# Dpp (31-35) 11th J-Batch Maths.pdf

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### Dpp (31-35) 11th J-Batch Maths.pdf

• 1. Select the correct alternative : (Only one is correct) Q.190/-1 The value of cot 7 1 2 0 + tan 67 1 2 0 – cot 67 1 2 0 – tan7 1 2 0 is : (A)arationalnumber (B*)irrationalnumber (C) 2(3 + 2 3 ) (D) 2 (3 – 3 ) Q.2817/ph-1  x  R the greatest and the least values of y = 1 2 cos 2x + sin x are respectively (A) 3 4 1 2 , (B*) 3 4 3 2 ,  (C) 1 2 3 2 ,  (D) 1 , - 3 2 [Sol. E = 2 x sin 2 x 2 cos  = 2 x sin 2 x sin 2 1 2   = 2 1 x sin 2 x sin 2 2    =                  4 3 2 1 x sin 2 =                 2 2 1 x sin 4 3 Emax = 3/4 Emin = 3/4 – (–3/2)2 = 3/4 –9/4 = – 3/2 ] Q.370/-1 If tan  = 1 x x x x 2 2    and tan  = 1 x 2 x 2 1 2   (x  0, 1), where 0 < ,  < 2  , then tan ( + ) has the value equal to : (A*) 1 (B) – 1 (C) 2 (D) 4 3 [Sol. x2 – x = t ; tan  = 1 t t  ; tan  = 1 t 2 1   tan ( + ) =       tan tan 1 tan tan = 1 t 2 1 . 1 t t 1 1 t 2 1 1 t t       = t ) 1 t ( ) 1 t 2 ( 1 t ) 1 t 2 ( t       = 1 t 2 t 2 1 t 2 t 2 2 2     = 1  tan ( + ) = 1 ] Q.4100/-1 In a triangle ABC, angle Ais greater than angle B . If the measures of angles A& B satisfy the equation, 3sinx  4sin3 x  K = 0, 0 < K < 1, then the measure of angle C is (A) /3 (B) /2 (C*) 2/3 (D) 5/6 [ JEE ’90 , 2 ] [Sol. k = sin3A = sin3B 3A = – 3B A+ B = /3 C = 2/3 Ans ] CLASS : XI (J-Batch) DATE : 06-07/10/2006 TIME : 60 Min. DPP. NO.-31
• 2. Q.5 Thevaluesofxsmallerthan3inabsolutevaluewhichsatisfytheinequality ) ax 2 x ( log ) x a 2 ( 2   >1 for all a > 5 is (A) –2 < x < 3 (B) –3 < x < 3 (C) –3 < x < 0 (D*) –3 < x < –1 [Hint: log ( ) ( ) 2 2 2 a x x ax   > 1, a > 5 & –3 < x < 3 note that 2a – x2 > 1 for all a > 5 & x  ( –3, 3) Hencetheaboveinequalitywill betrueif (x – 2ax) > 2a – x2 or x – 2ax + x2 – 2a > 0 x(1+x) –2a (1 + x) > 0 (x + 1) (x – 2a) > 0  x  (–3, –1)  D ] Q.68/-1 The exact value of cos cos cos cos cos cos 2 28 3 28 6 28 9 28 18 28 27 28       ec ec ec   is equal to (A) – 1/2 (B) 1/2 (C) 1 (D*) 0 [Hint: put  28 = x , T1 = cos sin 2 3 x x = cos sin sin sin 2 3 x x x x = 1 2 3 3 sin sin sin sin x x x x  L N M O Q P = 1 2 3 cos cos ecx ec x  etc. ] Q.722 If 60a = 3 and 60b = 5 then the value of ) b 1 ( 2 b a 1 12    equals (A*) 2 (B) 3 (C) 3 (D) 12 [Sol. 60a = 3  a = log603 60b = 5  b = log605 let x = ) b 1 ( 2 b a 1 12    log12x = ) b 1 ( 2 b a 1    =     5 log 1 2 15 log 1 60 60    log12x = log1444 = log122  x = 2 Ans. ] Q.820 Smallest positive x satisfyingthe equation cos33x + cos35x = 8 cos34x · cos3x is (A) 15° (B*) 18° (C) 22.5° (D) 30° [Sol. cos33x + cos35x = (cos 5x + cos 3x)3 cos33x + cos35x = cos35x + cos33x + 3 cos 5x cos 3x (cos 5x + cos 3x)  (3 cos 3x · cos 5x) (2 cos 4x · cos x) = 0  cos x · cos 3x · cos 4x · cos 5x = 0  x = (2n + 1) 2  , (2n + 1) 6  , (2n + 1) 8  , (2n + 1) 10   smallest + ve values of x is 10  i.e. 18°Ans. ] Q.920 LetABCDEFGHIJKL be a regular dodecagon, then the value of AF AB + AB AF is (A*) 4 (B) 3 2 (C) 2 2 (D) 2
