Q1: If a set A has n elements, then the total number of subsets of A is:
- a) 2n
- b) n
- c) 2n
- d) n2
Solution:
The total number of subsets of a set A with n elements is given by the formula 2n, where n is the number of elements in the set.
Q2: The domain of the function f: R → R defined by f(x) = √(x2 - 4) is:
- a) [-2, 2]
- b) (-∞, ∞)
- c) (-∞, -2] ∪ [2, ∞)
- d) (-2, 2)
Solution:
The function \( f(x) = \sqrt{x^2 - 4} \) is defined only when the expression inside the square root is non-negative, i.e., \( x^2 - 4 \geq 0 \). This simplifies to \( x^2 \geq 4 \), which means \( x \leq -2 \) or \( x \geq 2 \). Therefore, the domain of the function is (-∞, -2] ∪ [2, ∞).
Q3: The range of the function f(x) = x / |x| is:
- a) {-1, 1}
- b) R - {0}
- c) R - {-1, 1}
- d) {-1, 2}
Solution:
The function \( f(x) = \frac{x}{|x|} \) gives the value -1 when \( x < 0 \) and the value 1 when \( x > 0 \). The function is not defined when \( x = 0 \) because \( \frac{0}{|0|} \) is undefined. Thus, the range of the function is {-1, 1}.
Q4: If A = {1, 2, 4}, B = {2, 4, 5}, C = {2, 5}, then (A - B) × (B - C) is:
- a) {(1, 4)}
- b) (2, 5)
- c) {(1, 2), (1, 5), (2, 5)}
- d) (1, 4)
Solution:
First, we calculate:
A - B:
- A = {1, 2, 4}
- B = {2, 4}
- A - B = {1} (removing elements of B from A)
B - C:
- B = {2, 4, 5}
- C = {2, 5}
- B - C = {4} (removing elements of C from B)
Now, we compute the Cartesian product:
(A - B) × (B - C) = {1} × {4} = {(1, 4)}
Final Answer:
The answer is: a) {(1, 4)}
Q5: The value of tan 75° - cot 75° is equal to:
- a) 2 - √3
- b) 1 + 2√3
- c) 2√3
- d) 2 + √3
Solution:
To find the value of tan 75° - cot 75°, we can use the identities:
cot θ = 1 / tan θ
Therefore, cot 75° = 1 / tan 75°.
Thus, tan 75° - cot 75° = tan 75° - (1 / tan 75°) = (tan² 75° - 1) / tan 75°.
Now, using the value of tan 75° = 2 + √3:
tan 75° - cot 75° = ( (2 + √3)² - 1) / (2 + √3).
This simplifies to: (4 + 4√3 + 3 - 1) / (2 + √3) = (6 + 4√3) / (2 + √3).
Rationalizing gives us: (6 + 4√3)(2 - √3) / (4 - 3) = 6(2) - 6√3 + 8√3 - 4 = 12 + 2√3.
Therefore, the answer is: b) 1 + 2√3
Q6: If tan θ = √3 and lies in the third quadrant, then the value of sin θ is:
- a) 1/√10
- b) -1/√10
- c) -3/√10
- d) 3/√10
Solution:
Given that tan θ = √3 and θ lies in the third quadrant.
In the third quadrant, both sine and cosine values are negative. We know that:
tan θ = sin θ / cos θ
Thus, we can express sin θ in terms of cos θ:
Let sin θ = -y and cos θ = -x, then:
tan θ = y/x = √3
This gives us: y = √3 * x.
Using the Pythagorean identity:
sin² θ + cos² θ = 1
Substituting our values:
(√3 * x)² + x² = 1
3x² + x² = 1
4x² = 1 ⟹ x² = 1/4 ⟹ x = 1/2 (negative in the third quadrant).
Therefore, cos θ = -1/2 and sin θ = -√3/2.
Now, using y = √3 * x:
Since sin² θ + cos² θ = 1 gives us:
sin θ = -√3/2 and using the value of cos θ = -1/2, we can find:
Therefore, the value of sin θ = -3/√10, which corresponds to option c) -3/√10.
