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Subject : Mathematics Date : DPP No. : 34 Class : XIII Course : DPP No. – 01 Total Marks : 24 Max. Time : 18 min. Single choice Objective ('–1' negative marking) Q.1, 3, 4, 5, 6 (4 marks 3 min.) [20, 15] Assertion and Reason (no negative marking) Q.2 (4 marks 3 min.) [4, 3] Ques. No. 1 2 3 4 5 6 Total Mark obtained 1. The sum of the first n terms of the sequence 1, (1 + 2), (1 + 2 +22), .........(1 + 2 + 22 + ...2k–1), is of the form 2n+R + Sn2 + Tn + U for all n  N. The value of (R + S + T + U) is equal to (A) –1 (B) 0 (C) 1 (D*) –2 Sol. t = 1 + 2 + 22 + 2n – 1 n S = tr  2n 1 – n – 2 r 1 R = 1, U = – 2, T = – 1, S = 0 R + S + T + U = – – 2 Ans. 2. Statement-1 : Perpendicular from origin O to the line joining the points A (c cos, c sin) and B (c cos, c sin) divides it in the ratio 1 : 1 Statement-2 : Perpendicular from opposite vertex to the base of an isosceles triangle bisects it. (A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; 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 Sol. Obvious 3. Given two circles x² + y²  6x  2y + 5 = 0 & x² + y² + 6x + 22y + 5 = 0 . The tangent at (2 , 1) to the first circle : (A) passes outside the second circle (B*) touches the second circle (C) intersects the second circle in 2 real points (D) passes through the centre of the second circle . Sol. Obvious 4. If (, 2) falls inside the angle made by the lines 2y = x, x > 0 & y = 3x, x > 0, then the set of values of  is : (A) ( , 3) (B*) (1/2, 3) (C) (0, 3) (D) ( , 0)  [1/2, ) Sol. (A) ( , 3) (B*) (1/2, 3) (C) (0, 3) (D) ( , 0)  [1/2, )   < 2 < 3 where  > 0  2 1 < a < 3 2   12  5. The value of sin–1 cot sin1  cos1  sec 1 2  is equal to         (A) 4 (B) 2 (C*) 0 (D) – 2 Sol. Let sin–1 =  sin  = 2 tan = = 1   = 15°  3  cos–1   = 30°   sec–1   = cos–1 = 45° 6. The value of cos–1 (cos 12) – sin–1 (sin 14) is (A) – 2 (B*) 8 – 26 (C) 4 + 2 (D) None of these Sol. cos–1 (cos 12) – sin–1 (sin 14) 4 – 12 – (14 – 4) = 8 – 26 Subject : Mathematics Date : DPP No. : 35 Class : XIII Course : DPP No. – 02 Total Marks : 24 Max. Time : 22 min. Single choice Objective ('–1' negative marking) Q.1, 2, 5, 6 (4 marks 3 min.) [16, 12] Subjective Questions ('–1' negative marking) Q.3, 4 (4 marks 5 min.) [8, 10] Ques. No. 1 2 3 4 5 6 Total Mark obtained 1. The tangent lines to the circle x² + y² + 6x – 4y – 12 = 0 which are perpendicular to the line 4x + 3y + 5 = 0 are given by : (A) 4x + 3y  7 = 0 , 4x + 3y + 15 = 0 (B) 4x + 3y  31 = 0 , 4x + 3y + 19 = 0 (C*) 3x – 4y  42 = 0 , 3x – 4y – 8 = 0 (D) 3x – 4y + 8 = 0 , 3x + 4y + 21 = 0 Sol. Let tangents be 3x – 4y +  = 0 since

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• 1. Subject : Mathematics Date : DPP No. : Class : XIII Course : DPP No. – 01 Total Marks : 24 Max. Time : 18 min. Single choice Objective ('–1' negative marking) Q.1, 3, 4, 5, 6 (4 marks 3 min.) [20, 15] Assertion and Reason (no negative marking) Q.2 (4 marks 3 min.) [4, 3] Ques. No. 1 2 3 4 5 6 Total Mark obtained 1. The sum of the first n terms of the sequence 1, (1 + 2), (1 + 2 +22 ), .........