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### Dpp (25-30) 11th J-Batch Maths.pdf

1. CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-25 Select the correct alternative : (Only one is correct) Q.189/QE Number of ordered pair(s) (a, b) for each of which the equality, a (cos x  1) + b2 = cos (ax + b2)  1 holds true for all x  R are : (A) 1 (B*) 2 (C) 3 (D) 4 [Hint: Put x = 0  b2 = cos b2  1  cos b2 = 1 + b2  b = 0 when b = 0 we have a (cos x  1) = cos a x  1  2 a sin2 x 2 = 2 sin2 a x 2  a = 0 or a = 1. Hence a = 0 & b = 0 or a = 1 & b = 0 ] Q.280/QE For every x  R, the polynomial x8  x5 + x2  x + 1 is : (A*) positive (B) never positive (C) positve as well as negative (D) negative [Hint: for x  1 E = x5 (x3  1) + (x  1) + 1 > 0 for 1 < x < 0 , E = (1  x) + x2 (1  x3) + x8 > 0 For x < 0 , all terms are positive  > 0 Hence A] Q.396/QE Three roots of the equation, x4  px3 + qx2  rx + s = 0 are tanA, tanB & tanC whereA, B, C are theanglesof atriangle. Thefourthroot of thebiquadraticis : (A*) p r q s    1 (B) p r q s    1 (C) p r q s    1 (D) p r q s    1 [Hint: Let the fourth root be tan D Now tan ( A) =     tan tan tan tan tan tan tan A A B C A B A    1 tan D = p r q s    1 ] Q.4105/QE If the roots of the quadratic equation (4p – p2 – 5)x2 – (2p – 1)x + 3p = 0 lie on either side of unity thenthenumberofintegral valuesofpis (A) 0 (B) 1 (C*) 2 (D)infinite [Sol. note that a < 0 hence f(1) > 0 4p – p2 – 5 – 2p + 1 + 3p > 0 – p2 + 5p – 4 > 0  p2 – 5p + 4 < 0 (p – 4) (p – 1) < 0  1 < p < 4  p  {2, 3}] Q.5119/QE The inequalities y(1) 4, y(1) 0 & y(3) 5 are known to hold for y = ax2 +bx + c then the least value of 'a' is : (A)  1/4 (B)  1/3 (C) 1/4 (D*) 1/8 [Hint: a - b + c  4 ..... (i) and a + b + c  0  a  b  c  0 .... (ii) and 9a + 3b + c  5 .... (iii) (i) + (ii)   2b 4 ..... (iv) ; (ii) + (iii) + (iv)  8a  1  a 1/8 ] Q.64/-1 Given a2 + 2a + cosec2  2 ( ) a x  F H G I K J= 0 then, which of the following holds good? (A) a = 1 ; x I 2  (B*) a = –1 ; x I 2  (C) a  R ; x  (D) a , x are finite but not possible to find
2. [Sol. (a+1)2 + cosec2   a x 2 2  F H G I K J– 1 = 0 or (a+1)2 + cot2   a x 2 2  F H G I K J= 0 fromoption[B] If a = –1  tan2x/2 = 0 x/2 I ] Q.748/s&p For an increasing A.P. a1, a2, a3.....,an,.... if a1 + a3 + a5 = – 12 ; a1a3a5 = 80 then which of the followingdoesnothold? (A) a1= – 10 (B*) a2 = – 1 (C) a3 = – 4 (D) a5 = 2 [Hint: a1 = a – 2d, a2 = a – d, a3 = a ; a4 = a + d, a5 = a + 2d a1 + a3 + a5 = a – 2d + a + a + 2d = – 12 3a = – 12  a = – 4 a1a3a5 = (a – 2d)a(a + 2d) = 80 – 4(16 – 4d2) = 80  4d2 – 16 = 20  4d2 = 36  d2 = 9  d = ± 3 d = 3 series is increasing a1 = – 4 – 6 = – 10, a3 = – 4, a5 = – 4 + 6 = 2 ] Q.8 If p & q are distinct reals, then 2 {(x  p) (x  q) + (p  x) (p  q) + (q  x) (q  p)} = (p  q)2 + (x  p)2 + (x  q)2 issatisfiedby: (A) no value of x (B) exactlyone value of x (C) exactlytwo values of x (D*) infinitevalues of x . [Note x = p, q & o  infinite solutions ] Q.9 If the quadratic equation ax2 + bx + 6 = 0 does not have two distinct real roots, then the least value of 2a + b is (A) 2 (B*) –3 (C) –6 (D) 1 [Hint: no two distinct roots  f(x) > 0 or f(x) < 0. But f(0) = 6  f(x) > 0 V x  R hence f (2) > 