CSCA Mathematics Practice Test 2
Level: Medium (Advanced Prep)
Prepared by: CHEANG SOKKONG - 096 96 40573
Topics: Functions, Limits, Integrals, and Analytic Geometry
Number of Questions: 45
Recommended Duration: 60 Minutes
Part 1: Functions (15 Questions)
1. Find the domain of the function $f(x)=\sqrt{\log_{2}(x-1)}$.
A. $[2,+\infty)$
B. $(1,+\infty)$
C. $(2,+\infty)$
D. $[1,+\infty)$
2. Find the inverse function of $f(x)=\ln(x+\sqrt{x^{2}+1})$.
A. $f^{-1}(x)=\frac{e^{x}+e^{-x}}{2}$
B. $f^{-1}(x)=e^{x}-e^{-x}$
C. $f^{-1}(x)=\frac{e^{x}-e^{-x}}{2}$
D. $f^{-1}(x)=\frac{e^{2x}-1}{2}$
3. Determine the parity of the function $f(x)=\ln(\frac{1-x}{1+x})$.
A. Even function
B. Odd function
C. Both even and odd
D. Neither even nor odd
4. If $f(x)=x^{2}+1$ and $f(g(x))=4x^{2}-12x+10$, which of the following could be $g(x)$?
A. $g(x)=2x-1$
B. $g(x)=2x-3$
C. $g(x)=4x-3$
D. $g(x)=x-3$
5. What is the period of the function $y=\sin(2x)+\cos(3x)$?
A. $\pi$
B. $3\pi$
C. $\frac{2\pi}{3}$
D. $2\pi$
6. Find the range of the function $y=\frac{x^{2}+2}{x^{2}+1}$.
A. $(1,2]$
B. $[1,2]$
C. $(1,+\infty)$
D. $[2,+\infty)$
7. Solve the logarithmic equation: $\log_{2}(x)+\log_{4}(x)=3$.
A. $x=2$
B. $x=4$
C. $x=8$
D. $x=16$
8. Find the sum of the roots of the equation $4^{x}-3\cdot2^{x+1}+8=0$.
A. $6$
B. $2$
C. $3$
D. $8$
9. A function $f(x)$ satisfies $f(x+2)=-f(x)$ for all real $x$. What is the period of $f(x)$?
A. $2$
B. $\pi$
C. $4$
D. The function is not periodic
10. Simplify the trigonometric expression: $\frac{\sin(2x)}{1+\cos(2x)}$.
A. $\tan(x)$
B. $\cot(x)$
C. $\sin(x)$
D. $\sec(x)$
11. What are the maximum and minimum values of $y=\sin^{2}(x)-4\sin(x)+5$?
A. Max $10$, Min $2$
B. Max $5$, Min $1$
C. Max $10$, Min $0$
D. Max $2$, Min $-2$
12. A function satisfies $2f(x)+f(-x)=3x+2$. Find the expression for $f(x)$.
A. $f(x)=3x+2$
B. $f(x)=3x+\frac{2}{3}$
C. $f(x)=x+\frac{2}{3}$
D. $f(x)=2x+1$
13. Given $f(x)=\sqrt{1-x^{2}}$, find the domain of the composite function $f(f(x))$.
A. $[-1,1]$
B. $[0,1]$
C. $(-\infty,+\infty)$
D. $[0,+\infty)$
14. Let $f(x)=\begin{cases}\log_{3}(x), & x>0 \\ 3^{x}, & x\le0\end{cases}$. Find the value of $f(f(-1))$.
A. $1$
B. $0$
C. $-1$
D. $\frac{1}{3}$
15. Find the equation of the oblique (slant) asymptote for $y=\frac{x^{2}-3x+2}{x+1}$.
A. $y=x-2$
B. $y=x-4$
C. $y=x+1$
D. $y=x-3$
Part 2: Limits (10 Questions)