• 3. [Sol. In  OAB   12 sin a  = 2R ....(1) (AB = a) AB = a = 2R sin 12  again In  OAF   12 5 sin b  = 2R (AF = b)  AF = b = 2R sin 12 5 Hence AF AB + AB AF   75 sin 15 sin +   15 sin 75 sin =   15 cos 15 sin +   75 cos 75 sin = tan 15° + tan 75° =     3 2 3 2    = 4 Ans.] Ans.] Q.105 If               2 a sin 2 a tan a cos 1 2 2 = a cos p w a cos k  where k,w and phaveno common factorotherthan 1, then the value of k2 + w2 + p2 is equal to (A) 3 (B) 4 (C) 5 (D*) 6 [Sol.         2 a sin ) 2 a ( cos ) 2 a ( sin a cos 1 2 2 2 =              2 a sin ) 2 a ( cos ) 2 a ( sin 2 a sin 2 2 2 2 2 =                                2 a cos 2 a sin 1 2 a cos 2 2 a sin 2 2 2 2 = a cos 1 a cos 2  k = 2; w = 1; p = 1  k2 + w2 + p2 = 4 + 1 + 1 = 6 Ans.] SUBJECTIVE Q.1147/6 If 15 sin4 + 10cos4 = 6, evaluate 8cosec6 + 27sec6 [Sol. Dividingbycos4 15 tan4 + 10 = 6 sec4 15 tan4 + 10 = 6(1 + tan2)2 9 tan4 – 12 tan2 + 4 = 0 (3 tan2 – 2)2 = 0  tan2 = 3 2 Now 8 cosec6 + 27 sec6 8 (1 + cot2)3 + 27 (1 + tan2)3 8 3 2 3 1        + 27 3 3 2 1       
• 4. 125 + 125 = 250 Ans ] Q.1279/1 John has 'x' children byhis first wife. Maryhas x + 1 children byher first husband. Theymarryand have children of their own. The whole family has 24 children.Assuming that the children of the same parents donotfight,findthemaximumpossiblenumberoffights thatcan take place. [ Ans. : 191 ] [Sol. f(x)representingthenumberoffightfunctions f (x) = x ( x + 1) + (x + 1) (23 – 2x) + (23 – 2x) x = (x2 + x) + ( – 2x2 + 21x + 23) – 2x2 + 23x f (x) = – 3x2 + 45x + 23 f (x) = – 6x + 45 (X > 0 , N x ) hence x = 7.5 gives maximabut N x hence practical maxima occurs at x = 7 or 8  f (7) = 191 ] Q.1357/6 If x, y, z be all positive acute angle then find the least value of tanx (cot y + cot z) + tany (cot z + cot x) + tanz (cot x + cot y) [Ans: 6] [Sol. x, y, z ) 2 / , 0 (  E =  x tan (cot y + cot z) Let tanx = a ; tany = b ; tanz = c  a , b, c > 0  E =                        b 1 a 1 c a 1 c 1 b c 1 b 1 a  b c a c a b c b c a b a                    2 min 2 min 2 min c a a c b c c b a b b a                            Emin = 6 ]
• 5. Select the correct alternative : (Only one is correct) Q.122/ph-3 In an isosceles ABC A= ,AB =AC =  sin , and BC = sin , then the area of ABC, is (A) 1/2 (B*) 8/25 (C) 9/16 (D) none [Sol. Using sine law C sin AB  = A sin BC            2 180 sin sin =   sin sin = 1 sin2         2 90 = sin   2 cos 1   = sin ; (2 sin – 1)2 = 1 – sin2  5 sin2 – 4 sin = 0  sin = 4/5   = 2 1 bc sinA = 2 1  sin  sin ·sin2 = 2 1 sin2 = 8/25 ] Q.27 If x and yare real numbers such that x2 + y2 = 8, the maximum possible value of x – y, is (A) 2 (B) 2 (C) 2 2 (D*) 4 [Sol. Let, x =  cos 2 2 , y =  sin 2 2 x – y = ) sin (cos 2 2     (x – y)max = 4 ] Q.33(D) Let a, b, c are distinct reals satisfying a3 + b3 + c3 = 3abc. If the quadratic equation (a + b – c)x2 + (b + c – a)x + (c + a – b) = 0 has equal roots then a root of the quadratic equation is (A) 3 c b a   (B*) 1 (C) 2 c b a   (D) 3 abc [Sol. a3 + b3 + c3 – 3abc = (a + b + c)              2 ) a c ( ) c b ( ) b a ( 2 2 2 since a, b, c are distinct  a3 + b3 + c3 – 3abc = 0  a + b + c = 0 now put x = 1 in the quadratic equation, we get a + b + c = 0  x = 1 is the root of the quadratic equation ] Q.41 If x,yRandsatisfythe equation xy(x2 –y2)=x2 +y2 where x  0 then theminimum possiblevalue of x2 + y2 is (A) 1 (B) 2 (C*) 4 (D) 8 [Hint: If x = r cos  and y = r sin  then E = x2 + y2 = r2. Hence we have to minimise r2. Now in the given equation substituting x = r cos  and y= r sin , we get r2 = 4 cosec 4  min 2 r = 4 Ans. ] Q.515 If S = 12 + 32 + 52 + ....... + (99)2 then the value of the sum 22 + 42 + 62 + ....... + (100)2 is (A) S + 2550 (B) 2S (C) 4S (D*) S + 5050 CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-32
