Solution PERIODIC TEST -1 Class 12 - Mathematics 1. (b) 2 Explanation: Total possible pairs {(1, 1) (1, 3), (1, 5), (3, 3), (3, 1), (3, 5) (5, 5), (5, 1), (5, 3)} 1st equivalence relation R1 = {(1, 1,), (2, 2), (3, 3), (1, 3), (3, 1)} 2nd equivalence relation R2 = {(1, 1,), (2, 2), (3, 3), (1, 3), (3, 1), (3, 5), (5, 3)} no of possible equivalence relation =2 ∴ 2. (c) Reflexive, transitive but not symmetric Explanation: Since n divides n, ∀ n ∈ N, R is reflexive. R is not symmetric since for (3, 6) ∈ R but (6, 3)∉R. R is transitive since for n, m, r whenever n/m and m/r ⇒ n/r, i.e., n divides m and m divides r, then n will divide r. 3. (d) bijection Explanation: Given that f : [-1/2, 1/2] → [π /2, π /2] where f(x) = sin-1 (3x - 4x3) Put x = sinθ in f(x) = sin-1 (3x - 4x3) ⇒ f (x = sin θ) = sin −1 3 (3 sin θ − 4 sin θ ) f(x) = sin-1 (sin3θ) ⇒ f(x) = 3θ ⇒ ⇒ f(x) = 3 sin-1x If f(x) = f(y) Then 3 sin-1x = 3 sin-1y ⇒ x = y So, f is one-one. y = 3 sin-1x ⇒ x = sin y 3 x ∈ R also y ∈ R so f is onto. Hence, f is bijection. ∵ 4. (b) equivalence Explanation: Given that, R be a relation on T defined as aRb if a is congruent to b ∀ a, b ∈ T Now, aRa ⇒ a is congruent to a, which is true since every triangle is congruent to itself. ⇒ (a,a) ∈ R ∀ a ∈ T ⇒ R is reflexive. Let aRb ⇒ a is congruent to b b is congruent to a ⇒ bRa ∴ (a,b) ∈ R ⇒ (b,a) ∈ R ∀ a, b ∈ T ⇒ R is symmetric. Let aRb ⇒ a is congruent to b and bRc ⇒ b is congruent to c ⇒ a is congruent to c ⇒ aRc (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) ∈ R ∀ a, b,c ∈ T ⇒ R is transitive. Hence, R is an equivalence relation. ⇒ ∴ 1 / 10 5. (b) a bijection Explanation: Given that A = {x : −1 ≤ x ≤ 1} and f : A → A such that f(x) = x |x|. For x < 0, f(x) < 0 ⇒ y = -x2 − − − ⇒ x = √−y , which is not possible for x > 0. Hence, f is one-one and onto. ∴ the given function is bijective. 6. Let us take A = {2,4,6} Define a relation R on A as: A = {(2,2), (4,4), (6,6), (2,4), (4,2), (4,6), (6,4)} Relation of R is reflexive as for every a ∈ A, (a,a) ∈ R ⇒ (2,2), (4,4), (6,6) ∈ R, Relation R is symmetric as (a,b) ∈ R ⇒ (b,a) ∈ R for all a ,b ∈ R And Relation R is not transitive as (2,4), (4,6) ∈ R, but (2,6) ∉ R Therefore, relation R is reflexive and symmetric but not transitive. 