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1. CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-13 Fill in the blanks. Q.1827/log The expression   1 . 0 20 log ) 05 . 0 ( is a perfect square of the natural number ______. (where 0 1 . denotes 0.111111 ..... ) [Ans. 9] [Sol.             9 1 log 2 20 20 1 = 2 20 3 log 2 20   = 4 20 3 log 20 = 81  9 ] Q.2808/log The expression, xlogy  logz · ylogz  logx · zlogx  logy when simplified reduces to ______. [Ans. 1] [Sol. P = xlogy  logz · ylogz  logx · zlogx  logy log P = (log y– log z) log x + (log z – log x) log y+ (log x – log y) log z log P = 0  P = 1 ] Select the correct alternative : (Only one is correct) Q.3log If logx(logyz) = 0 and logy(logzx) = 0, where x, y, z > 1, then 2z – x – y equals (A*) 0 (B) 1 (C) xy (D)yz [Hint: x = y = z ] Q.4 Whichofthefollowingisthelargest? (A) 6 log5 2 (B) 5 log6 3 (C*) 6 log5 3 (D) 3 Q.5805/QE If sin  and cos  are the roots of the equation ax2 – bx + c = 0, then (A) a2 – b2 = 2ac (B) a2 + b2 = 2ac (C) a2 + b2 + 2ac = 0 (D*) b2 – a2 = 2ac Q.642/ph-1 Let a = cos x + cos         3 2 x + cos         3 4 x and b = sin x + sin         3 2 x + sin         3 4 x then whichoneof thefollowingdoesnot hold good? (A) a = 2b (B) b = 2a (C) a + b = 0 (D*) a  b [Sol. Using cos(A + B) = cosA cosB – sinA sinB a = cos x + 2 cos (x + )cos 3  (using C-Drelation) = cos x – cos x = 0 |||ly b = sin x + 2 sin x( + x)cos 3  = sin x – sin x = 0 = sin x – sin x = 0 Hence a = b = 0 ] Q.7log Suppose that log10(x – 2) + log10y = 0 and y x 2 y x     Then the value of (x + y), is (A) 2 (B) 2 2 (C*) 2 + 2 2 (D) 4 + 2 2
2. [Sol. y(x – 2) = 1  x = y 1 + 2 .....(1) x + 2 y  = y x  x + y – 2 + 2 ) 2 y ( x  = x + y 1 = ) 2 y ( x  x(y – 2) = 1 ....(2) using(1)in(2)          2 y 1 (y – 2) = 1 1 – y 2 + 2y – 4 = 1 y2 – 2y – 1 = 0 y = 2 4 4 2   y = 1 ± 2 ....(3) rejecting y = 1 – 2 y = 1 + 2 using(3)in(1) x = 1 + 2  x + y = 2 + 2 2 Ans. ] Q.829/log Let x, y, z are real numbers greater than 1 and 'w' is a positive real number. If logxw = 24, logyw = 40 and logxyzw =12 then logwz has the value equal to (A) 120 1 (B*) 120 2 (C) 120 3 (D) 120 5 [Sol. Given logwx = 24 1 ....(1); logwy= 40 1 ....(2), Let logwz = k ....(3) (1) + (2) + (3)  logw(x y z) = 24 1 + 40 1 + k  12 1 = 24 1 + 40 1 + k  k = 12 1 – 24 1 – 40 1 = 120 3 5 10   = 120 2 , hence logwz = 120 2 Ans. ] Q.949/log The value of the expression, log4         4 x2  2 log4 (4 x4) when x =  2 is : (A*)  6 (B)  5 (C)  4 (D) meaningless [Hint: (log4x2 – 1) –2(1 + log4x4)  when x = – 2 the equation becomes log44 – 3 – 2 log416 = 1 – 7 = – 6 Ans ]