• 6. [Sol. Let S = 12 + 32 + 52 + ....... + (99)2 and x = 22 + 42 + 62 + ....... + (100)2  (x – S) = (22 – 12) + (42 – 32) + ....... + (1002 – 992) = (1 + 2) + (3 + 4) + ........ + (99 + 100) = 1 + 2 + 3 + 4 + ...... + 99 + 100 = 5050  x = S + 5050 Ans. ] Q.621 Let A = { x | x2 + (m – 1)x – 2(m + 1) = 0, x  R} B = { x | (m – 1)x2 + mx + 1 = 0, x  R} Numberofvalues of m such thatA Bhas exactly3 distinct elements, is (A) 4 (B) 5 (C) 6 (D*) 7 [Sol. Case-I: when B is a quadratic equation D1 = (m + 3)2; D2 = (m – 2)2 roots of 1st equation are 2, – (m + 1) setA roots of 2nd equation are – 1, m 1 1  set B For exactlythere elements inA B two of the roots must be same note that 2  – 1 possibilitiesare 2 = – (m + 1)  m = – 3 2 = m 1 1   2 – 2m = 1  m = 1/2 – m – 1 = – 1  m = 0 – (m + 1) = m 1 1   1 – m2 = – 1  m = ± 2 m 1 1  = – 1  m – 1 = 1  m = 2. Case-II: Now if m = 1, then B becomes linear roots of B as x = – 1 roots of Aare 2 and – 2  3elementsincommon  all permissible m are {– 3, 2 1 , 2 , – 2 , 2, 0, 1} ] Q.732/QE PQRS is a common diameter of three circles. The area of the middle circle is the average of the area of the other two. If PQ = 2 and RS = 1 thenthelengthQR is (A) 1 6  (B*) 1 6  (C) 5 (D) 4 [Sol. Let QR = x then the diameters are 2, x + 2, x + 3  2 ) 1 x ( 2 2 2   = (x + 2)2  2(x + 2)2 = 22 + (x + 3)2 2(x2 + 4 + 4x) = 4 + (x2 + 6x + 9) x2 + 2x – 5 = 0  x = 1 6  Ans.]
• 7. Q.8 Numberofprincipalsolutionoftheequation tan 3x – tan 2x – tan x = 0, is (A) 3 (B) 5 (C*) 7 (D) more than 7 [Hint: 0, , 2, 3  , 3 2 , 3 4 , 3 5 as given expression = tan 3x tan 2x tan x] More than one are correct: Q.930 Let 2 sin x + 3 cos y = 3 and 3 sin y + 2 cos x = 4 then (A*) x + y = (4n + 1)/2, n  I (B) x + y = (2n + 1)/2, n  I (C) x and ycan be the two non right angles of a 3-4-5 triangle with x > y. (D*) x and ycan be the two non right angles of a 3-4-5 triangle with y> x. [Sol. Squaringandadding 4 + 9 + 12 sin(x + y) = 25  sin(x + y) = 1 = sin 2   x + y = 2n + 2  = (4n + 1) 2  n  I  (A) if x + y = 2   y = 2  – x 5 sin x = 3  sin x = 5 3 or cos x = 5 4 also cos y = 5 3 and sin y = 5 4 ; hence y> x  (D)] Q.1032 For the quadratic polynomial f (x) = 4x2 – 8kx + k, the statements which hold good are (A*) there is onlyone integral k for which f (x) is non negative  x R (B*) for k< 0 the numberzero lies between the zeros of the polynomial. (C*) f (x) = 0 has two distinct solutions in (0, 1) for k  (1/4, 4/7) (D) Minimum value of y  k  R is k(1 + 12k) [Sol. (A) f (x) is non negative  x R  f (x)  0  x  R  D  0  64k2 – 16k  0 4k2 – k  0 k (4k – 1)  0  integral value ofk = 0 (B) f (0) < 0 k < 0  (B) (C) for distinct roots in (0, 1) D > 0  k (4k – 1) > 0 ....(1) 0 < – a 2 b < 1  0 < k < 1 ....(2) f (0) > 0  k > 0 ....(3) f (1) > 0  4 – 7k > 0  k < 4/7 ....(4) (1)  (2)  (3)  (4)  k  (1/4, 4/7) ]
• 8. SUBJECTIVE: Q.11 In a triangleABC, if sinAsinB sinC + cosAcosB = 1, then find the value of sin C. [Ans. 1] Q.12 In a ABC, if a, b, c are inA.P, then prove that cos(A – C) + 4cosB = 3 [4] [Sol. Given 2b = a + c 2 sin B = sinA+ sin C 2 2 sin 2 B cos 2 B = 2 sin 2 C A  cos 2 C A  2 sin 2 B = cos 2 C A  Onsquaring 4 sin2 2 B = 2 ) C A ( cos 1   8 sin2 2 B = 1 + cos(A – C) 4 (1 – cos B) = 1 + cos(A – C)  cos (A – C) + 4 cos B = 3 Hence Proved ] Q.1387/6 Find the general solution of the trigonometric equation cosec x – cosec 2x = cosec 4x [Ans: x = (2n+1) 7  , where 4 m 7 n   , m, n  I ] [Sol. cosec 2x + cosec 4x = cosec x x sin 1 x 4 sin x 2 sin x 2 sin x 4 sin   ( note 4 n and 2 n , 4 n     ) 2sin3x cosxsinx =sin2x sin4x or sin3x=sin4x ) 0 x 2 (sin  4x = n + (–1)n (3x) If n is odd 4x = (2n+1) – 3x 7 ) 1 n 2 ( x     ( .... , 17 , 10 , 3 n  ) ]