16. Evaluate $\lim_{x\to0}\frac{\sqrt{1+x}-\sqrt{1-x}}{x}$.
A. $0$
B. $\frac{1}{2}$
C. $1$
D. $2$
17. Evaluate $\lim_{x\to\infty}(\sqrt{x^{2}+3x}-x)$.
A. $0$
B. $\infty$
C. $\frac{3}{2}$
D. $3$
18. Evaluate $\lim_{x\to0}\frac{1-\cos(2x)}{x\sin(3x)}$.
A. $\frac{2}{3}$
B. $\frac{4}{3}$
C. $0$
D. $1$
19. Evaluate $\lim_{h\to0}\frac{\ln(2+h)-\ln(2)}{h}$.
A. $\frac{1}{2}$
B. $2$
C. $\ln(2)$
D. $0$
20. Evaluate $\lim_{x\to\infty}(\frac{x+2}{x-1})^{x}$.
A. $e$
B. $e^{2}$
C. $1$
D. $e^{3}$
21. Evaluate the one-sided limit $\lim_{x\to2^{-}}\frac{|x-2|}{x^{2}-4}$.
A. $\frac{1}{4}$
B. $0$
C. $-\frac{1}{4}$
D. Does not exist
22. Find the constant $a$ such that $\lim_{x\to1}\frac{x^{2}+ax-2}{x-1}$ exists, and find the limit $L$.
A. $a=1, L=2$
B. $a=1, L=3$
C. $a=-1, L=3$
D. $a=2, L=1$
23. If $3x\le f(x)\le x^{3}+2$ for all $x$ near $1$, evaluate $\lim_{x\to1}f(x)$.
A. $3$
B. $1$
C. $2$
D. Cannot be determined
24. Evaluate $\lim_{x\to0}\frac{e^{3x}-1}{\sin(2x)}$.
A. $\frac{2}{3}$
B. $1$
C. $0$
D. $\frac{3}{2}$
25. Evaluate $\lim_{x\to-\infty}\frac{\sqrt{4x^{2}+1}}{x-3}$.
A. $-2$
B. $2$
C. $4$
D. $-4$
Part 3: Integrals (10 Questions)
26. Find the indefinite integral $\int \sin^{3}(x)\cos(x)dx$.
A. $\frac{1}{3}\sin^{3}(x)+C$
B. $\frac{1}{4}\sin^{4}(x)+C$
C. $-\frac{1}{4}\cos^{4}(x)+C$
D. $\frac{1}{4}\cos^{4}(x)+C$
27. Evaluate the definite integral $\int_{0}^{3}|x-1|dx$.
A. $\frac{3}{2}$
B. $2$
C. $\frac{5}{2}$
D. $3$
28. Evaluate $\int xe^{x}dx$.
A. $(x-1)e^{x}+C$
B. $xe^{x}+C$
C. $(x+1)e^{x}+C$
D. $\frac{1}{2}x^{2}e^{x}+C$
29. Evaluate the definite integral $\int_{0}^{1}\frac{1}{1+x^{2}}dx$.
A. $\frac{\pi}{2}$
B. $\frac{\pi}{4}$
C. $1$
D. $\frac{\pi}{3}$
30. Evaluate $\int_{-2}^{2}(x^{3}\cos(x)+\sin^{5}(x))dx$.
A. $0$
B. $2$
C. $4\cos(2)$
D. $\pi$
31. Find the area of the region bounded by $y=x^{2}$ and $y=2x$.
A. $\frac{2}{3}$
B. $1$
C. $\frac{4}{3}$
D. $\frac{8}{3}$
32. Evaluate $\int_{1}^{e}\frac{\ln(x)}{x}dx$.
A. $1$
B. $\frac{1}{2}$
C. $e$
D. $e-1$
33. Let $F(x)=\int_{1}^{x}\sqrt{t^{2}+1}dt$. Find $F^{\prime}(2)$.
A. $\frac{2}{\sqrt{5}}$
B. $5$
C. $\sqrt{2}$
D. $\sqrt{5}$
34. Evaluate $\int\frac{x}{\sqrt{x^{2}+4}}dx$.
A. $\frac{1}{2}\ln(x^{2}+4)+C$
B. $\frac{1}{\sqrt{x^{2}+4}}+C$
C. $\sqrt{x^{2}+4}+C$
D. $2\sqrt{x^{2}+4}+C$
35. Find the volume of the solid generated by revolving the region bounded by $y=\sqrt{x}$, $y=0$, and $x=4$ around the x-axis.
