Haryana Board (HBSE) Class 11 Maths SAT-1 Question Paper 2025 Answer Key. Get the Students Assessment Test (SAT) Class 11 Maths question paper with complete solution, accurate answer key, and expert preparation tips. The Haryana Board of School Education (HBSE) conducts SAT as an important assessment for Class 11 students. Best resource for Haryana Board Class 11 SAT Mathematics exam practice, quick revision, and scoring better marks.
HBSE Class 11 Maths SAT-1 Question Paper 2025 Answer Key
Instructions :
• All questions are compulsory.
• Questions (1-11) carry 1 mark each.
• Questions (12-14) carry 2 marks each.
• Questions (15-17) carry 3 marks each.
• Questions (18-19) carry 5 marks each.
• Question (20) case study, carry 4 marks.
1. Express the complex number i–39 in a + ib form.
(A) 0 – i
(B) 0 + i
(C) 1 + i
(D) 2 + i
Answer : (B) 0 + i
i–39 = 1/i39 × i/i = i/i40 = i/(i4)10 = i/(1)10 = i/1 = i = 0 + i
2. State which of the following set is empty :
(A) Set of even prime numbers.
(B) The set of letters in the English alphabet.
(C) The set of animals living on the earth.
(D) B = {x : x2 – 2 = 0 and x is rational number}.
Answer : (D) B = {x : x2 – 2 = 0 and x is rational number}
x2 = 2
x = ± √2 ∉ Q (rational number)
3. If A = {1, 2} and B = {3, 4}. How many subsets will A × B have?
(A) 16
(B) 8
(C) 4
(D) 2
Answer : (A) 16
No. of sunsets in A × B = 2m×n = 22×2 = 24 = 16
4. Solve: 3x – 7 > 5x – 1
(A) (– 3, ∞)
(B) (3, ∞)
(C) (– ∞, 3)
(D) (– ∞, – 3)
Answer : (D) (– ∞, – 3)
3x – 7 > 5x – 1
– 7 + 1 > 5x – 3x
– 6 > 2x
– 3 > x
x < – 3
x ∈ (– ∞, – 3)
5. Convert 40°20′ into radian measure.
(A) 540π/121
(B) 7π/540
(C) 121π/540
(D) 5π/121
Answer : (C) 121π/540
40°20′ = 40° + (20/60)° = 40° + (1/3)° = (121/3)° = 121/3 × π/180 = 121π/540
6. Which of the following is a set :
(A) The collection of ten most talented writers of India.
(B) A collection of the most dangerous animals of the world.
(C) The collection of all boys in your class.
(D) A team of eleven best-cricket batsmen of the world.
Answer : (C) The collection of all boys in your class.
7. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of (– 6 – 24i) is ………….
Answer : Let z = – 6 – 24i
Then conjugate of z = – 6 + 24i
As given, (x – iy) (3 + 5i) = (– 6 + 24i)
(3x + 5y) + (5x – 3y)i = – 6 + 24i
on comparing real and imaginary parts,
3x + 5y = – 6 …………(i)
5x – 3y = 24 …………(ii)
Multiply eqn.(i) by 3 and eqn.(ii) by 5, then add both equations, so we get
x = 3 and y = – 3
8. Find the range of the function f(x) = x2 + 2, x is a real number …………..
Answer : As given x is real number so square of a real no. is always +ve,
x2 ≥ 0
x2 + 2 ≥ 0 + 2
f(x) ≥ 2
so, the range of f(x) is [2, ∞)
9. Find the value of sin(31π/3).
Answer : sin(31π/3) = sin(10π + π/3) = sin(π/3) = √3/2
10. The marks obtained by a student of Class XI in first and second terminal examination are 62 and 48 respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.
Answer : Let marks obtained in annual exam = x, then marks are 62, 48 and x
Average of numbers is (62+48+x)/3 which must be at least 60
(62+48+x)/3 ≥ 60
62 + 48 + x ≥ 180
x ≥ 70
Minimum marks he should get is 70.
11. Assertion (A) : sin15° = (√3–1)/2√2
Reason (R) : sin(A + B) = sinA cosB + cosA sinB
(i) Both (A) and (R) are true and (R) is the correct explanation of (A).
(ii) Both (A) and (R) are true and (R) is not the correct explanation of (A).
(iii) (A) is true and (R) is false.
(iv) (A) is false and (R) is true.
Answer : (ii) Both (A) and (R) are true and (R) is not the correct explanation of (A).
sin15° = sin(45° – 30°) = sin45° cos30° – cos45° sin30° = (1/√2 × √3/2) – (1/√2 × 1/2) = (√3–1)/2√2 [using sin(A – B) = sinA cosB – cosA sinB]
sin(A + B) = sinA cosB + cosA sinB
12. Write the following in set builder form :
(i) {2, 4, 8, 16, 32}
Answer : A = {x : x = 2n, 1 ≤ n ≤ 5, where n ∈ N}
(ii) {1/2, 2/3, 3/4, 4/5, 6/7}
Answer : A = {x : x = n/(n+1) where n = 1, 2, 3, 4, 6}
13. If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16}, D = {5, 10, 15, 20}. Evaluate the following :
(i) A – B
Answer : A – B = {3, 6, 9, 12, 15, 18, 21} – {4, 8, 12, 16, 20} = {3, 6, 9, 15, 18, 21}
(ii) C – D
Answer : C – D = {2, 4, 6, 8, 10, 12, 14, 16} – {5, 10, 15, 20} = {2, 4, 6, 8, 12, 14, 16}
14. Prove that : tan3x.tan2x.tanx = tan3x – tan2x – tanx
Answer : Taking tan3x = tan(2x + x) = (tan2x + tanx)/(1 – tan2x tanx)
using formula, tan(x + y) = (tanx + tany) / (1 – tanx tany)
tan3x = (tan2x + tanx)/(1 – tan2x tanx)
tan3x (1 – tan2x tanx) = tan2x + tanx
tan3x – tan3x tan2x tanx = tan2x + tanx
tan3x – tan2x – tanx = tan3x tan2x tanx
tan3x tan2x tanx = tan3x – tan2x – tanx
Hence Proved.
