# Ph-1,2,3 & Binomial(F).pdf

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(n – 2)2 = n (n – 1) – 4n + 10 n2 – 4n + 4 = n2 – 5n + 10 n = 6 Ans ] Q.138107/bin The sum of the series aC0 + (a + b)C1 + (a + 2b)C2 + + (a + nb)Cn is where Cr's denotes combinatorial coefficient in the expansion of (1 + x)n, n  N (A) (a + 2nb)2n (B) (2a + nb)2n (C) (a +nb)2n – 1 (D*) (2a + nb)2n – 1 Q.139109/bin The coefficient of the middle term in the binomial expansion in powers of x of (1 + x)4 and of (1 – x)6 is the same if  equals 5 (A) – 3 10 (B) 3 3 (C*) – 10 3 (D) 5 Q.14029/bin (2n + 1) (2n + 3) (2n + 5) ....... (4n  1) is equal to : (A) (4n) ! 2n . (2n) ! (2n) ! (B*) (4n) ! n ! 2n . (2n) ! (2n) ! (C) (4n) ! n ! (2n) ! (2n) ! (D) (4n) ! n ! 2n ! (2n) ! [Hint: E = (2 n + 1) (2 n + 3) (2 n + 5) ......(4 n  1) Multiply numerator and denominator by (2 n + 2) (2 n + 4) ...... (4 n) & also by (2 n ) ! . E = (2 n) ! (2 n  1) (2 n  2) (2 n  3) (4 n 1) . 4 n (2 n) ! (2 n 2) (2 n  4) (2 n  2 n) = (4 n) !  (n) ! = (n !) . (4 n) !  B ] (2 n) ! 2n (n  1) (n  2) (2 n)  n ! 2n . (2 n) !2 Q.141 110/bin If Sn n =  n r0 Cr and Tn n =  n r0 Cr Tn then n is equal to (A*) n 2 (B) n 1 2 (C) n – 1 (D) 2n 1 2 Q.14211/bin The coefficient of xr (0  r  n 1) in the expression : (x + 2)n1 + (x + 2)n2. (x + 1) + (x + 2)n3 . (x + 1)² + ...... + (x + 1)n1 is : (A) nCr (2r  1) (B*) nCr (2nr  1) (C) nCr (2r + 1) (D) nCr (2nr + 1)  x1  x1 2  x1 n1  [Hint : E = (x + 2)n  1 1   x2 x2 ......   x2       1   x  1 n   n  1  x  2  n (x  2)n  (x  1)n  n n = (x + 2)    x  1  x  2  = (x + 2)   (x  2)n  = (x + 2)   (x + 1) Now co-efficient of xr in  (2  x)n  (1  x)n  = nCr 2n  r  nCr = nCr (2n  r  1) ]  x  1 x  1  10 Q.12663/bin In the expansion of  x2/ 3  x1/3  1  x  x1/ 2  , the term which does not contain x is :   (A) 10C0 (B) 10C7 (C*) 10C4 (D) none x1/3 3  1 [Hint : x2/3  x1/3  1 = x1/3 + 1  (x1/3  x1/2)10 ]  1 8 Q.127 65/bin If the 6th term in the expansion of the binomial  x8/3  x2 log10 x is 5600, then x equals to (A) 5 (B) 8 (C*) 10 (D) 100  1  8 5 [Hint : T6 = 8C5  x8/3 . (x2 log10 x)5 = 100  x = 10 ] Q.12868/bin Co-efficient of t in the expansion of, ( + p)m  1 + ( + p)m  2 ( + q) + ( + p)m  3 ( + q)2 + ( + q)m  1 where    q and p  q is : (A) (C) m Ct pt  qt  p  q m Ct pt  qt  p  q (B*) (D) m Ct pm t  qm t  p  q m Ct pm t  qm t  p  q    q    q 2    q m1  [Hint : E = ( + p)m  1 1        ......        p    p    p   co-efficient of t =   pm    qm p  q p  m  q m or p  q = m t pm t  qm t  p q ] Q.12970/bin (1 + x) (1 + x + x2) (1 + x + x2 + x3) ...... (1 + x + x2 + ...... + x100) when

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### Ph-1,2,3 & Binomial(F).pdf