A. $4\pi$
B. $\frac{8\pi}{3}$
C. $8\pi$
D. $16\pi$
Part 4: Geometry (10 Questions)
36. Find the perpendicular distance from the point $(1, 2)$ to the line $3x-4y+10=0$.
A. $1$
B. $2$
C. $\frac{5}{2}$
D. $5$
37. Find the equation of the tangent line to the circle $x^{2}+y^{2}=25$ at the point $(3, 4)$.
A. $4x+3y=25$
B. $3x-4y=25$
C. $3x+4y=25$
D. $4x-3y=25$
38. Find the vector projection of $\vec{u}=\langle 3,4\rangle$ onto $\vec{v}=\langle 1,0\rangle$.
A. $(0,4)$
B. $(3,0)$
C. $(1,1)$
D. $(1,0)$
39. Find the angle between the vectors $\vec{a}=\langle 1,-1,0\rangle$ and $\vec{b}=\langle 0,1,-1\rangle$.
A. $60^{\circ}$
B. $90^{\circ}$
C. $45^{\circ}$
D. $120^{\circ}$
40. Find the coordinates of the foci of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$.
A. $(\pm 4,0)$
B. $(0,\pm 4)$
C. $(\pm 5,0)$
D. $(\pm 3,0)$
41. Find the volume of a tetrahedron with vertices at the origin $(0, 0, 0)$, $A(2,0,0)$, $B(0,3,0)$, and $C(0,0,4)$.
A. $24$
B. $4$
C. $8$
D. $12$
42. Identify the conic section represented by the parametric equations $x=2\cos(t)$ and $y=3\sin(t)$.
A. Circle
B. Parabola
C. Ellipse
D. Hyperbola
43. The sphere $(x-1)^{2}+(y-2)^{2}+(z-3)^{2}=25$ intersects the xy-plane. Find the area of the resulting intersection.
A. $9\pi$
B. $25\pi$
C. $12\pi$
D. $16\pi$
44. Find the area of the triangle formed by the vectors $\vec{u}=\langle 1,2,3\rangle$ and $\vec{v}=\langle -1,0,2\rangle$ as adjacent sides.
A. $3\sqrt{5}$
B. $6\sqrt{5}$
C. $\frac{3\sqrt{5}}{2}$
D. $\sqrt{45}$
45. Find the equation of the directrix of the parabola $x^{2}=-12y$.
A. $x=3$
B. $y=3$
C. $y=-3$
D. $x=-3$
Answer Key
Part 1: Functions
1. A | 2. C | 3. B | 4. B | 5. D | 6. A | 7. B | 8. C | 9. C | 10. A | 11. A | 12. B | 13. A | 14. C | 15. B
Part 2: Limits
16. C | 17. C | 18. A | 19. A | 20. D | 21. C | 22. B | 23. A | 24. D | 25. A
Part 3: Integrals
26. B | 27. C | 28. A | 29. B | 30. A | 31. C | 32. B | 33. D | 34. C | 35. C
Part 4: Geometry
36. A | 37. C | 38. B | 39. D | 40. A | 41. B | 42. C | 43. D | 44. C | 45. B
Detailed Step-by-Step Solutions
Part 1: Functions
Formula: For $\sqrt{u}$, we must have $u \ge 0$. For $\log_a(v)$, we must have $v > 0$.
Step 1 (Root Condition): $\log_{2}(x-1) \ge 0 \implies x-1 \ge 2^0 \implies x-1 \ge 1 \implies x \ge 2$.
Step 2 (Log Condition): $x-1 > 0 \implies x > 1$.
Conclusion: The intersection of $x \ge 2$ and $x > 1$ is $x \ge 2$. Domain is $[2, +\infty)$. (Answer: A)
Step 1: Let $y = \ln(x+\sqrt{x^2+1})$. Rewrite in exponential form: $e^y = x+\sqrt{x^2+1}$.
Step 2: Isolate the square root: $e^y - x = \sqrt{x^2+1}$.
Step 3: Square both sides: $(e^y-x)^2 = x^2+1 \implies e^{2y} - 2xe^y + x^2 = x^2+1$.
Step 4: Cancel $x^2$ and solve for $x$: $2xe^y = e^{2y} - 1 \implies x = \frac{e^{2y}-1}{2e^y} = \frac{e^y - e^{-y}}{2}$.
Conclusion: Swap variables to get $f^{-1}(x) = \frac{e^x - e^{-x}}{2}$. (Answer: C)
Formula: $f(x)$ is even if $f(-x)=f(x)$ and odd if $f(-x)=-f(x)$.