7. Given: ∀ a, b ∈ Z, a R b if and only if a – b is divisible by 5. Now, R is Reflexive if (a, a) ∈ R ∀ a ∈ Z a R a = (a-a) is divisible by 5. a - a = 0 = 0 × 5 [since 0 is multiple of 5 it is divisible by 5] = a - a is divisible by 5 = (a, a) ∈ R Thus, R is reflexive on Z. R is Symmetric if (a,b) ∈ R ⇒ (b,a) ∈ R ∀ a,b ∈ Z (a,b) ∈ R = (a-b) is divisible by 5 = (a - b) = 5z for some z ∈ Z = -(b - a) = 5z = b - a = 5(-z) [∵ z ∈ Z ⇒ -z ∈ Z] = (b - a) is divisible by 5 = (b,a) ∈ R Thus, R is symmetric on Z. R is Transitive if (a,b) ∈ R and (b,c) ∈ R = (a,c) ∈ R ∀ a,b,c ∈ Z (a,b) ∈ R = (a-b) is divisible by 5 = a-b = 5z1 for some z1∈ Z (b,c) ∈ R = (b-c) is divisible by 5 = b-c = 5z2 for some z2∈ Z Now, a-b = 5z1 and b-c = 5z2 = (a-b) + (b-c) = 5z1 + 5z2 = a-c = 5(z1 + z2 ) = 5z3 where z1 + z2 = z3 = a-c = 5z3 [∵ z1,z2 ∈ Z ⇒ z3∈ Z] = (a-c) is divisible by 5. = (a, c) ∈ R Thus, R is transitive on Z. Since R is reflexive, symmetric and transitive it is an equivalence relation on Z. 8. Given that, R = {(1, 39), (2, 37), (3, 35) .... (19, 3), (20, 1)} Domain = {1,2,3,.......,20} Range = {1,3,5,7......,39} 2 / 10 R is not reflexive as (2, 2) ∉ R as 2 × 2 + 2 ≠ 41 R is not symmetric as (1, 39) ∈ R but (39, 1) ∉ R R is not transitive as (11, 19) ∈ R, (19, 3) ∈ R But (11, 3) ∉ R Hence, R is neither reflexive, nor symmetric and nor transitive. 9. i. (d) (X,Y) ∉ R ii. (a) both (X,W) and (W,X) ∈ R iii. (a) (F1, F2) ∈ R, (F2, F3) ∈ R and (F1, F3) ∈ R iv. (c) Equivalence relation v. (a) All those eligible voters who cast their votes 10. (b) − 24 25 Explanation: put cos −1 sin 2θ = 2sinθcosθ = 11. 2.4 5 3 5 ) = θ ⟹ . (− 3 5 ) = (− cos θ = − 24 25 3 5 therefore the given expression is ). π (a) 4 Explanation: sin = sin 12. (− −1 (sin π 4 −1 )= (sin π 3π 4 ) = sin −1 (sin(π − π 4 ) . 4 π (d) 6 Explanation: Let the principle value be given by x Now, let x = cosec-1(2) ⇒ cosec x = 2 ⇒ cosec x = cosec ( (c) (− (b) [ 2 −π 2 ) (∵ cos( π 6 ) = 2) 6 π , π 2 ) Explanation: (− 14. 6 π ⇒ x = 13. π , π 2 ] π 2 , π 2 ) - {0} Explanation: To Find: The range of coses-1(x) Here, the inverse function is given by y = f-1(x) The graph of the function coses-1(x) can be obtained from the graph of Y = coses-1(x) by interchanging x and y axes.i.e, if a, b is a point on Y = cosec x then b, a is the point on the function y = coses1(x) Below is the Graph of the range of coses-1(x) From the graph, it is clear that the range of coses-1(x) is restricted to interval [ , ] - {0} −π 2 15. Let tan π 2 −1 – (− √3) = y – ⇒ tan y = − √3 3 / 10 π ⇒ tan y = − tan 3 π ⇒ tan y = tan(− ) 3 Since, the principal value branch of tan-1 is [− Therefore, principal value of tan 16. sin −1 ∘ (sin(− 600 )) = sin = sin = sin = sin −1 −1 −1 10π {sin(− 3 π 3 π 3 – (− √3) {sin(−600 × )} = sin {− sin(3π + (sin −1 −1 −1 (− sin )} = sin −1 10π 3 π is − . 