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1. CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-7 Fill in the blanks : Q.1866/log If x = 7 5 2 3   1 7 5 2 3  , then the value of x3 + 3x  14 is equal to ______. [Ans. 0] [Hint : a =   7 5 2 1 3  / , b =   3 1 2 5 7 1  , then x = a  b , x3 = 14  3x ] Select the correct alternatives : (More than one are correct) Q.2509/log If p, q  N satisfy the equation   x x x x  then p & q are : (A*) relativelyprime (B) twin prime (C*) coprime (D*) if logq p is defined then logp q is not & vice versa [Sol.   x x x x  p, q  N x log x = x log x log x        2 x x = 0  log x = 0 or        2 x x = 0 x = 1 or x = 0 or x = 4 ; x = 0 (rejected) ] Q.3505/log The expression, logp logp ...... p p p p p n radical sign        where p  2, p  N, when simplified is : (A*) independent of p, but dependent on n (B) independent of n, but dependent on p (C) dependent on both p & n (D*) negative . [Hint: ...... p p p p = ppn 1 . Therefore logp logp ppn 1 = logp 1 pn       . logp (pn) =  n ] Q.4510/log Which ofthefollowingwhensimplified,reduces tounity? (A*) log105 . log1020 + 2 log2 10 (B*) 2 2 3 48 4 log log log log   (C*)  log5 log3 9 5 (D) 1 6 log 3 2 64 27       [Hint: D = – 1 ] *Q.5508/log The number N =   1 2 2 1 2 2 3 3 6 2 2    log log log when simplified reduces to : (A) a prime number (B) anirrational number (C*) a real which is less than log3 (D*) a real which is greater than log7 6 [Hint: 1 2 2 1 2 2 1 2 1 2 1 2 1 3 3 2 3 2 3 2 3 2 3 2         log ( log ) (log ) ( log ) ( log ) ( log ) ] Subjective : Q.69/06 If tanA& tan B are the roots of the quadratic equation, ax2 + bx + c = 0 then evaluate a sin2 (A + B) + b sin (A + B). cos (A + B) + c cos2 (A + B). [Ans:c] [Sol. tanA + tan B =  b a ; tanA. tan B = c a ; tan (A + B) =   b a c a 1 = b c a 
2. Now E = cos2 (A + B) [a tan2 (A + B) + b tan (A + B) + c ] = 1 1 2 2   b c a ( ) a b c a b c a c 2 2 2 ( )           = ( ) ( ) c a b c a    2 2 2 b c a a c a c 2 1                 = ( ) ( ) c a b c a    2 2 2 b c c a c 2 2 ( )         E = c ] Q.718/06 If cos + cos = a and sin + sin= b then prove that, cos2 + cos2 =      a b a b a b 2 2 2 2 2 2 2     [Sol. squaringand adding, cos(–) = a b 2 2 2 2   ....(1) using C – D relations & dividing, tan     2 b a  cos( + ) = 1 1 2 2 2    b a b a = a b a b 2 2 2 2   ....(2) now cos2 + cos2 = 2cos( + ) cos ( – ) ....(3) use (1) & (2) in (3) to get the result ] Q.8 Establishtricotomyineachofthisfollowingpairsofnumbers (i) 2 log 3 log 4 27 2 and 3 (ii) ) 25 / 1 ( log and 5 log 16 / 1 4 (iii) 4 and 81 log 10 log 10 3  (iv) ) 5 / 1 ( log and ) 7 / 1 ( log 7 / 1 5 / 1 [(i)(Hint: >); (ii) (Hint:=); (iii) (Hint:<); (iv) (Hint: >) ] [Hint: log log 3 10 10 4 3 4 4 4      y y = y y           2 0 2 ] Q.96(sub)/log Given, log712 = a & log1224 = b. Show that, log54168 = 1 8 5   ab a b ( ) . [Sol. log712 = a  2 log72 + log73 = a ; log1224 = b  log724 = ab 3 log72 + log73 = ab Now log54168 = log log 7 7 168 54 = 3 2 3 1 3 3 2 7 7 7 7 log log log log    . Substituting the value of, log72 & log73 we get the result ] Q.10 If b a c log a c b log c b a log      , show that aa . bb . cc = 1. Q.11 Prove that the expression, cos2 8  + cos2 8 3 + cos2 8 5 + cos2 8 7 is not irrational. [Sol. cos2 8  + cos2 8 3 + cos2 8 5 + cos2 8 7 cos2 8  +cos2          8 8 4 + cos2          8 8 4 + cos2          8           8 sin 8 cos 2 2 +          8 cos 8 sin 2 2 = 1 + 1 = 2 Ans. ]
3. CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-8 Fill in the blanks : Q.1832/log If logx log18   2 8  = 1 3 . Then the value of 1000 x is equal to _____. [Ans. 125] [Hint: logx log18   2 2 2  = 1/3 logx log18 18 = 1/3  x = 1/8  1000x = 125 ] Q.2864/log The solution set of the equation 27 log 2 log x log 3 9 9 2 x . 6 4   = 0 is ______ . [Ans {9, 81}and note that x log9 2 = 2 log9 x ] [Hint: put 2 9 log x = y we get y2 – 6y + 8 = 0  y = or 2  2 9 log x = 22  log9x = 2  x = 81 or 2 9 log x = 21  log9x = 1 x = 9 ] Select the correct alternative : (Only one is correct) Q.3502/log Which oneof the followingwhensimplified does not reduceto an integer? (A) 2 6 12 3 log log log  (B) 243 log 32 log 3 2 (C*) log log log 5 5 5 16 4 128  (D) log1/4 1 16 2        [ Hint : A=1, B = 2, C = 2/7, D = – 4] Q.4 Let u = (log2x)2 – 6 log2x + 12 where x is a real number. Then the equation xu = 256 has (A) no solution for x (B*) exactlyonesolution for x (C)exactlytwodistinct solutions for x (D)exactlythreedistinct solutions forx [Hint: 12 x log 6 ) x (log 2 2 2 x   = 256 ; (t2 – 6t + 12) t = 8 where t = log2x ;  (t – 2)3 = 0  t = 2  x = 4 ] Q.55/log The equation, log2 (2x2) + log2 x .   1 x log log 2 x x  + 1 2 log4 2 (x4) +   x log log 3 2 2 / 1 2 = 1 has: (A)exactlyonereal solution (B)tworeal solutions (C)3real solutions (D*)no solution. [Hint: y3 + 3y2 + 3y + 1 = 1 where y = log2 x  y(y2+3y+3) = 0  x = 0 (rejected) ; y2 + 3y + 3 (complex roots) ] Select the correct alternative : (More than one are correct) Q.6513/log The equation 2 8 2 8 ) x (log ) x 8 ( log = 3 has : (A)nointegralsolution (B*)onenatural solution (C*) tworeal solutions (D)oneirrationalsolution [Hint: 3 y y 2 1 2   (log8x = y) 3y2 + 2y – 1 = 0  3y2 + 3y – y – 1 = 0  3y (y + 1) – 1(y + 1) log8x = y = 1/3, – 1  x = 2 , 1/8 ] Subjective Q.750/06 Find the exact value of tan2 16  + tan2 16 3 + tan2 16 5 + tan2 16 7 [Ans : 28]
4. [Sol. Let 16  =  tan2 + tan23 + tan25 + tan27 = (tan2  + cot2  ) + (tan23 + cot23) [ Note that tan7 = tan(8 – ) = cot and tan5 = tan(8 –3) = cot3 ] = (cot – tan)2 + (cot3 – tan3)2 + 4 = 4 [cot22 + cot26] + 4 = 4 [ cot22 + tan22 ] + 4 = 4 [ (cot2 – tan2)2 + 2] + 4 = 4 (cot2 – tan2)2 + 12 = 4 . 4 cot2 4 + 12 = 16 × 12 + 12 = 28 Ans ] Q.816/06 In anytriangle, if (sinA+ sin B + sin C) (sinA+ sin B  sin C) = 3 sinAsin B, find the angle C. [Sol. (sinA+ sin B)2  sin2 C = 3 sinAsin B [Ans. C = 60º] sin2 A sin2 C + sin2 B = sinAsin B sin (A+ C) sin (A C) + sin2 B = sinAsin B sin B [ sin (A C) + sin (A+ C)]= sinAsin B (Using sin(A+C)= sinB) 2 sinAcos C = sinA (sinB  0) cos C = 1/2  C = 60º ] Q.9 Whichissmaller? 80 1 log 3 1 or        2 15 1 log 2 1 [Hint: 80 1 log 3 1 < 81 1 log 3 1 = 4 or 80 1 log 3 1 < 4 ....(1) || ly 2 15 1 log 2 1  > 16 1 log 2 1 = 4; hence 2 15 1 log 2 1  > 4 ....