Anzeige
Anzeige
Anzeige
Nächste SlideShare
Dpp (31-35) 11th J-Batch Maths.pdf
Wird geladen in ... 3
1 von 11
Anzeige

### Dpp (16-18) 11th J-Batch Maths.pdf

1. CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-16 Fill in the blanks : Q.1844/log The solution set of the equation x x   3 1 4 = x x   3 2 3 is ______. [Ans.2,4,11] [Hint: Takinglogonboth the sides 4 1 x  log | x – 3 | = 3 2 x  log | x – 3 |  log | x – 3 |                      3 2 x 4 1 x = 0  log | x – 3 | = 0 or                      3 2 x 4 1 x = 0  x = 4 , 2 or x = 11 ] Q.2846/log If (a2 + b2)3 = (a3 + b3)2 and ab  0 then the numerical value of a b b a  is equal to ______ [Hint: a6 + b6 + 3a2b2 (a2 + b2) = a6 + b6 + 2a3b3 [Ans. 2/3 ]  3 2 ab b a 2 2    3 2 3 1 3 1 a b b a     Ans ] Q.347/log 3 5 log 1 ) 1 . 0 ( log 1 10 7 5   simplifies to ________. [Ans. 2] [Hint: 3 10 7 5 log 10 / 1 log 1 5   = 3 1 1 7  = 3 8 = 2 Ans ] Select the correct alternative : (Only one is correct) Q.4 The tangents oftwo acute angles are 3 and 2. Thesine of twice theirdifference is : (A) 7/24 (B) 7/48 (C) 7/50 (D*) 7/25 [Hint: tan  = 3; tan  = 2. Now sin 2() = sin2 cos2 cos2 sin2 = 4 1 4 1 . 9 1 3 . 2     4 1 2 . 2 . 9 1 9 1    = 25 7  D] Q.5 log3 1 1 3        + log3 1 1 4        + log3 1 1 5        + .............. + log3 1 1 242        when simplified has the value equal to (A) 1 (B) 3 (C*) 4 (D) 5 *Q.6 Whichofthesestatements is false? (A)Arectangleis sometimes arhombus. (B)Arhombusis always a parallelogram. (C*) Thedigonals of a parallelogram always bisect theangles at the vertices. (D) The diagonals of a rectanglearealways congurent. Q.7104/QE If f (x) = x2 + 6x + c, where 'c' is an integer, then f (0) + f (–1) is (A) aneveninteger (B) anodd integer always divisibleby3 (C) an odd integer not divisibleby3 (D*) an odd integer may or not be divisible by3 [Sol. f (0) + f (–1) = c + c – 5 = 2c – 5 Now 2c is always even so 2c – 5 is odd. mayor not be divisible by 3.]
2. *Q.85/ph-1 Theminimumvalueofthefunction f (x) = (3sin x – 4 cos x – 10)(3 sin x + 4 cos x – 10), is (A*) 49 (B) 2 2 60 195 (C) 84 (D) 48 [Sol. f (x) = 9 sin2x – 16 cos2x – 10(3 sin x – 4 cos x) – 10(3 sin x + 4 cos x) + 100 = 25 sin2x – 60 sin x + 84 = (5 sinx – 6)2 + 48  f (x)min occurs when sin x = 1 minimumvalue=49 ] Select the correct alternative : (More than one are correct) Q.9154/log Which of the following are correct ? (A*) log3 19 . log1/7 3 . log4 1/7 > 2 (B*) log5 (1/23) lies between –2 & – 1 (C*) log10 cosec(160º) is positive (D) log sin . log sin 5 5 5 5         simplifiestoanirrationalnumber [Hint: (A) log419 ( since log416 = 2 ) hence log419 > 2 (B) log51/5 = – 1 and log51/25 = – 2 so log51/23 loes between – 1 and – 2 (C) since sin1600 < 1 (cosec1600 > 1) Hence numberand base on sameside of unity (D) 2log5sin(/5) . 