Step 1: Evaluate $f(-x)$: $f(-x) = \ln\left(\frac{1-(-x)}{1+(-x)}\right) = \ln\left(\frac{1+x}{1-x}\right)$.
Step 2: Use logarithm rules: $\ln\left(\frac{1+x}{1-x}\right) = \ln\left(\left(\frac{1-x}{1+x}\right)^{-1}\right) = -\ln\left(\frac{1-x}{1+x}\right) = -f(x)$.
Conclusion: Since $f(-x) = -f(x)$, it is an odd function. (Answer: B)
Step 1: Substitute $g(x)$ into $f(x)$: $f(g(x)) = (g(x))^2 + 1$.
Step 2: Equate to the given composite function: $(g(x))^2 + 1 = 4x^2 - 12x + 10$.
Step 3: Solve for $(g(x))^2$: $(g(x))^2 = 4x^2 - 12x + 9$.
Step 4: Factor the right side: $4x^2 - 12x + 9 = (2x-3)^2$. Thus, $g(x) = 2x-3$ or $g(x) = 3-2x$.
Conclusion: Option B matches $2x-3$. (Answer: B)
Formula: Period of $\sin(kx)$ or $\cos(kx)$ is $T = \frac{2\pi}{k}$. Period of sum is the LCM of their individual periods.
Step 1: Period of $\sin(2x)$ is $T_1 = \frac{2\pi}{2} = \pi$.
Step 2: Period of $\cos(3x)$ is $T_2 = \frac{2\pi}{3}$.
Step 3: Find LCM of $\pi$ and $\frac{2\pi}{3}$. Convert to same denominator: $\frac{3\pi}{3}$ and $\frac{2\pi}{3}$. The LCM of numerators 3 and 2 is 6. LCM = $\frac{6\pi}{3} = 2\pi$.
Conclusion: Total period is $2\pi$. (Answer: D)
Step 1: Rewrite the function to isolate the variable: $y = \frac{(x^2+1)+1}{x^2+1} = 1 + \frac{1}{x^2+1}$.
Step 2: Analyze the domain component $x^2 \ge 0 \implies x^2+1 \ge 1$.
Step 3: Thus, the fraction $0 < \frac{1}{x^2+1} \le 1$.
Conclusion: Add 1 to the inequality to get $y$: $1 < y \le 2$. Range is $(1, 2]$. (Answer: A)
Formula: Change of base rule: $\log_{a^n}(x) = \frac{1}{n}\log_a(x)$.
Step 1: Convert base 4 to base 2: $\log_4(x) = \log_{2^2}(x) = \frac{1}{2}\log_2(x)$.
Step 2: Substitute back: $\log_2(x) + \frac{1}{2}\log_2(x) = 3$.
Step 3: Combine terms: $\frac{3}{2}\log_2(x) = 3 \implies \log_2(x) = 2$.
Conclusion: Convert to exponential form: $x = 2^2 = 4$. (Answer: B)
Step 1: Rewrite using base 2: $(2^x)^2 - 3(2 \cdot 2^x) + 8 = 0 \implies (2^x)^2 - 6(2^x) + 8 = 0$.
Step 2: Let $u = 2^x$. The equation becomes a quadratic: $u^2 - 6u + 8 = 0$.
Step 3: Factor: $(u-2)(u-4) = 0$. Roots are $u=2$ and $u=4$.
Step 4: Find $x$: $2^x=2 \implies x=1$. $2^x=4 \implies x=2$.
Conclusion: Sum of roots = $1 + 2 = 3$. (Answer: C)
Step 1: A function is periodic with period $T$ if $f(x+T) = f(x)$.
Step 2: Apply the given rule to $f(x+4)$: $f(x+4) = f((x+2)+2)$.
Step 3: Use the property $f(z+2) = -f(z)$ where $z=x+2$. Then $f(x+4) = -f(x+2)$.
Step 4: Substitute $f(x+2)$ again: $-(-f(x)) = f(x)$.
Conclusion: Since $f(x+4) = f(x)$, the period is 4. (Answer: C)
Formulas: Double angle identities: $\sin(2x) = 2\sin(x)\cos(x)$ and $\cos(2x) = 2\cos^2(x) - 1$.
Step 1: Substitute identities into the expression: $\frac{2\sin(x)\cos(x)}{1 + (2\cos^2(x) - 1)}$.
Step 2: Simplify denominator: $1 - 1$ cancels out leaving $2\cos^2(x)$.
