Answer key
✅Section A – 1 Mark Questions with Explanations
1. Let A = {x ∈ ℝ, x > 4} and B = {x ∈ ℝ, x < 5}. Then A ∩ B is:
Solution:
Set A = {x ∈ ℝ, x > 4} → All real numbers greater than 4
Set B = {x ∈ ℝ, x < 5} → All real numbers less than 5
Intersection A ∩ B → All numbers greater than 4 and less than 5
Thus, A ∩ B = (4, 5)
Correct answer: (b) (4,5)
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2. If A and B are two sets, then A ∩ (A ∪ B)’ is equal to:
Explanation:
A ∪ B is the union of sets A and B
(A ∪ B)’ is the complement of A ∪ B → elements not in A ∪ B
Intersection A ∩ (A ∪ B)’ → elements that are in A and also not in A ∪ B → such elements do not exist
So, the result is the empty set ∅
Correct answer: (a) ∅
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3. If there are p elements in set A and q elements in set B, the number of elements in A × B is:
Definition:
The Cartesian product A × B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B.
Solution:
The total number of pairs = p × q
Correct answer: (a) p × q
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4. The range of |x| / x, where x ≠ 0 is:
Solution:
|x| / x = 1 when x > 0
|x| / x = -1 when x < 0
So, the range is {1, -1}
Correct answer: (c) {-1, 0, 1} ❌→ but 0 is excluded because x ≠ 0
So, the correct range is {-1, 1}
Correct answer: (d) (-1, 1) ❌
Only option matching {-1, 1} → Not given explicitly, but (c) is the closest, assuming typo.
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5. The radius of the circle whose arc length 15π cm makes an angle 3π/4 rad at the centre is:
Formula:
Arc length = rθ
Where r is radius and θ is angle in radians.
Solution:
15π = r × (3π / 4)
r = 15π / (3π / 4)
r = 15π × (4 / 3π)
r = (15 × 4) / 3
r = 60 / 3
r = 20 cm
Correct answer: (b) 20 cm
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6. If tan x = -1/√5 and x lies in the 4th quadrant, then the value of cos x is:
Explanation:
In the 4th quadrant, cosine is positive and tangent is negative
Let’s write tan x = -1/√5
tan x = opposite / adjacent
Let adjacent = √5, opposite = -1
Then hypotenuse = √(√5² + (-1)²) = √(5 + 1) = √6
So, cos x = adjacent / hypotenuse = √5 / √6
Correct answer: (a) √5 / √6
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7. tan x is periodic with period:
Definition:
The function tan x has period π → tan(x + π) = tan x
Correct answer: (b) π
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8. If x + iy = (3 + 5i) / (7 - 6i), then y is:
Solution:
Multiply numerator and denominator by conjugate of denominator:
= (3 + 5i) × (7 + 6i) / [(7 - 6i)(7 + 6i)]
= (3×7 + 3×6i + 5i×7 + 5i×6i) / (7² + 6²)
= (21 + 18i + 35i + 30i²) / (49 + 36)
Since i² = -1 → 30i² = -30
= (21 + 53i - 30) / 85
= (-9 + 53i) / 85
So, x = -9/85, y = 53/85
Correct answer: (c) 53/85
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9. The modulus of z = √5 + 3i is:
Definition:
The modulus of z = a + bi is |z| = √(a² + b²)
Solution:
|z| = √[(√5)² + (3)²]
= √(5 + 9)
= √14
Correct answer: (b) √14
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10. The value of i⁹⁹⁹ is:
Definition:
i = √-1
i powers cycle every 4 terms:
i¹ = i
i² = -1
i³ = -i
i⁴ = 1
Then repeat.
Find remainder of 999 ÷ 4:
4 × 249 = 996 → remainder 3
So, i⁹⁹⁹ = i³ = -i
Correct answer: (b) -i
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11. If (5 - 2x) / 3 ≤ (x / 6) - 5, then the value of x is:
Solution:
Multiply both sides by 6 to clear denominators:
6 × [(5 - 2x)/3] ≤ 6 × [(x/6) - 5]
2 × (5 - 2x) ≤ x - 30
10 - 4x ≤ x - 30
Bring all terms to one side:
10 + 30 ≤ x + 4x
40 ≤ 5x
x≥8
Correct answer: (b) x ≥ 8
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12. The value of cos 1° × cos 2° × cos 3° × ... × cos 179° is:
Observation:
cos(180° - θ) = -cos(θ)
Thus, the terms cancel in pairs, and the product becomes zero because cos(90°) = 0 is included.
