1. If f(x) = 2x + 3, find f(4).
A) 5 B) 9 C) 11 D) 14
Answer: C) 11
Explanation: Step 1: Substitute x = 4 into f(x): f(4) = 2×4 + 3.
Step 2: Calculate 2×4 = 8.
Step 3: Add 3 to get 8 + 3 = 11.
So, f(4) = 11.
2. If f(x) = x + 1, what is f(2)?
A) 1 B) 2 C) 3 D) 4
Answer: C) 3
Explanation: Step 1: Substitute x = 2 into f(x): f(2) = 2 + 1.
Step 2: Add 2 and 1 to get 3.
So, f(2) = 3.
3. Domain of f(x) = √(x - 2) is:
A) [2, ∞) B) (2, ∞) C) (-∞, 2] D) ℝ
Answer: A) [2, ∞)
Explanation: Step 1: The square root function is defined only when the expression inside is non-negative.
Step 2: Set x - 2 ≥ 0, so x ≥ 2.
Step 3: Therefore, the domain is all x such that x ≥ 2, or [2, ∞).
Number line:
────────●====================>
2
All values to the right of 2 (including 2) are allowed.
4. Range of f(x) = |x| is:
A) [0, ∞) B) (-∞, 0] C) ℝ D) [1, ∞)
Answer: A) [0, ∞)
Explanation: Step 1: The absolute value of any real number is always zero or positive.
Step 2: So, f(x) can take any value from 0 upwards.
Step 3: Therefore, the range is [0, ∞).
Number line:
0●====================>
All values from 0 to the right are possible outputs.
5. If f(x) = x² - 4 and g(x) = x + 2, find (f ∘ g)(1).
A) -3 B) 0 C) 5 D) 12
Answer: A) -3
Explanation: Step 1: Find g(1): g(1) = 1 + 2 = 3.
Step 2: Now, find f(g(1)) = f(3).
Step 3: f(3) = 3² - 4 = 9 - 4 = 5.
So, (f ∘ g)(1) = 5.
6. Inverse of f(x) = 3x - 6 is:
A) f⁻¹(x) = x/3 + 2 B) f⁻¹(x) = (x + 6)/3 C) f⁻¹(x) = 3x + 6 D) f⁻¹(x) = x/3 - 2
Answer: B) (x + 6)/3
Explanation: Step 1: Let y = 3x - 6.
Step 2: Solve for x: y + 6 = 3x ⇒ x = (y + 6)/3.
Step 3: Replace y with x for the inverse: f⁻¹(x) = (x + 6)/3.
7. Which is even?
A) x³ B) sin x C) x² + cos x D) x + 1
Answer: C) x² + cos x
Explanation: Step 1: A function is even if f(-x) = f(x).
Step 2: f(-x) = (-x)² + cos(-x) = x² + cos x = f(x).
Step 3: So, x² + cos x is even.
8. Period of f(x) = sin(3x) is:
A) π B) 2π C) 2π/3 D) π/3
Answer: C) 2π/3
Explanation: Step 1: The period of sin(bx) is 2π/b.
Step 2: Here, b = 3, so period = 2π/3.
9. If f(x) = 1/x, find f(f(x)).
A) x B) 1/x² C) 0 D) x/(x-1)
Answer: A) x
Explanation: Step 1: f(x) = 1/x.
Step 2: f(f(x)) = f(1/x) = 1/(1/x) = x.
So, f(f(x)) = x.
10. If f(x) = 2^x, then f(0) + f(1) =
A) 1 B) 2 C) 3 D) 4
Answer: C) 3
Explanation: Step 1: f(0) = 2⁰ = 1.
Step 2: f(1) = 2¹ = 2.
Step 3: Add: 1 + 2 = 3.
So, f(0) + f(1) = 3.
11. Which is one-to-one?
A) x² B) |x| C) e^x D) cos x
Answer: C) e^x
Explanation: Step 1: A function is one-to-one if every output comes from only one input.
Step 2: e^x is strictly increasing, so each value of x gives a unique output.
Step 3: Therefore, e^x is one-to-one.
12. If f(x) = x² + 2x + 1, find f(-1).
A) 0 B) 1 C) 2 D) 4
Answer: A) 0
Explanation: Step 1: Substitute x = -1 into f(x): f(-1) = (-1)² + 2×(-1) + 1.
Step 2: Calculate (-1)² = 1.
Step 3: 2×(-1) = -2.
Step 4: Add: 1 + (-2) + 1 = 0.
So, f(-1) = 0.
13. Range of f(x) = 1/(x² + 1) is:
A) (0, 1] B) [0, 1] C) (1, ∞) D) ℝ
Answer: A) (0, 1]
Explanation: Step 1: x² + 1 is always at least 1, so the smallest value of the denominator is 1.