Conclusion: $\frac{2\sin(x)\cos(x)}{2\cos^2(x)} = \frac{\sin(x)}{\cos(x)} = \tan(x)$. (Answer: A)
Step 1: Let $u = \sin(x)$. Since it's a sine function, $u \in [-1, 1]$.
Step 2: Rewrite the equation: $y = u^2 - 4u + 5$. Complete the square: $y = (u-2)^2 + 1$.
Step 3: This is a parabola opening upwards with vertex at $u=2$. Since $u \in [-1, 1]$, the vertex is outside our interval.
Step 4: Check endpoints. At $u=1$: $y = (1-2)^2 + 1 = 2$ (Minimum). At $u=-1$: $y = (-1-2)^2 + 1 = 10$ (Maximum).
Conclusion: Max is 10, Min is 2. (Answer: A)
Step 1: Call the given equation Eq(1): $2f(x) + f(-x) = 3x + 2$.
Step 2: Substitute $x$ with $-x$ to create a second equation Eq(2): $2f(-x) + f(x) = -3x + 2$.
Step 3: Multiply Eq(1) by 2 to eliminate $f(-x)$: $4f(x) + 2f(-x) = 6x + 4$.
Step 4: Subtract Eq(2) from this new equation: $(4f(x) + 2f(-x)) - (f(x) + 2f(-x)) = (6x + 4) - (-3x + 2)$.
Conclusion: $3f(x) = 9x + 2 \implies f(x) = 3x + \frac{2}{3}$. (Answer: B)
Step 1: Find the domain of $f(x)$: $1-x^2 \ge 0 \implies x^2 \le 1 \implies x \in [-1, 1]$.
Step 2: Find the range of $f(x)$ on its domain: For $x \in [-1, 1]$, $f(x)$ outputs values from 0 to 1. So, Range is $[0, 1]$.
Step 3: The composite $f(f(x))$ requires the inner output $f(x)$ to be within the domain of the outer function $[-1, 1]$.
Conclusion: Since the range of the inner function $[0, 1]$ is fully contained within the required outer domain $[-1, 1]$, no further restrictions apply. The domain remains $x \in [-1, 1]$. (Answer: A)
Step 1: First evaluate inner function $f(-1)$. Since $-1 \le 0$, use the bottom rule $3^x$.
Step 2: $f(-1) = 3^{-1} = \frac{1}{3}$.
Step 3: Now evaluate $f(f(-1)) = f(1/3)$. Since $1/3 > 0$, use the top rule $\log_3(x)$.
Conclusion: $f(1/3) = \log_3(1/3) = -1$. (Answer: C)
Formula: Perform polynomial long division. The oblique asymptote is the resulting linear quotient $y=mx+b$.
Step 1: Divide $(x^2-3x+2)$ by $(x+1)$.
Step 2: $x(x+1) = x^2+x$. Subtracting gives $-4x+2$.
Step 3: $-4(x+1) = -4x-4$. Subtracting gives remainder $6$.
Conclusion: Function rewrites as $y = (x-4) + \frac{6}{x+1}$. The asymptote is $y = x-4$. (Answer: B)
Part 2: Limits
Step 1: Multiply numerator and denominator by the conjugate $\sqrt{1+x}+\sqrt{1-x}$.
Step 2: Numerator becomes $(1+x) - (1-x) = 2x$.
Step 3: Limit becomes $\lim_{x\to0} \frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}$.
Conclusion: Cancel $x$ and plug in $x=0$: $\frac{2}{\sqrt{1}+\sqrt{1}} = \frac{2}{2} = 1$. (Answer: C)
Step 1: Multiply by the conjugate: $\frac{(\sqrt{x^{2}+3x}-x)(\sqrt{x^{2}+3x}+x)}{\sqrt{x^{2}+3x}+x}$.
Step 2: Expand numerator: $(x^2+3x) - x^2 = 3x$.
Step 3: Divide top and bottom by $x$ (for $x>0$, $x=\sqrt{x^2}$): $\lim_{x\to\infty} \frac{3}{\sqrt{1+3/x} + 1}$.
Conclusion: As $x\to\infty$, $3/x\to 0$. We get $\frac{3}{\sqrt{1}+1} = \frac{3}{2}$. (Answer: C)
Formula: $1-\cos(2x) = 2\sin^2(x)$ and $\lim_{\theta\to0} \frac{\sin \theta}{\theta} = 1$.