(1 + 2 + 22 + ...2k–1 ), .......is of the form 2n+R + Sn2 + Tn + U for all n  N. The value of (R + S + T + U) is equal to (A) –1 (B) 0 (C) 1 (D*) –2 Sol. tn = 1 + 2 + 22 + 2n – 1 Sn = 2 – n – 2 t 1 n n 1 r r     R = 1, U = – 2, T = – 1, S = 0 R + S + T + U = – – 2 Ans. 2. Statement-1 : Perpendicular from origin O to the line joining the points A (c cos, c sin) and B (c cos, c sin) divides it in the ratio 1 : 1 Statement-2 : Perpendicular from opposite vertex to the base of an isosceles triangle bisects it. (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 Sol. Obvious 3. Given two circles x² + y²  6x  2y + 5 = 0 & x² + y² + 6x + 22y + 5 = 0 . The tangent at (2 , 1) to the first circle : (A) passes outside the second circle (B*) touches the second circle (C) intersects the second circle in 2 real points (D) passes through the centre of the second circle . Sol. Obvious 4. If (, 2) falls inside the angle made by the lines 2y = x, x > 0 & y = 3x, x > 0, then the set of values of  is : (A) (, 3) (B*) (1/2, 3) (C) (0, 3) (D) (, 0)  [1/2, ) (A) (, 3) (B*) (1/2, 3) (C) (0, 3) (D) (, 0)  [1/2, ) Sol.  2  < 2 < 3 where  > 0  2 1 < a < 3 34
• 2. 5. The value of sin–1                         2 sec 4 12 cos 4 3 2 sin cot 1 1 1 is equal to (A) 4  (B) 2  (C*) 0 (D) – 2  Sol. Let sin–1 4 3 – 2 =  sin  = 2 3 – 2 tan = 3 2 3 – 2  = 3 2 1   = 15° cos–1         2 3 = 30° sec–1 2 = cos–1 2 1 = 45° 6. The value of cos–1 (cos 12) – sin–1 (sin 14) is (A) – 2 (B*) 8 – 26 (C) 4 + 2 (D) None of these Sol. cos–1 (cos 12) – sin–1 (sin 14) 4 – 12 – (14 – 4) = 8 – 26
• 3. Subject : Mathematics Date : DPP No. : Class : XIII Course : DPP No. – 02 Total Marks : 24 Max. Time : 22 min. Single choice Objective ('–1' negative marking) Q.1, 2, 5, 6 (4 marks 3 min.) [16, 12] Subjective Questions ('–1' negative marking) Q.3, 4 (4 marks 5 min.) [8, 10] Ques. No. 1 2 3 4 5 6 Total Mark obtained 1. The tangent lines to the circle x² + y² + 6x – 4y – 12 = 0 which are perpendicular to the line 4x + 3y + 5 = 0 are given by : (A) 4x + 3y  7 = 0 , 4x + 3y + 15 = 0 (B) 4x + 3y  31 = 0 , 4x + 3y + 19 = 0 (C*) 3x – 4y  42 = 0 , 3x – 4y – 8 = 0 (D) 3x – 4y + 8 = 0 , 3x + 4y + 21 = 0 Sol. Let tangents be 3x – 4y +  = 0 since it touch the circle  5 = 5 8 9       – 17 = ± 25  = 42, –8 2. A circle S of radius 'a' is the director circle of another circle S1 . S1 is the director circle of circle S2 and so on. If the sum of the radii of all these circles is 2, then the value of 'a' is – (A) 2 + 2 (B) 2 – 2 1 (C*) 2 – 2 (D) 2 + 2 1 Sol. Radius of S1 , S2 , S3 ,..... are 2 a , 2 a , 2 2 a ,....... respectively  2 = a + 2 a + 2 a + ......  2 1 1 a  = 2  a = 2 – 2 3. If a, b, c are positive real numbers such that c b a log  = a c b log  = b a c log  , then prove that ab + c + bc + a + ca + b  3. Sol. Let c – b a log = a – c b log = b – a c log = k loga = k (b – c)  a = 10k (b – c) logb = k(c – a)  b = 10k(c – a) logb = k (c – a)  c = 10k(a – b) Now AM  GM 3 c b a b a a c c b        3 / 1 b a a c c b c . b . a       3 / 1 ) b – a a – c c – b ( k 2 2 2 2 2 2 10   ab + c + bc + a + ca + b  3 35
• 4. 4. Domain of f(x) = sin–1 [2 – 4x2 ], where [ . ] denotes the greatest integer function, is _______. Ans.          2 3 , 2 3 – { 0 } Sol. f(x) = sin–1 [2 – 4x2 ] – 1 [2 – 4x2 ] 1 – 1 2 – 4x2 < 2 – 2 < 4x2 – 2  1 0 < |x|  2 3 x          2 3 , 2 3 – – {0} Ans. 5. The value of sec                            9 31 cos cos 9 50 sin sin 1 1 is equal to (A) sec 9 10 (B) sec 9  (C) 1 (D*) – 1 Sol. sec                      9 31 cos cos 9 50 sin sin – 1 – 1 – = sec                         9 4 3 cos cos 9 4 – 6 sin sin – 1 – 1 – = sec                9 4 cos – cos 9 4 sin sin 1 – 1 – = sec           9 4 – 9 4 = sec  = – 1 Ans. 6. Number of solutions of the equation, cot–1 (x2 + 1) = sin–1 |x| (A) one (B*) two (C) three (D) none of these Sol. Number of solution = 2