Step 2: The largest value of f(x) is 1/(1) = 1 (when x = 0).
Step 3: As x becomes very large, x² + 1 increases, so f(x) approaches 0 but never reaches it.
Number line for range:
0(====]1
So, the range is (0, 1].
14. If f(x) = 5x - 3, solve f(x) = 12.
A) 2 B) 3 C) 4 D) 5
Answer: B) 3
Explanation: Step 1: Set f(x) = 12: 5x - 3 = 12.
Step 2: Add 3 to both sides: 5x = 15.
Step 3: Divide by 5: x = 3.
So, the solution is x = 3.
15. Which is odd?
A) x⁴ B) sin(2x) C) e^x D) cos x
Answer: B) sin(2x)
Explanation: Step 1: A function is odd if f(-x) = -f(x).
Step 2: sin(2(-x)) = sin(-2x) = -sin(2x).
Step 3: So, sin(2x) is odd.
16. If f(x) = ⎣x⎦ (floor), find f(3.7).
A) 3 B) 3.7 C) 4 D) 0
Answer: A) 3
Explanation: Step 1: The floor function ⎣x⎦ gives the greatest integer less than or equal to x.
Step 2: For x = 3.7, the greatest integer ≤ 3.7 is 3.
Number line:
... 2 3 [====) 4 ...
↑
3.7
So, f(3.7) = 3.
17. Difference quotient [f(x+h) - f(x)]/h for f(x) = x² is:
A) 2x B) 2x + h C) x + h D) h
Answer: B) 2x + h
Explanation: Step 1: f(x) = x², so f(x+h) = (x+h)² = x² + 2xh + h².
Step 2: f(x+h) - f(x) = (x² + 2xh + h²) - x² = 2xh + h².
Step 3: Divide by h: (2xh + h²)/h = 2x + h.
So, the difference quotient is 2x + h.
18. Domain of f(x) = √(9 - x²) is:
A) [-3, 3] B) (-3, 3) C) (0, 3) D) ℝ
Answer: A) [-3, 3]
Explanation: Step 1: The square root is defined when 9 - x² ≥ 0.
Step 2: 9 - x² ≥ 0 ⇒ x² ≤ 9 ⇒ -3 ≤ x ≤ 3.
Number line:
...====●=======●====...
-3 3
All x between -3 and 3 (inclusive) are allowed.
19. If f(x) = (x-1)/(x+1), find f(0).
A) -1 B) 0 C) 1 D) Undefined
Answer: A) -1
Explanation: Step 1: Substitute x = 0: f(0) = (0-1)/(0+1) = -1/1 = -1.
So, f(0) = -1.
20. Which is periodic?
A) x³ B) tan x C) e^x D) ln x
Answer: B) tan x
Explanation: Step 1: A function is periodic if it repeats its values at regular intervals.
Step 2: tan x repeats every π units.
Number line for period:
...|---π---|---π---|---π---|...
So, tan x is periodic with period π.
21. If f(x) = x², g(x) = √x, domain of g ∘ f is:
A) ℝ B) [0, ∞) C) (-∞, 0] D) ∅
Answer: B) [0, ∞)
Explanation: Step 1: f(x) = x², so f(x) is always ≥ 0 for any real x.
Step 2: g(x) = √x is defined for x ≥ 0.
Step 3: So, g(f(x)) = √(x²) is defined for all real x.
Number line for domain:
────────●====================>
0
All real x are allowed, so the domain is ℝ.
22. Inverse of f(x) = e^{2x} is:
A) (1/2)ln x B) ln(2x) C) e^{-2x} D) ln√x
Answer: A) (1/2)ln x
Explanation: Step 1: Let y = e^{2x}.
Step 2: Take the natural log of both sides: ln y = 2x.
Step 3: Divide by 2: x = (1/2)ln y.
Step 4: Replace y with x for the inverse: f⁻¹(x) = (1/2)ln x.
23. If f(x) = 1/(1-x), find f(f(f(x))).
A) x B) 1/x C) 1-x D) x/(x-1)
Answer: A) x
Explanation: Step 1: f(x) = 1/(1-x).
Step 2: f(f(x)) = 1/(1-1/(1-x)).
Step 3: f(f(f(x))) = x (after simplifying).
So, f(f(f(x))) = x (the function is an involution of order 3).
24. Range of f(x) = 2 sin x - 3 is:
A) [-5, -1] B) [-1, 5] C) [-3, 3] D) ℝ
Answer: A) [-5, -1]
Explanation: Step 1: sin x ranges from -1 to 1.
Step 2: 2 sin x ranges from -2 to 2.
Step 3: 2 sin x - 3 ranges from -2-3 = -5 to 2-3 = -1.
Number line for range:
-5●====●-1
So, the range is [-5, -1].
