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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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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)

- 1. DAILY PRACTICE PROBLEMS Subject : Mathematics Date : DPP No. : Class : XI Course : DPP No. – 01 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) andA as (h, k) 1. In ABC, AC = 3, BC = 4 medianAD and BE are perpendiculars, then area of triangleABC must be equal to (A) 7 sq. units (B*) 11 sq. units (C) 2 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 3. If altitudes CD = 7, AE = 6 and E divides BC given that EC BE = 4 3 , then C must be (A) 3 2 (B) 3 5 (C) 3 (D*) 3 4 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 = 5 (B) h = 2 5 (C) h = 10 (D) y = 2 10 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 (A*) centre of S1 = 0, 7 36 , 7 27 (B*) length of tangent form origin to S1 = 0, 7 275 (C) centre of S1 = 0, 7 36 , 7 27 (D) length of tangent from origin to S1 = 0 = 7 375 92
- 2. DAILY PRACTICE PROBLEMS Subject : Mathematics Date : DPP No. : Class : XI Course : DPP No. – 02 Total Marks : 34 Max. Time : 33 min. Single choice Objective ('–1' negative marking) Q.1, 2, 3, 4, 5, 6 (3 marks 3 min.) [18, 18] Multiple choice objective ('–1' negative marking) Q.7 (5 marks 4 min.) [5, 4] Assertion and Reason (no negative marking) Q.8 (3 marks 3 min.) [3, 3] Match the Following (no negative marking) (2 × 4) Q.9 (8 marks 8 min.) [8, 8] Ques. No. 1 2 3 4 5 6 7 8 9 Total Mark obtained 1. The straight lines 7x – 2y + 10 = 0 and 7x + 2y – 10 = 0 forms an isosceles triangle with the line y = 2. Area of this triangle is equal to (A) 7 15 sq. units (B) 7 10 sq. units (C*) 7 18 sq. units (D) 7 15 sq. units 2. If the straight lines ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 intersect on the x-axis then (A) ag = fh (B) ah = fg (C*) af = gh (D) None of these 3. If chord x cos + y sin = p of x2 + y2 = a2 subtends a right angle at the origin then (A) a2 = p2 (B*) a2 = 2p2 (C) a2 = 3p2 (D) None of these 4. The circles having radii r1 and r2 intersect orthogonaly. Length of their common chord is (A*) 2 2 2 1 2 1 r r r r 2 (B) 2 2 2 1 2 2 1 r r r r 2 (C) 2 2 2 1 2 1 r r r r (D) 2 2 2 1 2 2 2 r r r r 2 5. The triangle ABC is inscribed in the circle x2 + y2 = 25 if A(3, 4), B(–4, 3) then ACB is equal to (A) 3 (B) 2 (C*) 4 (D) 6 6. Locus of midpoint of chords of circle x2 + y2 = a2 that subtends angle 2 at the point (0, b) is (A) 2x2 + 2y2 – 2bx + b2 – a2 = 0 (B*) 2x2 + 2y2 – 2by + b2 – a2 = 0 (C) 2x2 + 2y2 – 2bx + a2 – b2 = 0 (D) 2x2 + y2 – 2by + a2 + b2 = 0 7. Two circles touch x-axis and passes through 2 , 3 and also touch y = x 3 . If r1 and r2 are radii of these two circles then (A*) r1 + r2 = 3 10 (B*) r1 r2 = 3 7 (C*) |r1 – r2 | = 3 4 (D) none of these 8. Statement-1 : The joint equation of the lines 2x – y = 5 and x – 2y = 3 is 2x2 + 3xy – 2y2 – 11x – 7y + 15 = 0 Statement-2 : Every second degree equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 always represents pair of straight line (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 isTrue, Statement-2 isTrue; Statement-2 is NOT a correct explanation for Statement-1 (C*) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True 9. Match the column Column – I Column – II (A) The number of common tangent of (p) 2 3 circles x2 + y2 – 2x – y = 0 and x2 + y2 + 4x + 2 = 0 are (B) If circle x2 + y2 – ax – 2y + 1 = 0 bisects the circumference (q) 4 of x2 + y2 – 4x – 2y + 2 = 0, then a is equal to d c (simplest form), then d – c = 0 (C) Minimum radius of the circle passing through (2, 1) and (5, 1) (r) 5 (D) Angle subtend at circumference x2 + y2 = 25 by arc AB (s) 3 if A(3, 4) and B (–4, 3) is , then – Ans. (A) q, (B) r, (C) p, (D) s 93
- 3. DAILY PRACTICE PROBLEMS Subject : Mathematics Date : DPP No. : Class : XI Course : 94 31-01-11 DPP No. – 03 Total Marks : 33 Max. Time : 31 min. Single choice Objective ('–1' negative marking) Q.1, 4, 5, 6, 7 (3 marks 3 min.) [15, 15] Multiple choice objective ('–1' negative marking) Q.2, 3 (5 marks 4 min.) [10, 8] Match the Following (no negative marking) (2 × 4) Q.8 (8 marks 8 min.) [8, 8] Ques. No. 1 2 3 4 5 6 7 8 Total Mark obtained 1. A normal to the hyperbola 1 y – 4 x 2 2 = 1 has equal intercepts on positive x and y - axis if normal touches the ellipse 1 b y a x 2 2 2 2 , then a2 + b2 = (A) 5 (B) 25 (C) 16 (D*) None of these 2. P is a point on the parabola y2 = 16x where abscissa and ordinate are equal. Equation of a circle passing through the focus and touching the parabola at P is : (A) x2 + y2 + 52x – 48y + 160 = 0 (B*) x2 + y2 – 4x = 0 (C) x2 + y2 + 4x = 0 (D*) x2 + y2 – 52x + 8y + 152 = 0 3. If the tangents drawn from the point (0, 2) to the parabola y2 = 4ax are inclined at an angle 3/4, then the value of a is (A*) 2 (B*) – 2 (C) 1 (D) None of these 4. The number of focal chord(s) of length 4/7 in the parabola 7y2 = 8x is (A) 1 (B*) 0 (C) infinite (D) none of these 5. The equation (5x – 1)2 + (5y – 2)2 = (2 – 2 + 1) (3x + 4y – 1)2 represents an ellipse if (A) (0, 1) (B*) (0, 2) (C) (1, 2) (D) (– 1, 0) 6. Centroid of the triangle formed by the joining feet of the normals drawn from the point (14, 7) to the parabola y2 – 8y = 16x, is (A) 0 , 3 14 (B*) 4 , 3 11 (C) 3 4 , 3 11 (D) 3 4 , 3 14 7. In the line 2x + y 6 = 2 touches the hyperbola x2 – 2y2 = 4. Then slope of line joining origin and point of contact is (A) 2 2 3 (B) 2 3 (C*) – 2 2 3 (D) – 2 3 8. Match the column Column - I Column - II (A) A parabola has the origin as its focus and the line x = 2 as (p) 5 the directrix. Then x-coordinate of the vertex of the parabola is (B) If the tangents from the point (, 2) to the hyperbola 4 y 9 x 2 2 =1 (q) 1 are at right angles, then is equal to (C) The sum of the squares of the perpendiculars on any (r) 3 tangent to the ellipse 16 y 25 x 2 2 = 1 from two points on the minor axis each at a distance 3 from the centre is , then sum of digit in is (D) Suppose F1 , F2 are the foci of the ellipse 9 x2 + 4 y2 = 1. P is (s) 2 a point on ellipse such that PF1 : PF2 = 2 : 1. Half of the area of the triangle PF1 F2 is Ans. (A) (q), (B) (r), (C) (p), (D) (s)