Correct answer: (b) 0
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13. If (x + 3) / (x + 5) > 3, then x belongs to:
Solution:
(x + 3) / (x + 5) > 3
Subtract 3:
[(x + 3) - 3(x + 5)] / (x + 5) > 0
= [x + 3 - 3x - 15] / (x + 5) > 0
= (-2x - 12) / (x + 5) > 0
Factor numerator:
= -2(x + 6) / (x + 5) > 0
Multiply both sides by -1 (reverse inequality):
2(x + 6) / (x + 5) < 0
So, (x + 6) / (x + 5) < 0
Find zeros:
x = -6 → numerator zero
x = -5 → denominator zero
Test intervals:
1. x < -6 → both negative → positive → not valid
2. -6 < x < -5 → numerator positive, denominator negative → negative → valid
3. x > -5 → both positive → positive → not valid
So, x ∈ (-6, -5)
Correct answer: (d) (6,12) ❌
Given options don't match exactly → correct interval is (-6, -5)
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14. How many ways can letters of the word BANANA be arranged?
Explanation:
Word BANANA → 6 letters
Repeating letters:
A → 3 times
N → 2 times
Formula:
= 6! / (3! × 2!)
= 720 / (6 × 2)
= 720 / 12
= 60
Correct answer: (b) 60
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15. In how many ways 5 boys can sit at a round table?
Definition:
Circular arrangements = (n - 1)!
Solution:
= (5 - 1)! = 4!
=4×3×2×1
= 24
Correct answer: (c) 24
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16. If α + β = π/4, then the value of (1 + tan α)(1 + tan β) is:
Identity:
tan(α + β) = (tan α + tan β) / (1 − tan α tan β)
Since α + β = π/4 → tan(π/4) = 1
So,
1 = (tan α + tan β) / (1 − tan α tan β)
Cross multiply:
1 − tan α tan β = tan α + tan β
Bring all to one side:
1 − tan α tan β − tan α − tan β = 0
1 − tan α − tan β − tan α tan β = 0
Factor:
(1 − tan α)(1 − tan β) = 0
But that is not immediately helpful.
Let's directly expand (1 + tan α)(1 + tan β):
= 1 + tan α + tan β + tan α tan β
= 1 + (tan α + tan β) + tan α tan β
From earlier, we have:
1 − tan α tan β = tan α + tan β
Thus, tan α + tan β = 1 − tan α tan β
Plugging in:
= 1 + [1 − tan α tan β] + tan α tan β
= 1 + 1 − tan α tan β + tan α tan β
=2
Correct answer: (b) 2
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17. Find x if (1/8! + 1/9!) / (1/10!) = x
Solution:
First, find numerator:
= 1/8! + 1/9!
= 1/8! + 1/(9×8!)
= (1×9 + 1)/(9×8!)
= (9 + 1)/(9×8!)
= 10 / (9×8!)
Now, denominator:
1/10! = 1 / (10×9×8!)
So, dividing:
x = [10 / (9×8!)] ÷ [1 / (10×9×8!)]
= [10 / (9×8!)] × [(10×9×8!) / 1]
= (10 × 10 × 9 × 8!) / (9 × 8!)
= (10 × 10) / 1
= 100
Correct answer: (d) 100
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18. In how many ways can 6 boys be arranged in a row?
Definition:
For linear arrangement of n objects → n!
Solution:
= 6!
=6×5×4×3×2×1
= 720
Correct answer: (b) 6!
✅Section B – Very Short Answer Type Questions (2 Marks each)
21. Find all possible proper subsets of {1, 2}?
Definition:
A proper subset is a subset that is not equal to the original set.
For set {1, 2}, the total subsets are 2ⁿ where n = 2 → 2² = 4 subsets:
Subsets:
1. ∅ (empty set)
2. {1}
3. {2}
4. {1, 2}
The proper subsets are all subsets except the set itself, i.e., excluding {1, 2}.
Proper subsets:
∅, {1}, {2}
Total = 3 proper subsets
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22. Find the domain of the real function f(x) = √(x² − 4).
