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ALP Smart Syllabus Grade 11 Math Solved Model Paper 2027 – Punjab Board
Math Smart Syllabus ALP 2027 Model Paper
Prepare for the Grade 11 Mathematics Annual Examination 2027 with the latest ALP Smart Syllabus Mathematics Model Paper 2027, designed according to the new syllabus issued by PECTAA Punjab. This Mathematics Test Series for Class 11 includes a Class Test 1, Full Book Test, and Chapter Wise Test Series to help students prepare effectively for Punjab Board examinations. These Punjab Board 2027 Model Papers are designed to support exam preparation, improve mathematical concepts, and familiarise students with the expected examination pattern. Download and practise the Grade 11 Mathematics Model Papers 2027 to strengthen your preparation for the Annual Examination.
MODEL PAPER OF MATHEMATICS FOR GRADE-11
FOR (ANNUAL EXAM-2027) — Objective Type
Time Allowed: 30 MinutesMaximum Marks: 20
1. Note: Four possible answers A, B, C and D to each question are given. The choice which you think is correct, fill that circle with marker or ink pen in the answer-book. Cutting or filling two or more circles will result in zero mark in that question.
i. The multiplicative identity in complex number is:
(A) (1, 1)(B) (0, 0)(C) (1, 0)(D) (0, 1)
Ans: (C) — (a,b)(1,0) = (a,b).
ii. The sum of all the three cube roots of unity is:
(A) 1(B) 2(C) 3(D) 0
Ans: (D) — 1 + ω + ω² = 0.
iii. The domain of the function f(x) = √(4 − x²) is:
(A) (−2, 2)(B) (−2, ∞)(C) (−∞, 2)(D) [−2, 2]
Ans: (D) — 4 − x² ≥ 0 ⇒ −2 ≤ x ≤ 2.
iv. What is the maximum value of f(x) = −2x² + 4x + 3?
(A) 1(B) 3(C) 5(D) 7
Ans: (C) — vertex x = −b/2a = 1; f(1) = −2 + 4 + 3 = 5.
v. If order of matrix A is m×n and order of matrix B is n×p then order of AB is:
(A) n×p(B) m×n(C) m×p(D) p×n
Ans: (C)
vi. If C is a matrix of order 3×3 and k ∈ R is any scalar then |kC| is equal:
(A) k|C³|(B) k³|C³|(C) k³|C|(D) (kC)³
Ans: (C) — |kC| = kn|C|, n = 3.
vii. An expression of the form p(x)q(x), q(x) ≠ 0 is called:
(A) Partial fraction(B) Rational fraction(C) Irrational fraction(D) Polynomial
Ans: (B)
viii. The Arithmetic mean between 3√3 and √3 is:
(A) 4(B) 4√3(C) 2√3(D) √3
Ans: (C) — (3√3 + √3)/2 = 2√3.
ix. The next term of the sequence 1, 3, 9, 27,… is:
(A) 30(B) 60(C) 81(D) 99
Ans: (C) — r = 3; 27 × 3 = 81.
x. 0! is equal to
(A) 0(B) 1(C) 2(D) 3
Ans: (B)
xi. The number of diagonals of 6-sided figure is:
(A) 9(B) 18(C) 20(D) 35
Ans: (A) — n(n−3)/2 = 6·3/2 = 9.
xii. The remainder when f(x) = x&sup4; + x³ + x² + 1 is divided by x + 1 is:
(A) 1(B) 2(C) 3(D) 4
Ans: (B) — f(−1) = 1 − 1 + 1 + 1 = 2.
xiii. Cosec(π2 + α) is equal to:
(A) cosα(B) sinα(C) secα(D) cosecα
Ans: (C) — sin(π/2 + α) = cosα, so cosec = secα.
xiv. The value of cos(π2 + π6) is:
(A) −15(B) 12(C) −12(D) √32
Ans: (C) — cos(90° + 30°) = −sin30° = −1/2.
xv. The range of y = tan(x) is:
(A) (−∞, 0)(B) (−∞, ∞)(C) (−∞, −1) ∪ (1, ∞)(D) (−1, 1)
Ans: (B)
xvi. limx→a xp is equal to:
(A) p(B) a(C) pa(D) ap
Ans: (D)
xvii. The notation used by Newton to denote derivative of y = f(x) is:
(A) dydx(B) ḟ(x)(C) f′(x)(D) Df(x)
Ans: (B) — Newton: dot notation; Leibniz: dy/dx; Lagrange: f′; Cauchy: D.
xviii. The derivative of f(x) = x² − 3 at x = 2?