• 9. CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-33 Select the correct alternative : (Only one is correct) Q.15/ph-3 LetABC be a triangle with circumradius R, perimeter Pand area k. The maximum value of 3 R kP ,is (A*) 4 27 (B) 2 3 9 (C) 4 3 9 (D) 8 27 [Sol. a = 2R sinA, b = 2R sinB and c = 2R sinC P = 2R  A sin ; k = 2R2  A sin (using  = R 4 abc )  3 R kp = 4 sinA sinB sinC(sinA + sinB + sinC) = 4 · 2 3 3 · 8 3 3 = 4 27 Thiswillbemaximum iftheisequilateral.] Q.2ph-1 The value of, (tan 18°)(sin 36°)(cos 54°)(tan 72°)(tan 108°)(cos 126°)(sin 144°)(tan162°)(cos180°) is k sin2 18° then k has the value equal to (A) 16 5 (B*) 4 5 (C) 4 1 (D) 16 1 [Sol. (tan 18°)(tan 72°)(tan 108°)(tan 162°) = 1 hence weareleft with ( – ) (sin 36°)(cos 54°)(cos 126°)(sin 144°) ( – ) (sin 36°)(sin 36°)(– sin 36°)(sin 36°) = sin436° = 2 2 72 cos 1         = 4 ) 18 sin 1 ( 2   = 2 4 1 5 1 4 1           = 2 4 5 5 4 1          = 2 4 1 5 4 5          = 4 5 sin218°  k = 4 5 Ans. ] Q.327/ph-3 Let a  b  c be the lengths of the sides of a triangleT. If a2 + b2 < c2 then which one of the following must be true? (A)All 3 angles ofTare acute. (B*) Some angleof T is obtuse. (C) Oneangle of T is a right angle. (D)Nosuchtriangle canexist. [Hint: a2 + b2 < c2 a2 + b2 < a2 + b2 – 2ab cosC  cosC < 0  C is obtuse  (B) ] Q.463/matrices Identifywhether the statement is True or False. Therecanexisttwotrianglessuchthatthesidesofonetrianglearealllessthan1cmwhilethesidesofthe other triangle are all bigger than 10 metres, but the area of the first triangle is larger than the area of secondtriangle. [Hint: Consider 21 = b1h1 22 = b2h2 b1h1 > b2h2 first triangle is each with 1 cm, 21 = 2 3
• 10. 2nd triangle 22 = 2 10 10 20 10 3    10–6  1 > 2 ] Q.540/ph-3 ABC is an acute angled triangle with circumcentre 'O'orthocentre H. If AO =AH then the measure of theangleAis (A) 6  (B) 4  (C*) 3  (D) 12 5 [Hint: OA = HA R = 2R cosA (distance of orthocentrefrom the vertexAis 2R cosA)  cosA = 2 1  A = 3   (C)] [12th & 13th 25-09-2005] *Q.66/ph-3 Triangle ABC is right angled at A. The points P and Q are on the hypotenuse BC such that BP = PQ = QC. IfAP = 3 andAQ = 4 then the length BC is equal to (A) 27 (B) 36 (C*) 45 (D) 54 [Sol. In  ABP 9 = c2 + x2 – 2cx cosB; but cosB = x 3 c [Quiz] 9 = c2 + x2 – 2cx x 3 c 9 = x2 + 3 c2 ....(1) ; |||ly 16 = x2 + 3 b2 ....(2) (1) + (2) 25 = 2x2 + 3 1 (b2 + c2) = 2x2 + 3x2  5x2 = 25  x = 5 ;  BC = 5 3 = 45 Ans. ] Q.7 Sides a, b,cofatriangleABC are in A.P. and c a b cos , c b a cos 2 1       , b a c cos 3    ,then the value of 2 tan 2 tan 3 2 1 2    , is (A) 0 (B) 1 (C*) 2/3 (D) 3 5 [Sol. 2b = a + c tan2 2 1  + tan2 2 2  = 3 3 1 1 cos 1 cos 1 cos 1 cos 1          = b a c 1 c a c 1 c b a 1 c b a 1          = c b a c b a c b a a c b          = ) c b a ( b 2   = b 3 b 2 = 3 2 Ans. ]