3 π 180 )} ) {− (− sin ⇒ ( 2 cos x 2 sin 1− cos 2 2 = ) = tan −1 [∵ 1 − cos sin x x x π 3 )} 3 −1 ⇒ tan . π ) = 2 cos x ] 2 π 17. Here, we are required to find the value of x, Now, the given equation is 2 tan (cos x) = tan −1 π , 2 ( 2 2 sin x ) [∵ 2 tan x = sin 2 −1 −1 (2cosecx) x = tan −1 ( 2x 2 1−x ); −1 < x < 1] x] sinx cosx - sin2x = 0 ⇒ sinx(cosx - sinx) = 0 ⇒ sinx = 0 or cosx = sinx ⇒ sin x = sin 0 or cot x = 1 = cot π/4 ⇒ π ∴ x = 0 or 4 But here at x = 0, the given equation does not exist. π Hence, x = 18. is the only solution. 4 (b) 3 × 3 ⎡ 1 −1 Explanation: ⎢ 0 ⎣ ⎡ 1 = ⎢0 ⎣ ⎡ = ⎢ ⎣ 19. 2 −1 2 3 2 2 3 ⎤ ⎤ −1 ⎦ 2 ] 2 ⎦ 0 1 − 2×3 0 1 23 1 0 21 [ ] } 2×3 3×2 −1 −1 −21 1 0 −20 [ ⎥ 0 {[ ⎥ ] 2×3 3×2 −2 −1 2 0 1 −2 −1 ⎤ −40 ⎥ −102 ⎦ 3×3 (d) B2 = B and A2 =A Explanation: AB = A ...(i) BA = B ....(ii) From equation (ii) B × (AB) = B B2A= B From equation (ii) B2A = BA B2 = B From equation (i) A × (BA) = A A2B= A From equation (i) A2B = AB A2 = A Hence, A2 = A & B2 = B 20. (c) A skew-symmetric matrix Explanation: Given A and B are symmetric matrices A'= A ...(i) 4 / 10 B'= B ...(ii) Now (AB – BA)'= (AB)'– (BA)’ = B'A'– A’B’ [∵ (AB)' = B’A'] = BA – AB [Using (i) and (ii)] ∴ (AB – BA)' = - (AB - BA) AB - BA is a skew symmetric matrix. 21. (c) square matrix Explanation: We know that as matrix is said to be square matrix if the number of rows is equal to the number of columns. 0 0 4 Therefore, the given matrix P = ⎢ 0 4 0⎥ ⎡ ⎣ 4 0 0 ⎤ is a square matrix. ⎦ (c) A + At is symmetric Explanation: For every square matrix (A + A’ ) is always symmetric. Since (A+A')'=A'+A=A+A' 22. 1 2 −1 3 1 2 23. We have A = [ A + (B + C) = [ 5 0 1 2 2 24. Given: A = ⎢ 2 1 2⎥ 2 1 2 2 2 1 6 0 ]+ [ 8 0 1 −2 ] ] 3 2 0 7 ]= [ 2 5 2 ],C = [ 6 3 and (A + B) + C = [ ⎣ 0 1 ]+ [ −1 ⎡ 4 ],B = [ 7 2 1 6 ]= [ 1 −2 ] = A + (B + C) ⎤ ⎦ To prove A2 – 4A – 5 I2 = O2 Here I2 is the identity matrix of order 2 and O2 is the zero matrix of order 2. Now, we will find the matrix for A2, we get 1 2 2 1 2 2 = A× A= ⎢2 1 2⎥⎢2 1 2⎥ 2 1 2 1 ⎡ 2 A ⎣ ⇒ 2 ⎤⎡ ⎦⎣ 2 ⎤ ⎦ A2 1 × 1 + 2 × 2 + 2 × 2 1 × 2 + 2 × 1 + 2 × 2 1 × 2 + 2 × 2 + 2 × 1 =⎢ 2 × 1 + 1 × 2 + 2 × 2 2 × 2 + 1 × 1 + 2 × 2 2 × 2 + 1 × 2 + 2 × 1⎥ 2 × 2 + 2 × 1 + 1 × 2 2 × 2 + 2 × 2 + 1 × 1 ⎡ ⎣ 2 × 1 + 2 × 2 + 1 × 2 ⎤ ⎦ [as cij = ai1b1j + ai2b2j + … + ainbnj] ⇒ A2 = 1 + 4 + 4 2 + 2 + 4 2 + 4 + 2 ⎢2 + 2 + 4 4 + 1 + 4 4 + 2 + 2⎥ 4 + 2 + 2 4 + 4 + 1 ⎡ ⎣ 9 8 8 A2 =⎢ 8 9 8⎥ ⎡ ⇒ 2 + 4 + 2 ⎣ 8 8 9 ⎤ ⎦ ⎤ .........(1) ⎦ So, Substitute corresponding values from eqn(i) in equation A2 – 4A – 5 I2, we get 9 8 8 1 2 2 1 0 0 = ⎢8 9 8⎥ − 4⎢2 1 2⎥ − 5⎢0 1 0⎥ 8 8 9 2 1 0 1 9 8 8 4 × 1 4 × 2 4 × 2 5 0 0 = ⎢8 9 8⎥ − ⎢4 × 2 4 × 1 4 × 2⎥ − ⎢0 5 0⎥ 8 8 9 4 × 2 4 × 1 0 5 9 8 8 4 8 8 5 0 0 = ⎢8 9 8⎥ − ⎢8 4 8⎥ − ⎢0 5 0⎥ 8 9 8 4 0 5 ⎡ ⇒ ⎣ ⎡ ⇒ ⎣ ⎡ ⇒ ⎣ 8 ⎤ ⎦ ⎤ ⎡ ⎣ ⎡ ⎦ ⎣ ⎤ ⎡ ⎦ ⎣ 2 4 × 2 8 ⎤ ⎡ ⎦ ⎤ ⎦ ⎣ ⎡ ⎣ 0 0 ⎤ ⎦ ⎤ ⎦ ⎡ ⎣ 0 ⎤ ⎦ ⎤ ⎦ 5 / 10 9 − 4 − 5 8 − 8 − 0 8 − 8 − 0 = ⎢8 − 8 − 0 9 − 4 − 5 8 − 8 − 0⎥ 8 − 8 − 0 9 − 4 − 5 ⎡ ⇒ ⎣ 8 − 8 − 0 ⎤ ⎦ [as rij = aij + bij + cij], 0 0 0 ⇒= ⎢ 0 0 0⎥ ⎡ ⎣ 0 0 = O2 ⎦ 0 A2 – Hence the ⎤ 4A – 5I = O2 (Proved) 25. Given equations are 2x + 3y = [ 2 3 4 0 ...... (i) ] 2 −2 −1 5 3x + 2y = [ ] .....(ii) eqn. (i) × 3 - eqn. (ii) × 2 6x+9y = 6x+4y = − − 9 0 ] 4 −4 −2 10 [ ] = 5y y = 6 12 [ − = 1 5 2 13 14 −10 [ ] 2 13 14 −10 [ ] eqn. (i) × 2 - eqn. (ii) × 3 4 6 8 0 4x+6y=[ ] 6 −6 −3 15 9x+6y =[ − ] − = −5x x = − −2 12 11 −15 =[ 1 5 ] 2 −12 −11 15 [ ] 26. i. (a): Let F be the matrix representing the number of family members and R be the matrix representing the requirement of calories and proteins for each person. Then Men W omen Children F= F amily A 4 4 4 2 2 2 [ F amily B ] Calories M an R= ⎡ woman P roteins 2400 ⎢ 1900 ⎣ C hildren 1800 45 ⎤ 55 ⎥ 33 ⎦ ii. (b): The requirement of calories and proteins for each of the two families is given by the product matrix FR. FR = [ 4 4 ⎡ 2400 ] ⎢ 1900 2 =[ 4 2 2 ⎣ 1800 45 ⎤ 55 ⎥ 33 ⎦ 4(2400 + 1900 + 1800) 4(45 + 55 + 33) 2(2400 + 1900 + 1800) 2(45 + 55 + 33) ] Calories P roteins FR = [ 24400 532 F amily A ] 12200 266 F amily B iii. (c) iv. (c) Since, AB = B ...(i) and BA = A ..(ii) ∴ A2 + B2 = A⋅ A + B⋅ B = A(BA) + B(AB) [using (i) and (ii)] - (AB)A + (BA)B [Associative law] = BA + AB [using (i) and (ii)] =A+B 6 / 10 v. (a) A = (aij)m × n, B = (bij)n × p, C = (cij)p × q BC = (bij)n × p × (Cij)p × q = (dij)n × q (BC)A = (dij)n × q × (aij)m × M Hence, (BC)A is possible only when m = q 27. (c) 1 2 2 4 3 −5 [ ] Explanation: A = BX B-1A = B-1BX X = B-1A Using Adjoint method of inverse −1 B 1 = 2 2 0 0 1 [ ] X = B-1A X = X = 28. 