• 1. (n – 2)2 = n (n – 1) – 4n + 10 n2 – 4n + 4 = n2 – 5n + 10 n = 6 Ans ] Q.138107/bin The sum of the series aC0 + (a + b)C1 + (a + 2b)C2 + ..... + (a + nb)Cn is where Cr's denotes combinatorialcoefficient in the expansion of (1 + x)n, n N (A) (a + 2nb)2n (B) (2a + nb)2n (C) (a +nb)2n – 1 (D*) (2a + nb)2n – 1 Q.139109/bin The coefficient ofthe middle term in the binomialexpansion in powers of x of(1 + x)4 and of (1 – x)6 is the same if  equals (A) – 3 5 (B) 3 10 (C*) – 10 3 (D) 5 3 Q.14029/bin (2n+1) (2n+3) (2n+5) ....... (4n 1) is equal to : (A) ( ) ! . ( ) ! ( )! 4 2 2 2 n n n n (B*) ( ) ! ! . ( ) ! ( )! 4 2 2 2 n n n n n (C) ( ) ! ! ( ) ! ( ) ! 4 2 2 n n n n (D) ( ) ! ! ! ( ) ! 4 2 2 n n n n [Hint: E = (2 n + 1) (2 n + 3) (2 n + 5) ......(4 n  1) Multiply numerator and denominator by (2n + 2) (2n + 4) ...... (4n) & also by (2n )!. E = ( ) ! ( ) ( ) ( ) ........ ( ) . ( ) ! ( ) ( ) ........ ( ) 2 2 1 2 2 2 3 4 1 4 2 2 2 2 4 2 2 n n n n n n n n n n n        =   ( ) ! ( ) ! ( ) ! ( ) ( ) ...... ( ) ! 4 2 2 1 2 2 n n n n n n n n    =   ( !) . ( )! . ( ) ! n n n n 4 2 2 2  B ] Q.141110/bin If Sn =   n 0 r r n C 1 and Tn =   n 0 r r n C r then n n S T is equal to (A*) 2 n (B) 1 2 n  (C) n – 1 (D) 2 1 n 2  Q.14211/bin The coefficient of xr (0  r  n1) in the expression : (x+ 2)n1 + (x+ 2)n2. (x+ 1) + (x+ 2)n3 . (x+ 1)² + ...... + (x+ 1)n1 is : (A) nCr (2r  1) (B*) nCr (2nr  1) (C) nCr (2r + 1) (D) nCr (2nr + 1) [Hint: E = (x + 2)n  1                                   1 n 2 2 x 1 x ...... 2 x 1 x 2 x 1 x 1 = (x + 2)n  1 1 1 2 1 1 2                         x x x x n = (x + 2)n ( ) ( ) ( ) x x x n n n           2 1 2 = (x + 2)n  (x + 1)n Now co-efficient of xr in   ( ) ( ) 2 1    x x n n = nCr 2n  r  nCr = nCr (2n  r  1) ]
• 2. Q.12663/bin In the expansionof x x x x x x             1 1 1 2 3 1 3 1 2 10 / / / , the term which does not contain x is : (A) 10C0 (B) 10C7 (C*) 10C4 (D) none [Hint:   x x x 1 3 3 2 3 1 3 1 1 / / /    = x1/3 + 1  (x1/3  x1/2)10 ] Q.12765/bin Ifthe 6th termin the expansion of the binomial 1 8 3 2 10 8 x x x / log        is 5600, then xequals to (A) 5 (B) 8 (C*) 10 (D) 100 [Hint: T6 = 8C5 1 8 3 8 5 x /        . (x2 log10 x)5 = 100  x = 10 ] Q.12868/bin Co-efficient of t in the expansion of, ( + p)m  1 + ( + p)m  2 ( + q) + ( + p)m  3 ( + q)2 + ...... ( + q)m  1 where  q and p  q is : (A)   m t t t C p q p q   (B*)   m t m t m t C p q p q     (C)   m t t t C p q p q   (D)   m t m t m t C p q p q     [Hint: E = ( + p)m  1 1 2 1                                      q p q p q p m ......  co-efficient of t =           p q p q m m or     p q p q m m       =   m t m t m t C p q p q     ] Q.12970/bin (1 + x) (1 + x + x2) (1 + x + x2 + x3) ...... (1 + x + x2 + ...... + x100) when written in the ascending power of x then the highest exponent of x is ______ . (A) 4950 (B*) 5050 (C) 5150 (D) none [Hint: Highest exponent inthe product offirst two is 3 = 1 + 2 Highest exponent inthe product offirst three is 6 = 1 + 2 + 3 Similarly, Highest exponent inthe product offirst hundred = 1 + 2 + ...... + 100 = 5050 ] Q.13073/bin Let   5 2 6  n = p + f where n  N and p  N and 0 < f < 1 then the value of, f2  f + pf  p is (A) a naturalnumber (B*) a negative integer (C) a prime number (D) are irrationalnumber [Hint: value =  1 ] Q.13175/bin Number ofrational terms inthe expansion of   2 3 4 100  is : (A) 25 (B*) 26 (C) 27 (D) 28
