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### Dpp (19-21) 11th J-Batch Maths.pdf

1. Daily Practice Problems CLASS : XI (J-Batch) TIME : 50 Min. DPP. NO.-19 Select the correct alternative : (Only one is correct) Q.148/QE The equation, x =  2x2 + 6x  9 has: (A*)nosolution (B)onesolution (C)twosolutions (D)infinitesolutions [Hint: Left always +ve& right always –ve] Q.2129/QE If a > b > 0 are two real numbers, the value of , a b a b a b a b a b a b a b        ( ) ( ) ( ) ....... is : (A*) independent of b (B) independent of a (C) independent of both a & b (D) dependent on both a & b . [Hint: We have, y = ab a b y   ( ) y2 = ab + (a – b)y  y2 – (a – b)y – ab = 0  (y – a) (y + b) = 0  y = a or y = –b (not possible)  A ] Q.3 Numberofintegerswhichsimultaneouslysatisfiestheinequalities | x | + 5 < 7 and | x – 3 | > 2, is (A) exactly1 (B*) exactly2 (C)more than2but finite (D)infinitelymany [Hint: The onlyone integer that satisfy| x | + 5 < 7 are the ones that satisfy| x | < 2, namely, – 1, 0, 1. The integer that satisfy | x – 3 | > 2 are 6, 7, 8, ..... and 0, – 1, –2, ..... So, the integer that satisfyboth are 0, –1, and there are 2 of them. ] Q.473/ph-1 If tan A & tan B are the roots of the quadratic equation x2  ax + b = 0, then the value of sin2 (A +B) is : (A*) a a b 2 2 2 1   ( ) (B) a a b 2 2 2  (C) a b a 2 2 ( )  (D) a b a 2 2 2 1 ( )  [Sol. tanA + tanB = a ; tanA.tanB = b tan(A+B) = b 1 a  , hence sin2(A+B) = a a b 2 2 2 1   ( ) Ans: ] *Q.566/QE If themaximum and minimum valuesof y= c x 3 x c x 3 x 2 2     are 7 and 7 1 respectivelythen the value of c is equal to (A) 3 (B*) 4 (C) 5 (D) 6 [Hint: y = c x 3 x c x 3 x 2 2     = 7 or 7 1 must give two coincident values of x  c = 4 ] Subjective *Q.6 Find the value of x satisfying the equation 3 2 4 x x2     = x2 + x – 12. [Ans. 11/2] [Hint: 3 2 4 x x2     = x2 + x – 12
2. 3 2 x x2    = x2 + x – 12 1 x x2   = x2 + x – 12 2x = 11  x = 11/2 Ans. ] *Q.721/1 Find all values of 'a' for which 1 x x 2 ax x 2 2     lies between –3 and 2 for all real values of x. [Ans: ) 2 , 1 ( a   ] [Sol. – 3 < 1 x x 2 ax x 2 2     < 2  – 3x2 – 3x – 3 < x2 – ax – 2 < 2x2 + 2x + 2 R x  x2 + x(a + 2) + 4 > 0  (a + 6) (a – 2) < 0 ....(1) and 4x2 + x(3 – a) + 1 > 0 (3 – a)2 – 16 < 0  (a + 1) (a – 7) < 0 ....(2) from (1) and (2) ) 2 , 1 ( a   Ans. ] Q.8 If un = sinn  + cosn , prove that u u u u u u 3 5 1 5 7 3    [Sol. consider         u u u u 5 7 3 5 5 5 7 7 3 3 5 5          sin cos sin cos sin cos sin cos         =         sin sin cos cos sin sin cos cos 5 2 5 2 3 2 3 2 1 1 1 1               =     sin cos sin cos sin cos sin cos 2 2 3 3 2 2           = u u 3 1 ] *Q.929/6 Show that the triangleABC isright angles if andonlyif sinA+ sinB + sinC= cosA+cosB+ cosC+ 1. [Sol. 