Explanation:
The expression under the square root must be ≥ 0.
x² − 4 ≥ 0
(x − 2)(x + 2) ≥ 0
Critical points: x = 2 and x = -2
Test intervals:
1. x < -2 → (-)(-) → + → valid
2. -2 < x < 2 → (+)(-) → - → invalid
3. x > 2 → (+)(+) → + → valid
Domain: (-∞, -2] ∪ [2, ∞)
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23. Find the value of sin 15°.
Using identity:
sin(15°) = sin(45° − 30°)
sin(A − B) = sin A cos B − cos A sin B
= sin 45° cos 30° − cos 45° sin 30°
= (√2/2)(√3/2) − (√2/2)(1/2)
= (√6 / 4) − (√2 / 4)
= (√6 − √2) / 4
Final answer: sin 15° = (√6 − √2) / 4
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OR
If [(1 − i) / (1 + i)]¹⁰⁰ = a + ib, find the value of a and b.
Step 1 – Rationalize the denominator:
Multiply numerator and denominator by the conjugate of the denominator (1 − i):
= [(1 − i)(1 − i)] / [(1 + i)(1 − i)]
= [(1 − i)²] / [1² − i²]
= [(1 − 2i + i²)] / (1 − (-1))
= [(1 − 2i − 1)] / 2
= (-2i) / 2
= -i
So, [(1 − i) / (1 + i)] = -i
Step 2 – Find the 100th power:
(-i)¹⁰⁰
The powers of -i cycle every 4 terms:
(-i)¹ = -i
(-i)² = (-i)(-i) = -1 × -1 × i × i = (-1)(-1)(-1) = -1
Wait: better approach →
(-i)² = (-i) × (-i) = (-1)(-1)(i)(i) = 1 × -1 = -1
(-i)³ = (-i)² × (-i) = -1 × (-i) = i
(-i)⁴ = i × (-i) = i × (-i) = -i² = -(-1) = 1
So the cycle is:
1 → -i → -1 → i → 1
Cycle length = 4
Find remainder of 100 ÷ 4:
4 × 25 = 100 → remainder 0
So, (-i)¹⁰⁰ = 1
Thus, a = 1, b = 0
Final answer:
a = 1, b = 0
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24. Solve the inequality (5x − 8)/3 ≥ (4x − 7)/2 where x ∈ ℝ.
Step 1 – Eliminate denominators by multiplying both sides by 6 (LCM of 3 and 2):
6 × [(5x − 8)/3] ≥ 6 × [(4x − 7)/2]
= 2 × (5x − 8) ≥ 3 × (4x − 7)
= 10x − 16 ≥ 12x − 21
Step 2 – Bring all terms to one side:
10x − 16 − 12x + 21 ≥ 0
= -2x + 5 ≥ 0
Step 3 – Solve for x:
-2x ≥ -5
x ≤ 5/2
Final answer: x ≤ 5/2
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25. In how many ways 4 boys and 5 girls can sit in a row such that no two girls sit together?
Step 1 – Arrange the boys first:
There are 4 boys → can be arranged in 4! ways
= 4 × 3 × 2 × 1 = 24
Step 2 – Place the girls:
To ensure no two girls sit together, place the boys first and then insert the girls in the gaps.
For example:
B_B_B_B_
There are 5 gaps (before, between, and after the boys)
So, the 5 girls must be placed in these 5 gaps → 5! ways
= 5 × 4 × 3 × 2 × 1 = 120
Step 3 – Multiply both:
Total = 4! × 5!
= 24 × 120
= 2880
Final answer: 2880 ways
✅Section C – Short Answer Type Questions (3 Marks each)
26. Find the domain and range of
f(x) = \frac{1}{\sqrt{9 - x^2}}.
✅Step 1 – Find the domain
For the square root to be defined and the denominator to be non-zero:
1. The expression inside the square root must be ≥ 0:
2. The denominator must not be zero:
So:
Taking square roots:
-3 < x < 3
Domain:
(-3, 3)
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✅Step 2 – Find the range
The square root term must be positive:
\sqrt{9 - x^2} > 0
Since , the smallest value of approaches but never reaches it, and the largest value occurs when :
\sqrt{9 - 0} = \sqrt{9} = 3
Thus, the denominator varies between (not included) and .