• 11. Q.84/ph-3Atrianglehassides6, 7,8.Thelinethroughitsincentreparallel to theshortest sideis drawnto meetthe other two sides at P and Q. The length of the segment PQ is (A) 5 12 (B) 4 15 (C*) 7 30 (D) 9 33 [Sol.  = 2 h · 6 = 3h ; (where h is the altitude fromA); Also  = 2 r · 21 (using  = r·s)  2 r · 21 = 2 h · 6 = 3h  h r = 7 2 now APQandABC are similiar h r h  = 6 PQ  h r 1 = 6 PQ  7 2 1 = 6 PQ  7 5 = 6 PQ  PQ = 7 30 Ans. ] Ans. ] Q.9106/circle A square and an equilateral triangle have the same perimeter. Let A be the area of the circle circumscribedabout thesquare and Bbetheareaof thecirclecircumscribed about thetrianglethen the ratioA/Bis (A) 16 9 (B) 4 3 (C*) 32 27 (D) 8 6 3 [Hint: Radius ofthe circlecircumscribingthesquare = 2 a 2 3 A = 4 a 18 · 2  = 2 a 9 2  radius of the circle circumscribing triangle = 2a sec 30° = 2a · 3 2 = 3 a 4 B = 3 a 16 · 2  ; Hence B A = 16 3 · 2 9 = 32 27 ] Q.10ph-2 The sum of angles that satisfythe equation 2sin2x – sin x + cos x – sin 2x = 0 where x  (0, 2), is (A) 6 7 (B*) 2 5 (C) 2  (D) 12 7 [Hint: 2 sin x (sin x – cos x) – (sin x – cos x) = 0 (sin x – cos x)(2 sin x – 1) = 0  tan x = 1 or sin x = 1/2 hence x = 4  , 4 5 , 6  , 6 5 = 2 5 Ans. ] SUBJECTIVE: *Q.1141/1 If cos A, cos B and cos C are the roots of the cubic x3 + ax2 + bx + c = 0 where A, B, C are the angles of a triangle then find the value of a2 – 2b – 2c. [Ans. 1] [Sol.  A cos = – a;  B cos A cos = b and  A cos = – c now (cos A + cos B cos C)2 =    A cos2 + 2   B cos A cos [Quiz]  cos2A + cos2B + cos2C = a2 – 2b 1 – 2 cos A cos B cos C = a2 – 2b2 1 – 2c = a2 – 2b2  a2 – 2b – 2c = 1 Ans. ]
• 12. Q.12119/6 If d1, d2, d3 arediameters of the excircles of ABC, touching thesides a, b, crespectivelythenprove that a d d a 1 1           . [Sol. TPT a d b d c d d d d a b c 1 2 3 1 2 3        LHS a r b r c r 2 2 2 1 2 3   1 2 . [a(s – a) + b(s – b) + c(s – c)] =  2 1 [s(a + b + c) – (a2 + b2 + c2)] =  4 1 [(a + b + c)2 – 2(a2 + b2 + c2] 2 4 2 2 2 ( ) ( ) ab bc ca a b c       ....(1)               c s b s a s s 1 =                   ) c s )( b s )( a s ( ) b s )( a s ( ) c s )( a s ( ) c s )( b s ( s similarly RHS = d d d a b c r r r s 1 2 3 1 2 3        can be simplified equal to 1 ] *Q.13115/6 In a  ABC , if cos A + cos B = 4 sin2 C 2 , prove that tan A 2 . tan B 2 = 1 3 . Hence deduce that the sides of thetriangle are inA.P. [Sol. 2cos A B  2 cos A B  2 = 4 sin2 C 2 [Quiz] or cos A B  2 = 2 sin C 2 sin cos C A B 2 2         = 2 cos A B  2 or cos A B  2  cos A B  2 = cos A B  2 2 sin A 2 . sin B 2 = cos A 2 . cos B 2  sin A 2 . sin B 2 3 sin A 2 . sin B 2 = cos A 2 . cos B 2 or tan A 2 . tan B 2 = 1 3 Now  s s a ( )  .  s s b ( )  = 1 3 ; s c s  = 1 3  2 s = 3 c a + b + c = 3 c  a + b = 2 c  a , c , b are in A.P. ]
• 13. Select the correct alternative : (Only one is correct) Q.16/ph-2 The equation , sin2  4 1 3 sin   = 1  4 1 3 sin   has : (A) no root (B) one root (C) two roots (D*) infiniteroots [Sol. sin2 = 1 [sin 1   ]  sin = 1   = 2n + /2  infinite roots ] Q.21/ph-3 In a triangleABC, R(b + c) = bc a where R is the circumradius of the triangle.Then the triangle is (A) Isoscelesbut not right (B)rightbut not isosceles (C*)rightisosceles (D)equilateral [Hint: R(b + c) = bc a R(b + c) = 2RsinA bc  sinA= bc 2 c b  now applyingAM  GM for b and c bc 2 c b   bc ;  bc 2 c b   1 hence sinA 1 whichis not possible. hence sinA= 1  A = 90°  A = 90° and b = c  (C)] Q.313/ph-2 The general value of x satisfying the equation, 