25. If f(x) = x³ + x, g(x) = x - 1, find (f ∘ g)(2).
A) 8 B) 9 C) 10 D) 12
Answer: C) 10
Explanation: Step 1: g(2) = 2 - 1 = 1.
Step 2: f(1) = 1³ + 1 = 2.
So, (f ∘ g)(2) = 2.
26. Function f(x) = x/|x| at x = 0 is:
A) 0 B) 1 C) -1 D) Undefined
Answer: D) Undefined
Explanation: Step 1: At x = 0, the denominator |x| = 0.
Step 2: Division by zero is undefined.
Number line:
...-1====0====1...
At x = 0, the function is undefined.
27. Which is bijective from ℝ to ℝ?
A) x² B) x³ C) |x| D) cos x
Answer: B) x³
Explanation: Step 1: A function is bijective if it is both one-to-one and onto.
Step 2: x³ is strictly increasing and covers all real numbers.
Step 3: So, x³ is bijective from ℝ to ℝ.
28. Period of f(x) = sin²x is:
A) π B) 2π C) π/2 D) 4π
Answer: A) π
Explanation: Step 1: The period of sin x is 2π.
Step 2: The period of sin²x is half that, so π.
Number line for period:
...|---π---|---π---|---π---|...
So, sin²x is periodic with period π.
29. If f(x) = (ax+b)/(cx+d), f(f(x)) = x, then:
A) a = d B) a = -d C) a = c D) b = c
Answer: B) a = -d
Explanation: Step 1: For f(f(x)) = x, the function must be an involution.
Step 2: This requires a + d = 0, so a = -d.
30. Domain of f(x) = log(log x) is:
A) (0, ∞) B) (1, ∞) C) (0, 1) D) [1, ∞)
Answer: B) (1, ∞)
Explanation: Step 1: log x is defined for x > 0.
Step 2: log(log x) is defined when log x > 0, so x > 1.
Number line for domain:
────────○====================>
1
All x greater than 1 are allowed (not including 1).
31. If f(x) = 2x + 1, find f(x + h) - f(x).
Answer: 2h
Explanation: Step 1: f(x + h) = 2(x + h) + 1 = 2x + 2h + 1.
Step 2: f(x) = 2x + 1.
Step 3: Subtract: (2x + 2h + 1) - (2x + 1) = 2h.
So, f(x + h) - f(x) = 2h.
32. Range of f(x) = x/(x² + 1) is:
A) [-1, 1] B) [-1/2, 1/2] C) ℝ D) (0, ∞)
Answer: B) [-1/2, 1/2]
Explanation: Step 1: For large |x|, x/(x² + 1) approaches 0.
Step 2: The maximum and minimum occur at x = 1 and x = -1: f(1) = 1/2, f(-1) = -1/2.
Number line for range:
-1/2●====●1/2
So, the range is [-1/2, 1/2].
33. If f(x) = 3^x, then f(x+y) =
A) f(x) + f(y) B) f(x)f(y) C) f(xy) D) f(x/y)
Answer: B) f(x)f(y)
Explanation: Step 1: f(x) = 3^x.
Step 2: f(x+y) = 3^{x+y} = 3^x × 3^y = f(x)f(y).
34. Function f(x) = x^{2/3} is:
A) Even B) Odd C) Neither D) Constant
Answer: A) Even
Explanation: Step 1: f(-x) = (-x)^{2/3} = [(-x)^2]^{1/3} = (x^2)^{1/3} = x^{2/3} = f(x).
Step 2: So, f(x) is even.
35. If f(x) = ⎡x⎤ (ceiling), find f(-0.5).
A) -1 B) 0 C) -0.5 D) 1
Answer: B) 0
Explanation: Step 1: The ceiling function ⎡x⎤ gives the smallest integer greater than or equal to x.
Step 2: For x = -0.5, the smallest integer ≥ -0.5 is 0.
Number line:
...-1(====]0 1...
↑
-0.5
So, f(-0.5) = 0.
36. Asymptote of f(x) = 2x/(x-3) is:
A) x = 3 B) y = 2 C) y = 3 D) x = 2
Answer: A) x = 3, B) y = 2
Explanation: Step 1: The vertical asymptote is where the denominator is zero: x - 3 = 0 ⇒ x = 3.
Step 2: The horizontal asymptote is the ratio of leading coefficients: y = 2.
Number line for vertical asymptote:
...====| 3 |====...
At x = 3, the function is undefined (vertical asymptote).
37. If f(x) = 1/(x-1), find f⁻¹(x).
Answer: (1/x) + 1
Explanation: Step 1: Let y = 1/(x-1).
Step 2: Solve for x: y(x-1) = 1 ⇒ x-1 = 1/y ⇒ x = 1/y + 1.
Step 3: Replace y with x for the inverse: f⁻¹(x) = (1/x) + 1.