(A) 4(B) 2(C) 3(D) 5
Ans: (A) — f′(x) = 2x; f′(2) = 4.
xix. Angle between vectors p and q, if p·q = 35 and |p×q| = 35 is:
(A) 30°(B) 45°(C) 60°(D) 90°
Ans: (B) — tanθ = |p×q| / p·q = 1 ⇒ θ = 45°.
xx. Magnitude of a vector perpendicular to both u = 2i − 3j + k and v = i + 4j − 2k is:
(A) √50(B) √150(C) √500(D) √1550
Ans: (B) — u×v = (6−4)i − (−4−1)j + (8+3)k = 2i + 5j + 11k; magnitude = √(4+25+121) = √150.
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Subjective Type (Part-I)
Time allowed: 2:30 hoursMax. Marks: 80
2. Write short answers to any EIGHT (8) questions:2 × 8 = 16
(i) If z1 = (4, 2) and z2 = (3, −1), then find z1/z2.
z1/z2 = (4+2i)/(3−i) = (4+2i)(3+i)/(9+1) = (12+4i+6i−2)/10 = (10+10i)/10 = 1 + i = (1, 1).
(ii) Show that: in+1 + in+2 + in+3 + in+4 = 0, for all n ∈ N.
LHS = in+1(1 + i + i² + i³) = in+1(1 + i − 1 − i) = in+1(0) = 0 = RHS.
(iii) Find the square root of complex number 5 + 12i.
Let √(5+12i) = x+iy ⇒ x²−y² = 5, 2xy = 12. Since (x²+y²)² = 25 + 144 = 169, x²+y² = 13. So x² = 9, y² = 4. Since xy > 0: ±(3 + 2i).
(iv) If ω is an imaginary cube root of unity, prove that a + bω² + cωaω + bω²… — i.e. (a + bω² + cω)/(aω² + bω + c) = ω.
Multiply denominator by ω: ω(aω² + bω + c) = aω³ + bω² + cω = a + bω² + cω (as ω³ = 1). Hence numerator = ω × denominator, so ratio = ω.
(v) Given f(x) = x³ − ax² + bx + 1. If f(2) = −3 and f(−1) = 0, find a and b.
f(2) = 8 − 4a + 2b + 1 = −3 ⇒ 2a − b = 6. f(−1) = −1 − a − b + 1 = 0 ⇒ a + b = 0. Adding: 3a = 6, a = 2, b = −2.
(vi) Find the domain and range of f(x) = √(x² − 9).
x² − 9 ≥ 0 ⇒ Domain = (−∞, −3] ∪ [3, ∞). Range = [0, ∞).
(vii) Find the inverse of f(x) = 3x² − 2x + 6, x ≥ 5.
y = 3x² − 2x + 6 ⇒ 3x² − 2x + (6 − y) = 0 ⇒ x = [1 ± √(3y − 17)]/3. As x ≥ 5, take +. f−1(x) = [1 + √(3x − 17)]/3.
(viii) Solve |x² + 1| = 5.
x² + 1 > 0, so x² + 1 = 5 ⇒ x² = 4 ⇒ x = ±2.
(ix) If A and B are square matrices of the same order, explain why in general (A+B)² ≠ A² + 2AB + B².
(A+B)² = A² + AB + BA + B². Matrix multiplication is not commutative (AB ≠ BA in general), so AB + BA ≠ 2AB.
(x) Find the values of x if the determinant |2 1 x; −1 −4 −3; x 1 0| = 5.
Expand along R1: 2(0 + 3) − 1(0 + 3x) + x(−1 + 4x) = 6 − 3x − x + 4x² = 4x² − 4x + 6 = 5 ⇒ 4x² − 4x + 1 = 0 ⇒ (2x−1)² = 0 ⇒ x = ½.
(xi) If A is a square matrix of order 3, show that |kA| = k³|A|.
kA multiplies each of the 3 rows of A by k. Taking k common from each row gives k·k·k|A| = k³|A|.
(xii) Resolve 2x² − 1 into partial fractions.
2/[(x−1)(x+1)] = A/(x−1) + B/(x+1). 2 = A(x+1) + B(x−1). x = 1: A = 1; x = −1: B = −1. = 1/(x−1) − 1/(x+1).