• 3. Q.11219/bin Let n ) 3 4 7 (  = p +  when n and p are positive integers and   (0, 1) then (1 – ) (p + ) is (A) rationalwhichis not aninteger (B) a prime (C) a composite (D*) none ofthese [Hint: (1 – ) (p + ) = 1  (D) ] Q.11324/bin If(11)27 + (21)27 when divided by 16 leaves the remainder (A*) 0 (B) 1 (C) 2 (D) 14 [Hint: an + bn = (a + b) (Q (a, b) ) if n is odd alternatively : interpret from (16 – 5)27 + (16 + 5)27 ] Q.11425/bin Last three digits ofthe number N = 7100 – 3100 are (A) 100 (B) 300 (C) 500 (D*) 000 [Hint: consider (5 + 2)100 – (5 – 2)100 = 2 [100C1 599 · 2 + 100C3 597 · 23 + ....... + 100C995 · 299] = 2 [1000 · 598 + 100C3 594 + ....... + 1000 · 298]  minimum 000 as last three digits.  (D) ] Q.11527/bin The last two digits of the number 3400 are : (A) 81 (B) 43 (C) 29 (D*) 01 [Hint : 3400 = 81100 = (1 + 80)100 = 100C0 + 100C1 80 + ....... + 100C100 80100  Last two digits are 01 ] Q.11628/bin If (1 + x + x²)25 = a0 + a1x + a2x² + ..... + a50 . x50 then a0 + a2 + a4 + ..... + a50 is : (A*) even (B) odd & ofthe form 3n (C) odd & of the form (3n1) (D) odd & ofthe form (3n+1) [Hint: putting x = 1 and 1 and adding a0 + a2 + ...... + a50 = 3 1 2 25  =   1 2 1 2 25   ( 2 even 2 1 odd   ) = 25 2 2 2 1 2 0 25 1 25 2 2 25 25 25 C C C C     . . . =   2 1 2 2 2 25 1 25 2 25 25 24     C C C . ...... . = 2 [13 + 25C2 + ...... + 25C25 . 223]  even ] Q.11731/bin The sum of the series (1² + 1).1! + (2² +1).2! + (3² +1). 3! + ..... + (n²+ 1). n! is : (A) (n+1). (n+2)! (B*) n.(n+1)! (C) (n+1). (n+1)! (D) none of these [Hint: Tn = [n (n + 1)  (n  1) ] n! = n. (n+1)!  (n  1). n! Now put n = 1, 2, 3, ...... , n and add ] Q.11835/bin Let Pm stand for nPm . Then the expression 1 . P1 + 2 . P2 + 3 . P3 + ..... + n. Pn = (A*) (n+ 1) !  1 (B) (n+ 1) ! + 1 (C) (n+ 1)! (D) none of these [Hint: Tn = n . n ! = n! [(n + 1)  1 ] = (n + 1) !  n! Now put n = 1, 2, 3 , ....... and add ]
• 4. y = ) 2 / cos( ) 2 / sin( 2 2 / cos 2 sin cos 1 2        = cot /2 x = 2 / tan 1 2 / tan 1     = 1 y 1 y 1 2 / cot 1 2 / cot         (C) Applying C/D y 1 y 1 y 1 y 1 y 1 x 1 x            (B) Also , y =    x 1 1 x y – xy – x–1 = 0  (D) ] Q.102 If2 cos + sin = 1, then the value of 4 cos + 3sin is equalto (A*) 3 (B) –5 (C*) 7 5 (D) –4 [Hint: (2cos)2 = (1 – sin)2  5sin2 – 2sin– 3 = 0  sin = 1 or  3 5 , now proceed If sin = 1  cos = 0  E = 3; if sin = – 3 5  cos = 4 5 or – 4 5 but cos – 4 5 (think!) hence E = 7 5 ] Q.103 Ifsint + cost = 1 5 then tan t 2 is equal to : (A) 1 (*B) – 1 3 (C*) 2 (D)  1 6 [Hint: sint & cost in tangent of half the angle  (3y+ 1)(y  2) = 0  y = 2 or 1/3 where y= tan(t/2) ] BINOMIAL