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Subject : Mathematics Date : DPP No. : 92 DPP No. – 01 Class : XI Course : Total Marks : 32 Max. Time : 31 min. Comprehension ('–1' negative marking) Q.1 to Q.3 (3 marks 3 min.) [9, 9] Single choice Objective ('–1' negative marking) Q.4, 5, 6, 7, 8, 9 (3 marks 3 min.) [18, 18] Multiple choice objective ('–1' negative marking) Q.10 (5 marks 4 min.) [5, 4] Ques. No. 1 2 3 4 5 6 7 8 9 10 Total Mark obtained Comprehension (1 to 3) : At time of methods of coordinate becomes effective in solving problems of properties of triangle. We may choose one vertex of the triangle and one side passing through this vertex as x-axis. Thus, without loss of generality, we can assume that every triangle ABC has a vertex B situated at B(0, 0) vertex C at (a, 0) and A as (h, k) 1. In ABC, AC = 3, BC = 4 median AD and BE are perpendiculars, then area of triangle ABC must be equal to (A) sq. units (B*) sq. units (C) 2 sq. units (D) none of these 2. Suppose the bisector AD of the interior angle A of ABC divide side BC into segment BD = 4, DC = 2, then we must have (A) b > C and C < 4 (B) 2 < b < 6 and C < 1 (C*) 2 < b < 6 and C = 2b (D) none of these BE 3. If altitudes CD = 7, AE = 6 and E divides BC given that EC (A) 2 (B) 5 (C) = 3 , then C must be 4 (D*) 4. Consider the straight line ax + by = c where a, b, c  R+. This line meets the coordinate axes at 'P' and 'Q' respectively. If the area of triangle OPQ. 'O' being origin, does not depend upon a, b and c then (A) a, b, c are in G.P. (B*) a, c, b are in G.P. (C) a, b, c are in A.P. (D) a, c, b are in A.P. 5. ABC is a right angled isosceles triangle, right angled at A(2, 1). If the equation of side BC is 2x + y = 3, then the combined equation of lines AB and AC is (A) 3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0 (B) 3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0 (C) 3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0 (D*) None of these 6. The straight line x cos  + y sin  = 2 will touch the circle x2 + y2 – 2x = 0 if (A)  = n, n   (B)  = (2n + 1), n   (C*)  = 2n, n   (D) None of these 7. If the circles x2 + y2 – 8x + 2y + 8 = 0 and x2 + y2 – 2y – 6y + 10 – a2 = 0 have exactly two common tangents, then (A) 1 < |a| < 8 (B*) 2 < |a| < 8 (C) 3 < |a| < 8 (D) 4 < |a| < 8 8. Equation of the circle that cuts the circles x2 + y2 = a2, (x – b)2 + y2 = a2 and x2 + (y – c)2 = a2 orthogonally is (A) x2 + y2 – bx – cy – a2 = 0 (B) x2 + y2 + bx + cy – a2 = 0 (C) x2 + y2 + bx + cy + a2 = 0 (D*) x2 + y2 – bx – cy + a2 = 0 9. A circle passes through the points A(1, 0), B(5, 0) and touches the y-axis at C(0, h). If ACB is maximum then (A*) h = (B) h = 2 (C) h = (D) y = 2 10. If Line L : (3x – 4y – 25 = 0) touches the circle S : (x2 + y2 – 25 = 0) at P and L is common tangent of circles S = 0 and S1 = 0 at P and S1 = 0 passes through (5, –6), then  27 ,  36  (A*) centre of S1 = 0,    7  (B*) length of tangent form origin to S1 = 0,  27 ,  36  (C) centre of S1 = 0,    7  (D)

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