38. Which is transcendental?
A) x² + 1 B) sin x C) √x D) |x|
Answer: B) sin x
Explanation: Step 1: A transcendental function is not algebraic (not a root of a polynomial equation with rational coefficients).
Step 2: sin x is transcendental.
39. If f(x) = {x², x < 0; x+1, x ≥ 0}, find f(-1) + f(1).
Answer: 3
Explanation: Step 1: For x = -1, use the first part: f(-1) = (-1)² = 1.
Step 2: For x = 1, use the second part: f(1) = 1 + 1 = 2.
Step 3: Add: 1 + 2 = 3.
Number line for piecewise:
...x²|0|x+1...
Left of 0: use x²; right of 0: use x+1.
40. Domain of f(x) = √(log₂ x) is:
A) (0, ∞) B) [1, ∞) C) (1, ∞) D) [2, ∞)
Answer: B) [1, ∞)
Explanation: Step 1: log₂ x is defined for x > 0.
Step 2: √(log₂ x) is defined for log₂ x ≥ 0 ⇒ x ≥ 1.
Number line for domain:
────────●====================>
1
All x ≥ 1 are allowed.
41. If f(x+1) = x² - 3x + 2, find f(3).
Answer: 0
Explanation: Step 1: Set x + 1 = 3 ⇒ x = 2.
Step 2: f(3) = 2² - 3×2 + 2 = 4 - 6 + 2 = 0.
42. Range of f(x) = sec⁻¹(x) is:
A) [0, π] B) [0, π/2) ∪ (π/2, π] C) [0, π/2] D) [-π/2, π/2]
Answer: B) [0, π/2) ∪ (π/2, π]
Explanation: Step 1: sec⁻¹(x) is defined for |x| ≥ 1.
Step 2: The range is [0, π/2) ∪ (π/2, π].
Number line for domain:
...●====-1 1====●...
sec⁻¹(x) is defined for x ≤ -1 or x ≥ 1.
43. If f(x) = ln(sin x), domain is:
A) ℝ B) (0, π/2) C) ⋃(nπ, (n+1)π) D) ⋃(2nπ, (2n+1)π)
Answer: D) ⋃(2nπ, (2n+1)π)
Explanation: Step 1: ln(sin x) is defined when sin x > 0.
Step 2: sin x > 0 in intervals (2nπ, (2n+1)π).
Number line for domain:
...|====π|====2π|====3π|...
Defined between even and odd multiples of π.
44. f(x) = (x³ - 8)/(x² - 4) at x = 2 is:
A) 0 B) 3 C) Undefined D) Limit exists
Answer: D) Limit exists
Explanation: Step 1: At x = 2, denominator is zero, so f(2) is undefined.
Step 2: Check the limit as x approaches 2: (x³ - 8)/(x² - 4) = [(x-2)(x²+2x+4)]/[(x-2)(x+2)] = (x²+2x+4)/(x+2).
Step 3: Substitute x = 2: (4+4+4)/4 = 12/4 = 3.
So, the limit exists and is 3.
Number line:
...====| 2 |====...
At x = 2, the function is undefined, but the limit exists.
45. If f(x) = e^{2x}, g(x) = ln√x, find (f ∘ g)(e²).
Answer: e²
Explanation: Step 1: g(e²) = ln√(e²) = ln(e) = 1.
Step 2: f(1) = e^{2×1} = e².
So, (f ∘ g)(e²) = e².
46. Number of solutions of f(x) = f⁻¹(x) for f(x) = 1 - x is:
A) 0 B) 1 C) 2 D) Infinite
Answer: D) Infinite
Explanation: Step 1: f(x) = 1 - x, and its inverse is also 1 - x.
Step 2: Set f(x) = f⁻¹(x): 1 - x = 1 - x, which is true for all x.
Step 3: So, there are infinitely many solutions.
Number line:
────────●====================>
-∞ ∞
All real x are solutions.
47. If f(x) = ∫₀ˣ e^{t²} dt, then f'(x) =
Answer: e^{x²}
Explanation: Step 1: By the Fundamental Theorem of Calculus, the derivative of the integral from 0 to x of e^{t²} dt is just the integrand evaluated at x.
Step 2: So, f'(x) = e^{x²}.
48. Period of f(x) = sin 3x + cos 5x is:
A) π B) 2π C) 4π D) 30π
Answer: B) 2π
Explanation: Step 1: The period of sin 3x is 2π/3, and the period of cos 5x is 2π/5.
Step 2: The period of the sum is the least common multiple (LCM) of these two periods.
Step 3: LCM of 2π/3 and 2π/5 is 2π.
Number line for period:
...|---2π---|---2π---|---2π---|...
So, the period is 2π.
49. If f(x) = (1 - x)/(1 + x), find f(f(cos 2θ)).
A) tan²θ B) cot²θ C) cos 4θ D) sin 2θ
Answer: A) tan²θ
Explanation: Step 1: f(cos 2θ) = (1 - cos 2θ)/(1 + cos 2θ).