There are 39 questions in this question bank. Q.1042/bin Given that the termofthe expansion(x1/3  x1/2)15 whichdoes not contain xis 5mwhere m  N , then m = (A) 1100 (B) 1010 (C*) 1001 (D) none [Hint: Tr + 1 = 15Cr (x1/3)15  r (–x 1/2) r  15 3  r  r 2 = 0  r = 6 Hence T7 is independent of x and T7 = 15C6 = 5005 = 5m  m = 1001 ] Q.1053/bin In the binomial (21/3 + 31/3)n, if the ratio of the seventh term from the beginning of the expansion to the seventh term fromits end is 1/6, then n = (A) 6 (B*) 9 (C) 12 (D) 15 [Hint: Tr + 1 = nCr an  r . br where a = 21/3 and b = 3 1/3 T7 from beginning = nC6 an  6 b6 and T7 from end = nC6 bn  6 a6  a b n n   12 12 = 1 6  3 12 n 3 12 n 3 . 2   = 6 1  n – 12 =  3  n = 9 ]
• 5. [Sol. sin = sin  = n +(–1)n n = 0  =  sin/3 = sin/3  (A) n = 1  = – sin/3 = sin(/3–/3)  (B) n = –1  = –– sin/3 = sin(–/3–/3) = –sin(/3+/3)  (D) ] Q.95 Choose the INCORRECT statement(s). (A)iii sin 82 1 2  . cos 37 1 2  and sin 127 1 2  . sin 97 1 2  have the same value. (B*)v If tanA= 3 4 3  & tanB = 3 4 3  then tan(A B) must be irrational. (C*)viii The sign ofthe product sin2 . sin3 . sin5 is positive. (D*)xiv There exists a value of  between 0 & 2 which satisfies the equation ; sin4  – sin2  – 1 = 0. [Sol. (A) 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 ] (B) 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 ] (C) [Sol. sin2 = + ; sin3 = + ; sin5 = – ] (D) [Sol. sin2 = 2 5 1  sin2 = 2 5 1 (not possible) sin2 = 1 2 5 1    not possible ] Q.96 Whichofthefollowing functions have the maximumvalueunity? (A*) sin2 x  cos2 x (B*) sin cos 2 2 2 x x  (C*)  sin cos 2 2 2 x x  (D*) 6 5 1 2 1 3 sin cos x x       
• 6. Q.85 Minimum value of 8cos2x + 18sec2x  x  R wherever it is defined, is : (A) 24 (B) 25 (C*) 26 (D) 18 [Sol. y = 8 cos2x + 18 sec2x = 8 (cos2x + sec2x) + 10 sec2x = 8 [ (cos x – sec x )2 + 2 ] + 10 sec2x where cosx = secx  x = 0 ymin = 16 + 10 = 26 Ans ] Q.86 In a ABC           C sin c B sin b A sin a 2 2 2 . sin 2 A sin 2 B sin 2 C simplifies to (A) 2 (B*)  (C) 2  (D) 4  where  is the area ofthe triangle [Hint: 2R (a + b + c)  (sin 2 A ) 4 R s · R 4 r = r s = ] Q.87 If  is eliminated fromthe equations x = a cos( – ) and y = b cos ( – ) then ) cos( ab xy 2 b y a x 2 2 2 2      is equalto (A) cos2 (  – ) (B*) sin2 ( – ) (C) sec2 (  – ) (D) cosec2 ( – ) [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     ) cos( ab xy 2 b y a x 2 2 2 2      = sin2 ( – ) ] Q.88 Thegeneralsolutionofthetrigonometric equation tan x + tan 2x + tan 3x = tan x · tan 2x · tan 3x is (A) x = n (B) n ± 3  (C) x = 2n (D*) x = 3 n where n  I [Hint: tan x + tan 2x + tan 3x = tan 3x – tan 2x – tan x  tan x + tan 2x = 0  tan 2x = tan (– x)