2logsin(/5)5 = 4 ] Q.1076/ph-1 It is known that sin  = 5 4 and 0 <  <  then the value of            sin ) cos( 6 cos 2 ) sin( 3 is: (A*) independent of  for all  in (0, /2) (B*) 3 5 for tan  > 0 (C*) 3 7 24 15 ( cot )   for tan  < 0 (D) none [Sol. E =         sin 3 ) cos( 4 ) sin( 3 = 3(sin  · cos  + cos  · sin ) – 4(cos  · cos  + sin  · sin ) = 3 sin  · 5 3 + 3 cos  · 5 4 – 4 cos  · 5 3 + 4 sin  · 5 4 = 5 3 for 0 <  < /2 & to 3 7 24 15 ( cot )   for /2 <  < ] Subjective : Q.1148/06 If cos ( y – z ) + cos ( z – x ) + cos ( x – y ) = – 2 3 , prove that cosx + cosy + cosz = 0 = sinx + siny + sinz [Sol. Given 2[cos (x – y) + cos (y – z) + cos (z – x) ] = – 3 or 2[cos (x – y) + cos (y – z) + cos (z – x) + (sin2x + cos2x) + (sin2y + cos2y) + (sin2z + cos2z) ] = 0 or  x cos2 + 2 y cos x cos  +  x sin2 + 2 y sin x sin  = 0 ( cosx + cosy + cosz )2 + ( sinx + siny + sinz)2 = 0   x cos = 0 =  x sin ]
3. Q.1251/06 If 1 a x sin = 3 a x 3 sin = 5 a x 5 sin then show that 3 5 3 1 a a a 2 a   = 1 1 3 a a 3 a  . [Hint: both ratios simplifies to 2(cos 2x  1) ] [Alternate: 5 3 1 a x 5 sin a x 3 sin a x sin   3 5 3 1 a a a 2 a   = 1 1 3 a a 3 a  or 3 a a 2 a a a a 1 3 3 5 3 1     or 1 a a a a a a 1 3 3 5 3 1    = 0 0 1 x sin x 3 sin x 3 sin x 5 sin x 3 sin x sin     x 3 sin x 5 sin x sin  = x sin x sin x 3 sin  x 3 sin x 2 cos x 3 sin 2 = x sin x sin . x 2 cos 2 = 2 cos2x = 2 cos2x ] Q.135/01 Solve the following equation for x : 6 5 . aA – 3B = 9C whereA= loga x. log10 a . loga 5, B = log10(x/10) & C = log100 x + log4 2. [Ans. x = 100] [ Sol. 2 log x log 10 x log 5 log . a log . x log 4 100 10 a 10 a 9 3 a . 5 6                2 1 x log 2 1 x log 5 log . x log 10 10 a 10 3 3 a . 5 6 x log 1 x log x log x log 10 10 10 10 3 . 3 3 3 3 1 5 . 5 6     x log x log 10 10 3 3 10 5 . 5 6  2 x log 3 5 3 5 10              Hence log10 2 x   x = 100 ]
4. CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-17 Select the correct alternative : (Only one is correct) *Q.19/ph-1 If the expression, 2 cos10° + sin 100° + sin 1000° + sin 10000° issimplified,thenitsimplifiesto (A*) cos 10° (B) 3 cos 10° (C) 4 cos 10° (D) 5 cos 10° Q.2 If x2 + 2 x 1 = 7 then the value of 3 3 x 1 x  equals (x > 0) (A*) 18 (B) 21 (C) 24 (D) 27 [Hint: 2 x 1 x        – 2 = 7  2 x 1 x        = 9  x + x 1 = 3,  – 3 (why?) x > 0 3 x 1 x        = 27  x3 + 3 x 1 + 3        x 1 x = 27  x3 + 3 x 1 = 27 – 9 = 18 ] *Q.380(i)/aucNumber of positive integers x for which f (x) = x3 – 8x2 + 20x – 13, is a prime number, is (A) 1 (B) 2 (C*) 3 (D) 4 [Hint: f (x) = (x – 1)(x2 – 7x + 13) for f (x) to be prime at least one of the factors must be prime. Hence x – 1 = 1  x = 2 or x2 – 7x + 13 = 1  x2 – 7x + 12 = 0  x = 3 or 4  x = 2, 3, 4  (C) ] *Q.47/ph-1 In the inequality below the value of the angle is expressed in radian measure. Which one of the inequalitiesbelowistrue? (A) sin 1 < sin 2 < sin 3 (B) sin 3 < sin 2 < sin 1 (C) sin 2 < sin 1 < sin 3 (D*) sin 3 < sin 1 < sin 2 [Hint: sin 1 – sin 2 = – 2 cos 2 3 · sin 2 1 < 0  sin 1 < sin 2 sin 1 – sin 3 = – 2 cos 2 sin 1 > 0  sin 1 > sin 3 ] Q.5 Aparticleismovingalongastraightlineso thatitsvelocityattimet0isv(t)=3t2.Atwhattimetduring the interval from t = 0 to t = 9 is its velocitythe same as the average velocityoverthe entire interval? (A) 3 (B) 4.5 (C*) 3(3)1/2 (D) 9 [Sol. 3t2 = 9 1  9 0 2 dt t 3 = 9 1  9 0 3 t = 81 t2 = 27  t = 3 3  (C) ]
5. *Q.68/log If log10sinx + log10cosx = – 1 and log10(sin x + cos x) = 2 1 ) n (log10  then the value of 'n' is (A) 24 (B) 36 (C) 20 (D*) 12 [Sol. Given log10       2 x 2 sin = – 1  2 x 2 sin = 10 1  sin 2x = 5 1 ....(1) Also log10(sin x + cos x) = 2 10 n log10        log10(sin x + cos x)2 = log10       10 n  1 + sin 2x = 10 n  1 + 5 1 = 10 n  5 6 = 10 n  n = 12  (D) ] *Q.715/ph-1 The value of x satisfying the equation, x = x 2 2 2    is (A) 2 cos 10° (B) 2 cos 20° (C*) 2 cos 40° (D) 2 cos 80° [Sol. Note that x  [–2, 2] Let x = 2 cos  where   [0, ] x =     cos 2 2 2 2 2 cos  = 2 cos 2 2 2    =          2 cos 1 2 2 = 4 sin 2 2   =           4 2 cos 2 2 =                   4 2 cos 1 2 2 cos  = 2 cos          8 4 ;   = 8 4     8 9 = 4    = 9 2 Hence x = 2 cos 9 2 = 2 cos 40°  (C) Ans. ] *Q.859/ph-1 If  is eliminated from the equations x = a cos( – ) and y= b cos ( – ) then ) cos( ab xy 2 b y a x 2 2 2 2      is equal to (A) sec2 (  – ) (B) cosec2 ( – ) (C) cos2 (  – ) (D*) sin2 ( – ) Sol. ( – ) = ( – ) – ( – ) cos( – ) = cos ( – ) cos ( – ) + sin ( – ) sin( – ) cos( – ) = 2 2 2 2 b y 1 . a x 1 a x . b y                                  2 2 2 2 2 b y 1 a x 1 ) cos( ab xy  ) cos( ab xy 2 ) ( cos b a y x 2 2 2 2 2         = 2 2 2 2 2 2 2 2 b a y x a x b y 1   
6.  ) cos( ab xy 2 b y a x 2 2 2 2      = sin2 ( – ) ] Q.985/QE The quadratic equation X2 – 9X + 3 = 0 has roots r and s. If X2 + bX + c = 0 has roots r2 and s2, then (b, c) is (A) (75, 9) (B*) (–75, 9) (C) (–87, 4) (D) (–87, 9) *Q.10112/ph-1 The sum             134 sin 133 sin 1 ...... 50 sin 49 sin 1 48 sin 47 sin 1 46 sin 45 sin 1 is equal to (A) sec (1)° (B*) cosec (1)° (C) cot (1)° (D) none [Sol. T1 =  1 sin 1            46 sin 45 sin ) 45 46 sin( =  1 sin 1 [cot45° – cot46°] |||ly T2 =  1 sin 1            47 sin 48 sin ) 47 48 sin( =  1 sin 1 [cot47° – cot48°]  Tl =  1 sin 1            134 sin 133 sin ) 134 133 sin( =  1 sin 1 [cot133° – cot134°] Onadding   l 1 r r T =  1 sin 1 [{cot45° + cot47° + cot49° + ... + cot133°}– {cot46° + cot48° + cot50° + ... + cot134°}] = cosec1° [all terms cancelledexcept cot45° remains] ] SUBJECTIVE *Q.1116/1 If log10(15) = a and log20(50) = b then find the value of log9(40). [Ans. 2 b 4 a 2 ab 2 b 5     ] [Sol. Since log20(50) = b,it follows that 20 log 2 100 log       = b  2 log 1 2 log 2   = b  log102 = b 1 b 2   ....(1) Since log10(15) = a, we have log 3 + log 5 = a  log 3 + log       2 10 = a  log 3 – log 2 = a – 1  log 3 = log 2 + a – 1. Taking (1) into account, we find that log 3 = b 1 b 2   + a – 1. ....(2) Taking (1)and (2) into consideration,we obtained log9(40) = 3 log 2 2 log 2 1 = 2 b 4 a 2 ab 2 b 5     Ans ]