3. Write short answers to any EIGHT (08) questions:2 × 8 = 16
(i) Which term of the A.P. 3, 8, 13, … is 123?
a = 3, d = 5: 3 + (n−1)5 = 123 ⇒ n − 1 = 24 ⇒ n = 25.
(ii) Find the sum of the first 100 positive integers.
S = n(n+1)/2 = 100×101/2 = 5050.
(iii) Find the 12th term of 1 + i, 2i, −2 + 2i, …
r = 2i/(1+i) = 1 + i (G.P.). a12 = (1+i)(1+i)11 = (1+i)12 = [(1+i)²]6 = (2i)6 = 64i6 = −64.
(iv) If 5 is the harmonic mean between 2 and b, find b.
H = 2ab/(a+b): 5 = 4b/(2+b) ⇒ 10 + 5b = 4b ⇒ b = −10.
(v) Evaluate 9!6! 3!.
= (9·8·7)/(3·2·1) = 504/6 = 84.
(vi) How many signals can be given by 6 flags of different colours, using 2 flags at a time?
6P2 = 6 × 5 = 30.
(vii) How many diagonals and triangles can be formed by joining the vertices of a polygon having 15 sides?
Diagonals = n(n−3)/2 = 15×12/2 = 90. Triangles = 15C3 = 15·14·13/6 = 455.
(viii) When 4x&sup4; + 2x³ + kx² + 13 is divided by x + 1, the remainder is 16. Find k.
f(−1) = 4 − 2 + k + 13 = 16 ⇒ k = 1.
(ix) For P(x) = 3x³ − 4x² + 2x − 5, predict the value at x = −1.
P(−1) = −3 − 4 − 2 − 5 = −14.
(x) Prove: sin 210° + cos 240° + tan 225° + cot 225° = 1.
sin210° = −½, cos240° = −½, tan225° = 1, cot225° = 1. Sum = −1 + 2 = 1. ✓
(xi) Prove: cos(α + 45°) = 1√2(cosα − sinα).
cosαcos45° − sinαsin45° = (1/√2)cosα − (1/√2)sinα = (1/√2)(cosα − sinα). ✓
(xii) Express sin 5x + sin 7x as a product.
sinP + sinQ = 2 sin[(P+Q)/2] cos[(P−Q)/2] = 2 sin 6x cos x.
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4. Write short answers to any NINE (09) questions:2 × 9 = 18
(i) Find the period of 12 sin(3x2 − π2).
Period = 2π/(3/2) = 4π/3.
(ii) Find the maximum and minimum values of 32 + cos(x − π4).
−1 ≤ cos(·) ≤ 1 ⇒ Max = 3/2 + 1 = 5/2, Min = 3/2 − 1 = 1/2.
(iii) Evaluate limx→3 (x − 3)/(√x − √3).
x − 3 = (√x − √3)(√x + √3), so limit = lim (√x + √3) = 2√3.
(iv) Evaluate limn→+∞ (1 + 4/n)n.
Using lim(1 + a/n)n = ea: e&sup4;.
(v) Define divergent sequence.
A sequence which is not convergent (has no finite limit) is called divergent, e.g. 1, 2, 3, 4, …
(vi) s(t) = 4t² + 2t + 1 (miles). Find the average velocity over [2, 5].
s(2) = 21, s(5) = 111. Avg velocity = (111 − 21)/(5 − 2) = 90/3 = 30 miles/hour.
(vii) Find the gradient of f(x) = 3x² + 2x at x = 1.
f′(x) = 6x + 2; f′(1) = 8.
(viii) Find dy/dx if y = 34x&sup4; + 23x³ + 12x² + 2x + 5.
dy/dx = 3x³ + 2x² + x + 2.
(ix) Differentiate (2x − 3)/(2x + 1) w.r.t. x.
= [(2x+1)(2) − (2x−3)(2)]/(2x+1)² = (4x+2−4x+6)/(2x+1)² = 8/(2x+1)².
(x) If u = 2i+3j+k, v = 4i+6j+2k, w = −6i−9j−3k, show that u, v, w are parallel.
v = 2u and w = −3u, so all are scalar multiples of u ⇒ parallel.
(xi) Find t so that |2i + (t−1)j + tk| = √13.