3. Q.10 The solution set of the equation, 3 log10 x + 2 log10 1 x = 2 is : (A) {10, 102} (B) {10, 103} (C*) {10, 104} (D) {10, 102, 104} [Sol. 3 log10 x – 2 log10 x = 2 log10 x = y 3y – 2y2 = 2 2y2 – 3y + 2 = 0 2y2 – 2y – y + 2 = 0 2y(y – 1) – (y – 1) = 0 (y – 1)(2y – 1) = 0 y = 1; y = 1/2 if log10 x = 1  x = 10 log10 x = 1/2  x = 104] Subjective : Q.1113/6 If  and  are the roots of the quadratic equation (sin 2a)x2 – 2(sin a +cos a)x + 2 = 0, findthem and hence prove that 2 + 2 = 2 · 2. [Ans. sec a , cosec a]  [Sol. x = a 2 sin 2 a 2 sin 8 ) a cos a (sin 4 ) a cos a (sin 2 2     = a 2 sin a 2 sin 2 ) a cos a (sin ) a cos a (sin 2     = a 2 sin a 2 sin 2 a 2 sin 1 ) a cos a (sin     = a 2 sin ) a 2 sin 1 ( ) a cos a (sin    = a cos · a sin 2 ) a cos a (sin ) a cos a (sin    with+vesign a cos · a sin 2 a sin 2 = sec a with–vesign a cos · a sin 2 a cos 2 = cosec a now 2 + 2 = sec2a + cosec2a = a cos 1 2 + a sin 1 2 = a cos · a sin 1 2 2 = sec2a · cosec2a = 2 · 2 hence proved] Q.12 Findallintegral solutionof theequation,       3 x 2 2 x 4 2 x x log 3 x log 2 x log 4   .  [Ans. x = 1, x = 4] [Sol.   2 x log x log 4 2 2 + ) x 4 ( log ) x ( log 2 2 2 2 = ) x 2 ( log ) x ( log 3 2 3 2 1 x log ) x ( log 2 1 · 4 2 2  + ) x ( log 2 ) x ( log 4 2 2  = ) x ( log 1 ) x ( log 9 2 2 
4. let log2x = t 1 t t 2  + 2 t t 4  = 1 t t 9   1 t 2  + 2 t 4  = 1 t 9  ) 2 t )( 1 t ( 4 t 4 4 t 2      = 1 t 9   6t (t + 1) = 9(t2 + t – 2)  6t2 + 6t = 9t2 + 9t – 18  3t2 + 3t – 18 = 0  t2 + t – 6 = 0  (t + 3)(t – 2) = 0 t = 0, t = 2 & t = – 3, x = 1, x = 4, x = 8 1 (rejected  it is not integral value)]
5. Select the correct alternative : (Only one is correct) Q.1log Let N = 2 2 12 22 2 12 22          then log2N equals (A) 2 (B) 3 (C*) 4 (D) none [Sol. N = 2 2 2 ) 2 2 3 ( ) 2 2 3 (          = 1  2 ) 2 2 3 ( ) 2 2 3 (    = 16 hence log2N = log216 = 4 Ans. ] Q.2log The sum of all valuesof x so that ) 2 x 3 x ( ) 1 x 3 x ( 2 2 8 16      ,is (A) 0 (B) 3 (C*) – 3 (D) – 5 [Sol. ) 2 x 3 x ( ) 1 x 3 x ( 2 2 8 16       ) 2 x 3 x ( 3 ) 1 x 3 x ( 4 2 2 2 2       6 x 9 x 3 ) 4 x 12 x 4 2 2 2 2      4x2 + 12x – 4 = 3x2 + 9x + 6  x2 + 3x – 10 = 0  (x + 5)(x – 2) = 0 x = – 5 or x = 2  the sum of values of x is (– 5) + 2 = – 3 Ans. ] Q.3 The equation, | sin x | = sin x + 3 in [0, 2] has (A*) no root (B) onlyone root (C) two roots (D) more than two roots Q.420/log Given log2(log8x) = log8(log2x) then (log2x)2 has the value equal to (A) 9 (B) 12 (C*) 27 (D) 3 3 [Sol. Given log2       x log 3 1 2 = 3 1 log2(log2x) let log2x = 3y  log2y= 3 1 log23y  3log2y= log23 + log2y  2 log2y = log23  y2 = 3  9 1 (log2x)2 = 3  (log2x)2 = 27 Ans. ] Q.530/log The reals x and ysatisfy log8x + log4(y2) = 5 and log8y+ log4(x2) = 7 thenthe valueofxyis (A) 1024 (B*) 512 (C) 256 (D) 81 [Sol. equation(1)  3 1 log2x + log2y= 5 ....(3) and equation(2)  3 1 log2y+ log2x = 7 .....(4) (3) + (4)  3 1 log2(xy) + log2(x y) = 12  3 4 log2(xy) = 12 log2(xy) = 9  xy = 29 = 512 Ans.] CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-14