0  4a + 2b + 6 > 0  2a + b > –3  B ] Q.1011/QE Let a > 0, b > 0 & c > 0. Then both the roots of the equation ax2 + bx + c = 0. (A) are real & negative (B*) have negative real parts (C) are rational numbers (D) none [Hint:  = – b a 2 + b ac a 2 4 2  ; consider the examples x2 + x + 1 = 0 and x2 + 3x + 2 = 0 ] SUBJECTIVE: Q.1123/6 Prove that : 5 sin x = sin(x + 2y)  2 tan(x + y) = 3 tan y. [Sol. Given 1 5 x sin ) y 2 x sin(   or 1 5 1 5 x sin ) y 2 x sin( x sin ) y 2 x sin(        = 3 2 3 2 y cos ) y x sin( 2 y sin ) y x cos( 2    ) y x tan( y tan  = 3 2 ;  2 tan(x + y) = 3 tan y ] Q.12 Solve the inequality, log2x (x2  5x + 6) < 1. [ Ans. : (0, 1/2)  (1, 2)  (3, 6)]
3. CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-26 Select the correct alternative : (Only one is correct) Q.1109/QE If the quadratic polynomial, y= (cot )x2 + 2   sin  x + 1 2 tan ,   [0, 2] can take negative values for all x R, then the value of  must in the interval : (A*)  5 6   ,       (B)  5 6   ,        11 6 2   ,       (C)   6 2 ,   (D)   0 6 ,   5 6   ,       [Sol. b2 – 4ac < 0 and a < 0 hence cot  < 0 i.e.   2nd and 4th quadrant and 4sin  – 2 tan  cot  < 0 2 sin  < 1  sin  < 2 1    2nd and 4th quadrant (Also sin  > 0  ( onlyin 2nd quadrant) hence           , 6 5 ] Q.264/QE Numberof integral values of x satisfyingthe inequality 64 27 4 3 2 x 10 x 6          is (A) 6 (B*) 7 (C) 8 (D)infinite [Hint: 6x + 10 – x2 > 3  x2 – 6x – 7 < 0 (x + 1) (x – 7) < 0  0, 1, 2, 3, 4, 5, 6  B ] Q.3 The equation a sin x + cos 2x = 2a – 7 has a solution, if (A) a > 2 (B) a < 2 (C*) 2 < a < 6 (D) a < 2 or a < 6 [Sol. a sin x + 1 – 2 sin2x = 2a – 7 2 sin2x – 9 sin x + (2a – 8) = 0  sin x = 4 ) 8 a 2 ( 8 a a 2    ] Q.413 log Numberofpositivesolution whichsatisfythe equation log2x · log4x · log6x =log2x ·log4x + log2x · log6x + log4x ·log6x? (A) 0 (B) 1 (C*) 2 (D)infinite [Hint: log2x · log4x · log6x          1 x log 1 x log 1 x log 1 2 4 6 = 0  either x = 1 or logx48 = 1  x = 48 ]
4. Q.5 Numberofintegralvaluesofxsatisfyingtheinequality 3 x 1 x    4 x 2 x   (A) 1 (B) – 2 (C) – 1 (D*) 0 Q.6 The quadratic equations 2006 x2 + 2007 x + 1 = 0 and x2 + 2007x + 2006 = 0 have a root in common. Then theproduct of theuncommonroots is (A*) 1 (B) – 2 (C) – 1 (D) 0 [Sol. 2006 x2 + 2007 x + 1 = 0 2006 x2 + 2006 x + x + 1 = 0 2006 x(x + 1) + (x + 1) = 0 (x + 1)(2006 x + 1) = 0  x = – 1 or x = – 2006 1 obviouslyx = – 1 satisfies both the quadratic equations hence it is thecommon root if ,  are the roots of 1st then  = 2006 1 (–1) = 2006 1   = – 2006 1 and |||ly  = 2006 (from 2nd equation) (–1) = 2006  = – 2006  product of uncommon roots is 1 Ans. ] Q.7 Suppose sin  – cos  = 1 then the value of sin3 – cos3 is ( R) (A*) 1 (B) – 2 (C) – 1 (D) 0 Q.841/s&p Let an = 16,4, 1, .... be a geometric sequence. Define Pn as the product of the first n terms. The value of   1 n n n P (A) 8 (B) 16 (C*) 32 (D) 64 [Sol. For the G.P. a, ar, ar2, .......... Pn = a(ar)(ar2) ........(arn–1) = anrn(n – 1)/2  S =   1 n n n P =     1 n 2 / ) 1 n ( ar now,     1 n 2 / ) 1 n ( ar = a[1 + r + r + r r + ...... + ] = r 1 a  Given a = 16 and r = 1/4  S = ) 2 1 ( 1 16  = 32 Ans. ] Q.9 Consider anA.P. t1, t2, t3, ........ If 5th, 9th and 16th terms of thisA.P. form three consecutive terms of a G.P. with nonzero common ratio q, then thevalue of q is (A) 4/7 (B) 2/7 (C*) 7/4 (D) none