The function will be largest when the denominator is smallest and smallest when the denominator is
largest.
So:
The smallest value →
The largest value → approaches as approaches
Range:
\left(\frac{1}{3}, +\infty\right)
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✅Final Answer:
Domain:
Range:
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✅OR
Let and be two sets. By using the properties of sets, prove that:
A - (A \cap B) = A - B
✅Step 1 – Expand both sides using definitions
Definition of set difference:
A - B = \{x \in A : x \notin B\}
Intersection:
A \cap B = \{x : x \in A \text{ and } x \in B\}
Left-hand side:
A - (A \cap B) = \{x \in A : x \notin (A \cap B)\}
x \in A \text{ and } x \notin (A \cap B)
Thus:
A - (A \cap B) = \{x \in A : x \notin B\} = A - B
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✅Step 2 – Final conclusion
We have shown that:
A - (A \cap B) = A - B
Hence, proved.
✅27. Find the number of signals that can be sent by 4 flags of different colors taking one or more at a
time.
✅Step 1 – Understand the problem
Each flag can either be used or not used in a signal.
For each flag → 2 choices → included or excluded.
So, for 4 flags → total possible combinations = .
However, the case where no flag is used is not allowed.
✅Step 2 – Subtract the empty set
Total signals =
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✅Final Answer:
The number of signals = 15
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✅28. The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? (π = 3.14)
✅Step 1 – Find the distance traveled by the minute hand
The minute hand traces the circumference of a circle when it rotates.
The full circumference is:
C = 2\pi r
C = 2 \times 3.14 \times 1.5 = 9.42 \text{ cm}
This is the distance the tip travels in 60 minutes.
✅Step 2 – Find the distance in 40 minutes
The distance traveled in 40 minutes is:
\text{Distance} = \frac{40}{60} \times 9.42
= \frac{2}{3} \times 9.42
= 6.28 \text{ cm}
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✅Final Answer:
The tip moves 6.28 cm in 40 minutes.
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✅29. Solve the linear inequality . Represent it on the number line.
✅Step 1 – Expand the right-hand side
3x + 17 \leq 2(1 - x)
3x + 17 \leq 2 - 2x
✅Step 2 – Bring like terms together
Add to both sides:
3x + 2x + 17 \leq 2
5x + 17 \leq 2
Subtract from both sides:
5x \leq -15
✅Step 3 – Solve for
x \leq \frac{-15}{5}
x \leq -3
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✅Final Answer:
x \leq -3
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✅30. Find the value of .
✅Step 1 – Express in degrees
\frac{\pi}{8} = \frac{180^\circ}{8} = 22.5^\circ
So we are to find .
✅Step 2 – Use the half-angle formula
\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}
Let’s choose , so .
We know:
\cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \sin 45^\circ = \frac{\sqrt{2}}{2}
Plugging into the formula:
\tan 22.5^\circ = \frac{1 - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}
Multiply numerator and denominator by :
= \frac{2 - \sqrt{2}}{\sqrt{2}}
Divide each term by :
= \frac{2}{\sqrt{2}} - \frac{\sqrt{2}}{\sqrt{2}}
= \frac{2\sqrt{2}}{2} - 1
= \sqrt{2} - 1
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✅Final Answer:
\tan \frac{\pi}{8} = \sqrt{2} - 1
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✅31. Find the modulus of the complex number .
✅Step 1 – Rationalize the denominator
Multiply the numerator and denominator by the conjugate of the denominator:
(1 - 3i)(1 + 3i) = 1^2 - (3i)^2 = 1 - (-9) = 10
So:
Z = \frac{(1 + 2i)(1 + 3i)}{10}
✅Step 2 – Multiply the numerators
= \frac{1(1 + 3i) + 2i(1 + 3i)}{10}
= \frac{1 + 3i + 2i + 6i^2}{10}
Since :
= \frac{1 + 3i + 2i - 6}{10}
= \frac{-5 + 5i}{10}
= \frac{-1 + i}{2}
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✅Step 3 – Find the modulus
|Z| = \left| \frac{-1 + i}{2} \right|
= \frac{\sqrt{(-1)^2 + (1)^2}}{2}
= \frac{\sqrt{1 + 1}}{2}
= \frac{\sqrt{2}}{2}
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✅Final Answer:
|Z| = \frac{\sqrt{2}}{2}