6. Q.6 If sin 2x = 2025 2024 , where 4 5 < x < 4 9 , the value of the sin x – cos x is equal to (A*) – 45 1 (B) 45 1 (C) ± 2025 1 (D) none [Sol. (sin x – cos x)2 = 1 – sin 2x = 1 – 2025 2024 = 2025 1 sin x – cos x = 45 1 or – 45 1 in 4 5 to 4 9 , sin x – cos x < 0 Hence answer is B ] Q.7log The equation ln           ) 1 k ( 1 k 1 ) 1 k ( k = F(k) ·                k n k 1 1 k 1 1 n l l is true for all k wherever defined. F(100) has the value equal to (A) 100 (B*) 101 1 (C) 5050 (D) 100 1 [Sol. L.H.S. = k n k 1 l – 1 k 1  ln(k + 1) = ) 1 k ( k ) 1 k ( n k k n ) 1 k (     l l R.H.S. = k ) k ( F [k ln k – k ln(k + 1) + ln k] = k ) k ( F [(k + 1) ln k – k ln(k + 1)]  L.H.S. = R.H.S.  ) 1 k ( k ) 1 k ( n k k n ) 1 k (     l l = k ) k ( F [(k + 1) ln k – k ln(k + 1)]  F (k) = 1 k 1  F(100) = 101 1 Ans. ] Q.848/log If x1 and x2 are two solutions of the equation log3 2x  7 = 1 where x1 < x2 , then the number of integer(s) between x1 and x2 is/are : (A*) 2 (B) 3 (C) 4 (D) 5 [Hint: log3| 2x – 7| = 1  |2x – 7| = 3 Case 1 : x > 7/2  2x – 7 = 3  x = 5 Case 2 : x < 7/2  – 2x + 7 = 3  x = 2 number of integers between 2 and 5 are 3 and 4 = 2 ]
7. Select the correct alternative : (More than one are correct) Q.9522/log The equation 1 27  logx log3 x + 1 = 0 has : (A*) no integral solution (B) oneirrationalsolution (C) two real solutions (D*) no primesolution [ Hint : x log x log 2 3 1 3 3          + 1 = 0 let log3x = y y y 2 3 1          = – 1  2 y 1 y 2 3 1            2 y 1 y 2 3 y 2   2y2 + 3y – 2 = 0  2y2 + 4y – y – 2 = 0  ( y +2) (2y – 1) = 0 y = 1/2 or y = – 2  x = 31/2 (rejected) or x = 1/9 ] Subjective : Q.1012/1 Find the set of values of ‘x’ satisfying the equation x 64 – x 3 x 3 2  + 12 = 0. [Ans: x = 3 ] [Hint: Here x  N, therefore the 2nd value x = 6 log 3 2 has to be rejected given equation reduced to  2 x 3 2 – 8 · x 3 2 + 12 = 0 ; put x 3 2 = y and proceed. ] Q.11 Let aandbaretworealnumbers suchthat, sina+sinb= 2 2 andcos a+cosb = 2 6 .Findthevalueof (i) cos(a – b) and (ii) sin(a + b).  [Sol. squaringandadding [Ans. (i) 0, (ii) 2 3 ] 2 + 2 cos(a – b) = 2 1 + 2 3 = 2 cos(a – b) = 0 again 2 sin 2 b a  cos 2 b a  = 2 1 2cos 2 b a  cos 2 b a  = 2 3  tan 2 b a  = 3 1  sin(a + b) = 2 b a tan 1 2 b a tan 2 2    = 3 1 1 3 1 2   = 4 3 3 2  = 2 3 ] Q.12 For any 3 angle ,  and , prove that sin  + sin  + sin  = sin( +  + ) + 4 sin          2 ·sin          2 ·sin          2 . 