5. [Sol. q = d 4 a d 8 a   = d 8 a d 15 a   = d 4 d 7  q = 4 7 Ans. ] Q.10 If the mth, nth and pth terms of G.P. form three consecutive terms of another G.P. then m, n and p are in (A*)A.P. (B) G.P. (C) H.P. (D)A.G.P. SUBJECTIVE: Q.11 A quadratic polynomial f (x)= x2 + ax + b is formed with one of its zeros being 3 2 3 3 4   where aand b are integers. Also g (x) = x4 + 2x3 – 10x2 + 4x – 10 is a biquadratic polynomial such that           3 2 3 3 4 g = d 3 c  where c and d are also integers. Find the value of a, b, c and d. [Ans. a = 2; b = – 11, c = 4; d = – 1] [Sol. x2 + ax + b x = 3 2 3 3 4   ; x =    3 2 3 3 4   x1 = 8 – 3 4 + 3 6 – 9 = 3 2 – 1  x1 = – 1 + 3 2 ; x2 = – 1 – 3 2 sum = – 2 product = 1 – 12 = – 11  equation is x2 + 2x – 11 = 0  a = 2, b = – 11 given g (x) = x4 + 2x3 – 10x2 + 4x – 10           3 2 3 3 4 g = d 3 c  ; x = 3 2 3 3 4   = 3 2 – 1  (x + 1)2 =  2 3 2  x2 + 1 + 2x = 12  x2 + 2x – 11 = 0 x2(x2 + 2x – 11) + 1(x2 + 2x – 11) + 2x + 1 0 + 0 + 2x + 1 = 2( 3 2 – 1) + 1 = 3 4 – 1]
6. Q.12 If  and  be the roots of the equation x2 + 3x + 1 = 0 then find the value of 2 2 1 1                      . [Ans. 18] [Sol.  +  = – 3;  = 1, also 2 + 3 + 1 = 0 and 2 + 3 + 1 = 0 where 2 = – (3 + 1) & 2 = – (3 + 1) E = 2 2 2 2 ) 1 ( ) 1 (        E = 2 2 2 1      + 2 2 2 1      =              ) 1 3 ( +            ) 3 1 ( y =    3 1 +    3 1 = ) 3 1 ( ) 3 1 (        (as  = 1) = 3(2 + 2) + ( + ) = 3[9 – 2] + (–3) = 21 – 3 = 18 Ans. ]
7. CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-27 Select the correct alternative : (Only one is correct) Q.153/log Smallest integral x satisfyingthe inequality, x log 1 x log 1 2 4    2 1 ,is (A) 2 (B*) 2 (C) 3 (D) 4 [Sol. Let log2x = t (obviouslyx > 0)   t 1 2 t 1    2 1  t 1 t 2    1  1 t 1 t 2     0  t 1 t 1 t 2      0  1 t t 2 1    0  1 t 1 t 2    0  t  2 1 or t < – 1 Hence,either log2x  2 1  x  2 or log2x < – 1  0 < x < 2 1 0 < x < 2 1 . Hence the solution set is           , 2 2 1 , 0  smallest integer is 2  (B) ] Q.2 Consider the ten numbers, ar, ar2, ar3, .......... ar10. If their sum is 18 and the sum of their reciprocals is 6 then the product of these ten numbers, is (A) 324 (B) 343 (C*) 243 (D) 729 [Sol. Given 1 r ) 1 r ( ar 10   = 18 ....(1) Also r 1 1 r 1 1 ar 1 10         = 6  1 r r ) 1 r ( · ar 1 10 11   = 6 1 r ) 1 r ( ar · r a 1 10 11 2   = 6 ....(2) from (1) and (2) 11 2 r a 1 · 18 = 6  a2r11 = 3 now P = a10r55 = (a2r11)5 = 35 = 243 Ans. ] Q.383/ph-1 IfA= 3400 then 2 2 sin A is identical to (A) 1 1    sin sin A A (B)     1 1 sin sin A A (C) 1 1    sin sin A A (D*)     1 1 sin sin A A [Hint: A/2 = 1700 hence 2sinA/2 > 0 now 3400 lies in IV quadrant. Hence sinA <0. So 1+ sinA < 1 – sinA. Hence B & C are rejected because theygive – values. Now we will checkA& D. A: | sinA/2 + cosA/2 | + | sinA/2 – cosA/2 |
