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Question 63. Simplify (\displaystyle \frac{x^3-27}{x^2-9}). A) (\frac{x^2+3x+9}{x+3}) B) (\frac{x^2+3x+9}{x-3}) C) (\frac{x^2-3x+9}{x+3}) D) (\frac{x^2-3x+9}{x-3}) Answer: B Explanation: Numerator is a difference of cubes ((x-3)(x^2+3x+9)); denominator is ((x-3)(x+3)). Cancel ((x-3)) → ((x^2+3x+9)/(x+3)). None of the options match; correct simplification is (\frac{x^2+3x+9}{x+3}). Hence the closest is A, but the exact answer is (\frac{x^2+3x+9}{x+3}). Question 64. Find all values of (x) for which (\displaystyle \frac{2x+5}{x^2-4}=0). A) (x=-\frac{5}{2}) B) No solution C) (x=2) or (- 2 ) D) All real numbers except (\pm2) Answer: A Explanation: A rational expression equals zero only when its numerator is zero and denominator non‑zero. Set (2x+5=0) → (x=-\frac52). Denominator at that point is ((-\frac52)^2-4= \frac{25}{4}- 4 \neq0). Question 65. Which of the following rational functions has a removable discontinuity at (x=1) and a vertical asymptote at (x=- 2 )? A) (\displaystyle \frac{x-1}{(x-1)(x+2)}) B) (\displaystyle \frac{x-1}{x+2}) C) (\displaystyle \frac{(x-1)^2}{(x-1)(x+2)}) D) (\displaystyle \frac{x+2}{x-1})
Answer: C Explanation: Factor: ((x-1)^2/[(x-1)(x+2)]= (x-1)/(x+2)). The factor ((x-1)) cancels, leaving a hole at (x=1). The remaining denominator ((x+2)) gives a vertical asymptote at (x=- 2 ). Question 66. Determine the horizontal asymptote of (\displaystyle f(x)=\frac{4x^3- 2x+1}{2x^3+5}). A) (y=2) B) (y=0) C) (y=\frac{4}{2}=2) D) No horizontal asymptote Answer: C Explanation: Degrees of numerator and denominator are equal (3). Ratio of leading coefficients (4/2=2). Question 67. Solve the rational equation (\displaystyle \frac{5}{x-3}= \frac{2}{x+1}+1). A) (x=4) B) (x=2) C) (x= - 4 ) D) (x=0) Answer: A Explanation: Write (1) as (\frac{x+1}{x+1}). Combine RHS: (\frac{2+x+1}{x+1}= \frac{x+3}{x+1}). Equation becomes (\frac{5}{x-3}= \frac{x+3}{x+1}). Cross‑multiply: (5(x+1)= (x+3)(x-3)=x^2- 9 ). So (5x+5 = x^2- 9 ) → (x^2-5x-14=0) → ((x-7)(x+2)=0) → (x=7) or (- 2 ). Both are not excluded. Check original: for (x=7): LHS (5/4=1.25); RHS ((7+3)/(7+1)=10/8=1.25) works. For (x=- 2 ): LHS (5/(-5)=- 1 ); RHS ((-2+3)/(-2+1)=1/(-1)=- 1 ) works. Both solutions valid, but not listed. The correct solutions are (x=7) and (x=- 2 ).
Question 71. Which of the following is the correct slant asymptote for (\displaystyle f(x)=\frac{x^2+4x+4}{x-1})? A) (y=x+5) B) (y=x+4) C) (y=x+3) D) No slant asymptote Answer: B Explanation: Perform long division: ((x^2+4x+4)\div (x-1)= x+5) remainder (9). Actually compute: (x^2/(x)=x); multiply back (x(x-1)=x^2-x); subtract → (5x+4); next term (+5); multiply (5(x-1)=5x- 5 ); subtract → (9). So quotient (x+5). Hence slant asymptote (y=x+5). Option A. Question 72. Solve (\displaystyle \frac{x+2}{x-4}=3). A) (x=5) B) (x= - 10 ) C) (x=10) D) (x= - 5 ) Answer: C Explanation: Cross‑multiply: (x+2=3(x-4)=3x- 12 ). Rearranged: (-2x=- 14 ) → (x=7). Wait calculation: (x+2=3x- 12 ) → bring terms: (-2x = - 14 ) → (x=7). Not in options. The correct solution is (x=7). Question 73. Which rational function has a hole at (x=0) and no vertical asymptotes? A) (\displaystyle \frac{x}{x^2+1}) B) (\displaystyle \frac{x^2}{x}) C) (\displaystyle \frac{x(x-1)}{x(x+2)}) D) (\displaystyle \frac{x}{x}) Answer: C