4 + (t−1)² + t² = 13 ⇒ 2t² − 2t − 8 = 0 ⇒ t² − t − 4 = 0 ⇒ t = (1 ± √17)/2.
(xii) Work done by F = 2i + 5j + 3k moving from P1(2, −3, 1) to P2(7, 5, 3).
d = (5, 8, 2). W = F·d = 10 + 40 + 6 = 56 units.
(xiii) Show that |a×b|² = |a|²|b|² − (a·b)².
|a×b|² = |a|²|b|²sin²θ = |a|²|b|²(1 − cos²θ) = |a|²|b|² − (|a||b|cosθ)² = |a|²|b|² − (a·b)². ✓
Subjective (Part-II)
Note: Attempt any three questions.3 × 10 = 30
5. (a) If z1 = x + yi, z2 = a + bi, find x, y, a, b such that z1 + z2 = 10 + 4i and z1 − z2 = 6 + 2i. (5)
Step 1: (x+a) + (y+b)i = 10 + 4i ⇒ x + a = 10, y + b = 4.
Step 2: (x−a) + (y−b)i = 6 + 2i ⇒ x − a = 6, y − b = 2.
Step 3: Add: 2x = 16 ⇒ x = 8; so a = 2. Add: 2y = 6 ⇒ y = 3; so b = 1.
Ans: x = 8, y = 3, a = 2, b = 1.
Step 2: (x−a) + (y−b)i = 6 + 2i ⇒ x − a = 6, y − b = 2.
Step 3: Add: 2x = 16 ⇒ x = 8; so a = 2. Add: 2y = 6 ⇒ y = 3; so b = 1.
Ans: x = 8, y = 3, a = 2, b = 1.
(b) Solve |x² − 3x + 2| > 4. (5)
Step 1: |A| > 4 ⇒ A > 4 or A < −4.
Step 2: x² − 3x + 2 > 4 ⇒ x² − 3x − 2 > 0; roots (3 ± √17)/2 ⇒ x < (3−√17)/2 or x > (3+√17)/2.
Step 3: x² − 3x + 2 < −4 ⇒ x² − 3x + 6 < 0; discriminant 9 − 24 < 0 ⇒ no real solution.
Solution set: (−∞, (3−√17)/2) ∪ ((3+√17)/2, ∞).
Step 2: x² − 3x + 2 > 4 ⇒ x² − 3x − 2 > 0; roots (3 ± √17)/2 ⇒ x < (3−√17)/2 or x > (3+√17)/2.
Step 3: x² − 3x + 2 < −4 ⇒ x² − 3x + 6 < 0; discriminant 9 − 24 < 0 ⇒ no real solution.
Solution set: (−∞, (3−√17)/2) ∪ ((3+√17)/2, ∞).
6. (a) Show that det[ a−b−c, 2a, 2a ; 2b, b−c−a, 2b ; 2c, 2c, c−a−b ] = (a+b+c)³. (5)
Step 1: R1 → R1+R2+R3: every entry of row 1 becomes (a+b+c). Take it common: (a+b+c)·|1 1 1; 2b b−c−a 2b; 2c 2c c−a−b|.
Step 2: C2 → C2−C1, C3 → C3−C1: row 1 = (1, 0, 0); row 2 = (2b, −(a+b+c), 0); row 3 = (2c, 0, −(a+b+c)).
Step 3: Expand along R1: 1·[(a+b+c)(a+b+c)] = (a+b+c)².
Step 4: Result = (a+b+c)(a+b+c)² = (a+b+c)³. ✓
Step 2: C2 → C2−C1, C3 → C3−C1: row 1 = (1, 0, 0); row 2 = (2b, −(a+b+c), 0); row 3 = (2c, 0, −(a+b+c)).
Step 3: Expand along R1: 1·[(a+b+c)(a+b+c)] = (a+b+c)².
Step 4: Result = (a+b+c)(a+b+c)² = (a+b+c)³. ✓
(b) Resolve (x² − 10x + 13)/[(x−1)(x² − 5x + 6)] into partial fractions. (5)
Step 1: x²−5x+6 = (x−2)(x−3). Let = A/(x−1) + B/(x−2) + C/(x−3).
Step 2: x²−10x+13 = A(x−2)(x−3) + B(x−1)(x−3) + C(x−1)(x−2).
