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Match the items of Column I with those of Column IIColumn IColumn II(A) If the line segment joining the points P(1, 3) and Q(5, 7) subtends a right angle at a point R, such that the area of ΔPQR is 2 sq. unit, then the number of such points R is(P) 2(B) If A(1, 2), B(4, 6), C(5, 7) and S(a, b) are the vertices of a parallelogram in the given order, then the value of a + b is (q) 1(C) If (P/q , r/s) is the centroid of ΔABC given in (B), then the value of p + r/q + s + r is (r) 4(D) Let p = lim n → ∞, lim m→ ∞ cos2n Δm πx (s)3where x x rational and q = lim n → ∞ limlim n → ∞ cos2m Δn x, where x is irrational. Then the area of the triangle with vertices (p, q), (2, 1) and ( 2, 1) is(t) 5 |
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Answer» (A) Since ΔPRQ = 90° , in general, the locus represented by R is a circle with P and Q as ends of the diameter. Because area of ΔPQR is 2 sq. unit, there will be four positions for R (two each in the two semicircles for which PQ is a diameter). Answer: (A) → (r) (B) It is known that a = 1 + 5 - 4 = 2 and b = 2 + 7 - 6 = 3 Therefore a + b = 5 Answer: (B) → (t) (C) Centroid (10/3,15/3) P + r/q + s - 1 = 25/5 = 5 Answer: (C) → (t) (D) We have p = x Δm is even and cos Δm π = 1) Similarly, q = x. Since p = x is rational and q = x is irrational, we have p = q = 0. Therefore, (p,q) = (0, 0). Hence the area of the triangle is 1/2|2(1) - (-2)(1)| = 2 Answer: (D) → (P) |
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