8. [Sol. TPT sin  + sin  + sin  – sin( +  + ) = 4 sin          2 ·sin          2 ·sin          2 LHS =                                        2 sin 2 cos 2 2 cos 2 sin 2 =                                     2 2 cos 2 cos 2 sin 2 =                                  2 sin 2 sin 2 2 sin 2 = 4 sin          2 ·sin          2 ·sin          2 = RHS hence proved. ]
9. CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-15 Fill in the blanks : Q.1803/log Solution set of the equation, x log2 10 + log10x2 = 2 log2 10  1 is ______ . [Ans. 1 20 , 1 5 ] [Sol. (log10x)2 + 2log10x + 1 = 2 log2 10 = (log10x + 1)2 = 2 log2 10 log10x + 1 = + log102 log10x + 1 = – log102 log10x + 1 = log102 log10(2x) = – 1 log10   x 2 = 1 2x = 10–1   x 2 = 10 = 5 1 x = 20 1 ] Select the correct alternative : (Only one is correct) Q.2 Let m denotesthe number of digits in 264 and n denotes the number of zeroes between decimal point and the first significant digit in 2–64, then theordered pair (m, n) is (you mayuse log102=0.3) (A) (20, 21) (B) (20, 20) (C) (19, 19) (D*) (20, 19) [Sol. x = 264 y = 2–64 log10x = 64 log10x log10y = – 64 log102 = 64 × 0.3 = – 64 × 0.3 log10x = 19.2 log10y = – 19.2 m = 19 + 1 log10y = 8 . 20 m = 20 n = 19  (20, 19) ] Q.326/log The ratio 1 a 7 a 2 3 2 a log 4 log a log 49 3 2 4 / 1 ) ( 1 a 27 2      simplifies to : (A) a2  a  1 (B) a2 + a  1 (C) a2  a + 1 (D*) a2 + a + 1 [Hint: N D r r = a a a a a 4 2 2 2 1 1       D ] Q.4 Whichoneof the followingdenotesthegreatest positive properfraction? (A) 6 log 2 4 1       (B) 5 log3 3 1       (C*) 2 log3 3  (D)          2 log 1 3 8 [Hint: A: a = 6 log 2 2 2 1 = 36 log 2 2 1 = 1 36 ; B: b = 5 log 3 3 1 = 1 5 C : c = 2 1 3 1 2 log 3  ; D: d = 3 log 2 8  = 27 log 2 2  = 27 1 2 1 27 log 2  ] Select the correct alternative : (More than one are correct) Q.5152/log The solution set of the system of equations, log12x 1 2 2 log log x y        = log2 x and log2 x . (log3 (x + y)) = 3 log3 x is: (A*) x = 6; y = 2 (B) x = 4; y = 3 (C*) x = 2; y = 6 (D) x = 3; y = 4 [Hint: log12x(log2xy) = log2x  log2xy = log212 xy= 12 ....(1) log3( x + y) = 3. log32  x + y = 8 ....(2) from (1) and (2) x = 6 , y = 2 or y = 6 , x = 2 ]