1 2 1 2 2 0 [ 1 2 3 −5 ][ 0 1 2 4 3 −5 [ ] ] (b) none of these Explanation: 3A3 + 2A2 + 5A + I = 0 By pre-operating by A-1, we get 29. ⇒ 3A-1 A3 + 2A-1 A2 + 5A-1 A + A-1 I = A-1 0 ⇒ 3A2 + 2A + 5 I + A-1 = 0, By making A-1 as the subject, we get ⇒ A-1 = -(3A2 + 2A + 5I) (c) Idempotent Explanation: clearly for given matrix A Therefore idempotent 2 30. (b) = A 1 det(A) Explanation: We know that, A-1 = So, ∣∣A −1 ∣ ∣ = ∣∣ = = 1 |A| 1 | Adj(A)| n |A| n−1 1 |A| Adj (A) ∣ Adj(A) ∣ n |A| 1 |A| = 1 1 |A| 1 = 1 |A| {since adj(A) is of order n and |Adj(A)| = |A|n-1} 31. (d) A Explanation: A2 = I A-1A2 = A-1I A = A-1 32. a 11 = 2, a12 = −3, a13 = 5 A31 = −12, A32 = 22, A33 = 18 L.H.S = a 11 A31 + a12 A32 + a13 A33 = 2 (−12) + (−3) (22) + 5 (18) = 0 Hence proved. 33. Given set of lines are:x+y-z=1 3x + y - 2z = 3 x-y-z=-1 Converting the following equations in matrix form, AX = B 7 / 10 1 1 −1 ⎢3 1 −2 ⎥ ⎢ y ⎥ = ⎢ ⎡ ⎣ 1 −1 x ⎤⎡ ⎦⎣ −1 z ⎤ 1 ⎡ ⎦ ⎣ ⎤ ⎡ ⎤ 3 ⎥ −1 ⎦ R2 - 3R1 R3 - R1 1 ⎡ ⎢0 ⎣ 0 1 −1 −2 1 −2 0 x ⎤⎡ 1 ⎥⎢y ⎥ = ⎢ ⎦⎣ z ⎦ ⎣ ⎤ 0 ⎥ −2 ⎦ Again converting into equations we get x+y-z=1 -2y + z = 0 -2y = - 2 ⇒ y = 1 -2 + z = 0 ⇒z = 2 x+1-2=1 ⇒ x = 2 ∴ x = 2, y = 1, z = 2 ∣ 1 34. We have, ∣ cos C 1 ∣ cos B ∣ ∴ LHS = ∣ ∣ cos B ∣ cos C ∣ ∣ cos A cos A ∣ 1 1 cos C cos C 1 cos A cos A 1 ∣ cos B = 0 ∣ cos B ∣ ∣ ∣ = 0 ∣ Expanding along R1 2 = 1 (1 − cos A) − cos C (cos C − Cos A. cos B) + cos B (cos C . cos A − cos B) 2 2 2 2 2 = sin A − cos C + cos A. cos B. cos C + cos A. cos B. cos C − cos B 2 = sin A − cos B + 2 cos A. cos B. cos C − cos C 2 = − cos(A + B). cos(A − B) + 2 cos A. cos B. cos C − cos C 2 2 [∵ cos B − sin A = cos(A + B). cos(A − B)] = − cos(−C ). cos(A − B) + cos C (2 cos A cos B − cos C ) [∵ cos(−θ = cos θ)] = − cos C (cos A. cos B = sin A. sin B − 2 cos A. cos B + cos C ) = cos C (cos A. cos B − sin A. sin B − cos C ) = cos C (cos(A + B) − cos C ) = cos C (cos C − cos C ) = 0 = RHS Hence proved. 35. x − y + z = 4 x − 2y − 2z = 9 2x + y + 3z = 1 1 −1 Let A = ⎢ 1 −2 ⎡ ⎣ 2 1 ⎤ ⎡ x ⎤ ⎡ 4 ⎤ −2 ⎥ X = ⎢ y ⎥ C = ⎢ 9 ⎥ 1 3 ⎦ ⎣ z ⎦ ⎣ 1 ⎦ Then, given system of equations can be rewritten as, AX = C 1 −1 Now, AB = ⎢ 1 −2 ⎡ ⎣ 2 8 0 0 = ⎢0 8 0⎥ 0 8 ⎡ ⎣ 0 1 1 −4 4 4 −2 ⎥ ⎢ −7 1 3 −3 −1 3 ⎤⎡ ⎦⎣ 5 ⎤ ⎥ ⎦ ⎤ ⎦ AB = 8I −1 −1 A = 1 8 B[ ∵ A −1 AB = 8A −1 I ] B = 8A 8 / 10 −1 1 1 2 2 2 ⎢ ⎢ ⎢ −7 1 3 8 8 8 ⎣ 5 −3 −1 8 8 8 ⎡ −1 ⇒ A = 1 8 ⎤ ⎥ ⎥ ⎥ ⎦ Now, AX = C, ⇒ X = A-1C ⎡ −1 1 1 2 2 2 1 3 ⎡ x ⎤ ⎢ ⇒ ⎢ y ⎥ = ⎢ ⎢ ⎣ ⎦ z ⎣ −7 8 8 8 5 −3 −1 8 8 8 −4 ⎡ ⎢ = ⎢ ⎢ ⎣ 2 −28 8 20 8 9 + 2 9 + 8 −27 + 8 1 + + 4 ⎡ ⎤ ⎥ ⎥⎢9⎥ ⎥ ⎣ ⎦ 1 ⎦ ⎤ 2 + ⎤ 3 8 −1 3 ⎡ ⎤ ⎥ ⎥ = ⎢ −2 ⎥ ⎥ ⎣ ⎦ −1 ⎦ 8 ⇒ x = 3, y = −2, z = −1 36. If a set P has m elements and set Q has n elements then the number of functions possible from P to Q is nm. So, number of functions from A to B = 62 37. As the total number of Relations that can be defined from a set P to Q is the number of possible subsets of P × Q. If n(P) = m and n(Q) = n then n(P × Q) = mn and the number of subsets of P × Q = 2mn. So number of relations possible from A to B = 22× 6 = 212 If n(A) = P and n(B) = q then n(A × B) = pq and the number of subsets of A × B = 2pq. 38. R = {(1, 2), (2, 2),(1, 3), (3, 4), (3, 1), (4, 3), (5, 5),} R is not reflexive. (3, 3) ∉ R R is not symmetric. Because for (1, 2) ∈ R there (2, 1) ∉ R. R is not transitive. Because for all element of B there does not exist, (a, b) (b, c) ∈ R and (a, c) ∈ R. 39. R is reflexive, since every element of B i.e, B = {1, 2, 3, 4, 5, 6} is divisible by itself. i.e, (1, 1), (2, 2), (3, 3), (4, 4,), (5, 5), (6, 6) ∈ R Further, (1, 2) ∈ R But (2, 1) ∉ R Moreover, (1, 2), (2, 4) ∈ R ⇒ (1, 4) ∈ R ⇒ R is transitive. Therefore, R is reflexive and transitive but not symmetric. ∣ 40. Required Area = ∣ ∣ ∣0 1 ∣ 2 ∣ 3 0 1 ∣∣ – √3 1 – ∣3 − √3 – – [1(−3√3 − 3√3)]∣ ∣ =∣ = 1 2 6√3 2 – = 3√3 ∣∣ ∣∣ 1 ∣∣ sq units 41. Since, a face of the Pyтamid consists of 25 smaller equilateral triangles. – – ∴ Area of a face of the Pyramid = 25 × 3√3 = 75√3 sq. units. 42. Area of equilateral triangle = – ∴ 3√3 = √3 4 √3 4 (side)2 (side)2 [As calculated above area of equilateral triangle is 3√3 sq. units] – (side)2 = 12 – ⇒ side = 2√3 (units) Let h be the length of the altitude of a smaller equilateral triangle. – Then, × base × height = 3√3 ⇒ 1 2 9 / 10 or, 1 2 × – side × height = 3√3 or, height = 3√3×2 2√3 = 3 units. 43. Let the third point on the line be (x, y). The area of triangle with vertices (x, y), (1, 2), (3, 6) ∣x = y 1∣ 1 2 1 ∣3 6 1∣ 1 ∣ 2 ∣ ∣ ∣ Since the three points are collinear, the area formed will be zero. ∣x ⇒ ⇒ ⇒ y 1∣ 1 2 1 ∣3 6 1∣ ∣ ∣ ∣ ∣ =0 x(2 - 6) - y(1 - 3) + 1(6 - 6) = 0 -4x + 2y = 0 2x - y = 0 Hence, the equation of line joining (1, 2) and (3, 6) is 2x - y = 0 ⇒ 10 / 10