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5.[A,B,D] We have adj A = |A| A –1adj (AB) = |AB| (AB) –1 = |A| |B| (AB) –1= |A| |B| (B –1 A –1 )= ( |B| B –1 ) (|A| A –1 )= (adj B) (adj A)6.[A,B,C] We have⎡cosα − sin α 0 ⎤A(α, β)′ =⎢⎥⎢sin α cosα0⎥β⎢⎣0 0 e ⎥⎦⎡ cos( −α)sin( −α)0 ⎤=⎢⎥⎢− sin( −α)cos( −α)0⎥β⎢⎣0 0 e ⎥⎦= A (–α, β)Also, A (α, β) A( – α, –β)⎡ cosαsin α 0 ⎤ ⎡cosα − sin α=⎢⎥⎢− sin α cosα0⎢⎥ ⎢sin α cosαβ⎢⎣0 0 e ⎥⎦⎢⎣0 0⇒ A(α, β) –1 = A(–α, –β)Next, adj A(α, β) = |A (α, β)| A (α, β) –1= e β A(– α, – β).0 ⎤0⎥⎥= I⎥⎦−βe7.[A,B,C,D]n⎛ 5 4 ⎞Let (r + 1) th term of ⎜ + x ⎟ be2⎝ x ⎠independent of x. We haven−rT r + 1 = n ⎛ 5 ⎞C r ⎜ ⎟ (x 4 ) r2⎝ x ⎠= n C r 5 n – r 6r – 2nxFor this term to be independent of x, 6r – 2n = 0or n= 3r. As each of 18, 21, 27 and 99 is divisibleby 3, each of this can be a possible value of n.8.[A,B]Let k = 2n + 1, then 2n + 1 C r is maximum whenr = n. Also 2n+1 C n = 2n + 1 C n+1Thus, k C r is maximum whenr = 21 (k – 1) or r = 21 (k + 1).9.[A,B,C,D]As a 912 , a 951 and a 480 are divisible by 3, none ofthem is prime. For a 91 , we have1 1a 91 = (99....9) = (109 1424391 – 1) 991 times= 91 [(10 7 ) 13 – 1]⎡7 13(10 ) −1⎤⎡10 7 −1⎤= ⎢7⎥ ⎢ ⎥⎢⎣10 −1⎥⎦⎢⎣10 −1⎥⎦= [(10 7 ) 12 + (10 7 ) 11 + ….+ 10 7 + 1]× [10 6 + 10 5 + ….. + 10 + 1]= a 91 is not prime.Column Matching :10. [A] → p, r; [B] → p,q, r; [C] → t ; [D] → p, q, r3 πsin θ = = sin2 3⇒ θ = nπ + (–1) n π π = 2nπ +3 34 sin θ cos θ – 2 sinθ – 2 3 cos θ + 3 = 0⇒ (2 sin θ – 3 ) (2 cos θ – 1) = 03⇒ sin θ =2⇒ θ = nπ + (–1) n π 1 , cosθ =3 2π⇒ θ = 2nπ ± 3(C) sin 2θ + cos 2θ + 4 sin θ = 1 + 4 cos θ⇒ 2 sin θ cos θ + 1 – 2 sin 2 θ + 4 sin θ = 1+ 4cos θ⇒ 2 sin θ (cos θ– sin θ) – 4(cosθ– sinθ) = 0⇒ (2 sin θ –4) (cos θ – sin θ) = 0⇒ sin θ = 2 or sin θ = cos θ⇒ tan θ = 1 ⇒ θ = nπ + π/4, n ∈ I(D) cos 2 θ = 1/4= cos 2 π π⇒ θ = 2nπ ±3311. [A] → r ; [B] → r ; [C] → p ; [D] → q(A)We have b – a = c – b and (c – b) 2 = a(b – a)⇒ (b – a) 2 = a(b – a) ⇒ b = 2a and so c = 3a.Thus a : b : c = 1 : 2 : 3a + b(B) If the numbers are a and b, then x = and213b = ar 3 ⎛ b ⎞⇒ r = ⎜ ⎟⎠⎝ aNow,y3 3 3 3 3 6+ z a r + a r=2xyz x (ar)(ar )a (1 + r ) a + b= = = 2x a + b2(C) c > 4b –3a ⇒ ar 2 + > 4ar –3a⇒ r 2 – 4r + 3 > 0 ⇒ r < 1 or r > 3, But the termsare positive so r ∈ (0, 1) ∪ (3, ∞)(D) tan –1 ⎛ 1 ⎞⎜ ⎟ = tan –1 ⎛ 2 ⎞⎜ ⎟2⎝ 2r ⎠2⎝ 4r ⎠= tan –1 (2r + 1) − (2r −1)1+(2r + 1)(2r −1)= tan –1 (2r +1) –tan –1 (2r–1)∴n∑r=1−1⎛1 ⎞tan ⎜ ⎟ (2n+1) – tan –1 (1)2⎝ 2r ⎠= tan –1 (2n + 1) – 4π3XtraEdge for IIT-JEE 97APRIL 2010