10. Subjective : Q.6830/log If log18 36 = a & log2472 = b, then find the value of 4 (a +b)  5 ab. [Ans. 2] [Hint: a = y 2 y 2 2 2 3 log 2 3 log 18 log 36 log 2 3 2 2 3       where log32 = y b = 1 y 3 2 y 3 3 2 log 3 2 log 24 log 72 log 3 3 2 3 3       and henceresult ] ] Q.711/log It is known that x = 9 is aroot of the equation log (x2 + 15a2) – log(a2) = log 8 2 ax a  . Find theother root(s) ofthis equation. [Sol. 2 a ax 8 log 2 a a 15 x log 2 2       .........(1)  x2 + 15a2 = 8ax or x2 – 8ax + 15a2 = 0 .........(2) (x – 5a) (x – 3a) = 0  a = x 5 or a = x 3  x = 9 satisfies (1) hence a = 9 5 or a = 3 but a = 9 5 is not possible (think!)  a = 3 substitutinga = 3 in (2) x2 – 24x + 225 = 0  x = 9 or x = 15  other root is x = 15 ] Q.849/06 If  = 1 2n   , prove that 2n cos cos2 cos22 ........ cos2n–1 = 1. What the value of the product whould be if  = 1 2n   . [Ans. –1] [Sol. We have 2n ·        sin 2 cos . .......... . 2 cos . 2 cos . cos . sin 1 n 2 =        sin 2 cos ...... .......... 2 cos . 2 cos . 2 sin . 2 1 n 2 1 n =       sin 2 cos ....... .......... 2 cos . 2 sin . 2 1 n 2 2 2 n =       sin 2 cos ....... .......... 2 cos . ) . 2 sin( . 2 1 n 3 3 3 n Thisexpressionfinallyconvergerto =   sin 2 sin n =          sin 1 2 2 sin n n =            sin 1 2 2 sin n n =              sin 1 2 2 sin n n =                 1 2 sin 1 2 sin n n = 1 ] Q.9 Evaluate the product     1 n 2 1 r ) r tan( where 4n = . [Ans : 1] [Sol. P = tan  × tan 2 × tan 3 × ........... × tan(2n – 1) = tan  · tan(2n – 1) n 4  tan (2n – 1) n 4  = 1
11.          n 4 2 tan = cot n 4  tan n = tan 4  = 1 Ans. ] Q.1059/06 Find the exact value of cosec10° + cosec50° – cosec70° [Ans : 6] [Sol:      70 sin 1 50 sin 1 10 sin 1      20 cos 1 40 cos 1 80 cos 1            80 cos . 40 cos . 20 cos 80 cos 40 cos 20 cos 80 cos 20 cos 40 cos = 8 [cos20°(cos40° + cos80°) – cos40° cos80° ] = 8 [2cos20° cos60° cos20° – cos40° cos80°] = 4 [2cos220° – 2cos40° cos80°] = 4 [1 + cos40° – (cos120° + cos40°)] = 4 . 2 3 = 6 Ans ] Q.1153/06 If sin x + cos x + tanx + cot x + secx + cosecx = 7 then sin2x = a  b 7 where a, b  N. Find the ordered pair (a, b). [Ans. a = 22 , b = 8] [Sol. S  sin x ; C  cos x S + C + C S + S C + C 1 + S 1 = 7  SC C S C S S C C S 2 2 2 2      = 7 SC (S + C) + 1 + S + C = 7 S C  (S + C) (1 + SC) = 7SC – 1 (S + C)2 (1 + SC)2 = (7 SC – 1)2 (1 + 2 SC) (1 + SC)2 = (7 SC – 1)2 put sin x · cos x = y (1 + 2y) (1 + y)2 = (7y – 1)2  (1 + 2y) (1 + y2 + 2y) = 49y2 – 14y + 1 1 + y2 + 2y + 2y + 2y3 + 4y2 = 49y2 – 14y + 1  2y3 – 44y2 + 18y = 0 2y [y2 – 22y + 9] = 0  2y = 36 484 22   (as y  0) y = 11 ± 9 121 = 11 ± 7 4 sin x · cos x = 11 ± 7 4 2sin x · cos x = 22 – 7 8 + ve is rejected as sin 2x can have max. value as 1 ]
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