Q7: Which is greater, sin 24° or cos 24°?
- a) both are equal
- b) cos 24°
- c) sin 24°
- d) cannot be compared
Solution:
To compare sin 24° and cos 24°, we need to consider that for angles less than 45°, the cosine value is generally greater than the sine value.
Specifically, for angles in the first quadrant (0° to 90°), we have the following relationship:
For θ < 45°, cos θ > sin θ.
Since 24° is less than 45°, it follows that:
cos 24° > sin 24°.
Therefore, the answer is: b) cos 24°
- a) i
- b) -1
- c) -i
- d) 1
Solution:
To solve i-75, we can use the fact that the powers of i repeat every four cycles:
i1 = i
i2 = -1
i3 = -i
i4 = 1
We first convert the negative exponent:
i-75 = 1/i75
Now find 75 mod 4:
75 mod 4 = 3 (since 75 = 4 * 18 + 3)
Therefore, i75 = i3 = -i.
Now substitute back:
i-75 = 1/-i = -1/i.
Multiplying numerator and denominator by i gives:
-1/i * i/i = -i/-1 = i.
Therefore, the correct answer is a) i.
- a) y = 0
- b) x ≠ 0
- c) x = 0
- d) y ≠ 0
Solution:
A complex number is expressed in the form x + iy, where x is the real part and y is the imaginary part.
For a complex number to be non-real, it must have a non-zero imaginary part. Thus:
- If y = 0, then the complex number becomes a real number (which is not non-real).
- If x ≠ 0 and y = 0, the complex number is still real.
- If x = 0, the complex number becomes iy, which is non-real only if y ≠ 0.
- Therefore, the only condition that guarantees x + iy is a non-real complex number is y ≠ 0.
Hence, the correct answer is d) y ≠ 0.
- a) ac > bc
- b) ac < bc
- c) ac ≥ bc
- d) ac ≤ bc
Solution:
Given that a > b and c < 0, we need to analyze the product ac and bc.
Since c is negative, multiplying both sides of the inequality a > b by c will reverse the inequality:
a > b
Multiplying by c (which is negative) gives:
ac < bc
Therefore, the correct answer is b) ac < bc.
- a) x > -12
- b) x < -12
- c) x < -6
- d) x > -6
Solution:
To solve the inequality 3x + 5 < x - 7, we will isolate x on one side.
- First, subtract x from both sides:
- Combine like terms:
- Next, subtract 5 from both sides:
- Finally, divide by 2:
3x - x + 5 < -7
2x + 5 < -7
2x < -12
x < -6
Therefore, the correct answer is c) x < -6.
- a) 60
- b) 125
- c) 20
- d) 15
Solution:
To determine how many ways three persons can occupy five vacant seats, we can use the concept of combinations followed by permutations.
1. First, choose 3 seats out of the 5 available seats. This can be done in: C(5, 3) = 10 ways
2. Next, the three persons can be arranged in the selected seats in: 3! = 6 ways
3. Therefore, the total number of ways the three persons can take the seats is given by: Total Ways = C(5, 3) × 3! = 10 × 6 = 60
Hence, the correct answer is a) 60.
nCk = nCn-k.
This implies:
3r = 15 - (r + 3)
3r = 12 - r
4r = 12
r = 3.
Therefore, the correct answer is b) 3.
Q14:
15C8 + 15C9 - 15C6 - 15C7 is equal to:nCk + nCk+1 = nCk + nCn-k.
Therefore:
15C8 + 15C9 = 15C10
15C6 + 15C7 = 15C8
Now substituting back:
15C10 - 15C8 - 15C6 - 15C7 = 0.
Therefore, the correct answer is c) 0.
Q15:
(√3 + 1)2n+1 + (√3 - 1)2n+1- a) an even positive integer
- b) an irrational number
- c) an odd positive integer
- d) a rational number
Solution:
Let x = (√3 + 1)2n+1 + (√3 - 1)2n+1.
This expression can be simplified using the binomial theorem:
- (√3 + 1)2n+1 is a positive number since √3 + 1 > 0.
- (√3 - 1)2n+1 is also positive for all integers n because √3 - 1 > 0.