8. –ve +ve –sinA/2 – cosA/2 + sinA/2 – cosA/2 = – 2 cosA/2 Hence D is the answer ] Q.431/s&p If 3 + 1 4 (3 + d) + 1 42 (3 + 2d) +...... + upto  = 8, then the value of d is : (A*) 9 (B) 5 (C) 1 (D) none of these [Sol. 8 = 3 + 1 4 (3 + d) + 1 42 (3 + 2d) +...... + upto       ....... .......... 4 d 3 4 3 4 8 2  8 – 2 = 3 +     ...... .......... 4 d 4 d 4 d 3 2 6 = 3 + 4 / 1 1 4 / d   d = 9 ] Q.5179/QE The sum of the roots of the equation (x + 1) = 2 log2(2x + 3) – 2 log4(1980 – 2–x) is (A) 3954 (B*) log211 (C) log23954 (D)indeterminate Directions for Q.6 to Q.9: Considertwodifferent infinitegeometricprogressions with theirsums S1 and S2 as S1 = a + ar + ar2 + ar3 + ..........  S2 = b + bR + bR2 + bR3 + ..........  If S1 = S2 = 1, ar = bR and ar2 = 1/8 then answer the following: Q.693/s&pThe sum oftheircommonratios is (A) 2 1 (B) 4 3 (C*) 1 (D) 2 3 Q.7 Thesumoftheirfirsttermsis (A*) 1 (B) 2 (C) 3 (D) none Q.8 Common ratio ofthefirst G.P. is (A) 2 1 (B) 4 5 1 (C) 4 1 5  (D*) 4 1 5  Q.9 Common ratio of the second G.P. is (A) 4 5 3 (B*) 4 5 3 (C) 2 1 (D) none [Sol. Let the two GP's are a ar ar2 .......... | r | < 1 b br br2 .......... | R | < 1 now r 1 a  = 1 ; R 1 b  = 1 a = 1 – r .....(1) and b = 1 – R ....(2)  a + r = b + Q also ar = bR ....(3) and ar2 = 8 1 ....(4) (1) – (2) a – b = R – r a + r = b + R
9. From (1), (2) and (3), ar = bR (1 – r) r = (1 – R)R (substituting a and b for (1) and (2) in (3))  (r – R)(r + R) = r – R  r + R = 1 (as r  R) Ans. (Note : if r = R then a = b  G.P. will become identical) hence r + R = 1 ....(5) hence (1) + (2)  a + b = 2 – (r + R)  a + b = 1 now, from (4) 8ar2 = 1 8(1 – r)r2 = 1 8r2 – 8r3 – 1 = 0  8r3 – 8r2 + 1 = 0 (2r – 1)(4r2 – 2r – 1) = 0 (as r1/2asinthiscaseRwillalsobe1/2whichisnotpossible)  4r2 – 2r – 1 = 0 r = 8 20 2  = 4 5 1 or 4 5 1 if r = 4 5 1 then R = 1 – 4 5 1 = 4 5 3 > 1 hence r = 4 5 1 is rejected  r = 4 5 1 Ans. and R = 1 – 4 5 1 = 4 5 3 Ans. a = 1 – r = 4 5 3 ; b = 4 5 1 a + b = 1 Ans. ] More than one alternative are correct. Q.10124/QE If ax2 + bx + c = 0 , b  1 be an equation with integral co-efficients and  > 0 be its discriminant, then the equation b2 x2  x  4 ac = 0 has : (A) twointegral roots (B*) two rational roots (C) two irrational roots (D*) one integral root independent of a, b, c . [Hint: a, b, c  I ; b2 > 4ac ;  = b2 – 4ac  =     2 2 2 16 2 b ac b =     ( ) b ac b ac b 2 2 2 2 4 16 2 =    ( ) b ac b 2 2 2 4 2 = ( ) ( ) b ac b ac b 2 2 2 4 4 2    with +ve sign,  = 1 (independent of a, b & c)  = – 4 2 ac b which is rational  B & D]
10. SUBJECTIVE: Q.11 If mand n arepositive integers satisfying 1 + cos 2 + cos 4 + cos 6 + cos 8 + cos 10 =    sin n sin · m cos then find the value of (m + n). [Ans. 11] [Sol. Let S = cos 0° + cos 2 + cos 4 + .......... + cos 10 2 sin · S = 2 sin [cos0 + cos 2 + .......... + cos 10] = sin  + sin  = sin3 – sin  = sin5 – sin3 = sin7 – sin5 = sin9 – sin7 = sin11 – sin9 ————————— 2 sin · S = sin11 + sin 2 sin · S = 2 sin6 · cos5 S =    sin 2 5 cos 6 sin 2 =    sin m cos n sin  n = 6 and m = 5 Ans. ] Q.12 Suppose a cubic polynomial f (x) = x3 + px2 + qx + 72 is divisible by both x2 + ax + b and x2 + bx + a (where a, b, p, q are constants and a  b). Find the sum of the squares of the roots of the cubicpolynomial. [Ans. 146] [Sol. Sincecubicis divisiblebyboth x2 + ax + b and x2 + bx + a and  x2 + ax + b and x2 + bx + a must have a common roots. x2 + ax + b = 0 – x2 + bx + a = 0 subtract  x(a – b) = (a – b) x = 1  common root is 1 x2 + ax + b = 0 1 ·  = b   = b x2 + bx + a = 0 1 ·  = a   = a  roots of cubic be 1, a, b product of the roots be 1 · a · b = – 72 ....