• 7.  2b2 = a2 + c2 Þ a2, b2, c2 are in A.P. ] Q.75 The number ofsolution ofthe equation,   5 1 r ) x r cos( = 0 lying in (0, p) is : (A) 2 (B) 3 (C*) 5 (D) more than 5 [Hint: cos x + cos 2x + cos 3x + cos 4x + cos 5 x = 0 2 cos 3x cos 2x + 2 cos 3x cos x + cos 3x = 0 cos 3x [2 cos 2x + 2 cos x – 1] = 0 x = (2n – 1) 6   6  , 6 3 , 6 5 = 3      = 5 2nd equation gives cos x = 4 2 1 = 2 Q.76 If  = 3  and sin  = a a b 2 2  . The value of the expression, a cosec  b sec  is (A) 1 2 2 a b  (B*) 2 a b 2 2  (C) a + b (D) none [Sol. a cosec – bsec =    cos b sin a                 sin b a b cos b a a cos sin b a 2 2 2 2 2 2 Now sin3 = 2 2 b a a  gives                cos sin sin 3 cos cos 3 sin b a 2 2 = 2 2 b a 2  Ans ] Q.78 The value ofcot 7 1 2 0 + tan 67 1 2 0 – cot 67 1 2 0 – tan7 1 2 0 is : (A) arationalnumber (B*)irrationalnumber (C) 2(3 + 2 3 ) (D) 2 (3 – 3 ) Q.79 If in a triangle ABC 2 2 cos cos cos A a B b C c a bc b ca     then the value of the angle Ais : (A) 8  (B) 4  (C) 3  (D*) 2  Q.80 The value ofthe expression (sinx + cosecx)2 + (cosx + secx)2 – ( tanx+ cotx)2 wherever defined is equalto (A) 0 (B*) 5 (C) 7 (D) 9
• 8. [Hint: C sin b a 2 1 b r 2 1 a r 2 1      r (a + b) = 2  r = b a 2   ] Q.66 For eachnatural number k , let Ck denotes the circle withradius k centimeters and centre at the origin. Onthe circle Ck , a particle moves kcentimeters inthecounter- clockwise direction.Aftercompleting its motion on Ck , the particle moves to Ck+1 inthe radialdirection. Themotion ofthe particle continues in this manner .The particle starts at (1, 0).Ifthe particlecrosses the positive directionofthe x- axisfor the first time onthe circle Cn thenn equalto (A) 6 (B*) 7 (C) 8 (D) 9 [Hint: Totaldistance travelled = 35 cm;displacement at the instant it crosses the +ve x-axis first timeis 6cm Angular displacement oneachcircle is 1 radian.] Q.67 If in a ABC, cos cos cos A a B b C c   then the triangle is (A)right angled (B) isosceles (C*) equilateral (D) obtuse [Hint: cos sin cos sin cos sin A R A B R B C R C 2 2 2    tanA = tanB = tanC  A = B = C  (C) ] Q.68 If cos A+ cosB + 2cosC = 2 then the sides of the ABC are in (A*)A.P. (B) G.P (C) H.P. (D) none [Hint: cosA + cosB = 2(1-cosC) = 4 sin2 C 2 or 2cos A B  2 cos A B  2 = 4sin2 C 2 or cos A B  2 = 2sin C 2 or 2cos C 2 cos A -B 2 = 4sin C 2 cos C 2 = 2sinC 2sin A + B 2 cos A -B 2 = 2sinC or sinA+ sinB = 2sinC  a, c, b are inA.P. . ] Q.69 IfAand B are complimentaryangles, then : (A*) 1 2 1 2               tan tan A B = 2 (B) 1 2 1 2               cot cot A B = 2 (C) 1 2 1 2               sec cos A ec B = 2 (D) 1 2 1 2               tan tan A B = 2 [Sol. A = /2 – B    A B 2 4 2  Hence 1 + tanA/2 = 1 + 1 2 1 2   tan / tan / B B = 2 1 2  tan B HenceAis correct ]