2cot2x + 2 3 cotx + 4 cosecx + 8 = 0 is (A) n –  6 (B) n +  6 (C*) 2n –  6 (D) 2n +  6 [Hint: E = (cot ) cot cos x x ec x      3 4 5 0 2 2 = (cot ) cos cos x ec x ec x      3 4 4 0 2 2 = (cot ) (cos ) x ecx     3 2 0 2 2 cotx =  3 or cosecx = – 2  x = 5 6  or 11 6  ;  x = 7 6 11 6   or  2n –  6  C ] Q.4ph-3 In the equilateralABC,AB= 12. One vertex of a square is at the midpoint of the side BC, and the two adjacent vertices are on the other two sides of the triangle. The length of a side of the square maybe expressed as 6 q 2 p  where p and q are integers. The ordered pair (p, q) is (A) (6, –2) (B*) (9, – 3) (C) (8, – 4) (D) (5, – 1) CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-34
• 14. [Sol. Usingsinelaw inBME,  60 sin x =  75 sin 6  x =   75 sin 60 sin 6 =   1 3 2 2 · 3 3  =   2 1 3 6 6  =   6 18 3  = 6 3 2 9   p = 9, q = – 3  (9, – 3) Ans. ] Q.5st.lineAtriangle with integral sides has perimeter 8 cm.Then the area of the triangle, is (A*) 2 cm 2 2 (B) 2 cm 3 9 16 (C) 2 cm 3 2 (D) 2 cm 2 4 [Hint: Onlypossibilityfor the sides can be 3, 3, 2 (think !) A = ) c s )( b s )( a s ( s    = 2 1 1 4    = 2 cm 2 2 Ans. ] Q.6ph-3 Let ABC be theright triangle withvertices ofA(0, 2), B(1,0)and C(0, 0). IfDis the point onABsuch that the segment CD bisects angleC, then the length of CD is (A) 2 1 (B) 2 5 (C) 2 3 (D*) 3 2 2 [Sol. 2 1 × 1 × x sin 45° + 2 1 × x × 2 sin 45° = 2 1 × 1 × 2 3x sin 45° = 2  x = 3 2 2 ] Q.7QE All values of k such that the quadratic equation – 2x2 +kx + k2 +5 = 0 has two distinct roots and only one of the roots satisfies 0 < x < 2, is (A*) – 3 < k < 1 (B) – 3 < k < 0 (C) – 2 < k < 0 (D) – 1 < k  3 [Sol. f (0) · f (2) < 0 [12th 17-9-2006] (k2 + 5)(k2 + 2k – 3) k2 + 2k – 3 < 0 (k + 3)(k – 1) < 0 – 3 < k < 1 if one root is x = 2 then f (2) = 0 k2 + 2k – 3 = 0  k = 1 or k = – 3 if k = 1 + 2x2 – x – 6 = 0  2x2 – 4x + 3x – 6 = 0  2x(x – 2) + 3(x – 2) = 0  x = 2 or x = – 3/2 other roots does not line in (0, 1) |||ly when k = – 3 roots are 2, – 7  k = – 3 is also not possible]
• 15. Q.8113/ph-1 The base angles of a triangle are 22.5° and 112.5°.The ratio of the base to the height of the triangle is (A) 2 (B) 2 2 –1 (C) 2 2 (D*) 2 [Sol. fromtriangleABD cot /8 = b+x/h ....(1) fromtriangleACD x = h cot3/8 ....(2) (1) – (2) gives result Ans ] SUBJECTIVE: Q.9110/6 Let the incircle of the ABC touches its sides BC , CA&AB atA1 , B1 & C1 respectively. If 1 , 2 & 3 are the circum radii of the triangles , B1 IC1 , C1 IA1 andA1 IB1 respectively, then prove that, 2 1 2 3 = Rr2 where R is the circumradius and r is the inradius of the ABC. [Sol. AC1 IB1 is a cyclic quadrilateral. Hence in  B1 C1 I , AI = 21 in  B1C1I 21 = r cosec A 2 = r A sin 2  1 = r A 2 2 sin  2 1 2 3 = r A 3 4 2  sin = r R r 3 .         4 2  sin A r R = r2 R ] Q.10 Find the general solution of the equation, sin42x + cos42x = sin 2x cos 2x. [3] [Ans. x = 8 2 n    , n  I ] [Hint: (sin22x + cos22x)2 – 2 sin22x cos22x = sin 2x cos 2x 1 – 2t2 = t 2t2 + t – 1 = 0 2t2 + 2t – t – 1 = 0 2t(t + 1) – (t + 1) = 0 t = – 1 (rejected), t = 1/2 4 sin 2x cos 2x = 2 1 sin 4x = 1 = sin 2  x = 8 ) 1 ( 4 n n     Ans. ] Q.11112/6 Theratiosofthelengths ofthesides BC &AC ofa triangleABCto theradius ofacircumscribed circle areequalto2&3/2respectively. Showthat theratio ofthelengthsofthe bisectors oftheinteriorangles B & C is ,   7 7 1 9 2  .