Step 2: Use the identity 1 - cos 2θ = 2sin²θ and 1 + cos 2θ = 2cos²θ.
Step 3: f(cos 2θ) = (2sin²θ)/(2cos²θ) = tan²θ.
Step 4: f(f(cos 2θ)) = f(tan²θ).
Step 5: Substitute tan²θ into f(x): f(tan²θ) = (1 - tan²θ)/(1 + tan²θ).
Step 6: Use the identity (1 - tan²θ)/(1 + tan²θ) = cos 2ϕ if tan ϕ = tan θ, so f(f(cos 2θ)) = cos 4θ.
Number line for θ:
0●====θ====π/2
θ is between 0 and π/2 for tan²θ to be defined.
50. For f(x) = ⎣x⎦ + ⎣-x⎦, f(x) =
A) 0 B) -1 C) -x D) 0 if x ∈ ℤ, -1 otherwise
Answer: D) 0 if x ∈ ℤ, -1 otherwise
Explanation: Step 1: For integer x, ⎣x⎦ + ⎣-x⎦ = x + (-x) = 0.
Step 2: For non-integer x, ⎣x⎦ + ⎣-x⎦ = -1.
Number line:
...●---|---●---|---●...
At integer points, the value is 0; between integers, the value is -1.
51. If f(x) = |x-2| + |x-5|, what is the minimum value of f(x)?
Answer: 3
Explanation: Step 1: The sum of absolute values is minimized at the endpoints.
Step 2: Try x = 2: f(2) = 0 + 3 = 3.
Step 3: Try x = 5: f(5) = 3 + 0 = 3.
Number line:
2●====|====●5
Minimum occurs at x = 2 or x = 5.
52. If f(x) = x + 1/x, x > 0, what is the minimum value?
Answer: 2
Explanation: Step 1: By AM ≥ GM, x + 1/x ≥ 2 for x > 0.
Step 2: The minimum is at x = 1: f(1) = 1 + 1 = 2.
Number line:
0●====1====>
Minimum at x = 1.
53. If f(x) = x^2 - 6x + 10, what is the minimum value?
Answer: 1
Explanation: Step 1: The minimum of a quadratic ax² + bx + c is at x = -b/(2a).
Step 2: Here, a = 1, b = -6, so x = 6/2 = 3.
Step 3: f(3) = 9 - 18 + 10 = 1.
Number line:
...====3====...
Minimum at x = 3.
54. If f(x) = 2^x + 2^{-x}, what is the minimum value?
Answer: 2
Explanation: Step 1: By AM ≥ GM, 2^x + 2^{-x} ≥ 2 for all x.
Step 2: The minimum is at x = 0: f(0) = 1 + 1 = 2.
Number line:
...====0====...
Minimum at x = 0.
55. If f(x) = [x] + [-x], where [.] is the greatest integer, what is the range for x ∈ (0,1)?
Answer: -1
Explanation: Step 1: For 0 < x < 1, [x] = 0 and [-x] = -1.
Step 2: So, f(x) = 0 + (-1) = -1.
Number line:
0(====)1
For all x in (0,1), f(x) = -1.
56. If f(x) = x^2 for x ≤ 1, 2x-1 for x > 1, find f⁻¹(3).
Answer: x = -√3, √3, or 2
Explanation: Step 1: For x ≤ 1, set x² = 3 ⇒ x = ±√3 (but only x ≤ 1 allowed).
Step 2: For x > 1, set 2x - 1 = 3 ⇒ x = 2.
Number line for piecewise:
...x²|1|2x-1...
Left of 1: use x²; right of 1: use 2x-1.
57. If f(x) = x/(x+1), x ≠ -1, find f⁻¹(x).
Answer: x/(1-x)
Explanation: Step 1: Let y = x/(x+1).
Step 2: Solve for x: y(x+1) = x ⇒ yx + y = x ⇒ y = x - yx ⇒ y = x(1 - y) ⇒ x = y/(1 - y).
Step 3: Replace y with x for the inverse: f⁻¹(x) = x/(1-x).
58. Let f(x) = |x-3| + |x+1|. For what value of x is f(x) minimum?
Answer: x = 3
Explanation: Step 1: The sum of absolute values is minimized at the median of the points -1 and 3.
Step 2: Try x = 3: f(3) = 0 + 4 = 4.
Step 3: Try x = -1: f(-1) = 4 + 0 = 4.
Number line:
-1●====|====●3
Minimum occurs at x = -1 or x = 3.
59. If f(x) = [x] + [2x], where [.] is the greatest integer function, find the number of integers x in [0,2) for which f(x) = 2x - 1.
Answer: 1
Explanation: Step 1: Try integer values in [0,2): x = 0, 1.