(2) From (1) and (2) 2 15 1 log 80 1 log 2 1 3 1   ] Q.10 Find the possible value(s) of           cot ec cos sec tan cos sin if tan = – 3 4 . [2.5] [Ans. in 2nd quadrant 5 23 , in 4th quadrant 35 34 ] [Sol. If  lies in the 2nd quadrant then tan = – 3 4  sin = 5 4 ; cos = – 5 3 hence E = 4 3 4 5 3 5 3 4 5 3 5 4      =         15 20 9 12          9 15 20 12 = 4 · 15 12 · 23 = 5 23 Ans. If  lies in the 4th quadrant tan = – 3 4  sin = – 5 4 ; cos = 5 3 E = 4 3 4 5 3 5 3 4 5 3 5 4      =            9 15 20 20 9 12       15 12 = 15 · 14 12 · 17 = 35 34 Ans. ]
5. Q.11 Solve theequation ) 1 x ( log ) 5 . 0 x ( log 5 . 0 x 1 x      . [Ans. x = 1] [5] [Sol. ) 1 x ( log ) 5 . 0 x ( log 2 2   = ) 5 . 0 x ( log ) 1 x ( log 2 2    [log2(x + 1)]2 = [log2(x – 0.5)2]] log2(x + 1) = log2(x – 0.5) or – log2(x – 0.5) if log2(x + 1) = log2(x – 0.5) x + 1 = x – 0.5  nosolution if log2(x + 1) = log(x – 0.5)–1 x + 1 = ) 2 1 ( x 1  = 1 x 2 2  (x + 1) (2x – 1) = 2 2x2 + x – 3 = 0 2x2 + 3x – 2x – 3 = 0 (x – 1)(2x + 3) = 0  x = 1 Ans. x = – 3/2 (rejected) ]
6. Q.9 Compute the value of 81 27 3 1 3 36 4 9 5 9 7 log log log   [Hint: T1 = 625; T2 = 216; T3 = 49; Ans : 890 ] [Sol. 5 log 4 3 log 1 3 5 3 81  = 54 = 625 ....(1) 36 log 3 36 log 2 3 9 3 27  = 2 3 2 ) 36 ( log 6 3 2 / 3 3   = 216 ....(2) 7 log 2 7 log 4 9 log 4 3 9 7 3 3 3   = 49 ....(3) 625 + 216 + 49 = 890 ] Q.2819/log The expression log .5 0 2 8 has the value equal to ______. [Ans. 3] [Hint: Note x2 = x, if x is negative ]
7. Q.1 Simplifywheneverdefined )) 180 ( cos ) sin( ) 90 sin( ) 540 ( sin ) 270 sin( ) 720 ( cos ) 270 sin( 2 3 3                       + ) 450 ( ec cos ) 270 cot( 2       where  is taken suchthat thedenominator appearing inanyfraction in the expression does not vanish. [Ans. 1 ] [2.5] [Sol. E =            2 3 3 cos sin cos ·sin cos ) )(cos cos ( +   2 sec tan =         2 3 4 cos cos sin sin cos cos +   sec sin = ) cos (sin cos ) cos (sin cos 3 3         + sin cos = (1 – sin cos) + sin cos = 1 ] Q.2 Given x2 + 4y2 =12xy, where x>0, y>0 then prove that, log(x + 2y) – 2log2 = 2 1 (log x + log y). [3] [Sol. x2 + 4y2 + 4xy = 12xy+ 4xy = 16 xy (x + 2y)2 = 16xy 2 log(x + 2y) = log216 + log x + log y = 4 log 2 + log x + log y log (x + 2y) = 2 log 2 + 2 1 (log x + log y) ] Q.3 Solve the equation, 2 x log ) x log(   . [Ans. x = – 1 and x = – 10] [3] [Sol. Since the equation can be satisfied onlyfor x < 0 hence | x | x2  = – x  ) x log( ) x log(     log(–x) =[log(–x)]2 log(–x)[1 –log(–x)] = 0 if log(–x) = 0  – x = 1  x = – 1 if log10(–x) = 1 – x = 10  x = – 10 ] Q.4 Let fn(x) = sinnx + cosnx. Find the number of values of x in [0, ] for which the relation 6f4(x) – 4f6(x) = 2f2(x) holds valid. [Ans. infinitelymany] [3] [Hint: fn(x) = (sin x)n + (cosx)n (given) now 6f4(x) – 4f6(x) = 2f2(x) 6[sin4x + cos4x] – 4[sin6x + cos6x] = 2(sin2x + cos2x) 6[1 – 2sin2x cos4x] – 4[1 – 3sin2x cos2x] = 2 2 = 2  it becomesanidentity   solutions ] Q.5 If 2cos =x + x 1 , find thevalues of the followingin terms of cosine of the multipleangle of . (i) x2 + 2 x 1 ; (ii) x3 + 3 x 1 and (ii) x4 + 4 x 1 Hence deduce the value of xn + n x 1 , n  N. [3] [Ans. (i) 2 cos 2, (ii) 2 cos 3, (iii) 2 cos 4 and 2 cos n ] CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-9