(1) and a + b + 1 = 0 ....(2) (from x2 + ax + b = 0 put x = 1)  a – b 72 = – 1  a2 + a – 72 = 0 (a + 9)(a – 8) = 0 a = – 9, 8  roots are 1, – 9, 8  sum of their squares = 1 + 81 + 64 = 146 Ans. Note: Iftheydon’thaveacommonrootsthentherewill befourroots ofthecubicwhichis notpossible.Ifthey have bothcommonroots then a=b which contradicts whichis given]
11. CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-28 Select the correct alternative : (Only one is correct) Q.1 If the solutions of the equation sin2 = k (0 < k < 1) in (0, 2) are inA.P. then the value of k is (A*) 2 1 (B) 2 1 (C) 4 3 (D) 4 1 [Hint: The solutions of sin2 = k are ,  – ,  +  and 2 –   – 2 = 2 hence,  = 4   = 4   solutions are 4  , 4 3 , 7 5 , 4 7 and k = 2 1 ] Q.2106/ph-1 Maximum value of the expression cos · sin          6    R, is (A) 2 1 (B) 4 3 (C*) 4 1 (D) 1 [Hint: y = cos · sin          6 = 2 1                  6 sin 6 2 sin  max if 2 – 6  = 2   2 = 6 4   = 3   maximum value = cos 3  sin 6  = 2 1 · 2 1 = 4 1  (C) ] Q.355/s&p Consider a decreasing G.P. : g1, g2, g3, ...... gn ....... such that g1 + g2 + g3 = 13 and 2 3 2 2 2 1 g g g   =91 thenwhichofthefollowingdoesnothold? (A) The greatest term of the G.P. is 9. (B) 3g4 = g3 (C*) g1 = 1 (D) g2 = 3 [Hint: G.P. : r a , a, ar, ......... a = 3 and r = 3 1 or 3 (rejected) G.P. is 9, 3, 1, 3 1 , ......  C is not correct.] Q.4 Supposex,y,z isageometricseries with a commonratio of'r'suchthat x y.Ifx,3y, 5zis anarithmetic sequence then thevalue of 'r' equals (A) 1/3 (B*) 1/5 (C) 3/5 (D) 2/3 [Hint: x = a ; y = ar ; z = ar2 (r  1)] Q.579/s&p Let S1 , S2 , S3 be the sums of the first n , 2n and 3n terms of an A.P. respectively. If
12. S3 = C (S2 – S1) then , 'C' is equal to (A) 4 (B*) 3 (C) 2 (D) 1 [Sol. ] d ) 1 n ( a 2 [ 2 n S1    ; ] d ) 1 n ( a 2 [ 2 n 2 S2    ; ] d ) 1 n ( a 2 [ 2 n 3 S3    Now, S3 = C(S2 – S1) ] d ) 1 n ( a 2 [ 2 n 3   =                  2 n 2 n 2 d ) 1 n ( a 2 C  C = 3 ] Q.6102/QE If the equation a (x – 1)2 + b(x2 – 3x + 2) + x – a2 = 0 is satisfied for all x  R then the number of ordered pairs of (a, b) can be (A) 0 (B*) 1 (C) 2 (D)infinite [Sol. equation is an identity  coefficient of x2 = 0= coefficient of x = constant term  a + b = 0 ....(1) – 2a – 3b + 1 = 0 ....(2) and a + 2b – a2 = 0 ....(3) from (1) and (2) a = – 1 and b = 1 whichalsosatisfies(3)  (a, b) = (–1, 1)  (B) ] Q.797/seq&prog Let f (x) = x2 +x4 + x6 + x8 + .......  for all real x such that the sum converges. Number of real x for which the equation f (x) – x = 0 holds, is (A) 0 (B) 1 (C*) 2 (D) 3 [Hint: f (x) = 2 2 x 1 x  when | x | < 1  2 2 x 1 x  = x  x = 0 or x = 1 – x2 x2 + x – 1 = 0  x = 2 5 1  , 2 5 1  is to be rejected ] Q.812/s&p The sum to n terms of the series, 1 2 + 3 4 + 7 8 + 15 16 +...... is equal to : (A) 2n  n  1 (B) 1  2n (C*) 2n + n  1 (D) 2n  1 [Sol. S = .... 8 1 1 4 1 1 2 1 1                         up to n terms = n –          terms n to up .... 2 1 2 1 1 2 1 2  result] Q.924/log Solution set of the inequality   4 log 2 3 x log x log 2 2 1 2 3 3   is (A) [3, 9] (B*)           , 9 3 1 , 0 (C)             , 9 3 1 , (D)   9 , 1 1 , 3 1        [Sol. Given:   4 log 2 3 x log x log 2 2 1 2 3 3   2 2 log 4 log 2 3 x log x log 2 2 2 3 3    2 x log x log 2 3 3    put log3x = y