1st part : LetABC is aright angle. Also let A = 90° C = 90 – B now LHS, 1 + sinB + cosB RHS = cos90° + cosB + sinB + 1 = 1 + sinB + cosB = LHS henceifanyoneangle is 90° thenthegivenrelation is true. 2nd Part : Let sinA + sinB + sinC = cosA + cosB + cosC + 1 2 B A cos 2 B A sin 2   + 2 C cos 2 C sin 2 = 2 B A cos 2 B A cos 2   + 2 C cos 2 2           2 B A cos 2 B A sin 2 B A cos –        2 C sin 2 C cos 2 C cos = 0         2 C sin 2 C cos 2 B A cos –        2 C sin 2 C cos 2 C cos = 0
3.                2 C cos 2 B A cos 2 C sin 2 C cos = 0 if 2 C sin 2 C cos  = 0  2 C tan = 1  2 C = 4   C = 90° ....(1) if 2 B A cos  = 2 C cos = cos        2 C then A – B = C or A – B = – C A = B + C A + C = B here A + B + C =  A = 2  ....(2)  B = 2  ....(3) from (1), (2)and (3) one of the angle must be 90° henceABC is a right angled triangle.] Q.106/6From the relation, 3 (cos 2 – cos 2) = 1 – cos 2 cos 2, we get tan = k tan where ,          2 , 0 . Find the value of k. [Ans. k = 2 ] [Sol. Fromgivenrelation, 3cos 2 – 3cos 2 = 1 – cos 2 cos 2 or (3 + cos2)cos 2 = 1 + 3cos 2 or cos 2 =     2 cos 3 2 cos 3 1 applycomponendoanddividendo,     2 cos 1 2 cos 1 =     2 cos 1 2 cos 1 · 2 1 or tan2 = 2 1 tan2. But given tan = k tan wherebyk2 = 2, and since  and  are positive acute angles k = 2 ]
4. Daily Practice Problems CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-20 Select the correct alternative : (Only one is correct) *Q.179/ph-1 The value of the expression (sinx + cosecx)2 + (cosx + secx)2 – ( tanx + cotx)2 wherever defined is equal to (A) 0 (B*) 5 (C) 7 (D) 9 Q.287/QE The numberofintegers 'n' such that the equation nx2 +(n + 1)x +(n + 2) =0 has rational roots only, is (A) 1 (B) 2 (C*) 3 (D) 4 [Hint: roots rational D must be perfect square and D > 0 D = (n+1)2 –4n(n+2) D = –3n2– 6n + 1 = 0  3n2 + 6n – 1 = 0 or D = 4 – 3(n + 1)2 (now interpret)  =    6 36 12 6 =   3 2 3 6 =     1 2 3 3 1 2 3 3 or Hence possible integral values are –2 , –1 , 0 and all of them make D a perfect square. ] Q.3 If cos + cos= a and sin + sin= b, then the value of cos·coshas the value equal to (A)     a b a a b 2 2 2 2 2 2 4 4    (B)     a b b a b 2 2 2 2 2 2 4 2    (C)     a b a a b 2 2 2 2 2 2 2 4 2    (D*)     a b b a b 2 2 2 2 2 2 4 4    [Hint: square and add, cos( – ) = a b 2 2 2 2   using C – D relation tan     2 b a Now E = 1 2 [cos( + ) + cos( – )] = 1 2 1 1 2 2 2 2 2 2 2 2                  b a b a a b simplifyto gettheresult] Select the correct alternative : (More than one are correct) Q.4511/QE The value(s) of 'p' for which the equation ax2  px + ab = 0 and x2  ax  bx + ab = 0 mayhave a common root, given a, b are non zero real numbers, is (A) a + b2 (B*) a2 + b (C*) a(1 + b) (D) b(1 + a) [Sol. x2  (a + b) x + a b = 0 or (x  a) (x  b) = 0  x = a or b if x = a is the root of other equation , a3  a p + a b = 0  p = a2 + b if x = b is the root of the other equation , then a b2  p b + a b = 0 p = a (1 + b) ] Subjective:
5. *Q.584/1 Findthe solutionset ofthe inequality 2 1 | x | 2 | x | 3    . [ Ans:(–  , – 1)          5 4 , 1  {0}        1 , 5 4  (1, )] [Sol. Let |x| = y ( note |x|  1 ) 2 1 y 2 y 3    Hence 2 1 y 2 y 3    ....(1) or 2 1 y 2 y 3     ....(2) From (1) 0 2 1 y 2 y 3     0 1 y 2 y 2 2 y 3      0 1 y y   ) , 1 ( ] 0 , ( y     But |x|  0 Hence |x|  (1, ) x (– , –1)  (1, ) ....(3) From (2) 0 2 1 y 2 y 3     0 1 y 4 y 5                     1 , 5 4 | x | 1 , 5 4 y                  1 , 5 4 5 4 , 1 x ....(4) For (3) & (4) together with the fact that x = 0 is the obv. solution as equalityholds hence x  ) , 1 ( 1 , 5 4 } 0 { 5 4 , 1 ) 1 , (                      ] Q.656/QE Let a, b, c be the three roots of the equation x3 + x2 – 333x – 1002 = 0 then find the value of a3 + b3 + c3. [Ans. 2006] [Sol. Let t be the root of the given cubic where t can take values a, b, c hence t3 + t2 – 333t – 1002 = 0 or t3 = 1002 + 333t – t2   3 t = 1002 + 333t –  2 t = 3006 + 333 t –   ] [ 2 1 2 t t 2 t    but t = – 1 ;  2 1t t = – 333  a3 + b3 + c3 = 3006 – 333 – [1 + 666] = 3006 – 333 – 667 = 3006 – 1000 = 2006 Ans. ] *Q.734/1 A polynomial in x of degree greater than 3 leaves the remainder 2, 1 and –1 when divided by (x–1);(x+2)&(x+1)respectively.Findtheremainder,ifthepolynomialisdivided by,(x2 1)(x +2). [ Ans. : 7 6 x2 + 3 2 x  2 3 ] [Sol. f (x) = Q1 (x  1) + 2 = Q2 (x + 2) + 1 = Q3 (x + 1)  1
6.  f (1) = 2 ; f ( 2) = 1 ; f ( 1) =  1 again f (x) = Qr (x2  1) (x + 2) + ax2 + bx + c Hence a + b + c = 2 ; 4 a  2 b + c = 1 and a  b + c =  1 ] Q.852/6 Prove the inequality sin sin x x   1 2 + 1 2  2 3   sin sin x x  x  R. Q.933/6 Show that the sum to n terms of the series : sin cos 3 + sin 3 cos 5 + ..... + sin(2n – 1). cos(2n + 1)= sin ( ) .sin sin 2 1 2 2 2 n n     – n 2 sin2 [Sol: We have, 2S = 2sin cos 3 + 2 sin3 cos 5 + ..... + 2 sin(2n – 1) × cos (2n + 1) Hence, sin4 – sin2 + sin8 – sin2 2S = + sin12 – sin2 +  sin4n – sin2 ________________________________________________ 2S = sin sin sin ........ sin sin 4 8 12 4 2 1           n n say S                  now, 2 sin2.S1 = 2 sin2 sin4 + 2 sin2 sin8 + ....... + 2 sin2 sin4n = cos2 – cos6 + = cos6 – cos10  + = cos(4n – 2) – cos(4n + 2) _________________________________________________________________ 2 sin2.S1 = cos 2 – cos (4n + 2) = 2 sin(2n + 2). sin2n S1 =    2 2 2 2 2 2 sin( ).sin sin n n     Hence, S = sin ( ) .sin sin 2 1 2 2 2 n n     – nsin2 2  ] Q.1045/6 Let ) sin( ) sin(       = b a & ) cos( ) cos(       = d c then prove that cos () = bc ad bd ac   . [Hint: a sin ( )    = b sin ( )    = k1 ; c cos ( )    = d cos ( )    = k2 = a c bd a d bc   = k k k k 1 2 1 2                                 ) ( cos ) ( sin ) ( cos ) ( sin ) ( cos ) ( sin ) ( cos ) ( sin =     ) ( sin ) ( 2 sin ) ( sin ) ( 2 sin ) ( 2 sin ) ( 2 sin                           =     ) ( 2 sin 2 ) ( cos ) ( 2 sin 2              = cos () ]