7. *Q.12 Provetheinequality, sin x + 2 1 sin 2x + 3 1 sin 3x > 0 for 0 < x < 180° [6] [Sol. LHS = 6 ) x sin 4 x sin 3 ( 2 x cos x sin 6 x sin 6 3    = 6 ] x sin 4 3 x cos 3 3 [ x sin 2 2    = 3 ] x sin 4 3 x cos 3 3 [ x sin 2    = 3 ] x cos 4 x cos 3 2 [ x sin 2   = 3 ] x cos 3 ) x cos 1 ( ) x cos 1 [( x sin 2 2     which is + ve for 0° < x < 180° ] Q.13 Let u = 10x3 – 13x2 + 7x and v = 11x3 – 15x2 – 3. Findthe integralvaluesofxsatisfyingtheinequality, dx dv dx du  . [3] [Sol. dx du = 30x2 – 26x + 7 dx dv = 33x2 – 30x  dx dv dx du  30x2 – 26x + 7 > 33x2 – 30x 3x2 – 4x – 7 < 0 (x + 1)(3x – 7) < 0 integral solution set is x {0, 1, 2} ]
8. CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-18 Select the correct alternative : (Only one is correct) Q.11/ph-1 a, b,c arethesidesof a triangleABC which is right angled atC, then theminimumvalueof 2 b c a c        is (A) 0 (B) 4 (C) 6 (D*) 8 [Hint: a = c sin  b = c cos  E = 2 b c a c        = 2 cos 1 sin 1          =    2 sin ) 2 sin 1 ( 4 2 =          2 sin 1 2 sin 1 4 2 where 0 <  < 2   Emin = 8 when 2 = 90°   = 45°] Q.2 Thevalueof m for thezerosof thepolynomial P(x) = 2x2 – mx – 8 differ by(m – 1) is (A) 4, – 3 10 (B) – 6, 3 10 (C) 6, 3 10 (D*) 6, – 3 10 [Hint: ( – )2 = (m – 1)2 ( + )2 – 4 = (m – 1)2 etc.] Q.322/ph-2 Number ofsolutions of the equation 2 sin3x + 6 sin2x – sin x – 3 = 0 in (0, 2), are (A) 6 (B*) 4 (C) 3 (D) 2 [Hint: 2 sin2x (sin x + 3) – (sin x + 3) = 0 (2 sin2x – 1)(sin x + 3) = 0  sin2x = 2 1 Hence 4 solutions  (B) ] *Q.4 Minimum vertical distance between the graphs of y= 2 + sinx and y=cos x is (A) 2 (B) 1 (C) 2 (D*) 2 – 2 [Hint: dmin = min(2 + sin x – cos x) = min[2 + 2 sin         4 x ] = 2 – 2 ] *Q.541/ph-3 Each sideof triangleABC is divided into 3 equal parts.The ratio of the area of hexagon UVWXYZ to the area of triangleABC is (A) 9 5 (B*) 3 2 (C) 2 1 (D) 4 3
9. [Hint: area of hexagon =  – 9 1         B sin ca 2 1 A sin bc 2 1 C sin ab 2 1 =  – 9 1 [3] = 3 2  (where  is the area of the triangleABC)  triangle of area heaxgon of area = 3 2 ] Q.688/QE If a, b, c are real numbers such that a2 + 2b = 7, b2 + 4c = – 7 and c2 + 6a = – 14 then the value of a2 + b2 + c2 is (A*) 14 (B) 21 (C) 28 (D) 35 [Hint: a2 + 2b = 7 b2 + 4c = – 7 c2 + 6a = – 14 —————— a2 + b2 + c2 + 2b + 4c + 6a = – 14 (a + 3)2 + (b + 1)2 + (c + 2)2 – 14 = – 14  a = – 3 ; b = – 1 and c = – 2  a2 + b2 + c2 = 14 ] *Q.719/ph-1 Find the smallest natural 'n' such that tan(107n)° =       96 sin 96 cos 96 sin 96 cos . (A) n = 2 (B*) n = 3 (C) n = 4 (D) n = 5 [Hint:       6 cos 96 cos 6 cos 96 cos = –     45 sin 51 sin 2 45 cos 51 cos 2 = – cot51° = tan(90° + 51°) = tan141° = tan(180° + 141°) = tan(321°) = tan(3·107°)  n = 3 ] SUBJECTIVE *Q.8 IfA+ B + C = 2  then prove that  A sin2 + 2 A sin = 1. [3] [Sol. LHS sin2A + sin2B + sin2C + 2 A sin 1 – (cos2A – sin2B) + sin2C + 2 A sin 1 – cos(A + B) cos(A – B) + sin2C + 2 A sin (A + B = C 2   ; cos(A + B) = sinC) 1 – sinC [cos(A – B) – sinC] + 2 A sin 1 – sinC [cos(A – B) – cos(A + B)] + 2 A sin 1 – sinC[2sinA sinB] + 2 A sin 1 – 2 A sin + 2 A sin = 1 Hence proved. ]
10. Q.9 The position vector of a point P in space is given by k̂ t cos 4 j ˆ t sin 5 î t cos 3 r     (a) Show that its speed is constant. (b) Show that its velocityvector v  , is perpendicular to r  . [3] [Sol.(a) dt r d v    = k̂ t sin 4 j ˆ t cos 5 î t sin 3    speed = | v |  = t sin 16 t cos 25 t sin 9 2 2 2   = 5 unit Ans. (b) r · v   = – 9 (sin t)(cos t) + 25(sin t)(cos t) – 16(sin t)(cos t) = 0 Hence v  , is perpendicular to r  ] Q.10 If x is eliminated from the equation, sin(a + x) = 2b and sin(a – x) = 2c, then find the eliminant. [Ans. a sin ) c b ( 2 2  + a cos ) c b ( 2 2  = 1] [3] [Sol. adding sin(a + x) + sin(a – x) = 2(b + c) 2 sin a cos x = 2(b + c)  cos x = a sin c b  ....(1) sub sin(a + x) – sin(a – x) = 2(b – c) 2 cos a sin x = 2(b – c)  sin x = a cos c b  ....(2) squaringand addingboth equation (1) and (2), we get a sin ) c b ( 2 2  + a cos ) c b ( 2 2  = 1 Ans. ] *Q.1133/1 If p, q are the roots of the quadratic equation x2 + 2bx + c = 0, prove that 2 log  q y p y    = log 2 + log           c by 2 y b y 2 [Sol. x2 + 2bx + c = 0 ; c pq b 2 q p        ....(1) TPT 2 log  q y p y    = log 2 + log           c by 2 y b y 2 LHS =   q y p y 2 q y p y log       =             pq y ) q p ( y 2 ) q p ( y 2 log 2 =           c by 2 y 2 b 2 y 2 log 2 = log 2 +           c by 2 y b y log 2 = RHS ]
11. Q.1294/6 Solve for '' satisfying cos() · cos () = 1. [Ans.  = 0 ] [Sol. cos() · cos () = 1 either cos() = 1 & cos () = 1 ....(1) or cos() = –1 & cos () = –1 ....(2) Now for (1) to hold, from cos() = 1   = 2m & cos () = 1   = 2k which is onlypossible when =0 Equation (2) is not possible for anyas for cos() = – 1,  should be odd multiple of   irrational and for cos() = – 1   should be odd integer  rational boththe conditions can not be satisfied   =0 is theonlysolution. ] Q.13 Find thevalue of kforwhich thegraphof thequadratic polynomial P (x) = x2 + (2x + 3)k + 4(x + 2) + 3k – 5 intersects the axis of x at two distinct points. [3] [Ans. R – {1}] [Sol. P (x) = x2 + 2(k + 2)x + 3(2k + 1) for distinct zeros, D > 0 4(k + 2)2 – 4 · 3(2K + 1) > 0 (k2 + 4k + 4) – (6k + 3) > 0 k2 – 2k + 1 > 0 (k – 1)2 > 0 which is always true except at k = 1 whenit touchestheaxis of x. Hence k = R – {1}]
Anzeige