Step 2: For x = 1: [1] + [2×1] = 1 + 2 = 3, 2×1 - 1 = 1.
Step 3: Only x = 1 satisfies the equation.
Number line:
0●====1====2
Only x = 1 in [0,2) works.
60. If f(x) = x/(x+1), x ≠ -1, and g(x) = 1/x, find (f ∘ g)(2).
Answer: 1/3
Explanation: Step 1: g(2) = 1/2.
Step 2: f(1/2) = (1/2)/(1/2+1) = (1/2)/(3/2) = 1/3.
So, (f ∘ g)(2) = 1/3.
61. If f(x) = x^2 for x ≤ 0, 2x+1 for x > 0, find the range of f(x).
Answer: [0, ∞)
Explanation: Step 1: For x ≤ 0, f(x) = x², which gives all values from 0 upwards.
Step 2: For x > 0, f(x) = 2x + 1, which starts at f(0) = 1 and increases.
Step 3: So, the range is [0, ∞).
Number line for range:
0●====================>
All values from 0 to infinity are possible outputs.
62. If f(x) = x^2 + 1 for x ≤ 1, 2x-1 for x > 1, find f⁻¹(3).
Answer: x = -√2, √2, or 2
Explanation: Step 1: For x ≤ 1, set x² + 1 = 3 ⇒ x² = 2 ⇒ x = ±√2 (but only x ≤ 1 allowed, so x = -√2 or √2 if √2 ≤ 1).
Step 2: For x > 1, set 2x - 1 = 3 ⇒ x = 2.
Number line for piecewise:
...x²+1|1|2x-1...
Left of 1: use x²+1; right of 1: use 2x-1.
63. If f(x) = x^2 - 2x + 2, and f(a) = f(b), show that a + b = 2.
Answer: True
Explanation: Step 1: Set f(a) = f(b): a² - 2a + 2 = b² - 2b + 2.
Step 2: Rearranged: a² - b² - 2a + 2b = 0 ⇒ (a-b)(a+b-2) = 0.
Step 3: So, either a = b or a + b = 2.
Number line:
...●====2====●...
If a ≠ b, then a + b = 2.
64. Let f(x) = x^3 - 3x + 1. Find all real solutions to f(x) = 0.
Answer: x ≈ 1.532, -1.879, 0.347
Explanation: Step 1: Set x³ - 3x + 1 = 0.
Step 2: This cubic has three real roots, which can be found using Cardano's method or numerically.
Number line:
...●-1.879●0.347●1.532...
Roots are approximately at these points.
65. Let f(x) = x^2 + 2x + 2, g(x) = x - 1, find (f ∘ g)(x).
Answer: (x-1)^2 + 2(x-1) + 2 = x^2 - 2x + 3
Explanation: Step 1: g(x) = x - 1.
Step 2: Substitute into f(x): f(g(x)) = (x-1)² + 2(x-1) + 2.
Step 3: Expand: (x-1)² = x² - 2x + 1, so f(g(x)) = x² - 2x + 1 + 2x - 2 + 2 = x² - 2x + 3.
66. Let f(x) = x^3 - 3x. Find all real roots of f(x) = 0.
Answer: x = 0, √3, -√3
Explanation: Step 1: Set x³ - 3x = 0.
Step 2: Factor: x(x² - 3) = 0.
Step 3: So, x = 0, x = √3, x = -√3.
Number line:
...●-√3●0●√3...
Roots at x = -√3, 0, √3.
67. If f(x) = e^x + e^{-x}, what is the minimum value?
Answer: 2
Explanation: Step 1: By AM ≥ GM, e^x + e^{-x} ≥ 2 for all x.
Step 2: The minimum is at x = 0: f(0) = 1 + 1 = 2.
Number line:
...====0====...
Minimum at x = 0.
68. If f(x) = tan^{-1}(x) + tan^{-1}(1/x), x > 0, what is the value of f(x)?
Answer: π/2
Explanation: Step 1: For x > 0, tan⁻¹(x) + tan⁻¹(1/x) = π/2.
Number line:
0●====>
For all x > 0, f(x) = π/2.
69. If f(x) = sin(x) + sin(2x), what is the period of f(x)?
Answer: 2π
Explanation: Step 1: The period of sin(x) is 2π, and the period of sin(2x) is π.
Step 2: The period of the sum is the least common multiple (LCM) of these two periods.
Step 3: LCM of 2π and π is 2π.
Number line for period:
...|---2π---|---2π---|---2π---|...
So, the period is 2π.
70. If f(x) = ln(x^2 + 1), what is the range of f(x)?
Answer: [0, ∞)
Explanation: Step 1: x² + 1 ≥ 1 for all real x.
Step 2: ln(x² + 1) ≥ ln(1) = 0.
Step 3: As x → ±∞, ln(x² + 1) → ∞.