13. y – y2  – 2 y2 – y – 2  0 (y – 2) (y + 1)  0 log3x  2 x  9 x  3 1  (B) ] Q.1089/s&p If abcd = 1 where a, b, c, d are positive reals then the minimum value of a2 + b2 + c2 + d2 + ab + ac + ad + bc + bd + cd is (A) 6 (B*) 10 (C) 12 (D) 20 [Hint: Use AM  GM between the given 10 numbers i.e. a2, b2, c2, d2, ab, ac, ad, bc, bd, and cd ] SUBJECTIVE Q.11 Solvethe logarithmic inequality,            ) 5 x )( 1 x ( ) 2 x ( 2 log x 1 1. [Ans. (1, 2) ] [3] [Sol. Let x 1 > 1 i.e. 0 < x < 1            ) 5 x )( 1 x ( ) 2 x ( 2 log x 1  1  x 1 ) 5 x )( 1 x ( ) 2 x ( 2      0 x 1 ) 5 x )( 1 x ( ) 2 x ( 2      0 ) 5 x )( 1 x ( x ) 5 x )( 1 x ( ) 2 x ( x 2         0 ) 5 x )( 1 x ( x 5 x2     Hencenosolutioninthis case nowlet 0 < x 1 < 1 ; x > 1 in this case 0 < x < 5 ....(1) its intersection with x > 1 and the domain of the equation gives x  (1, 2) ] Q.12 We inscribe a square in a circle of unit radius and shade the region between them.Then we inscribe another circle inthesquare and anothersquare in the new circle and shade the regionbetween the new circleandthesquare. Iftheprocessis repeated infinitelymanytimes,find the areaoftheshaded region. [Ans. 2( – 2) sq. units] [4] [Sol. Area of 1st shaded region = ( – 2) Area of 2nd shaded region =         1 2 Area of 3rd shaded region =         2 1 4 and so on Hence total area of shaded region =           ........ 4 1 2 1 1 –          ........ 2 1 1 2 =     2 1 1 2 2 1 1     = 2 – 4 = 2( – 2) Ans. ]
14. Select the correct alternative : (Only one is correct) Q.156/s&p The maximum value of the sum of the decreasingA.P. 50 , 48 , 46 , 44 , ............ is (A) 325 (B) 648 (C*) 650 (D) 652 [Sol. 4 , 2 , 0 ..... ,......... 46 , 48 , 50            a = 50 , d = 2 , n= 25 ] d ) 1 n ( a 2 [ 2 n Sn    25 [50 –24 ] 25 × 26 = 650 Ans ] Q.2513/ph-1 The roots of the equation 2 + cot x = cosec x always lie in the quadrant number (A) Ionly (B) I and II (C) II and IV (D*) IIonly [Hint: Solving cosec x = 5/4 and cot x = – 3/4 ] Q.374/s&p If Sn = 1 1 1 2 1 2 3 3 3    +...... + 1 2 3 1 2 3 3 3 3 3         ...... ...... n n , n = 1, 2, 3,...... Then Sn is not greater than (A) 1/2 (B) 1 (C*) 2 (D) 4 [Sol. Sn = ... .......... 3 2 1 3 2 1 2 1 2 1 1 1 3 3 3 3 3 3          Tn = 3 3 3 3 n . .......... 3 2 1 n ... .......... 5 4 3 2 1        = ) 1 n ( n 2 2 ) 1 n ( n 2 ) 1 n ( n 2           Sn =          1 n 1 n 1 2 ] Q.4103/QE The absolute term in the quadratic expression                   n 1 k 2 k 3 1 x 1 k 3 1 x as n   is (A) zero (B) 1 (C) 3 2 (D*) 3 1 [Hint: Tn = ) 2 n 3 )( 1 n 3 ( 1   ; Tn =          ) 1 n 3 ( 1 ) 2 n 3 ( 1 3 1 ] Q.545/s&p Consider the A.P. a1 , a2 ,..... , an ,.... the G.P. b1 , b2 ,....., bn ,..... such that a1 = b1 = 1 ; a9 = b9 and 369 a 9 1 r r    then (A) b6 = 27 (B*) b7 = 27 (C) b8 = 81 (D) b9 = 18 CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-29