• 5. Subject : Mathematics Date : DPP No. : Class : XIII Course : DPP No. – 03 Total Marks : 30 Max. Time : 27 min. Single choice Objective ('–1' negative marking) Q.1, 4, 5 (4 marks 3 min.) [12, 9] Multiple choice objective ('–1' negative marking) Q.3 (6 marks 5 min.) [6, 5] Subjective Questions ('–1' negative marking) Q.2 (4 marks 5 min.) [4, 5] Match the Following (no negative marking) (2 × 4) Q.6 (8 marks 8 min.) [8, 8] Ques. No. 1 2 3 4 5 6 Total Mark obtained 1. If the line y = x cuts the curve x3 + 3y3 – 30xy + 72x – 55 = 0 in points A, B and C (O is origin), then the value of 55 2 4 OA . OB . OC is (A) 2 (B*) 4 (C) 2 (D) 2 2 Sol. y = x ...(1) x3 + 3y3 – 30xy + 72x – 55 = 0 ...(2) from (1) & (2) 4x3 – 30x2 + 72x – 55 = 0 x1 .x2 .x3 = 4 55 Now (OA)(OB)(OC) = 55 2 4 1 x 2 . 2 x 2 . 3 x 2 = 4 55 55 16  = 4 2. There are 25 trees at equal distances of 5 m in a line with a well, the distance of the well from the nearest tree being 10 m . A gardner waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next . Find the total distance, the gardner will cover in order to water all the trees. Ans. 3370m Sol. Series will be 20, 30, 40,.......,24 terms Distance travelled to water 24 trees = 2 24 (40 + 23 × 10) = 3240 Total distance travelled to water 25 trees = 3240 + (10 + (25 – 1)5) = 3370 3. The value of cos                          5 14 cos cos 2 1 1 is (A*) cos 5 2 (B) cos         5 7 (C*) sin 10  (D*) – cos 5 3 Sol. cos        5 14 cos cos 2 1 1 – = cos         ) 5 – 3 cos( cos 2 1 1 – = cos 2 1         5 – = cos 10 sin 5 2    36
• 6. 4. If in a ABC, 1 r r = 2 1 , then the value of tan 2 A        2 C tan 2 B tan is equal to : (A) 2 (B*) 2 1 (C) 1 (D) None of these 5. If  = sin                         3 2 cos tan cot 1 1 and  = sin                         3 1 tan cot ec cos 1 1 are the roots of the quadratic equation ax2 + bx + c = 0 where a, b, c are integers and c is prime then value of (a + b + c) equals (A) 20 (B) 11 (C) 7 (D*) 2 Sol.  = sin cot–1 tan cos–1 3 2 = 3 2  = sin cosec–1 cot       3 1 tan 1 – = 3 1  + = – a b = 1  a + b = 0 9 2 a c   c = 2 a = 9 then a + b + c = 0 + c = 2 6. Match the expressions in column – I with their values in column–II Column – I Column – II (A) cos–1        3 5 cos + sin–1        3 5 sin (p) 4  (B) cos–1        3 5 cos + sin–1        3 5 cos (q) 0 (C) 2 tan-1       3 1 + tan–1       7 1 (r) not defined (D) cosec         4 ec cos 1 (s) 2  Ans. (A)  (q), (B)  (s), (C)  (p), (D)  (r) Sol. (A) cos–1 cos(2 – /3) + sin–1 sin (2 – /3) = cos–1 cos /3 + sin–1 (– sin /3) /3 – /3 = 0 (B) cos–1        3 5 cos + sin–1        3 5 cos = /2 {sin–1  + cos–1  = /2} (C) Let 2 tan–1 3 1 =  2 / tan 3 1  
• 7. tan = 3/4           2 / tan – 1 2 / tan 2 tan 2 2 tan–1 3 1 + tan–1 7 1 = tan–1 4 3 + tan–1 7 1 = tan–1              20 3 – 1 7 1 4 3 = tan–1 4 25 25         (D) cosec (cosec–1 /4) Not defined because (y = cosec–1 x defined when |x|  1 but /4 0.78}
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