7. Daily Practice Problems CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-21 Select the correct alternative : (Only one is correct) *Q.1 The set of values of 'p' for which theexpression x2  2px +3p + 4 is negative for atleast one real x is: (A)  (B) (1, 4) (C*) (,  1)  (4, ) (D) {1, 4} [Put D > 0 ] Q.2 The value of f (x) = x2 + (p – q)x + p2 + pq + q2 for real values of p, q and x (p + q  0) (A)isalways negative (B*)isalways positive (C) is sometimes zero for non zero value of x (D) none of these [Hint: f (x) = x2 + (p – q)x + p2 + pq + q2 D = (p – q)2 – 4(p2 + pq + q2) = p2 + q2 – 2pq – 4p2 – 4pq – 4q2 = – 3p2 – 3q2 – 6pq = – 3(p + q)2 D < 0 ] Q.360/QE For R x  , the difference between the greatest and the least value of y = 1 x x 2  is (A*) 1 (B) 2 (C) 3 (D) 2 1 Q.4 If the roots of the cubic, x3 + ax2 +bx + c = 0 arethreeconsecutive positive integers. Then the value of 1 b a2  is equal to (A*) 3 (B) 2 (C) 1 (D) none of these [Sol. n, n + 1, n + 2 sum = 3(n + 1) = – a  a2 = 9(n + 1)2 sum of the roots taken 2 at atime = + b  n(n + 1) + (n + 1)(n + 2) + (n + 2)n + 1 = b + 1 n2 + n + n2 + 3n + 2 + n2 + 2n + 1 = b + 1  b + 1 = 3n2 + 6n + 3 b + 1 = 3(n + 1)2 = 3 a2 ;  1 b a2  = 3  (A) ] Q.577/ph-1 If x sin  = y cos  then x sec2 + y cosec2 is equal to (A*) x (B) y (C) x2 (D) y2 [Hint: sec 2 = 2 2 2 2 y x y x   ; cosec 2 = xy 2 y x 2 2  ]
8. Q.6112/QE The equation (x  R) x x 2 2 1 1 5 3    = x (A) has no root (B*) exactlyone root (C) two roots (D) four roots [Hint: only x =  4/3 satisfies ; x = 4/3 is rejected ] [Sol. x x x 2 2 1 1 5 3     ....(1) or x x x 2 2 1 5 3     ....(2) (2) – (1) given 2x = x x 2 2 5 3 1 5 3    squaring 4x2 = x x 2 2 5 3 1 5 3 2     3 1 5 3 11 3 0 2 2 x x     ; 3 1 5 3 11 3 y y    = 0 3 5 3 11 3 5 3 1 0 y y y  F H G I K J  F H G I K J  3y2 – 5y + (11/3)y – (55/9) – 1 = 0 27y2 – 12y – 6y = 0 27y2 – 48y + 36y – 6y = 0 3y(9y–16) + 4(9y–16) = 0 y =  4 3 or y = 16 9  x2 = – 3 4 (rejected) or x2 = 9 16  x = 3 4 or – 3 4 . Only x = – 3 4 satisfies. ] *Q.7280/flcd The function f (x) is defined by f (x) = cos4x + K cos22x + sin4x, where K is a constant. If the functionf(x)is aconstantfunction, thevalueofk is (A) – 1 (B*) – 1/2 (C) 0 (D) 1/2 (E) 1 [Sol. Your first impulse, like mine, might be to complete the square to get the expression (cos2x +sin2x)2 somewhereinthe problem. Whilethis works, it iseasier to substitute intwo values for x andforcethentobe