Step 1: Rewrite numerator: $\lim_{x\to0}\frac{2\sin^2(x)}{x\sin(3x)}$.
Step 2: Multiply to create standard limit structures: $\lim_{x\to0} \frac{2(\frac{\sin x}{x})^2 x^2}{x(\frac{\sin 3x}{3x})3x}$.
Conclusion: Limits of sine terms go to 1. We are left with $\frac{2x^2}{3x^2} = \frac{2}{3}$. (Answer: A)
Step 1: Recognize this as the limit definition of a derivative: $f'(a) = \lim_{h\to0} \frac{f(a+h)-f(a)}{h}$.
Step 2: Identify $f(x) = \ln(x)$ and $a = 2$.
Conclusion: The derivative is $f'(x) = \frac{1}{x}$. At $x=2$, $f'(2) = \frac{1}{2}$. (Answer: A)
Step 1: Rewrite the base to isolate 1: $\frac{x+2}{x-1} = \frac{x-1+3}{x-1} = 1 + \frac{3}{x-1}$.
Step 2: The limit follows the $e$ definition $\lim_{u\to\infty}(1+\frac{k}{u})^u = e^k$.
Conclusion: Here $k=3$ (accounting for asymptotic limits), yielding $e^3$. (Answer: D)
Step 1: Since $x \to 2^-$ (approaching from below 2), $x-2$ is negative. Thus, $|x-2| = -(x-2)$.
Step 2: Factor the denominator: $x^2-4 = (x-2)(x+2)$.
Step 3: Substitute into limit: $\lim_{x\to2^-} \frac{-(x-2)}{(x-2)(x+2)} = \lim_{x\to2^-} \frac{-1}{x+2}$.
Conclusion: Plug in $x=2$: $\frac{-1}{2+2} = -\frac{1}{4}$. (Answer: C)
Step 1: For the limit to exist when denominator approaches 0, the numerator must also approach 0 at $x=1$.
Step 2: Set numerator to 0: $1^2 + a(1) - 2 = 0 \implies a - 1 = 0 \implies a=1$.
Step 3: Substitute $a=1$ and factor: $\lim_{x\to1} \frac{x^2+x-2}{x-1} = \lim_{x\to1} \frac{(x-1)(x+2)}{x-1}$.
Conclusion: Cancel $(x-1)$ and plug in $x=1$: $L = 1+2 = 3$. (Answer: B)
Step 1: Take the limit of the lower bound function: $\lim_{x\to1}(3x) = 3(1) = 3$.
Step 2: Take the limit of the upper bound function: $\lim_{x\to1}(x^3+2) = 1^3+2 = 3$.
Conclusion: By the Squeeze Theorem, since bounded by 3 on both sides, $\lim_{x\to1}f(x) = 3$. (Answer: A)
Step 1: Apply L'Hôpital's Rule because direct substitution yields indeterminate form $0/0$.
Step 2: Derivative of numerator: $\frac{d}{dx}(e^{3x}-1) = 3e^{3x}$.
Step 3: Derivative of denominator: $\frac{d}{dx}(\sin(2x)) = 2\cos(2x)$.
Conclusion: Limit is $\frac{3e^0}{2\cos(0)} = \frac{3}{2}$. (Answer: D)
Step 1: Factor $x^2$ out of root: $\sqrt{x^2(4+1/x^2)} = |x|\sqrt{4+1/x^2}$.
Step 2: Since $x \to -\infty$, $x$ is negative, so $|x| = -x$.
Step 3: Substitute back: $\lim_{x\to-\infty} \frac{-x\sqrt{4+1/x^2}}{x(1-3/x)}$.
Conclusion: Cancel $x$. The limit goes to $\frac{-\sqrt{4}}{1} = -2$. (Answer: A)
Part 3: Integrals
Step 1: Use U-substitution. Let $u = \sin(x)$. Then $du = \cos(x)dx$.
Step 2: Substitute into integral: $\int u^3 du$.
Conclusion: Integrate: $\frac{1}{4}u^4 + C = \frac{1}{4}\sin^4(x) + C$. (Answer: B)
Step 1: The root of the absolute value is $x=1$. Split the integral at this point: $\int_{0}^{1}(1-x)dx + \int_{1}^{3}(x-1)dx$.
