = n × f(x)
Putting n = 2 and x = 1, we get f(2) = 2 × f(1) = 2
Putting n = 3 and x = 1, we get f(3) = 3 × f(1) = 3
…
Putting n = 100 and x = 1, we get f(100) = 100 × f(1) = 100.
∴ f(1) + f(2) + f(3) + … + f(100) = 1 + 2 +3 + … + 100 = 5050.
Hence, option (c).
Workspace:
f(x) = x(x + 1) ; x = 1,2,3, ........ Find the value of S = f(1) + f(2) + f(3) + ............. + f(10).
- (a)
438
- (b)
455
- (c)
440
- (d)
490
Answer: Option C
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Explanation :
Given, f(x) = x(x + 1)
f(1) + f(2) + f(3) + ............. + f(10) = Σx(x + 1) = Σx2 + Σx
= (12 + 22 + 32 + … + 102) + (1 + 2 +3 + … + 10)
= 385 + 55 = 440.
Hence, option (c).
Workspace:
The function f(x) is defined as f(xy) = f(x) + f(y). Also, f(2) = 3 and f(3) = 2. Find the value of f(16/27).
- (a)
0
- (b)
6
- (c)
8
- (d)
None of these
Answer: Option B
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Explanation :
Logarithm is a function which satisfies f(xy) = f(x) + f(y) i.e.,
logx(mn) = logx(m) + logx(n)
∴ f(2) = 3 = logx(2), and
f(3) = 2 = logx(3)
Now, f(16/27) = logx(16/27) = logx(16) - logx(27)
= 4logx(2) - 3logx(3)
= 4 × 3 – 3 × 2 = 12 – 6 = 6
Alternately,
Given, f(xy) = f(x) + f(y)
f(2) = 3 and f(3) = 2
putting x = 2 and y = 2, we get
f(4) = f(2) + f(2) = 6.
putting x = 4 and y = 4, we get
f(16) = f(4) + f(4) = 12.
Similarly, f(27) = 6.
Now putting x = y = 1, we get
f(1) = f(1) + f(1) ⇒ f(1) = 0
Now, putting x = 27 and y = 1/27
f(1) = f(27) + f(1/27)
⇒ 0 = 6 + f(1/27)
⇒ f(1/27) = -6
Now, putting x = 16 and y = 1/27, we get
f(16/27) = f(16) + f(1/27) = 12 – 6 = 6.
Hence, option (b).
Workspace:
Answer the next 3 questions based on the information given below:
lt(x, y) = Least of (x, y)
md(x) = |x|
ma(x, y) = Maximum of (x, y)
Find the value of ma(x + md(lt(x, y)), md(x + ma(md(x), md(y))), at x = -2 and y = -3.
- (a)
1
- (b)
0
- (c)
5
- (d)
3
Answer: Option A
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Explanation :
Putting x = –2 and y = –3, we get
ma(–2 + md(lt(–2, –3)), md(–2 + ma(md(–2), md(–3)))
= ma(–2 + md(–3), md(–2 + ma(2, 3)))
= ma(–2 + 3, md(–2 + 3))
= ma(1, md(1))
= ma(1, 1) = 1.
Hence, option (a).
Workspace:
Which of the following must always be correct for x, y > 0?
- (a)
md(lt(x, y)) ≥ (ma(md(x), md(y))
- (b)
md(lt(x, y)) > (ma(md(x), md(y))
- (c)
md(lt(x, y)) < (lt(md(x), md(y))
- (d)
md(lt(x,y)) = lt(md(x), md(y))
Answer: Option D
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Explanation :
The best way to approach such questions is by verifying each option by putting few values.
Let us put x = 2, y = 3.
Option (a):
LHS = md(lt(2, 3)) = md(2) = 2.
RHS = (ma(md(2), md(3)) = (ma(2, 3)) = 3.
∴, LHS < RHS.
Option (b):
LHS = md(lt(2, 3)) = md(2) = 2.
RHS = (ma(md(2), md(3)) = (ma(2, 3)) = 3.
Hence, LHS < RHS.
Option (c):
LHS = md(lt(2, 3)) = md(2) = 2.
RHS = (lt(md(2), md(3)) = lt(2, 3) = 2.
Hence, LHS = RHS.
Option (d):
LHS = md(lt(2, 3)) = md(2) = 2.
RHS = (lt(md(2),md(3)) = lt(2, 3) = 2.
Hence, LHS = RHS.
∴ Only option (d) is correct.
Hence, option (d).
Workspace:
For what values of ‘x’ is ma(x2 - 3x, x - 3) < 0?
- (a)
x > 3
- (b)
0 < x < 3
- (c)
x < 0
- (d)
x = 3
Answer: Option B
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Explanation :
Let us verify by taking arbitrary values of x in the range specified.
Option (a): x > 3. Let x = 4.
∴ ma(x2 – 3x, x – 3) = ma(4,1) = 4 > 0.
Option (b): 0 < x < 3. Let x = 2.
∴ ma(x2 – 3x, x – 3) = ma(–2, –1) = –1 < 0.
Option (c): x < 0. Let x = –1.
∴ ma(x2 – 3x, x – 3) = ma(4, –4) = 4 > 0.
Option (d): x = 3,
∴ ma(x2 – 3x, x– 3) = ma(0, 0) = 0.
Only for option (b) we get ma(x2 - 3x, x - 3) < 0.
Hence, option (b).
Workspace:
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