equal,sincethefunction isto beconstant.Anytwo values will work,but wewant touseones forwhich weknowthesineandcosine.All multiples of/2resultin1+K, soweneedto put in one value that is not a multiple of /2.Try/4.This gives us cos4(/4) + Kcos2(/2) + sin4(/4) = 1/4 + K · 0 + 1/4 = 1/2. So now we know that 1 + K = 1/2, so K = – 1/2. This does in fact give us the function f (x) = (cos2x + sin2x)2 – 1/2 = 1/2 ]
9. Subjective: Q.839/6 Prove that : sin 7 2 + sin 7 4 – sin 7 6 = 4 sin 7  sin 7 3 sin 7 5 [Sol. RHS = 4 sin 7  sin 7 2 sin 7 4             7 4 sin 7 3 sin and 7 2 sin 7 5 sin g sin u 2 sin 7          7 4 sin 7 2 sin 2 or 2 sin 7           7 6 cos 7 2 cos =           7 cos 7 2 cos 7 sin 2 sin 7 2 + 2 sin 7  cos 7 2  sin 7 2 + sin 7 3 – sin 7  sin 7 2 + sin 7 4 – sin 7 6 = L.H. S. proved ] Q.927/6 Find the two smallest positive values of x for which sin x° = sin (xc) [Ans.    180 180 or    180 360 ] [Sol. Let x be the number now  radian = 180 degree 1 radian =        180 degree x radian =  x 180 degree now        x 180 sin = sin x  x = 0 (not possible) or        x 180 sin = sin(180 – x) ;  x 180 = 180 – x         1 180 x = 180  x =    180 180 Ans. or        x 180 sin = sin(360 + x) ;  x 180 = 360 + x         1 180 x = 360  x =    180 360 Ans. ] Q.10 If one root of the quadratic equation x2 + mx – 24 = 0 is twice a root of the equation x2 –(m + 1)x + m = 0 thenfind thevalue of m. [Ans. m = – 2, 2, 10] [Sol. 2nd equationgiven x = m or x = 1 now put x = 1 and then x = m to determine the value of m. ]
10. Q.1166/6 For  = 1°, prove that 2 sin2 + 4 sin4 + 6 sin6 + ......... + 180 sin180 = 90 cot [Sol. Let LHS =S, now multiplyingbothsides bysin sin · S = 2 sin2 sin + 4 sin4 sin + 6 sin6 sin + ....... + 180 sin180 sin T1 = (cos – cos3) T2 = 2(cos3 – cos5) T3 = 3(cos5 – cos7) T4 = 4(cos7 – cos9)   = 89(cos177 – cos179) T90 = 90(cos179 – cos181) ———————————— S sin = cos + cos3 + cos5 + ....... + cos(177) + cos(179) – 90 cos(181) put  = 1° S·sin1° =             1 cos 90 179 cos 177 cos ....... 5 cos 3 cos 1 cos terms 90                          S·sin1° =                            vanish vanish vanish 91 cos 89 cos ........ 177 cos 3 cos 179 cos 1 cos             + 90 cos1°  hence S sin 1° = 90 cos 1°  S = cot1° ] *Q.12 If y= 8 x 2 x 3 x 2 x 2 2     then findthe interval in which ycan lie for everyx  R wherever defined.  [Sol. y = ) 2 x )( 4 x ( ) 1 x )( 3 x (     [Ans. y ) , 1 ( 9 4 ,           ] x2(y – 1) + 2x(y – 1) – 8y + 3 = 0,  x  R (y – 1)2 – (y – 1)(3 – 8y)  0 or (y – 1)(9y – 4)  0  y  1 or y  9 4 for y = 1, x2 + 2x – 8 = x2 + 2x – 3  – 8 = – 3 (notpossible)  y  1 |||ly y = 9 4 ; we get, x2 + 2x + 1 = 0 x = – 1  y  ) , 1 ( 9 4 ,           ]
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