15. Write the relation R = {(x, x3) : x is a prime number less than 10} in roster form.
Answer : Prime numbers less than 10 are 2, 3, 5, 7
so, R = {(2, 23), (3, 33), (5, 53), (7, 73)} = {(2, 8), (3, 27), (5, 125), (7, 343)}
16. Prove that : 2sin2(3π/4) + 2cos2(π/4) + 2sec2(π/3) = 10
Answer : L.H.S. = 2sin2(3π/4) + 2cos2(π/4) + 2sec2(π/3)
= 2[sin(3π/4)]2 + 2[cos(π/4)]2 + 2[sec(π/3)]2
= 2[sin(π – π/4)]2 + 2(1/√2)2 + 2(2)2
= 2(1/√2)2 + 2(1/2) + 2(4)
= 2(1/2) + 1 + 8
= 1 + 1 + 8 = 10 = R.H.S.
17. Solve: 3(x – 2)/5 ≤ 5(2 – x)/3
Answer : 3(x – 2)/5 ≤ 5(2 – x)/3
Multiplying by 15 throughout, we get
9(x – 2) ≤ 25(2 – x)
9x – 18 ≤ 50 – 25x
9x + 25x ≤ 50 + 18
34x ≤ 68
x ≤ 2
x ∈ (– ∞, 2]
18. Prove that : sin3x + sin2x – sinx = 4sinx cos(x/2) cos(3x/2)
Answer : L.H.S. sin3x + sin2x – sinx
= (sin3x – sinx) + sin2x
Using sinC – sinD = 2cos(C + D)/2 sin(C – D)/2 and sin2x = 2sinx cosx
= 2cos(3x + x)/2 sin(3x – x)/2 + sin2x
= 2cos2x sinx + 2sinx cosx
= 2sinx(cos2x + cosx)
= 2sinx[2cos(3x/2) cos(x/2)]
Using cosC + cosD = 2cos(C + D)/2 cos(C – D)/2
= 4sinx cos(x/2) cos(3x/2) = R.H.S.
Hence Proved.
OR
Prove that : tan4x = 4tanx(1 – tan2x) / [1 – 6tan2x + tan4x]
Answer : L.H.S. tan4x = tan2(2x)
Using tan2A = 2tanA/(1–tan2A)
so, tan4x = 2tan2x/(1–tan22x) …….(i)
Now tan2x = 2tanx/(1–tan2x) …….(ii)
Put (ii) in (i), then solve, we get
= 4tanx(1 – tan2x) / [1 – 6tan2x + tan4x] = R.H.S.
Hence Proved.
19. If a + ib = (x + i)2 / (2x2 + 1), then prove that : a2 + b2 = (x2 + 1)2 / (2x2 + 1)
Answer : As given, a + ib = (x + i)2 / (2x2 + 1) = (x2 – 1 + 2x) / (2x2 + 1)
Compare real and imaginary parts,
a = (x2 – 1) / (2x2 + 1) …….(i)
b = (2x) / (2x2 + 1) ………(ii)
Squaring and adding (i) and (ii),
a2 + b2 = [(x2 – 1)/(2x2 + 1)]2 + [(2x) / (2x2 + 1)]2
= [(x2 – 1)2 + (2x)2] / (2x2 + 1)2
:. Using (a – b)2 + 4ab = (a + b)2
= (x2 + 1)2 / (2x2 + 1)
Hence Proved.
OR
If z1 = 2 – i, z2 = 1 + i, find |(z1 + z2 + 1) / (z1 – z2 + 1)|
Answer : |(z1 + z2 + 1) / (z1 – z2 + 1)|
= [(2–i) + (1+i) + 1] / [(2–i) – (1+i) + 1]
= (4) / (2–2i)
= 2 / (1–i) × (1+i) / (1+i)
= 2(1+i) / (12–i2) = 2(1+i) / (1+1) = 1 + i
|(z1 + z2 + 1) / (z1 – z2 + 1)| = |1 + i| = √12+12 = √2
[Here, z = a + ib so |z| = √a2+b2 ]
20. Mathematics teacher Mr. Srinivas Rao explained the concept of subsets of a set which includes subset, element of a set, null set etc. Based on these concepts, the teacher asked a few questions to the students. Try to answer the following :
Let A = {1, 2, {3, 4}, 5}, which of the following statements are correct and which are incorrect.
(i) {1, 2, 3} ⊂ A
Answer : incorrect (because 3 ∉ A)
(ii) {3, 4} ∈ A
Answer : correct
(iii) Φ ⊂ A
Answer : correct (because empty set is subset of every set)
(iv) { {3, 4} } ⊂ A
Answer : correct (because {3, 4} ∈ A)