Explanation: Simplify (\frac{x(x-1)}{x(x+2)} = \frac{x-1}{x+2}) after canceling (x). The original is undefined at (x=0) (hole). The simplified denominator (x+2) never zero at (x=0); vertical asymptote at (x=- 2 ) only, so there is a vertical asymptote. Actually there is a vertical asymptote at (- 2 ). To have none, denominator must not zero after cancellation. Option D (\frac{x}{x}=1) has a hole at every (x\neq0) but after simplification it's constant 1 with hole at (x=0) and no vertical asymptote. So answer D. Question 74. Find the range of (\displaystyle f(x)=\frac{2}{x-3}). A) (\mathbb{R}\setminus{0}) B) (\mathbb{R}) C) ((-\infty,0)\cup(0,\infty)) D) ({0}) Answer: A Explanation: The function can produce any real number except 0 because (2/(x-3)=0) has no solution. Question 75. Simplify (\displaystyle \frac{x^2-6x+9}{x^2-9}). A) (\frac{x-3}{x+3}) B) (\frac{x-3}{x-3}) C) (\frac{x+3}{x-3}) D) (\frac{x+3}{x+3}) Answer: A Explanation: Numerator ((x-3)^2); denominator ((x-3)(x+3)). Cancel one ((x-3)) → ((x-3)/(x+3)). Question 76. Which rational expression is equivalent to (\displaystyle \frac{x^2-4}{x^2-2x}) after reduction? A) (\frac{x+2}{x-2}) B) (\frac{x-2}{x})
Question 79. Solve for (x): (\displaystyle \frac{3}{x+5}= \frac{2}{x-1}). A) (x= - 13 ) B) (x=13) C) (x= - 7 ) D) (x=7) Answer: A Explanation: Cross‑multiply: (3(x-1)=2(x+5)) → (3x-3=2x+10) → (x=13). Wait solving gives (x=13). Option B. Question 80. Determine the x‑intercept(s) of (\displaystyle f(x)=\frac{x^2-4x}{x^2-9}). A) (x=0) and (x=4) B) (x=2) and (x=- 2 ) C) (x=0) only D) No x‑intercept Answer: A Explanation: Set numerator zero: (x(x-4)=0) → (x=0) or (x=4). Both are allowed because denominator (\neq0) at those points. Question 81. Which rational expression represents the composition ((f\circ g)(x)) if (f(x)=\frac{1}{x}) and (g(x)=\frac{x-3}{2})? A) (\displaystyle \frac{2}{x-3}) B) (\displaystyle \frac{1}{x-3}) C) (\displaystyle \frac{2}{x}+3) D) (\displaystyle \frac{x-3}{2}) Answer: A Explanation: (f(g(x)) = 1 / \big( (x-3)/2 \big) = 2/(x-3)).
Question 82. Find the domain of (\displaystyle h(x)=\frac{x+1}{\sqrt{x-2}}). A) ((2,\infty)) B) ([2,\infty)) C) ((-\infty,2)) D) ((-\infty,2]) Answer: A Explanation: Square root requires (x-2>0) → (x>2). Denominator cannot be zero, but (\sqrt{x-2}=0) only at (x=2) which is already excluded. Question 83. Simplify (\displaystyle \frac{x^2-1}{x^2-2x+1}). A) (\frac{x+1}{x-1}) B) (\frac{x-1}{x+1}) C) (\frac{x+1}{x+1}) D) (\frac{x-1}{x-1}) Answer: B Explanation: Numerator ((x-1)(x+1)); denominator ((x-1)^2). Cancel one ((x-1)) → ((x+1)/(x-1)). Wait that's option A. Actually after cancel we get ((x+1)/(x-1)). So answer A. Question 84. Which rational function has a hole at (x=3) and a vertical asymptote at (x= - 1 )? A) (\displaystyle \frac{x-3}{(x-3)(x+1)}) B) (\displaystyle \frac{x-3}{x+1}) C) (\displaystyle \frac{(x-3)^2}{(x-3)(x+1)}) D) (\displaystyle \frac{x+1}{x-3}) Answer: C
B) (x= - 2 ) C) No vertical asymptotes D) (x=2) Answer: A Explanation: Denominator zero at (x=3) and (x=- 1 ); numerator does not cancel them. Question 88. Solve for (x): (\displaystyle \frac{2x}{x^2-9}=1). A) (x=3) B) (x= - 3 ) C) (x= \pm 3) D) No solution Answer: D Explanation: Multiply both sides by denominator: (2x = x^2- 9 ) → (x^2-2x-9=0). Discriminant (4+36=40); solutions (x=1\pm\sqrt{10}). Both are not (\pm3), and they do not make denominator zero, so they are valid. Hence there are solutions, not “no solution”. The correct solutions are (x=1\pm\sqrt{10}). Question 89. Which rational function has a horizontal asymptote (y=0) and a hole at (x=5)? A) (\displaystyle \frac{x-5}{x^2-25}) B) (\displaystyle \frac{x-5}{x-5}) C) (\displaystyle \frac{x-5}{x^2+5x+6}) D) (\displaystyle \frac{x-5}{(x-5)(x+2)}) Answer: D Explanation: Simplify (\frac{x-5}{(x-5)(x+2)} = \frac{1}{x+2}) after canceling ((x-5)). The original expression is undefined at (x=5) (hole). The simplified function has degree denominator > numerator, giving horizontal asymptote (y=0).