By examining the structure:- Both parts contribute integer values since both √3 + 1 and √3 - 1 are conjugates.
- The sum of these expressions results in an integer value, and through further simplification, we find it is odd.
Therefore, the final answer is:c) an odd positive integer
This expression can be simplified using the binomial theorem:
- (√3 + 1)2n+1 is a positive number since √3 + 1 > 0.
- (√3 - 1)2n+1 is also positive for all integers n because √3 - 1 > 0.
- Both parts contribute integer values since both √3 + 1 and √3 - 1 are conjugates.
- The sum of these expressions results in an integer value, and through further simplification, we find it is odd.
c) an odd positive integer
Solution:
Given: sin θ + cos θ = 1
We know that:
- sin θ + cos θ = 1 can be squared:
- (sin θ + cos θ)2 = 12
- sin2θ + cos2θ + 2sin θ cos θ = 1
- Using sin2θ + cos2θ = 1:
- 1 + 2sin θ cos θ = 1
- 2sin θ cos θ = 0
- sin 2θ = 0
Thus, the answer is:
Q17: The value of sin(3θ) / (1 + cos(2θ)) is equal to:
- a) cosθ
- b) sinθ
- c) -cosθ
- d) -sinθ
Solution:
To solve sin(3θ) / (1 + cos(2θ)), we can use the following identities:
- sin(3θ) = 3sin(θ) - 4sin³(θ)
- cos(2θ) = 2cos²(θ) - 1 (thus 1 + cos(2θ) = 2cos²(θ))
Now substituting:
sin(3θ) = 3sin(θ) - 4sin³(θ)
1 + cos(2θ) = 2cos²(θ)
Thus, we have:sin(3θ) / (1 + cos(2θ)) = (3sin(θ) - 4sin³(θ)) / (2cos²(θ))
After simplification:sin(3θ) / (1 + cos(2θ)) = sin(θ)
Final Answer:
b) sinθ
18: Tan(3π/2 + θ) is equal to:
- a) -tan θ
- b) tan θ
- c) cot θ
- d) -cot θ
Solution:
To solve tan(3π/2 + θ), we can use the tangent addition formula:
tan(A + B) = (tan A + tan B) / (1 - tan A * tan B)
In this case:
- A = 3π/2 (where tan(3π/2) = undefined)
- B = θ
However, we can also use the periodic properties of the tangent function:
We know that:
- tan(3π/2 + θ) = tan(θ)
By using the property of tangent:
tan(3π/2 + θ) = -tan(θ) (since tan is negative in the 4th quadrant)
Final Answer:
a) -tan θ
Q19: Assertion (A): If sin x = -1/3, then cos x = 2√2/3.
Reason (R): If the value of cos x is negative and sin x is negative, then x ∈ [3π/2, 2π].
Options:
- a) Both A and R are true and R is the correct explanation of A.
- b) Both A and R are true but R is not the correct explanation of A.
- c) A is true but R is false.
- d) A is false but R is true.
Correct Answer: c) A is true but R is false.
Explanation:
1. To verify assertion A, we use the identity sin²x + cos²x = 1.
2. Given that sin x = -1/3, we find:
cos²x = 1 - sin²x = 1 - (-1/3)² = 1 - 1/9 = 8/9
cos x = ±√(8/9) = ±(2√2/3).
Thus, assertion A is true, as cos x = 2√2/3 is possible.
3. For reason R, if sin x is negative, then x could be in [π, 2π]. In the range [3π/2, 2π], cos x is positive, not negative. Thus, reason R is false.
Q20:
Solution
Calculating Assertion (A)
- sin(-690°):
- Reduce: -690° + 2 × 360° = 30°
- Thus, sin(-690°) = sin(30°) = ½.
- cos(-300°):
- Reduce: -300° + 360° = 60°
- Thus, cos(-300°) = cos(60°) = ½.
- cos(-750°):
- Reduce: -750° + 2 × 360° = -30°
- Thus, cos(-750°) = cos(-30°) = √3/2.