Step 3: x=1: 4 = 2A ⇒ A = 2. x=2: −3 = −B ⇒ B = 3. x=3: −8 = 2C ⇒ C = −4.
Ans: 2/(x−1) + 3/(x−2) − 4/(x−3).
Step 2: x²−10x+13 = A(x−2)(x−3) + B(x−1)(x−3) + C(x−1)(x−2).
Step 3: x=1: 4 = 2A ⇒ A = 2. x=2: −3 = −B ⇒ B = 3. x=3: −8 = 2C ⇒ C = −4.
Ans: 2/(x−1) + 3/(x−2) − 4/(x−3).
7. (a) For what value of n is (an + bn)/(an−1 + bn−1) the positive G.M. between a and b? (5)
Step 1: Set equal to √(ab): an + bn = an−½b½ + a½bn−½.
Step 2: an − an−½b½ = a½bn−½ − bn ⇒ an−½(a½ − b½) = bn−½(a½ − b½).
Step 3: Cancel (a½−b½) ⇒ an−½ = bn−½ ⇒ n − ½ = 0. n = ½.
Step 2: an − an−½b½ = a½bn−½ − bn ⇒ an−½(a½ − b½) = bn−½(a½ − b½).
Step 3: Cancel (a½−b½) ⇒ an−½ = bn−½ ⇒ n − ½ = 0. n = ½.
(b) How many numbers between 100 and 1000 can be formed using digits 0,1,2,3,4,5 without repetition? How many are divisible by 5? (5)
Step 1 (all): Hundreds: 5 ways (not 0); tens: 5; units: 4 ⇒ 5×5×4 = 100.
Step 2 (ends in 0): 5 × 4 = 20.
Step 3 (ends in 5): hundreds 4 ways (not 0, 5), tens 4 ways ⇒ 16.
Divisible by 5: 20 + 16 = 36.
Step 2 (ends in 0): 5 × 4 = 20.
Step 3 (ends in 5): hundreds 4 ways (not 0, 5), tens 4 ways ⇒ 16.
Divisible by 5: 20 + 16 = 36.
8. (a) Prove that sin(π/9) sin(2π/9) sin(π/3) sin(4π/9) = 3/16. (5)
Step 1: In degrees: sin20° sin40° sin60° sin80°.
Step 2: Identity sinθ sin(60°−θ) sin(60°+θ) = ¼ sin3θ. With θ = 20°: sin20° sin40° sin80° = ¼ sin60° = √3/8.
Step 3: Multiply by sin60° = √3/2: (√3/8)(√3/2) = 3/16. ✓
Step 2: Identity sinθ sin(60°−θ) sin(60°+θ) = ¼ sin3θ. With θ = 20°: sin20° sin40° sin80° = ¼ sin60° = √3/8.
Step 3: Multiply by sin60° = √3/2: (√3/8)(√3/2) = 3/16. ✓
(b) f(x) = 2x + 3 for x ≤ 1; −x + 4 for x > 1. Discuss limit and continuity at x = 1. (5)
Step 1: LHL = lim(2x+3) = 5. Step 2: RHL = lim(−x+4) = 3. Step 3: f(1) = 5.
LHL ≠ RHL ⇒ limit does not exist, hence f is discontinuous at x = 1.
LHL ≠ RHL ⇒ limit does not exist, hence f is discontinuous at x = 1.
9. (a) Differentiate (√x + 1)(x3/2 − 1)/(x3/2 − x1/2) w.r.t. x. (5)
Step 1: Denominator = √x(x − 1). Numerator: x3/2 − 1 = (√x − 1)(x + √x + 1), so (√x+1)(√x−1)(x+√x+1) = (x−1)(x+√x+1).
Step 2: y = (x + √x + 1)/√x = √x + 1 + x−1/2.
Step 3: dy/dx = 1/(2√x) − 1/(2x3/2).
Step 2: y = (x + √x + 1)/√x = √x + 1 + x−1/2.
Step 3: dy/dx = 1/(2√x) − 1/(2x3/2).
(b) If |a| = 3, |b| = 4 and |a + b| = 5, find the angle between a and b. (5)
Step 1: |a+b|² = |a|² + |b|² + 2a·b ⇒ 25 = 9 + 16 + 2a·b ⇒ a·b = 0.
Step 2: cosθ = 0 ⇒ θ = 90°.
Step 2: cosθ = 0 ⇒ θ = 90°.
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