Number line for range:
0●====================>
All values from 0 to infinity are possible outputs.
71. If f(x) = [x] + [-x], where [.] is the greatest integer, what is the range for x ∈ (0,1)?
Answer: -1
Explanation: Step 1: For 0 < x < 1, [x] = 0 and [-x] = -1.
Step 2: So, f(x) = 0 + (-1) = -1.
Number line:
0(====)1
For all x in (0,1), f(x) = -1.
72. If f(x) = |x| + |x-2|, what is the minimum value?
Answer: 2
Explanation: Step 1: The sum of absolute values is minimized at the median of the points 0 and 2.
Step 2: Try x = 1: f(1) = 1 + 1 = 2.
Number line:
0●====|====●2
Minimum occurs at x = 1.
73. If f(x) = x/(x+1), x ≠ -1, find f⁻¹(x).
Answer: x/(1-x)
Explanation: Step 1: Let y = x/(x+1).
Step 2: Solve for x: y(x+1) = x ⇒ yx + y = x ⇒ y = x - yx ⇒ y = x(1 - y) ⇒ x = y/(1 - y).
Step 3: Replace y with x for the inverse: f⁻¹(x) = x/(1-x).
74. Let f(x) = |x-2| + |x-5|. For what value(s) of x is f(x) minimum?
Answer: x = 2 or x = 5
Explanation: Minimum occurs at the endpoints of the interval between 2 and 5.
75. If f(x) = [x] + [-x], where [.] is the greatest integer, what is the range for x ∈ (0,1)?
Answer: -1
Explanation: [x]=0, [-x]=-1 for 0 < x < 1.
76. If f(x) = x^2 + 1, g(x) = √x, find the domain of (g ∘ f)(x).
Answer: All real x
Explanation: Step 1: f(x) = x² + 1, which is always ≥ 1 for any real x.
Step 2: g(x) = √x is defined for x ≥ 0.
Step 3: So, g(f(x)) = √(x² + 1) is defined for all real x.
Number line for domain:
────────●====================>
All real x are allowed.
77. If f(x) = x^2 + 2x + 2, is f(x) one-one? Justify.
Answer: No
Explanation: Step 1: f(x) is a quadratic function (parabola), which is not one-to-one over ℝ.
Step 2: For example, f(0) = 0² + 0 + 2 = 2 and f(-2) = (-2)² + 2×(-2) + 2 = 4 - 4 + 2 = 2.
Step 3: So, two different x values give the same output.
Number line:
...●-2====0●...
Both x = -2 and x = 0 give f(x) = 2.
78. If f(x) = x^3, what is the inverse function f⁻¹(x)?
Answer: f⁻¹(x) = x^(1/3)
Explanation: Step 1: To find the inverse, set y = x³.
Step 2: Solve for x: x = y^(1/3).
Step 3: Replace y with x for the inverse: f⁻¹(x) = x^(1/3).
79. If f(x) = x^2 + 4x + 8, what is the axis of symmetry?
Answer: x = -2
Explanation: Step 1: The axis of symmetry for a quadratic ax² + bx + c is x = -b/(2a).
Step 2: Here, a = 1, b = 4, so x = -4/2 = -2.
Number line:
...====-2====...
Axis of symmetry at x = -2.
80. If f(x) = 3x^2 - 12x + 7, what is the vertex of the parabola?
Answer: (2, -5)
Explanation: Step 1: The vertex of ax² + bx + c is at x = -b/(2a).
Step 2: Here, a = 3, b = -12, so x = 12/6 = 2.
Step 3: f(2) = 3×4 - 12×2 + 7 = 12 - 24 + 7 = -5.
Number line:
...====2====...
Vertex at x = 2.
81. Define a surjective (onto) function. Give an example from ℝ to ℝ.
Answer: A function f: A → B is surjective if every element of B has a pre-image in A. Example: f(x) = x³ is surjective from ℝ to ℝ.
Explanation: Step 1: For every y in ℝ, there exists an x in ℝ such that f(x) = y.
Step 2: For f(x) = x³, for any y, x = y^(1/3) gives f(x) = y.
Number line:
────────●====================>
All real y are covered by f(x) = x³.
82. What is an injective (one-to-one) function? Give an example from ℝ to ℝ.
Answer: A function f: A → B is injective if f(a₁) = f(a₂) ⇒ a₁ = a₂. Example: f(x) = 2x + 1 is injective from ℝ to ℝ.
Explanation: Step 1: If f(a₁) = f(a₂), then 2a₁ + 1 = 2a₂ + 1 ⇒ a₁ = a₂.
Step 2: So, f(x) = 2x + 1 is one-to-one.
Number line:
────────●====================>
Each x gives a unique output.
83. What is a bijective function? State its significance in finding inverses.
Answer: A function is bijective if it is both injective and surjective. Only bijective functions have well-defined inverses.
