Q. Determine the x-intercept of the line 4x - 5y + 20 = 0.
Solution
Setting y = 0 in the equation gives 4x + 20 = 0, thus x = -5.
Correct Answer: D — -4
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Q. Determine the x-intercept of the line 5x + 2y - 10 = 0.
Solution
Setting y = 0 in the equation gives 5x - 10 = 0, thus x = 2.
Correct Answer: B — 5
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Q. Determine the x-intercept of the line given by the equation 2x - 3y + 6 = 0.
Solution
Set y = 0 in the equation: 2x + 6 = 0 => x = -3.
Correct Answer: B — 3
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Q. Evaluate cos(tan^(-1)(1)).
-
A.
√2/2
-
B.
1/√2
-
C.
1
-
D.
0
Solution
Let θ = tan^(-1)(1). Then, cos(θ) = 1/√(1 + tan^2(θ)) = 1/√(1 + 1) = 1/√2.
Correct Answer: A — √2/2
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Q. Evaluate cos(tan^(-1)(3/4)).
-
A.
4/5
-
B.
3/5
-
C.
5/4
-
D.
3/4
Solution
Using the triangle with opposite = 3 and adjacent = 4, hypotenuse = 5. Thus, cos(tan^(-1)(3/4)) = 4/5.
Correct Answer: A — 4/5
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Q. Evaluate cos(tan^(-1)(5/12)).
-
A.
12/13
-
B.
5/13
-
C.
13/12
-
D.
5/12
Solution
Using the right triangle definition, cos(tan^(-1)(5/12)) = adjacent/hypotenuse = 12/13.
Correct Answer: A — 12/13
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Solution
cos^(-1)(0) = π, since cos(π) = 0.
Correct Answer: C — π
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Q. Evaluate sin(cos^(-1)(1/2)).
Solution
sin(cos^(-1)(1/2)) = √(1 - (1/2)^2) = √(3/4) = √3/2.
Correct Answer: A — √3/2
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Q. Evaluate sin(tan^(-1)(3/4)).
-
A.
3/5
-
B.
4/5
-
C.
1/5
-
D.
5/5
Solution
Using the right triangle definition, sin(tan^(-1)(3/4)) = opposite/hypotenuse = 3/5.
Correct Answer: A — 3/5
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Q. Evaluate sin(tan^(-1)(x)).
-
A.
x/√(1+x^2)
-
B.
√(1-x^2)
-
C.
1/x
-
D.
x
Solution
Using the identity, sin(tan^(-1)(x)) = x/√(1+x^2).
Correct Answer: A — x/√(1+x^2)
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Q. Evaluate sin^(-1)(-1/2) + cos^(-1)(1/2).
Solution
sin^(-1)(-1/2) = -π/6 and cos^(-1)(1/2) = π/3. Therefore, -π/6 + π/3 = π/6.
Correct Answer: B — π/2
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Q. Evaluate sin^(-1)(sin(5π/6)).
-
A.
5π/6
-
B.
π/6
-
C.
7π/6
-
D.
0
Solution
Since 5π/6 is in the range of sin^(-1), sin^(-1)(sin(5π/6)) = 5π/6.
Correct Answer: A — 5π/6
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Q. Evaluate sin^(-1)(sin(π/3)).
-
A.
π/3
-
B.
2π/3
-
C.
π/6
-
D.
0
Solution
sin^(-1)(sin(π/3)) = π/3, since π/3 is in the range of sin^(-1).
Correct Answer: A — π/3
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Q. Evaluate sin^(-1)(sin(π/4)).
-
A.
π/4
-
B.
3π/4
-
C.
π/2
-
D.
0
Solution
sin^(-1)(sin(π/4)) = π/4
Correct Answer: A — π/4
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Q. Evaluate sin^(-1)(√3/2) + cos^(-1)(1/2).
-
A.
π/3
-
B.
π/2
-
C.
π/4
-
D.
2π/3
Solution
sin^(-1)(√3/2) + cos^(-1)(1/2) = π/2
Correct Answer: B — π/2
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Q. Evaluate tan(sin^(-1)(1/√2)).
Solution
If sin(x) = 1/√2, then x = π/4, thus tan(sin^(-1)(1/√2)) = tan(π/4) = 1.
Correct Answer: A — 1
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Q. Evaluate tan(sin^(-1)(3/5)).
-
A.
3/4
-
B.
4/3
-
C.
5/3
-
D.
3/5
Solution
Let θ = sin^(-1)(3/5). Then sin(θ) = 3/5 and using the Pythagorean theorem, cos(θ) = 4/5. Therefore, tan(θ) = sin(θ)/cos(θ) = (3/5)/(4/5) = 3/4.
Correct Answer: B — 4/3
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Q. Evaluate tan^(-1)(1) + tan^(-1)(1).
-
A.
π/2
-
B.
π/4
-
C.
π/3
-
D.
0
Solution
tan^(-1)(1) = π/4, thus tan^(-1)(1) + tan^(-1)(1) = π/4 + π/4 = π/2.
Correct Answer: A — π/2
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Q. Evaluate tan^(-1)(1) + tan^(-1)(√3).
-
A.
π/3
-
B.
π/4
-
C.
π/2
-
D.
π/6
Solution
tan^(-1)(1) = π/4 and tan^(-1)(√3) = π/3. Therefore, π/4 + π/3 = π/2.
Correct Answer: C — π/2
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Q. Evaluate tan^(-1)(√3).
-
A.
π/3
-
B.
π/4
-
C.
π/6
-
D.
π/2
Solution
tan^(-1)(√3) = π/3, since tan(π/3) = √3.
Correct Answer: A — π/3
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Q. Evaluate the definite integral ∫(0 to 1) (3x^2)dx.
-
A.
1
-
B.
0.5
-
C.
0.33
-
D.
0.25
Solution
The integral evaluates to 1.
Correct Answer: A — 1
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Q. Evaluate the definite integral ∫(1 to 2) (3x^2)dx.
Solution
The integral evaluates to 6.
Correct Answer: B — 6
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Q. Evaluate the derivative of f(x) = e^x + ln(x) at x = 1.
Solution
f'(x) = e^x + 1/x. At x = 1, f'(1) = e + 1.
Correct Answer: A — 1
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Q. Evaluate the determinant \( \begin{pmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1 \end{pmatrix} \).
Solution
The determinant is calculated as \( 2(0*1 - 2*2) - 1(1*1 - 2*3) + 3(1*2 - 0*3) = 2(0 - 4) - 1(1 - 6) + 3(2) = -8 + 5 + 6 = 3 \).
Correct Answer: A — -12
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Q. Evaluate the determinant \( \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \)
Solution
The determinant of the identity matrix is 1.
Correct Answer: B — 1
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Q. Evaluate the determinant \( \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \end{vmatrix} \)
Solution
The determinant is 0 because the first column is repeated.
Correct Answer: A — 0
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Q. Evaluate the determinant \( \begin{vmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 4 & 1 \end{vmatrix} \)
Solution
The determinant is 0 because the rows are linearly dependent.
Correct Answer: A — 0
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Q. Evaluate the determinant \( \begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{vmatrix} \).
Solution
The determinant evaluates to -12.
Correct Answer: A — -12
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Q. Evaluate the determinant \( \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix} \)
Solution
The determinant is 0 because the rows are linearly dependent.
Correct Answer: A — 0
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Q. Evaluate the determinant \( \begin{vmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{vmatrix} \).
Solution
The determinant is 0 because the rows are linearly dependent.
Correct Answer: A — 0
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