• 9. ymax = 2 625 2 252  ] Q.59 4 sin50 sin550 sin650 has the values equalto (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] Q.60 If x, y and z are the distances of incentre from the vertices of the triangle ABC respectively then z y x c b a is equalto (A)  2 A tan (B*)  2 A cot (C)  2 A tan (D)  2 A sin [Similar to 48] [Sol. x = r cosec 2 A a = r        2 C cot 2 B cot 2 A sin . 2 C cot 2 B cot x a         = 2 C sin . 2 B sin 2 A cos . 2 A sin  2 C sin . 2 B sin 2 A sin 2 C cos . 2 B cos . 2 A cos xyz abc  = 2 C cot . 2 B cot . 2 A cot In a triangle  2 A cot =  2 A cot ] Q.61 The medians ofa ABC are 9 cm, 12 cm and 15 cm respectively . Thenthe area of the triangle is (A) 96 sq cm (B) 84 sq cm (C*) 72 sq cm (D) 60 sq cm [Hint: Produce the medianAM to D such that GM = MD . Join D to B and C . Now GBDC is a parallelogram. Note that the sides ofthe  GDC are 6, 8, 10 GDC = 90º
• 10. Q.49 Number of roots of the equation cos sin 2 3 1 2 3 4 1 0 x x      which lie in the interval [] is (A) 2 (B*) 4 (C) 6 (D) 8 [Sol. 1 – sin2x + 2 1 3  sinx – 4 3 – 1 = 0 sin2x – 2 1 3  sinx + 4 3 = 0 4sin2x – 2 3 sinx – 2sinx + 3 = 0 On solvingwe get sinx = 1/2 ; 2 3 = (/6 , 5/6 ; /3 , 2/3 ] Q.50 sec sec 8 1 4 1     is equal to (A) tan 2 cot 8 (B) tan 8 tan 2 (C) cot 8 cot 2 (D*) tan 8 cot 2 Q.51 In a ABC if b = a   1 3  and C = 300 then the measure of the angle A is (A) 150 (B) 450 (C) 750 (D*) 1050 [Hint: use tan 2 B A  = b a b a   cot 2 C to get A– B and A+ B = 1500 (given) ] Q.52 Number ofvalues of   [ , ] 0 2 satisfying the equation cotx – cosx = 1 – cotx. cosx (A) 1 (B*) 2 (C) 3 (D) 4 Q.53 The exact value of cos273º + cos247º + (cos73º . cos47º)is (A) 1/4 (B) 1/2 (C*)3/4 (D) 1 [Sol. 2 146 cos 1 0  + 2 94 cos 1 0  + 2 26 cos 120 cos 0 0  = 1 + 2 26 cos 4 1 2 94 cos 146 cos 0 0 0          = 3/4 Ans ] Q.54 In a ABC, a = a1 = 2 , b = a2 , c = a3 such that ap+1 =           p p p 2 p p 2 p a 5 2 p 4 2 a 3 5 where p = 1,2 then (A) r1 = r2 (B) r3 = 2r1 (C) r2 = 2r1 (D*) r2 = 3r1 [Hint: put p = 1 , we get a2 = 4  b = 4 put p = 2 , we get a3 = 4  c = 4 Hence the ABC is isosceles
• 11. = 9 sin 18 sin 6 cos 18 cos 6 sin 4             = 4 Ans ] Q.41 In a ABC if b + c = 3a then cot B 2 · cot C 2 has the value equal to : (A) 4 (B) 3 (C*) 2 (D) 1 [Hint: cot B 2 · cot C 2 =   s s b   .   s s c   .   s a s a   = s s a  = 2 2 2 s s a  but given that a + b + c = 4a  2s = 4a Hence cot B 2 · cot C 2 = 4 2 a a = 2 ] Q.42 The set of values of ‘a’for which the equation, cos 2x + a sin x = 2a  7 possess a solution is : (A) (, 2) (B*) [2, 6] (C) (6, ) (D) () [Sol. cos2x + a sinx = 2a – 7 i.e. 2sin2x – a sinx + 2a – 8 = 0 sinx = a a a a a       2 8 2 8 4 8 4 ( ) ( ) sinx = a or  4 2 2 Hence –1 < (a– 4)/ 2 < 1  the range of a ] Q.43 Inaright angled trianglethehypotenuse is2 2 timesthe perpendicular drawnfromthe oppositevertex. Thentheother acute angles ofthe triangle are (A)  3 &  6 (B*)  8 & 3 8  (C)  4 &  4 (D)  5 & 3 10  [Sol. p2sec2 + p2cosec2 =  2 2 2 p2  8 cos sin 1 2 2    sin22 = 1/2 = 2 2 1       2 = n + /4  = n/2 + /8 for n = 0   = /8 for n=1   = 3/8 ] Q.44 Let f, g, h bethe lengths oftheperpendiculars fromthe circumcentreoftheABC onthe sides a, band c respectively . If a f b g c h   =  a b c f g h then the value of  is : (A*) 1/4 (B) 1/2 (C) 1 (D) 2