15. [ Hint: a1 = b1 = 1 ; a9 = 1 + 8d = bn = 1. r8 now ) a 1 ( 2 9 a 9 9 1 r r     = 3 r 369 ) r 1 ( 2 9 8      b7 = b. r6 = 1 6 3 = 27 ] Q.692/s&p Startingwithaunit square,asequenceofsquareis generated.Each squarein thesequencehashalfthe side length of its predecessor and two of its sides bisected byits predecessor's sides as shown. This process isrepeatedindefinitely.Thetotalareaenclosed byallthe squares in limitingsituation, is (A*) 4 5 sq. units (B) 64 79 sq. units (C) 64 75 sq. units (D) 12 1 sq. units [Hint: Required area = 1 + 4 3           ....... 4 1 4 1 4 1 3 2 = 1 + 4 3 · 4 1 1 4 1  = 4 1 1 = 4 5  (A) ] Q.7 The sum     1 k k 2 k 3 2 equal to (A) 12 (B*) 8 (C) 6 (D) 4 [Hint: S =          1 k k 3 2 4 =                        ....... 3 2 3 2 3 2 4 3 2 =        ) 3 2 ( 1 3 2 4 = 8 ] Q.817/ph-1 If A, B, C and D denotes the interior angles of a quadrilateral then (A)     A tan A cot A tan (B*)        A cot A tan A tan (C)      A tan A tan A cot (D)      A cot A tan A tan [Hint: tan(A + B + C + D) = 4 2 3 1 S S 1 S S    ( A + B + C + D = 360°) S1 = S3 tanA + tanB + tanC + tanD =     A cot A tan  B ] Q.940/s&p Consider anA.P. with first term 'a'and the common differenced. Let Sk denote the sum of the first K terms. Let S S kx x is independent of x, then (A*) a = d/2 (B) a = d (C) a = 2d (D) none [Sol. ] d ) 1 x ( a 2 [ 2 K )] 1 Kx ( a 2 [ 2 Kx S S x Kx      =           xd d a 2 Kxd d a 2 K If 2a – d = 0 then        xd Kxd K S S K Kx = K2 which is possible when a = d/2 ]
16. Q.10104/ph-1 4 sin50 sin550 sin650 has the values equal to (A) 3 1 2 2  (B*) 3 1 2 2  (C) 3 1 2  (D) 3 3 1 2 2  d i [Sol. 2[2sin50 sin550] sin650  2[cos500 – cos600]sin650  2cos500 sin650 – sin650  sin(1150) + sin150 – sin650 = 2 2 1 3  Ans] SUBJECTIVE: Q.1135/1 Find all values ofk for which the inequality, 2x2  4k2x  k2 + 1 > 0 is valid for all real x which do not exceed unityintheabsolutevalue. [ Ans. :  1 2 < k < 1 2 ] [Hint: Case I: when D < 0 whichgives the result Case II : f (1)  0 ; D  0 ;  b a 2 > 1 or f ( 1)  0 ; D  0 ;  b a 2 <  1 ; No solution in case II ] Q.12 Find all x such that   1 k k x · k = 20. [Ans. x = 4/5] [3] [Sol. Let S = x + 2x2 + 3x3 + 4x4 + .........+  Sx = + x2 + 2x3 + 3x4 + 4x5 + .........+  ————————————————— S(1 – x) = x + x2 + x3 + x4 + .........+  = x 1 x  S = 2 ) x 1 ( x  = 20  20(1 + x2 – 2x) = x  20x2 – 41x + 20 = 0 x = 40 1600 1681 41   = 40 9 41 = 40 50 or 40 32 = 4 5 or 5 4 x = 4 5 is rejected as | x | < 1;  x = 5 4 Ans. ]
17. Q.1841/-1 Identifywhether the statement is True or False. (i)(iii) sin 82 1 2  . cos 37 1 2  and sin 127 1 2  . sin 97 1 2  have the same value. [ Ans. True ; 4 2 3  ] [Sol. A= 0 0 2 1 37 cos . 2 1 82 sin = 0 0 2 75 cos . 2 165 sin =   0 0 45 sin 120 sin 2 1  = 2 4 1 6  B= 0 0 2 1 97 sin . 2 1 127 sin =   0 0 225 cos 30 cos 2 1  =        2 1 2 3 2 1 = 2 4 2 6  = 4 2 3   A= B  True ] (ii)(v) If tanA= 3 4 3  & tanB = 3 4 3  then tan(A B) must be irrational. [Ans. False] [Sol. tan(A–B) = B tan A tan 1 B tan A tan   =    3 4 3 4 3 . 3 1 3 4 3 3 4 3       =   3 3 16 3 4 3 4 3      = 3/8  rational ] (iii)(x) If tanA= 1, tanB = 2 and tanC = 3 thenA, B, C can not be the angles of a triangle. [Ans. False] [Sol. In a  ABC tanA + tanB + tanC = tanA. tanB. tanC 1 + 2 + 3 = 1 . 2 . 3 6 = 6 Hence theyare angles of atriangle ] (iv)(xiv) There exists a value of  between 0 & 2 which satisfies the equation ; sin4  – sin2  – 1 = 0. [Ans. False ] [Sol. sin2 = 2 5 1  sin2 = 2 5 1 (not possible) sin2 = 1 2 5 1    not possible ] Select the correct alternative : (Only one is correct) Q.221/-1 The number of solutions of the equation cos        3 2 x = x2 + x 3 2 + 4 is (A) more than 2 (B) 2 (C) 1 (D*) 0 CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-30
