# XI Maths DPP (30) - Prev Chaps + P&C

.pdf
Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website : www.resonance.ac.in | E-mail : Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029 PAGE NO.-1 TARGET : JEE (Main + Advanced) 2017 Course : VIKAAS(JA) MATHEMATICS DPP DPP DPP DAILY PRACTICE PROBLEMS NO. 74 TO 76 This DPP is to be discussed in the week (26-10-2015 to 31-10-2015) DPP No. : 74 (JEE-Main) Total Marks : 30 Max. Time : 30 min. Single choice Objective ('-1' negative marking) Q.1 to Q.10 (3 marks, 3 min.) [30, 30] 1. The number of distinct terms in the expansion of 15 2 2 1 x 5 x is (A) 256 (B) 60 (C*) 31 (D) 61 15 2 2 1 x 5 x ds izlkj es a fofHkUu ink sa dh la[;k gS & (A) 256 (B) 60 (C*) 31 (D) 61 Sol. The number of distinct terms in the expansion of 15 2 2 1 x 5 x is the same as the number of distinct terms in the expansion of 15 2 2 1 x 2 x = 30 1 x x = 30 2 x 1 x . the number of distinct terms is 31 Hindi 15 2 2 1 x 5 x ds izlkj eas fofHkUu inksa dh la[;k] 15 2 2 1 x 2 x = 30 1 x x = 30 2 x 1 x ds izlkj e sa fofHkUu ink sa dh la[;k ds cjkcj gSA fofHkUu ink sa dh la[;k 31 gSA 2. The sum of the radii of inscribed and circumscribed circles for an n sided regular polygon of side 'a', is : n Hkqtkvksa okys lecgqHkqt ftldh ,d Hkqtk dh yEckbZ 'a' gks] ds fy, vUr% o`r rFkk ifjo`Ùk dh f=kT;kvks a dk ;k sx gksxkμ (A) a cot n (B*) a 2 cot 2n (C) a cot 2n (D) a 4 cot 2n ANSWER KEY DPP No. : 74 (JEE-Main) 1. (C) 2. (B) 3. (C) 4. (C) 5. (B) 6. (B) 7. (B) 8. (D) 9. (B) 10. (C) DPP No. : 75 (JEE-Advanced) 1. (C) 2. (A) 3. (C) 4. (B) 5. (C) 6. (A)(B)(C) 7. (B)(C) 8. 1 DPP No. : 76 (JEE-Advanced) 1. (B) 2. (A) 3. (A)(D) 4. (A)(C) 5. (A)(C)(D) 6. (A)(C) 7. (A)(D) 8. (A) (r), (B) (s), (C) (q), (D) (p)
Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website : www.resonance.ac.in | E-mail : Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029 PAGE NO.-2 Sol. tan n = a 2r ; sin n = a 2R r + R = a 2 cot cosec n n r + R = a 2 .cot 2n 3. In ABC, I is incentre , If a = 3 cm and A = 60° then circumradius of I BC is ABC e sa] I vUr% dsUnz gSA ;fn a = 3 ls-eh- vkSj A = 60° gS rks I BC dh ifjf=kT;k gS& (A) 2 cm ls-eh- (B) 3 cm ls-eh- (C*) 1 cm ls-eh- (4) 4 cm ls-eh- Sol. D I C= 90 - C 2 D I B= 90 - B 2 B I C= 180 - B C 2 = 180 - A 90 2 = 90 + A 2 I n I BC es a , I BC sin( B C) = 2R 2R = 3 sin(120 ) = 2 cm R = 1 cm. 4. If (1!) 2 + (2!) 2 + (3!) 2 + ..... + (99!) 2 + (100!) 2 is divided by 100, the remainder is ;fn (1!) 2 + (2!) 2 + (3!) 2 + ..... + (99!) 2 + (100!) 