• 12. [Hint: Note that thegiven expressionsimplifies to 3 cot3x ] [Sol. cotx + ) 60 x sin( ) 60 x cos( ) x 60 sin( ) x 60 cos(      = ) 60 x sin( ) 60 x sin( ) x 2 sin( x sin x cos    = 3 x sin 4 x cos x sin 8 x sin x cos 2   = x sin 3 x sin 4 x cos x sin 8 x cos 3 x cos x sin 4 3 2 2    = x sin ] x cos 4 x cos 3 [ 3 3 3  = 3 cot3x  x tan x tan 3 ] x tan 3 1 [ 3 3 2   Ans ] Q.34 In a  ABC, cos 3A+ cos 3B + cos 3C = 1 then : (A) ABC is right angled (B) ABC is acute angled (C*) ABC is obtuse angled (D) nothing definite can be said about the nature ofthe . [Hint:  cos 3A = 1 + 4 sin 3 2 A sin 3 2 B sin 3 2 C = 1  A = = 2 3  or B = 2 3  or C = 2 3  ] Q.35 The value of 3 76 16 76 16       cot cot cot cot is : (A*) cot44º (B) tan44º (C) tan2º (D) cot46º [Sol. Using 0 0 0 0 0 0 0 0 16 cos 76 sin 16 sin 76 cos 16 cos 76 cos 16 sin . 76 sin 3   = 0 0 0 0 0 0 0 92 sin ] 16 cos 76 cos 16 sin 76 [sin 16 sin 76 sin 2   =       92 sin 60 cos 92 cos 60 cos = 0 0 92 sin 92 cos 1 = 0 0 0 2 46 cos 46 sin 2 46 sin 2 = tan460 = cot440 Ans ] Q.36 Ifthe incircle ofthe ABC touches its sides respectivelyat L, M and N and ifx, y, zbe the circumradii ofthe triangles MIN, NIL and LIM where I is the incentre then theproduct xyz is equalto : (A) Rr2 (B) rR2 (C*) 1 2 R r2 (D) 1 2 rR2 [Hint: note thatANIM is a cyclic quandrilateral.  cosec A 2 = 2x r  2x = r cosec A 2 x = r A 2 2 sin  xyz = 2 R r r . 2 R . r 2 A sin 4 . 2 r 2 3 3    ]
• 13. Q.25 With usual notation in a  ABC 1 1 1 1 1 1 1 2 2 3 3 1 r r r r r r                      = K R a b c 3 2 2 2 where K has the value equalto : (A) 1 (B) 16 (C*) 64 (D) 128 [Hint: 1st term = ) b s a s ( 1     = c   L H S = a bc 3 . Use  = a bc R 4 to get the result ] Q.26 If 5 2 3     x , then the value ofthe expression 1 1 1 1       sin sin sin sin x x x x is (A) –cot x 2 (B) cot x 2 (C) tan x 2 (D*) –tan x 2 [Hint: Onrationalizing;we get x sin 1 x sin 1 | x cos | 2 x sin 1 x sin 1        =   ) x (sin 2 | x cos | 1 2   = ) x (sin x cos 1    (D) ] Q.27 If x sin = y sin          2 3 = z sin          4 3 then : (A) x + y + z = 0 (B*) xy + yz + zx = 0 (C) xyz + x + y + z = 1 (D) none [Hint:        sin sin . 3 / 2 cos cos . 3 / 2 sin y x =           sin sin cos 3 2 1 = 2 1 cot 2 3   ....(1) |||ly        sin 3 / 4 sin . cos 3 / 4 cos . sin z x =    cot 2 3 2 1 ....(2) 1 z x y x     xz + xy + yz = 0 ] Q.28 In a ABC, the value of a A b B c C a b c cos cos cos     is equal to : (A*) r R (B) R r 2 (C) R r (D) 2r R [Hint: LHS     R A B C R A B C sin sin sin sin sin sin 2 2 2 2     = 4 2 4 2 2 2 sin sin sin . cos cos cos A B C A B C = 4 sin A 2 sin B 2 sin C 2 = r R ] Q.29 The value of cos  10 cos 2 10  cos 4 10  cos 8 10  cos 16 10  is : (A) 1 32 (B) 1 16 (C)   cos /  10 16 (D*)  10 2 5 64 