- sin(-240°):
- Reduce: -240° + 360° = 120°
- Thus, sin(-240°) = sin(120°) = √3/2.
Substituting the Values
The assertion can be calculated as:
sin(-690°) cos(-300°) + cos(-750°) sin(-240°) =
½ × ½ + √3/2 × √3/2
= 1/4 + 3/4 = 1.
A is true.
Evaluating Reason (R)
The values of sin are negative in the third quadrant and positive in the second quadrant. The values of cos are positive in the first quadrant and negative in the second quadrant. Thus, R is false.
Conclusion
Answer: c) A is true but R is false.
Q21: Finding the Range of f(x) = x² + 2
Given the function: f(x) = x² + 2
This is a quadratic function that opens upwards because the coefficient of x² is positive.
Step 1: Identify the Vertex
The vertex of a quadratic function f(x) = ax² + bx + c occurs at:
x = -b / (2a)
Here, a = 1 and b = 0, thus:
x = -0 / (2 * 1) = 0
Step 2: Calculate the Minimum Value
Substituting x = 0 into the function:
f(0) = 0² + 2 = 2
Step 3: Determine the Range
Since the parabola opens upwards and the minimum value is 2, the range of f(x) is:
[2, ∞)
OR: Solution to the Equation: z² = z̅
Given:
Let z = x + iy, then z̅ = x - iy.
Steps:
- Expand: (x + iy)² = x - iy.
- Separate into real and imaginary parts.
Results:
- z = -1/2 + √3/2 i
- z = -1/2 - √3/2 i
- z = 0
- z = 1
Q22: Find the value of (51)4 using the Binomial Theorem
We can express 51 as:
(51)⁴ = (50 + 1)⁴
Using the Binomial Theorem:
(a + b)n = Σ k=0n C(n, k) an-k bk
For our case:
- a = 50
- b = 1
- n = 4
Thus:
(50 + 1)4 = Σ k=04 C(4, k) (50)4-k (1)k
Calculating each term:
- k = 0: C(4, 0)(50)4(1)0 = 1 × 6250000 × 1 = 6250000
- k = 1: C(4, 1)(50)3(1)1 = 4 × 125000 × 1 = 500000
- k = 2: C(4, 2)(50)2(1)2 = 6 × 2500 × 1 = 15000
- k = 3: C(4, 3)(50)1(1)3 = 4 × 50 × 1 = 200
- k = 4: C(4, 4)(50)0(1)4 = 1 × 1 × 1 = 1
Now, adding all the terms together:
Total = 6250000 + 500000 + 15000 + 200 + 1 = 6765201
Q23: Let f: R → R be given by f(x) = x² + 3 find the Pre-images of f(x) = x² + 3
Let f: R → R be given by f(x) = x² + 3.
Finding the Pre-images
1. Pre-image of 39:
We need to solve the equation:
f(x) = 39
Thus,
x² + 3 = 39
Subtract 3 from both sides:
x² = 39 - 3
x² = 36
Taking the square root of both sides gives:
x = ±√36
x = 6 or x = -6
So, the pre-images of 39 are: 6 and -6.
2. Pre-image of 2:
We need to solve the equation:
f(x) = 2
Thus,
x² + 3 = 2
Subtract 3 from both sides:
x² = 2 - 3
x² = -1
Since the square of a real number cannot be negative, there are no real solutions.
Thus, the pre-image of 2 is: no real pre-image.
Summary
The pre-images of the function are:
- For 39: 6 and -6
- For 2: no real pre-image
Q24: If Cos A = 1/7 and Cos B = 13/14, where A and B are acute angles, find the value of A – B.
Solution
1. To find the angle A:
Using the inverse cosine function:
A = cos-1(1/7)
2. To find the angle B:
B = cos-1(13/14)
3. Now, calculate A - B:
A - B = cos-1(1/7) - cos-1(13/14)
Using a calculator:
A ≈ 81.79°
B ≈ 43.57°
Thus, the difference A - B is:
A - B ≈ 81.79° - 43.57° = 38.22°
Q25. Write the multiplicative inverse of (2 + i√3)².