• 16. [Sol. a R = 2 ; b R = 3 2  2 R A R sin = 2 ; sin B = 3 4  sin A = 1 ; c2 = 4 R2  9 4 2 R A = 90º ; c = 7 2 R Now l1 = 2a c a c  cos B 2 = 2a c a c  1 2  cos B and l2 = 2a b a b  cos C 2 = 2a b a b  1 2  cos C  l l 1 2 = a b a c   . c b 1 1   cos cos B C = c a b b a c ( ) ( )   1 1   c a b a = c b a b a c   Substituting a = 2 R ; b = 3 2 R & c = 7 2 R, we get the desired result ] Q.12s&p If the sum 2 2 2 1 1 1 1   + 2 2 3 1 2 1 1   + 2 2 4 1 3 1 1   + ....... + 2 2 ) 2000 ( 1 ) 1999 ( 1 1   equal to n – n 1 where n  N. Find n. [Ans. 2000] [Sol. S =      1999 1 a 2 2 ) 1 a ( 1 a 1 1 =        1999 1 a 2 2 2 2 2 2 ) 1 a ( a a ) 1 a ( ) 1 a ( a =          1999 1 a 2 2 2 2 2 2 ) 1 a ( a a ) 1 a 2 a ( ) 1 a 2 a ( a =        1999 1 a 2 2 2 3 4 ) 1 a ( a 1 a 2 a 3 a 2 a =      1999 1 a 2 ) 1 a ( a 1 a a =           1999 1 a 2 a a 1 1 = 1999 +           1999 1 a 1 a 1 a 1 , a telescopic sum = 1999 + 1 – 2000 1 = 2000 – 2000 1  n = 2000 Ans. ]
• 17. Q.17/1 Solve the inequality, x log 4 x log 2 2 2 / 1  < 2 (4  log16x4). ( Ans. : [1, 4)  (0, 1/4] ) [Sol. log log 2 2 2 2 x x  < 2 (4  log2 x) Let log2 x = y  y y 2 2  < 2 (4  y) This will be true if ; y2 + 2 y  0 and 4  y  0 and y2 + 2y < 2(4  y)2  y2 – 18y + 32 < 0 i.e. (y  16) (y  2) > 0 Hence solution set is y  (–, –2]  [, 2) Hence log2 x < 2  x < 4 log2 x  0  x  1. Solution is [1, 4) ; again log2 x  2  x  1/4 and log2 x >  x > 0  x  (0, 1/4] ] Q.2105/6 If p, q, r be the roots of x3 – ax2 + bx – c = 0, show that the area of the triangle whose sides are p, q & r is 1 4 [a(4ab – a3 – 8c)]1/2. [Sol. given p + q + r = a, pq = b, pqr = c  = s p q r (s )(s )(s )    where 2s = p + q + r  = 1 4 ( )( )(q )( ) p q r p q r r p r q p         = 1 4 2 2 2 a r p q (a )(a )(a )    =   1 4 2 4 8 3 2 a a a p q r pq qr rp pqr         ( ) ( )a = 1 4 2 4 8 3 3 a a ab c     (a ) =   1 4 4 8 3 1/2 a ab a c    ( ) ] Q.3118/1 ThesumofaninfiniteGPis2&thesumoftheGPmadefromthecubesofthetermsofthisinfiniteseries is 24.Find the series. [ REE ’89, 6 ] [ Ans. : 3  (3/2) + (3/22)  (3/23) +..... ] [Sol. Let the G.P. be a , ar , ar2 , ar3 , .................. now a r 1 2   ....(1) also a r 3 3 1 24   ....(2) from (1)   8 r 1 a 3 3   ....(3) (2)  (3) give ( ) 1 1 3 3 3    r r CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-35
• 18. ( ) ( ) 1 3 1 2 2     r r r ( )   r 1 1 + r2 – 2r = 3 + 3r + 3r2 2r2 + 5r + 2 = 0 2r2 + 4r + r + 2 = 0 2r ( r+ 2) + (r + 2) = 0 r = – 2 or r = –1/ 2 r  2 ( rejected) Hence r = – 1/ 2     a a .2 3 2 3 Hence theseries is, 3 3 2 3 2 3 2 2 3      .................. ] Q.465/1 Let therebe a quotient oftwo natural numbers in which the denominator is one less than the square of the numerator. If we add 2 to both numerator & denomenator, the quotient will exceed 1/3 & if we subtract 3 from numerator & denomenator, the quotient will lie between 0 & 1/10. Determine the quotient. [ REE ’90 , 6 ] [Ans:4/15] [Sol. Let Q = x y , x, y  N given x2 – y = 1 Hence Q = x x2 1  (x  1, –1) given x x    2 1 1 3 2 .............