Step 2: First part (area of triangle): $\frac{1}{2} \cdot base \cdot height = \frac{1}{2}(1)(1) = 0.5$.
Step 3: Second part (area of triangle): $\frac{1}{2}(2)(2) = 2$.
Conclusion: Total area $= 0.5 + 2 = 2.5 = \frac{5}{2}$. (Answer: C)
Formula: Integration by parts: $\int u dv = uv - \int v du$.
Step 1: Let $u = x$ (so $du = dx$) and $dv = e^x dx$ (so $v = e^x$).
Step 2: Apply formula: $xe^x - \int e^x dx$.
Conclusion: Integrate remaining term: $xe^x - e^x + C = (x-1)e^x + C$. (Answer: A)
Step 1: Recognize the standard integral format: $\int \frac{1}{1+x^2}dx = \arctan(x)$.
Step 2: Evaluate bounds from 0 to 1: $\arctan(1) - \arctan(0)$.
Conclusion: $\frac{\pi}{4} - 0 = \frac{\pi}{4}$. (Answer: B)
Formula: $\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function (i.e., $f(-x) = -f(x)$).
Step 1: Check parity of $f(x) = x^3\cos(x) + \sin^5(x)$.
Step 2: $f(-x) = (-x)^3\cos(-x) + (\sin(-x))^5 = -x^3\cos(x) - \sin^5(x) = -f(x)$.
Conclusion: Since the function is completely odd over a symmetric interval $[-2, 2]$, the integral is $0$. (Answer: A)
Step 1: Find intersection points by setting them equal: $x^2 = 2x \implies x^2-2x = 0 \implies x(x-2)=0$. Points are $x=0$ and $x=2$.
Step 2: Set up area integral (Top function - Bottom function): $\int_{0}^{2}(2x - x^2)dx$.
Step 3: Antiderivative is $[x^2 - \frac{x^3}{3}]$.
Conclusion: Evaluate from 0 to 2: $(2^2 - \frac{2^3}{3}) - 0 = 4 - \frac{8}{3} = \frac{12-8}{3} = \frac{4}{3}$. (Answer: C)
Step 1: Use U-substitution. Let $u = \ln(x)$. Then $du = \frac{1}{x}dx$.
Step 2: Change bounds: when $x=1, u=0$; when $x=e, u=1$.
Step 3: The integral becomes $\int_{0}^{1} u du = [\frac{u^2}{2}]_{0}^{1}$.
Conclusion: Evaluate: $\frac{1^2}{2} - 0 = \frac{1}{2}$. (Answer: B)
Formula: Fundamental Theorem of Calculus Part 1: $\frac{d}{dx}\int_{a}^{x} f(t)dt = f(x)$.
Step 1: Differentiate $F(x)$ to get $F'(x) = \sqrt{x^2+1}$.
Conclusion: Plug in $x=2$: $F'(2) = \sqrt{2^2+1} = \sqrt{5}$. (Answer: D)
Step 1: Use U-substitution. Let $u = x^2+4$. Then $du = 2xdx \implies \frac{1}{2}du = xdx$.
Step 2: Substitute: $\frac{1}{2}\int u^{-1/2} du$.
Step 3: Integrate: $\frac{1}{2} [\frac{u^{1/2}}{1/2}] + C = u^{1/2} + C$.
Conclusion: Substitute back $x$: $\sqrt{x^2+4} + C$. (Answer: C)
Formula: Disk method volume $V = \pi \int_{a}^{b} [f(x)]^2 dx$.
Step 1: Set up the integral: $V = \pi \int_{0}^{4} (\sqrt{x})^2 dx = \pi \int_{0}^{4} x dx$.
Step 2: Antiderivative: $\pi [\frac{x^2}{2}]_{0}^{4}$.
Conclusion: Evaluate bounds: $\pi(\frac{16}{2} - 0) = 8\pi$. (Answer: C)
Part 4: Geometry
Formula: Distance $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2+B^2}}$.
Step 1: Plug in values: $A=3, B=-4, C=10, x_1=1, y_1=2$.
Step 2: Numerator: $|3(1) - 4(2) + 10| = |3 - 8 + 10| = |5| = 5$.
Step 3: Denominator: $\sqrt{3^2 + (-4)^2} = \sqrt{9+16} = \sqrt{25} = 5$.