18. Q.361/-1 The value of 10 tan 2 10 sec 3 10 cos 4      is equal to (A) 1 (B) 1 5  (C) 1 5  (D*) zero [Sol. 4 cos180 – 0 18 cos 3 – 2 tan180 0 0 0 2 18 cos 18 sin 2 3 18 cos 4   = 0 0 0 18 cos 3 18 sin 2 ) 36 cos 1 ( 2     0 0 0 18 cos 3 ) 18 sin 36 cos 1 ( 2    = 0 18 cos 3 2 1 1 2         = 0 ] Q.428/s&p The sum of the first 100 terms common to the series 17, 21, 25, ..... and 16, 21, 26, ...... is (A*) 101100 (B) 111000 (C) 110010 (D) 100101 [Sol. 17 ,21 , 25 , 29 , 33 ,37 ,... A.P. with d = 4 16 , 21 , 26 , 31 , 36 , 41 ,..... A.P. with d = 5 common 21 , 41 , 61 ,............. A.P. with d =20 S = ] 20 ) 99 ( ) 21 ( 2 [ 2 100  = 100 [21 + 990] = 101100 Ans ] Q.5 If x, y, z  N then the number of ordered triplets of (x, y, z) satisfying the equation x + y+ z = 102 is (A) 4950 (B*) 5050 (C) 5150 (D) None [Hint: if x = 1 ; y + z = 101  100 solutions eg. (1, 100) , (2, 99) etc if x = 2 ; y + z = 100  99 solutions eg. (1, 99) , (2, 98) etc finally x = 100 ; y + z = 2  1 solution Hence total = 1 + 2 + 3 + ...... + 100 = 5050 ] SUBJECTIVE: Q.655/1 Find all real values of x for which the expression         1 x x log 2 2 / 1 is a real number. . [Ans. x  1 5 2  or 1 5 2   x < 0 ] Q.76/1 Solve the inequality, 2 log1/2 (x  1)  1 3  1 8 2 logx x  . [Ans. : [2, ) ] [Hint: converting on base 2 log2 2 1 1 6 ( ) ( ) x x x    0 note that x > 1 2 1 5 ( ) x x   1 put x  1 = y  y > 0 2 1 5 y y   1  0 2 1 1 5 y y y     0 2 2 1 1 5 y y y y      0 2 1 1 1 4 y y y y ( )      0 ( ) ( ) [ ] ( ) y y y y y      1 2 1 1 1 1 2  0  y y   1 1  0  y  1  x  2 ]
19. Q.810/06 If  = 2 7  , prove that , sec  + sec2 + sec4 =  4. [Sol. E = sec + sec2 + sec4, where 7 = 2 = 1 1 2 1 4 cos cos cos      = cos cos cos cos cos cos cos cos cos 2 4 4 2 2 4              But cos · cos2 · cos4 = 1 8 where = 2 7  hence E = 4 [2cos2cos4 + 2cos4cos + 2coscos2] = 4 [cos6 + cos2 + cos5 + cos3+ cos3 + cos] =8 [cos + cos2 + cos3] as cos6 = cos, cos5 = cos2 Now Let S = cos +cos2 + cos3 2sin  2 S = 2sin  2 cos + 2sin  2 cos2+ 2sin  2 cos3 = sin 3 2  – sin  2 + sin 5 2  – sin 3 2  + sin 7 2  – sin 5 2  = – sin  2 as sin 7 2  = sin  = 0  S = – 1 2 ; Hence E = – 4 Ans ] Q.950/1 Find thevalues of 'p'forwhich the inequality, 2 1 2               log p p x2+2x 1 1 2         log p p  2 1 1 2         log p p > 0 isvalid forall realx. [Sol. (2  t) x2 + 2 (1 + t) x  2 (1 + t) > 0 when t = 1 p p log2  when t = 2 , 6 x  6 > 0 which is not true  x  R. Let t  2 ; t < 2  (1) and 4 (1 + t)2 + 8 (1 + t) (2  t) < 0 (for given inequality to be valid) or (t  5) (t + 1) > 0  t > 5 or t <  1 (2) From (1) and (2) ; t <  1  0 < log2 p p  1 <  1 p p  1 < 1 2 or p p   1 1 < 0   1 < p < 1 but p p  1 > 0  p > 0 or p <  1. common solution is p  (0, 1) ]
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