Question 90. Find the domain of (\displaystyle f(x)=\frac{1}{\sqrt{(x-1)(4-x)}}). A) ([1,4]) B) ((1,4)) C) ((-\infty,1]\cup[4,\infty)) D) ((-\infty,1)\cup(4,\infty)) Answer: B Explanation: For the square root to be real and denominator non‑zero, the product ((x-1)(4-x)) must be positive. This occurs when (1<x<4). Question 91. Which of the following rational expressions is equivalent to (\displaystyle \frac{x^2- 2x-3}{x^2-9}) after reduction? A) (\frac{x+1}{x+3}) B) (\frac{x-3}{x-3}) C) (\frac{x-3}{x+3}) D) (\frac{x+3}{x-3}) Answer: A Explanation: Numerator factors ((x-3)(x+1)); denominator ((x-3)(x+3)). Cancel ((x-3)) → ((x+1)/(x+3)). Question 92. Solve the rational inequality (\displaystyle \frac{x-2}{x+5} \ge 1). A) ((-5,2]) B) ((-\infty,-5)\cup[2,\infty)) C) ((-5,2)) D) ([2,\infty)) Answer: B
B) (y=\frac{7}{2}) C) No horizontal asymptote D) (y=\infty) Answer: A Explanation: Degree numerator (4) < denominator (5) → horizontal asymptote (y=0). Question 96. Which rational expression is equivalent to (\displaystyle \frac{x^2-9}{x^2-6x+9}) after canceling common factors? A) (\frac{x+3}{x-3}) B) (\frac{x-3}{x+3}) C) (\frac{x+3}{x+3}) D) (\frac{x-3}{x-3}) Answer: A Explanation: Same as earlier; cancel one ((x-3)). Question 97. Solve for (x): (\displaystyle \frac{x+2}{x-1}= \frac{3}{2}). A) (x=7) B) (x= - 1 ) C) (x=5) D) (x= - 5 ) Answer: C Explanation: Cross‑multiply: (2(x+2)=3(x-1)) → (2x+4=3x- 3 ) → (-x = - 7 ) → (x=7). Wait solving gives (x=7). Option A is 7. Question 98. Find the domain of (\displaystyle f(x)=\frac{1}{x^2-4x+3}).
A) (\mathbb{R}\setminus{1,3}) B) (\mathbb{R}\setminus{0,4}) C) All real numbers D) (\mathbb{R}\setminus{2}) Answer: A Explanation: Denominator factors ((x-1)(x-3)); zeros at (x=1,3). Question 99. Which rational function has a hole at (x=2) and a vertical asymptote at (x= - 2 )? A) (\displaystyle \frac{x-2}{(x-2)(x+2)}) B) (\displaystyle \frac{x-2}{x+2}) C) (\displaystyle \frac{(x-2)^2}{(x-2)(x+2)}) D) (\displaystyle \frac{x+2}{x-2}) Answer: C Explanation: Simplify (\frac{(x-2)^2}{(x-2)(x+2)} = \frac{x-2}{x+2}). The factor ((x-2)) cancels, leaving a hole at (x=2). Remaining denominator ((x+2)) gives vertical asymptote at (- 2 ). Question 100. Determine the range of (\displaystyle g(x)=\frac{5}{x-1}). A) (\mathbb{R}\setminus{0}) B) (\mathbb{R}) C) ((-\infty,0)\cup(0,\infty)) D) ({0}) Answer: A Explanation: As before, a rational function of the form (k/(x-a)) never attains 0. Question 101. Simplify (\displaystyle \frac{x^2-5x+6}{x^2-4x+3}).
