CAT — Functions

Step 1: Total number of functions
Each of the 6 elements of A can be mapped to any of the 3 elements of
B.
Total functions = 3⁶ = 729

Step 2: Subtract functions where one element of B is not used
Choose the missing element of B in 3 ways.
The remaining 2 elements are available for all 6 elements of
A.
Number of such functions = 3 × 2⁶ = 3 × 64 = 192
Remaining functions = 729 − 192 = 537

Step 3: Add back functions where two elements of B are not used
Choose the two missing elements in 3 ways.
In each case, all 6 elements map to the only remaining element.
Number of such functions = 3 × 1 = 3

Hence, required number of onto functions = 537 + 3 = 540

Case 1 (3 ≤ x < 4): [x] = 3, [x]² = 9. Need [x²] = 9 → 9 ≤ x² < 10 → x ∈ [3, √10).

Case 2 (4 ≤ x < 5): [x] = 4, [x]² = 16. Need [x²] = 16 → x ∈ [4, √17).

Case 3 (5 ≤ x < 6): [x] = 5, [x]² = 25. Need [x²] = 25 → x ∈ [5, √26).

Case 4 (x = 6): [x²] = [36] = 36 = [x]² = 36. So x = 6 is included.

Complete set S = [3, √10) ∪ [4, √17) ∪ [5, √26) ∪ {6}. Option A, (3, √10) ∪ [5, √26) ∪ {6}, is a valid subset. Options B and C incorrectly include √10 and √17/√26 (endpoints not in S). Option D contains values outside S.

Given:
A function f satisfies
f(xy) = f(x)f(y) + f(x) + f(y)
for all natural numbers x and y.
Also,
f(p) = 1 for every prime number p.
Find the value of
f(160000).

Step 1: Define a new function
Let
g(x) = f(x) + 1
Then,
g(xy)
= f(xy) + 1
= f(x)f(y) + f(x) + f(y) + 1
= (f(x) + 1)(f(y) + 1)
= g(x)g(y)
Thus,
g(xy) = g(x)g(y)

Step 2: Find g for prime numbers
Since
f(p) = 1,
we have
g(p) = 2
for every prime p.

Step 3: Prime factorize 160000
160000
= 160 × 1000
= (2⁵ × 5) × (2³ × 5³)
= 2⁸ × 5⁴
Since g is multiplicative,
g(160000)
= g(2)⁸ × g(5)⁴
= 2⁸ × 2⁴
= 2¹²
= 4096

Step 4: Find f(160000)
f(160000)
= g(160000) − 1
= 4096 − 1
= 4095

Answer:
A. 4095

Step 1: Determine the ranges
⌊√n⌋ = 1 for n = 1 to 3 (3 values)
⌊√n⌋ = 2 for n = 4 to 8 (5 values)
⌊√n⌋ = 3 for n = 9 to 15 (7 values)
⌊√n⌋ = 4 for n = 16 to 24 (9 values)
⌊√n⌋ = 5 for n = 25 to 35 (11 values)
⌊√n⌋ = 6 for n = 36 to 48 (13 values)
⌊√n⌋ = 7 for n = 49 to 50 (2 values)

Step 2: Compute the sum
= (1×3) + (2×5) + (3×7) + (4×9) + (5×11) + (6×13) + (7×2)
= 3 + 10 + 21 + 36 + 55 + 78 + 14
= 217

Step 1: Square both sides: f(x) + 1 = 9701²
f(x) = 9701² − 1 = 9700 × 9702

Step 2: Let y = x² + 3x.
y(y + 2) = 9700 × 9702
y² + 2y + 1 = 9701²
(y + 1)² = 9701²
y + 1 = ±9701 → y = 9700 or y = −9702

Step 3: Solve for x
Case 1: x² + 3x − 9700 = 0 → Discriminant = 9 + 38800 = 38809 = 197²
x = (−3 ± 197)/2 → x = 97 or x = −100

Case 2: x² + 3x + 9702 = 0 → Discriminant = 9 − 38808 < 0 (no real roots)

Step 4: Sum of real roots = 97 + (−100) = −3

Step 1: Let y = (2x − 3)/(2x² + 4x − 6).
Rearrange: 2yx² + (4y − 2)x + (3 − 6y) = 0

Step 2: For real x, discriminant ≥ 0.
(4y − 2)² − 4(2y)(3 − 6y) ≥ 0
16y² − 16y + 4 − 24y + 48y² ≥ 0
64y² − 40y + 4 ≥ 0
16y² − 10y + 1 ≥ 0

Step 3: Factor: (8y − 1)(2y − 1) ≥ 0
This holds when y ≤ 1/8 or y ≥ 1/2.

Step 4: Range = (−∞, 1/8] ∪ [1/2, ∞)

Step 1: Find the domain of f(x) and g(x). f(x) = x/(2x − 1) is undefined when 2x − 1 = 0 → x = 1/2. Hence, x ≠ 1/2. g(x) = x/(x − 1) is undefined when x − 1 = 0 → x = 1. Hence, x ≠ 1.

Step 2: Find the domain of f(g(x)). For f(g(x)) to exist: (1) g(x) must be defined → x ≠ 1. (2) g(x) ≠ 1/2. Solving g(x) = 1/2: x/(x − 1) = 1/2 → 2x = x − 1 → x = −1. So, x ≠ −1. Therefore, f(g(x)) is defined only when x ≠ 1 and x ≠ −1.

Step 3: Find the domain of g(f(x)). For g(f(x)) to exist: (1) f(x) must be defined → x ≠ 1/2. (2) f(x) ≠ 1. Solving f(x) = 1: x/(2x − 1) = 1 → x = 2x − 1 → x = 1. So, x ≠ 1. Therefore, g(f(x)) is defined only when x ≠ 1/2 and x ≠ 1.

Step 4: Find the domain of h(x). Since h(x) = f(g(x)) + g(f(x)), both expressions must be defined simultaneously. Hence, x cannot be −1, 1/2, or 1.

Given:
For every non-zero real number x,
f(x) + 2f(1/x) = 3x
Find the sum of all possible values of x for which
f(x) = 3.

Step 1: Form another equation
Replace x by 1/x.
Then,
f(1/x) + 2f(x) = 3/x
We now have the two equations:
f(x) + 2f(1/x) = 3x
2f(x) + f(1/x) = 3/x

Step 2: Solve for f(x)
Multiply the first equation by 2,
2f(x) + 4f(1/x) = 6x
Subtract the second equation,
3f(1/x) = 6x − 3/x
f(1/x) = 2x − 1/x
Substitute into the first equation,
f(x) + 2(2x − 1/x) = 3x
f(x) = 3x − 4x + 2/x
f(x) = 2/x − x

Step 3: Use the given condition
Since
f(x) = 3,
2/x − x = 3
Multiply throughout by x,
2 − x² = 3x
x² + 3x − 2 = 0

Step 4: Find the sum of the solutions
For the quadratic,
x² + 3x − 2 = 0,
the sum of the roots is
−3

Final Answer
Answer:
C. −3