8. [Sol. (i) LHS = x2 + 2 x 1 = 2 x 1 x        – 2 = 4cos2– 2 = 2(1 + cos2) – 2 = 2 cos2 (ii) LHS = x3 + 3 x 1 = 3 x 1 x        – 3        x 1 x = 8cos3 – 6cos = 2(4cos3 – 3 cos) = 2 cos3 (iii) LHS = x4 + 4 x 1 = 2 2 2 x 1 x        – 2 = 4cos22 – 2 [usingtheresult of(i)] = 2(1 + cos4) – 2 = 2 cos4 Since x2 + 2 x 1 = 2 cos2 ; x3 + 3 x 1 = 2 cos3 ; x4 + 4 x 1 = 2 cos4 Hence xn + n x 1 = 2 cos n ] [11th PQRS, 29-5-2005] Q.6 If a  b > 1, then find the largest possible value of the expression       b a loga +       a b logb . [3] [Ans. 0 ] [Sol. Given E =       b a loga +       a b logb = logaa – logab + logbb – logba E = 2 – (logba + logab) now Ewillbemaximumif logba+logabisminimum but logba + logab =  2 a b b log a log  + 2 Henceitsminimumvalueis2  Emax = 2 – 2 = 0 Ans. ] Q.7 Prove that solutionofthe equation, 3 x 27 x 9 4 4 1 log 1 2 1 2 log 2                                is an irrational number. . [Ans. – log23] [5] [Sol.                        4 4 1 log 3 1 2 1 2 log 2 x 27 x 9        1 2 1 · 2 log x 3 =        4 4 1 log x 3  1 2 2 x  = 4 4 1 x  ; let x 2 1 = y 2y – 1 = y2 – 4 y2 – 2y – 3 = 0  (y – 3)(y + 1) = 0  y = 3 or y = – 1 (rejected) x 2 1 = 3  2–x = 3 – x = log23  x = – log23 TPT log23isirrational; let log23 = q p where p  N, q  N, p > q 2p/q = 3; 2p = 3q LHS is even but RHS is odd. Which is not possible. Hence log23 can not be rational. Hence x = – log23 is irrational. ] Q.8843/log Solution set of the equation
9. 1 1 6  log x + 2 = 3 1 6  log x is _______. [Ans. 1 6 ,        ] [Sol. Let log1/6x = a | 1 – a | + 2 = | 3 – a | I When a < 1 1 – a + 2 = 3 – a log1/6x < 1 a = 1 x > 1/6 log1/6x = 1 x = 1/6 II When 1  a < 3 1  log1/6x < 3 – (1 – a) + 2 = 3 – a x  1/6, x > 1 / 216 – 1 + a + 2 = 3 – a 2a = 2  a = 1 x = 1/6 III a  3 – (1 – a) + 2 = – (3 – a) – 1 + a + 2 = – 3 + a – 1 + 2  – 3 ] Q.9 Find the value of sin 2  and cos 2  . If sin = – 325 323 and           2 3 , [3] [Ans. sin 2  = 26 5 19 ; cos 2  = – 26 5 17 ] [Sol. Given sin = – 325 323 ,           2 3 ,  325 36 cos    now sin2 2  = 2 cos 1   = 650 361 2 325 36 1    sin 2  = 26 5 19 as 2           4 3 , 2 |||ly cos2 2  = 2 cos 1   =        325 36 1 2 1 = 650 289  cos 2  = – 26 5 17 Hence sin 2  = 26 5 19 ; cos 2  = – 26 5 17 Ans. ] Q.10 Show that, tan          2 6 tan          2 6 = 1 cos 2 1 cos 2     [3] [Sol. LHS = 2 tan 3 1 1 2 tan 3 1 · 2 tan 3 1 1 2 tan 3 1         = 2 tan 3 2 tan 3 1 2 2     = 2 sin 2 cos 3 2 sin 3 2 cos 2 2 2 2       =                                       2 cos 2 sin 2 sin 2 cos 2 2 cos 2 sin 2 sin 2 cos 2 2 2 2 2 2 2 2 2 = 1 cos 2 1 cos 2     Hence proved ]
10. Q.11 Let y = x 5 cos x 4 cos x 2 cos x cos x 5 sin x 4 sin x 2 sin x sin       . Find the value of y where x = 36  . [3] [Ans. 2 – 3 ] [Sol. y = ) x 4 cos x 2 (cos ) x 5 cos x (cos ) x 4 sin x 2 (sin ) x 5 sin x (sin       = x cos x 3 cos 2 x 2 cos x 3 cos 2 x cos x 3 sin 2 x 2 cos x 3 sin 2   = ] x cos x 2 [cos x 3 cos ] x cos x 2 [cos x 3 sin   = tan 3x  y = = tan3x = 36 3 tan  = 12 tan  = tan15° = 2 – 3 Ans. ]