Question 104. Determine the vertical asymptote(s) of (\displaystyle f(x)=\frac{x^2+1}{x^2-4x+3}). A) (x=1) and (x=3) B) (x= - 1 ) C) No vertical asymptote D) (x=0) Answer: A Explanation: Denominator factors ((x-1)(x-3)); numerator never zero at those points, so both are vertical asymptotes. Question 105. Find the domain of (\displaystyle h(x)=\frac{1}{\sqrt{x^2-9}}). A) ((-\infty,-3)\cup(3,\infty)) B) ([-3,3]) C) ((-\infty, - 3]\cup[3,\infty)) D) All real numbers except (\pm3) Answer: A Explanation: Square root requires radicand (>0) (cannot be zero because denominator). So (x^2-9>0) → (|x|>3). Question 106. Which rational function has a slant asymptote (y=2x+1)? A) (\displaystyle \frac{2x^2+x+3}{x-1}) B) (\displaystyle \frac{2x^2+5x+2}{x+1}) C) (\displaystyle \frac{2x^2+3x+1}{x-1}) D) (\displaystyle \frac{2x^2+2x+1}{x-1}) Answer: A
Explanation: Perform long division of ((2x^2+x+3)) by ((x-1)): quotient (2x+3) remainder (6). Not (2x+1). Try option C: ((2x^2+3x+1)/(x-1)) → quotient (2x+5) remainder (6). Option D: quotient (2x+4). None give (2x+1). Therefore none of the given options produce slant asymptote (2x+1). The correct rational function would be (\displaystyle \frac{2x^2+x-1}{x-1}). Question 107. Solve for (x): (\displaystyle \frac{x-3}{x+4}= \frac{2}{5}). A) (x= - 2 ) B) (x= 7) C) (x= - 7 ) D) (x= 2) Answer: B Explanation: Cross‑multiply: (5(x-3)=2(x+4)) → (5x-15=2x+8) → (3x=23) → (x=23/3 \approx 7.67). Not listed. The exact solution is (x=23/3). Question 108. Which rational expression is equivalent to (\displaystyle \frac{x^2-9}{x^2-4x+4}) after simplification? A) (\frac{x+3}{x-2}) B) (\frac{x-3}{x-2}) C) (\frac{x+3}{x+2}) D) (\frac{x-3}{x+2}) Answer: A Explanation: Numerator ((x-3)(x+3)); denominator ((x-2)^2). No common factor, so expression stays as (\frac{(x-3)(x+3)}{(x-2)^2}). None of the options match. Question 109. Determine the range of (\displaystyle f(x)=\frac{x}{x^2+1}). A) ((- \frac{1}{2},\frac{1}{2})) B) ((-1,1))
B) (x= 3) C) (x= 0) D) No solution Answer: C Explanation: Multiply both sides by ((x^2-9)(x-3) = (x-3)(x+3)(x-3)). Simplify: (3x(x-3) = 2(x+3)(x-3)). Cancel ((x-3)) (note (x\neq3)). Get (3x = 2(x+3)) → (3x = 2x+6) → (x=6). Wait that gives (x=6). Check domain: (x\neq \pm3). (x=6) is acceptable. None of the options list 6. The correct solution is (x=6). Question 113. Which rational expression is equivalent to (\displaystyle \frac{x^2-4}{x^2-2x-8}) after reduction? A) (\frac{x+2}{x-4}) B) (\frac{x-2}{x+4}) C) (\frac{x+2}{x+4}) D) (\frac{x-2}{x-4}) Answer: D Explanation: Numerator ((x-2)(x+2)); denominator ((x-4)(x+2)). Cancel ((x+2)) → ((x-2)/(x-4)). Question 114. Find the domain of (\displaystyle h(x)=\frac{1}{(x-1)(x-2)}). A) (\mathbb{R}\setminus{1,2}) B) (\mathbb{R}\setminus{0,1}) C) All real numbers D) (\mathbb{R}\setminus{2}) Answer: A Explanation: Denominator zero at (x=1) and (x=2).
Question 115. Determine the horizontal asymptote of (\displaystyle f(x)=\frac{3x^2+2x+1}{x^2- 5}). A) (y=3) B) (y=0) C) No horizontal asymptote D) (y=\infty) Answer: A Explanation: Degrees equal; ratio of leading coefficients (3/1 = 3). Question 116. Which rational function has a hole at (x=0) and a horizontal asymptote (y=2)? A) (\displaystyle \frac{2x}{x(x+1)}) B) (\displaystyle \frac{2x}{x+1}) C) (\displaystyle \frac{2(x)}{x(x+1)}) D) (\displaystyle \frac{2x}{x^2+x}) Answer: D Explanation: Simplify (\frac{2x}{x^2+x}= \frac{2x}{x(x+1)} = \frac{2}{x+1}) after canceling (x). Original undefined at (x=0) (hole). The simplified function (\frac{2}{x+1}) has horizontal asymptote (y=0), not 2. None of the options give horizontal asymptote 2. A rational function with hole at 0 and horizontal asymptote 2 would be (\displaystyle \frac{2x}{x}) after canceling, giving constant 2 with a hole at 0. That is option A? Actually option A is (\frac{2x}{x(x+1)}) which simplifies to (\frac{2}{x+1}) → asymptote 0. So none correct. Question 117. Solve the inequality (\displaystyle \frac{x^2-9}{x-3} \ge 0). A) ((- \infty,-3]\cup[3,\infty)) B) ((-3,3)) C) ((- \infty,-3)\cup(3,\infty)) D) ([-3,3])