Explanation: Step 1: Bijective means both one-to-one and onto.
Step 2: Only bijective functions have an inverse that is also a function.
Number line:
────────●====================>
Every y in the range is hit exactly once by some x.
84. If f: ℝ → ℝ is defined by f(x) = x², is f invertible? Why or why not?
Answer: No
Explanation: Step 1: f(x) = x² is not one-to-one on ℝ (since f(2) = f(-2)).
Step 2: So, it does not have an inverse on ℝ.
Number line:
...●-2====2●...
Both x = -2 and x = 2 give f(x) = 4.
85. State the definition of a functional equation and give an example relevant to CAT exams.
Answer: A functional equation is an equation in which the unknowns are functions. Example: Find all f: ℝ → ℝ such that f(x+y) = f(x) + f(y).
Explanation: Step 1: The equation involves the function itself, not just values.
Step 2: For example, f(x+y) = f(x) + f(y) is solved by f(x) = kx.
Number line:
────────●====================>
The solution is a straight line for all x.
86. If f(x) = |x|, is f differentiable at x = 0? Justify.
Answer: No
Explanation: Step 1: The left derivative at x = 0 is -1, the right derivative is 1.
Step 2: Since the left and right derivatives are not equal, f(x) is not differentiable at x = 0.
Number line:
...====0====...
Sharp corner at x = 0.
87. If f(x) = [x], where [.] is the greatest integer function, is f continuous at x = 1? Why or why not?
Answer: No
Explanation: Step 1: At x = 1, the function jumps from 0 to 1.
Step 2: So, f(x) is not continuous at x = 1.
Number line:
...0[====)1[====)...
Jump at x = 1.
88. If f(x) = x² for x ≤ 0, f(x) = 2x+1 for x > 0, is f continuous at x = 0?
Answer: Yes
Explanation: Step 1: Left limit at x = 0: f(0-) = 0² = 0.
Step 2: Right limit at x = 0: f(0+) = 2×0 + 1 = 1.
Step 3: Since left and right limits are not equal, f(x) is not continuous at x = 0.
Number line:
...x²|0|2x+1...
Jump at x = 0.
89. If f(x) = x/(x+1), x ≠ -1, is f invertible? If yes, find f⁻¹(x).
Answer: Yes, f⁻¹(x) = x/(1-x)
Explanation: Step 1: Let y = x/(x+1).
Step 2: Solve for x: y(x+1) = x ⇒ yx + y = x ⇒ y = x - yx ⇒ y = x(1 - y) ⇒ x = y/(1 - y).
Step 3: Replace y with x for the inverse: f⁻¹(x) = x/(1-x).
90. If f(x) = sin x, what is the period of f(x)?
Answer: 2π
Explanation: Step 1: The period of sin x is 2π.
Number line for period:
...|---2π---|---2π---|---2π---|...
So, the period is 2π.
91. If f(x) = x² + 1, g(x) = √x, find the domain of (g ∘ f)(x).
Answer: All real x
Explanation: Step 1: f(x) = x² + 1, which is always ≥ 1 for any real x.
Step 2: g(x) = √x is defined for x ≥ 0.
Step 3: So, g(f(x)) = √(x² + 1) is defined for all real x.
Number line for domain:
────────●====================>
All real x are allowed.
92. If f(x) = x² - 2x + 2, and f(a) = f(b), show that a + b = 2.
Answer: True
Explanation: f(a) = f(b) ⇒ (a-b)(a+b-2) = 0 ⇒ a = b or a + b = 2.
93. If f(x) = x² + 2x + 2, is f(x) one-one? Justify.
Answer: No
Explanation: Quadratic functions are not one-one over ℝ.
94. If f(x) = x³, what is the inverse function f⁻¹(x)?
Answer: f⁻¹(x) = x^(1/3)
Explanation: Cube root is the inverse of cube.
95. If f(x) = x² + 4x + 8, what is the axis of symmetry?
Answer: x = -2
Explanation: Axis is x = -b/2a = -2.
96. If f(x) = 3x² - 12x + 7, what is the vertex of the parabola?
Answer: (2, -5)
Explanation: Vertex at x = -b/2a = 2, f(2) = -5.
97. If f(x) = |x-1| + |x-2| + |x-3|, what is the minimum value?
Answer: 2
Explanation: Minimum at x = 2 (median).
98. If f(x) = x² - 2x + 2, what is the minimum value?
Answer: 1
Explanation: Vertex at x = 1, f(1) = 1.
99. If f(x) = x² + 2x + 2, what is the minimum value?
Answer: 1
Explanation: Vertex at x = -1, f(-1) = 1.
100. If f(x) = x² - 4x + 5, what is the minimum value?
Answer: 1
Explanation: Vertex at x = 2, f(2) = 1.