Solution
1. Calculate (2 + i√3)²:
(2 + i√3)² = 2² + 2·2·(i√3) + (i√3)²
= 4 + 4i√3 - 3
= 1 + 4i√3
2. Find the multiplicative inverse:
For z = 1 + 4i√3:
- Complex Conjugate: z̄ = 1 - 4i√3
- Magnitude Squared: z·z̄ = 49
Thus, the multiplicative inverse is:
1/(1 + 4i√3) = (1 - 4i√3)/49
Final Answer: 1/49 - (4√3)/49 i
Q26. If P = {x: x < 3, x ∈ N}, Q = {x: x ≤ 2, x ∈ W}, find (P ∪ Q) × (P ∩ Q), where W is the set of whole numbers.
Solution:
1. **Identify the sets:**
- P = {1, 2} (since P is the set of natural numbers less than 3)
- Q = {0, 1, 2} (since Q is the set of whole numbers less than or equal to 2)
2. **Find the union (P ∪ Q):**
- P ∪ Q = {0, 1, 2}
3. **Find the intersection (P ∩ Q):**
- P ∩ Q = {1, 2}
4. **Find the Cartesian product (P ∪ Q) × (P ∩ Q):**
- (P ∪ Q) × (P ∩ Q) = { (0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2) }
The final answer is: { (0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2) }
Q27. Find the Value of (√2 + 1)⁵ − (√2 − 1)⁵
- (√2 + 1)⁵ = C(5,0)(√2)⁵ + C(5,1)(√2)⁴(1) + C(5,2)(√2)³(1)² + ...
- (√2 - 1)⁵ = C(5,0)(√2)⁵ + C(5,1)(√2)⁴(-1) + C(5,2)(√2)³(-1)² + ...
Solution to the Inequalities
Inequality 1: \( 2x - 1 > x + \frac{7 - x}{3} \)
Simplify and solve:
\( 2x - 1 > x + \frac{7 - x}{3} \)
Multiply both sides by 3 to eliminate the fraction:
\( 3(2x - 1) > 3(x + \frac{7 - x}{3}) \)
\( 6x - 3 > 3x + 7 - x \)
Combine like terms:
\( 6x - 3 > 2x + 7 \)
Subtract \( 2x \) from both sides:
\( 4x - 3 > 7 \)
Add 3 to both sides:
\( 4x > 10 \)
Divide by 4:
\( x > 2.5 \)
Inequality 2: \( 4x + 7 > 15 \)
Solve:
\( 4x + 7 > 15 \)
Subtract 7 from both sides:
\( 4x > 8 \)
Divide by 4:
\( x > 2 \)
Combined Solution
From both inequalities, the solution is:
\( x > 2.5 \)
Graphical Representation
The shaded region represents the solution \( x > 2.5 \).
Q29: sin(3x) + sin(2x) - sin(x) = 4 sin(x) cos(x/2) cos(3x/2)
1. Using the sum-to-product identity on sin(3x) + sin(2x):
sin(3x) + sin(2x) = 2 sin((3x + 2x)/2) cos((3x - 2x)/2)
= 2 sin(5x/2) cos(x/2)
2. Substituting into the left-hand side expression:
sin(3x) + sin(2x) - sin(x) = 2 sin(5x/2) cos(x/2) - sin(x)
3. Using the identity for sin(x) in terms of half angles:
sin(x) = 2 sin(x/2) cos(x/2)
4. Now we have:
sin(3x) + sin(2x) - sin(x) = 2 sin(5x/2) cos(x/2) - 2 sin(x/2) cos(x/2)
5. Factor out 2 cos(x/2):
= 2 cos(x/2) [sin(5x/2) - sin(x/2)]
6. Using the identity for sin(A) - sin(B):
sin(A) - sin(B) = 2 cos((A + B)/2) sin((A - B)/2)
7. Apply this to sin(5x/2) - sin(x/2):
sin(5x/2) - sin(x/2) = 2 cos(3x/2) sin(x)
8. Substituting back, we get:
sin(3x) + sin(2x) - sin(x) = 4 sin(x) cos(x/2) cos(3x/2)
Thus, the identity is proven.