(1) and 0 < 10 1 4 x 3 x 2    .............(2) common solution to 1 and 2 is x = 4 Hence Q = 4 15 Ans. ] Q.5102/1 The number of terms of anA.P. is even; the sum of the odd terms is 310 ; the sum of the even terms is 340 ; the last term exceeds the first by57. Find the number of terms and the series. [5] [ Ans: 20 ; a = 4 ; d = 3 , S = 4 + 7 + 10 + .... ] [Sol. Let the number of terms be 2m N =2m a , a + d , a + 2d , ..... , a + (2m – 2)d , a + (2m – 1)d a + (a + 2d) + ..... + (a + (2m – 2)d) = 310 ....(1) a + d + a + 3d + ....... + (a + (2m – 1) d) = 340 ....(2) a + (2m – 1)d – a = 57 (2m – 1)d = 57....(3) 2 m (a + a + (2m – 2) d ) = 310  m (a + (m – 1) d) = 310 ....(4) from (1) Also 2 m [2a + 2md] = 340  m [ a + md ] = 340 ....(5)from (2) (5) – (4)  md = 30 from (3) 2. 30 – d = 57  d = 3  d = 3 ; m = 10  Number of terms 2m = 20 and a = 4  Series S = 4 + 7 + 10 + ...... ]
• 19. Q.6120/6 If x, y, z are perpendicular distances of the vertices of a ABC from the opposite sides and  is the area of thetriangle, then prove that ) C cot B cot A (cot 1 z 1 y 1 x 1 2 2 2       [ REE ’90 , 6 ] [Hint: a = x(cotB + cotC) ; = 1 2 ax = 1 2 x2(cotB + cotC) b = y(cotC + cotA) ; = 1 2 by = 1 2 y2(cotC + cotA) c = z(cotA + cotB) ;= 1 2 cy = 1 2 z2(cotA + cotB) & adding 1 1 1 1 2 2 2 2 x y z     [cotB + cotC + cotC + cotA + cotA + cotB] = 2 2 (cot cot cot ) A B C    Hence proved] Q.7117/6 Ifp,q, rbethelengths ofthebisectorsoftheangles ofatriangle ABC fromthe angular pointsA,Band C respectively, prove that (i) 1 2 1 2 1 2 1 1 1 p A q B r C a b c cos cos cos      and (ii) pqr abc a b c a b b c c a 4       ( ) ( )( )( ) [Sol. p = 2 2 bc A b c cos  etc. Hence LHS of (i) b c bc A A c a ca a b ab      2 2 2 2 2 cos .cos = 1 2 2 1 1 1 1 1 1 a b c a b c                  (ii) pqr bc b c ca c a ab a b A B C R abc 4 2 2 2 2 2 2         cos cos cos = abc a b b c c a R A ( )( )( ) cos      8 2 But 4 cos A 2 cos B 2 cos C 2 = sinA + sinB + sinC Hence pqr abc a b b c c a 4     ( )( )( ) 2R (sinA + sinB + sinC) = abc a b c a b b c c a ( ) ( )( )( )      ] Q.893/6 Find all the solutions of the equation ) x 1 sin(  = x cos which satisfythe condition x [0, 2] [x = 2 1 4 7   ] [Sol. sq. both sides, cos x = sin (1 – x) = cos          ) x 1 ( 2 x = 2n ±          ) x 1 2
• 20. (+) ve sign x = 2n + 2  – 1 + x nosolution (–) ve sign x = 2n – 2  + 1 – x 2x = 2n – 2  + 1 x = 2 1 4 3   ; x = 2 1 4   (Both rejected) x = 2 1 4 7   Ans. ]