2 dk s 100 ls foHkkftr fd;k tk,] rks 'k s"kQy gS & (A) 27 (B) 28 (C*) 17 (D) 14 Sol. 1 + 4 + 36 + 576 + multiple of 100 Remainder = 17 5. Sum of all the 4-digit numbers which can be formed using the digits 0, 3, 6, 9 (without repetition of digits) is (A) 119988 (B*) 115992 (C) 3996 (D) none of these vadks a 0, 3, 6, 9 dks ysdj 4 - v adk sa dh lHkh la[;kvks a dk ;k sxQy ( tcfd v adk sa dh iqujko`fÙk u gks ) gS & (A) 119988 (B*) 115992 (C) 3996 (D) bues a ls dksb Z ugh a Sol. Sum at unit place = 0 × 6 + 3 × 4+ 6 × 4 + 9 × 4 = 72 Sum at tens place = 0 × 6 + 3 × 4+ 6 × 4 + 9 × 4 = 72 Sum at hundred place = 0 × 6 + 3 × 4+ 6 × 4 + 9 × 4 = 72 Sum at thousand place = 3 × 6 + 6 × 6 + 9 × 6 = 108 Total sum 72 × 1 + 72 × 10 + 72 × 100 + 108 × 1000 = 115992. Hindi. bdkbZ LFkku ij ;ksxQy = 0 × 6 + 3 × 4+ 6 × 4 + 9 × 4 = 72 ngkb Z LFkku ij ;ksxQy = 0 × 6 + 3 × 4+ 6 × 4 + 9 × 4 = 72 lSdM+k LFkku ij ;ksxQy = 0 × 6 + 3 × 4+ 6 × 4 + 9 × 4 = 72 gtkjosa LFkku ij ;ksxQy = 3 × 6 + 6 × 6 + 9 × 6 = 108 dqy ;ksx 72 × 1 + 72 × 10 + 72 × 100 + 108 × 1000 = 115992. 6. Maximum value of (27 sinx 81 1+cosx ) 1/9 is (27 sinx 81 1+cosx ) 1/9 dk vf/kdre eku gS& (A) 1 (B*) 3 (C) 9 (D) 3 1/9 Sol. (27 sinx . 81 1+cosx ) 1/9 = sin x 4 4cos x 3 9 3 . 3 = 3sinx 4cos x 4 9 3 - 5 3 sin x + 4 cos x 5 1 1 9 3 , 3 max. value vf/kdre eku = 3
Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website : www.resonance.ac.in | E-mail : Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029 PAGE NO.-3 7. Suppose a ray of light leaves the point (3, 4) reflects from the y-axis and moves towards the x-axis, then reflects from the x-axis, and finally arrives at the point (8, 2), then the value of x, is ekuk ,d izdk'k dh fdj.k fcUnq (3, 4) ls pydj y- v{k ls ijkofrZr gksdj x- v{k dh vk sj tkrh gS rFkk varr% x- v{k ls ijkofrZr gksdj fcUnq (8, 2) ls xqtjrh gS] rks x dk eku gS& x y (0,y) O (x,0) (8,2) (3,4) (A) x = 1 4 2 (B*) x = 4 1 3 (C) x = 2 4 3 (D) 5 1 3 Sol. tan (90 - ) = 2 0 8 x = 2 8 x cot = 2 8 x ............ (i) tan (90 + ) = y 0 0 x = y x cot = 4 y 3 ............ (3) 2 8 x = y x y x = 4 y 3 2x = 8y - xy 3y = 4x - xy xy = 8y - 2x xy = 4x - 3y 8y - 2x = 4x - 3y 11y = 6x y = 6x 11 x 6x 11 = 4x - 3 6x 11 6x 2 = 44x - 18x 3x 2 = 22x - 9x 3x 2 = 13x x = 0 or x = 13 3 = 4 1 3 8. If the line y = x cuts the curve x 3 + 3y 3 - 30xy + 72x - 55 = 0 in points A, B and C, then the value of 4 2 55 OA.OB.OC (where O is the origin), is ;fn js[kk y = x x 3 + 3y 3 - 30xy + 72x - 55 = 0 dks fcUnq A, B ,oa C ij izfrPNsn djrh gS] rc 4 2 55 OA . OB . OC dk eku gS& ( tgk ¡ O ewy fcUnq gS ) (A) 55 (B) 1 4 2 (C) 2 (D*) 4 Sol. Let A(r 1 , r 1 ) B(r 2 , r 2 ) , C(r 3 , r 3 ) OA .OB.OC = 2 2 r 1 r 2 r 3 on putting point (r, r) in the curve r 3 + 3r 3 - 30r 2 + 72r - 55 = 0 4r 3 - 302 + 72r - 55 = 0 it has roots r 1 , r 2 , r 3 r 1 r 2 r 3 = 55 4 OA.OB.OC = 55 2 4 2 OA.OB.OC 55 = 4
Page1of 11