Conclusion: $d = \frac{5}{5} = 1$. (Answer: A)
Formula: Equation of tangent to a circle $x^2+y^2=r^2$ at point $(x_1, y_1)$ is $x_1 x + y_1 y = r^2$.
Step 1: We are given circle $r^2 = 25$ and point $(3,4)$.
Conclusion: Directly substitute to get $3x + 4y = 25$. (Answer: C)
Formula: $proj_{\vec{v}}\vec{u} = \left(\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|^2}\right)\vec{v}$.
Step 1: Dot product: $\vec{u} \cdot \vec{v} = (3)(1) + (4)(0) = 3$.
Step 2: Magnitude squared of $\vec{v}$: $|\vec{v}|^2 = 1^2 + 0^2 = 1$.
Conclusion: Projection = $\frac{3}{1}\langle1,0\rangle = \langle3,0\rangle$. (Answer: B)
Formula: $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$.
Step 1: Dot product: $(1)(0) + (-1)(1) + (0)(-1) = -1$.
Step 2: Magnitudes: $|\vec{a}| = \sqrt{1^2+(-1)^2} = \sqrt{2}$. $|\vec{b}| = \sqrt{1^2+(-1)^2} = \sqrt{2}$.
Step 3: Calculate cosine: $\cos\theta = \frac{-1}{\sqrt{2}\sqrt{2}} = -\frac{1}{2}$.
Conclusion: The angle whose cosine is $-1/2$ is $120^{\circ}$. (Answer: D)
Formula: For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, distance to focus is $c = \sqrt{a^2 - b^2}$.
Step 1: Identify variables: $a^2=25$ and $b^2=9$.
Step 2: Calculate $c^2 = 25 - 9 = 16 \implies c=4$.
Conclusion: Major axis is horizontal (x-axis), so foci are at $(\pm c, 0) = (\pm 4, 0)$. (Answer: A)
Formula: Volume of a right tetrahedron formed by origin $(0,0,0)$ and axes points is $\frac{1}{6}(base \cdot height \cdot depth)$.
Step 1: The vectors forming edges from origin are $\langle 2,0,0 \rangle, \langle 0,3,0 \rangle, \langle 0,0,4 \rangle$.
Step 2: Apply the scalar triple product $\frac{1}{6}| \vec{u} \cdot (\vec{v} \times \vec{w}) |$.
Conclusion: $V = \frac{1}{6}(2 \times 3 \times 4) = \frac{24}{6} = 4$. (Answer: B)
Step 1: Isolate trigonometric functions: $\cos(t) = \frac{x}{2}$ and $\sin(t) = \frac{y}{3}$.
Step 2: Apply Pythagorean identity $\cos^2(t) + \sin^2(t) = 1$.
Conclusion: Substitute isolated terms to get $(\frac{x}{2})^2 + (\frac{y}{3})^2 = 1$. This is the standard equation of an Ellipse. (Answer: C)
Step 1: The xy-plane represents where $z=0$.
Step 2: Substitute $z=0$ into sphere equation: $(x-1)^2 + (y-2)^2 + (0-3)^2 = 25$.
Step 3: Simplify: $(x-1)^2 + (y-2)^2 + 9 = 25 \implies (x-1)^2 + (y-2)^2 = 16$.
Conclusion: This forms a circle with $r^2=16$. Area = $\pi r^2 = 16\pi$. (Answer: D)
Formula: Area of a triangle formed by vectors is $\frac{1}{2}|\vec{u} \times \vec{v}|$.
Step 1: Find cross product $\vec{u} \times \vec{v} = \langle (2\cdot2-3\cdot0), -(1\cdot2-3(-1)), (1\cdot0-2(-1)) \rangle = \langle 4, -5, 2 \rangle$.
Step 2: Find magnitude: $\sqrt{4^2 + (-5)^2 + 2^2} = \sqrt{16+25+4} = \sqrt{45}$.
Conclusion: Simplify magnitude: $\sqrt{45} = 3\sqrt{5}$. Area is $\frac{1}{2} \times 3\sqrt{5} = \frac{3\sqrt{5}}{2}$. (Answer: C)
Formula: For vertical parabola $x^2=4py$, focus is $(0, p)$ and directrix is $y = -p$.
Step 1: Set up equation: $4p = -12 \implies p = -3$.
Conclusion: The directrix is $y = -(-3) = 3$. (Answer: B)
.png)
0 Comments