• 14. Q.17 With usualnotations, in a triangleABC, a cos(B – C) + b cos(C –A) + c cos(A– B) is equal to (A*) 2 R abc (B) 2 R 4 abc (C) 2 R abc 4 (D) 2 R 2 abc [Sol. Here a(cosB cosC + sinB sinC) + ........ using C sin c B sin b A sin a   = 2R a (cosB cosC + 2 R 4 bc ) + ...... = 2 R 4 abc 3 + a cosB cosC + b cosC cosA + c cosA cosB = 2 R 4 abc 3 + c cosC + c cosA cosB = 2 R 4 abc 3 + c [cosA cosB – cos(A + B)] = 2 R 4 abc 3 + c sinA sinB = 2 R 4 abc 3 + 2 R 4 abc = 2 R abc  (A) ] Q.18 sin cos sin cos 3 3        cos cot   1 2   2 tan cot  =  1 if : (A)  0 2 ,        (B*)    2 ,       (C)    , 3 2       (D)  3 2 2   ,       [Hint: simplifies to –cos |sin| + sin cos = 0 provided sin  cos  Q.19 With usual notations in a triangleABC, ( I I1 ) · ( I I2 ) · ( I I3 ) has the value equalto (A) R2r (B) 2R2r (C) 4R2r (D*) 16R2r [Hint: BICI1 is a cyclic quadrilateral with I I1 as the diameter also  BI1C = 2 A 2   applying sine law in BCI1 1 I I 2 A cos a   I I1 = 2 A cos 2 A cos · 2 A sin 2 · R 2 = 4R sin 2 A   1 I I = 64R3 sin 2 A sin 2 B sin 2 C = 16R2 r ]
• 15. Q.8           tan . cos sin cos . tan x x x x x            2 3 2 7 2 2 3 2 3 whensimplified reduces to : (A) sinx cosx (B) sin2 x (C) sinx cosx (D*) sin2x [Sol.           tan . cos sin cos . tan x x x x x            2 3 2 7 2 2 3 2 3 = x cot . x sin x cos x sin . x cot 3    = x sin x cos . x sin x sin x cos . x sin x cos3   = sin2x Ans ] Q.9 If in a ABC, sin3A+ sin3B + sin3C = 3 sinA· sinB · sinC then (A) ABC may be a scalene triangle (B) ABC isa right triangle (C) ABC is anobtuse angled triangle (D*) ABC isan equilateraltriangle [Hint: Use : a3+b3+c3–3abc = (1/2) (a+b+c) [(a–b)2+(b–c)2+(c–a)2 ] [Hint: either sinA + sinB + sin C = 0 (which isnot possible) or sinA = sinB = sinC  equilateral ] Q.10 In a triangleABC, CH and CM are the lengths of the altitude and median to the base AB. If a = 10, b = 26, c = 32 then length (HM) (A) 5 (B) 7 (C*) 9 (D) none [Hint: cos B = 32 10 26 2 3210 7 10 7 2 2 2      . . BH  MH = BM  BH = 16  7 = 9 ] Q.11 The value of 1 tan cos sin cos sin sin 2 2           for allpermissible vlaues of (A) is less than – 1 (B) is greater than 1 (C) lies between – 1 and 1 including both (D*) lies between – 2 and 2 Q.12 sin 3 = 4 sin  sin 2 sin 4 in 0  has : (A) 2 realsolutions (B) 4 realsolutions (C) 6 realsolutions (D*) 8 realsolutions. [Sol. given equation canbe written as 3 sin  - 4sin3 = 4sin sin 2 sin4 hence either sin  = 0   = n or 3 – 4sin2 = 4 sin 2 sin 4 3 – 2 (1 – cos 2) = 2 (cos 2 – cos 6) or 1 = – 2 cos 6 cos 6 = – 2 1 = cos 3 2 6 = 2n ± 3 2
• 16. Question bank on Compound angles, Trigonometric eqn and ineqn, Solutions of Triangle & Binomial There are 142 questions in this question bank. Select the correct alternative : (Only one is correct) Q.1 If x + y = 3 – cos4 and x – y = 4 sin2 then (A) x4 + y4 = 9 (B) 16 y x   (C) x3 + y3 = 2(x2 + y2) (D*) 2 y x   [Sol. On addingand subtracting x = 2 2 sin 4 4 cos 3     ; y = 2 2 sin 4 4 cos 3     x = 2 ) 4 cos 1 ( ) 2 sin 1 ( 4      ; y = 2 ) 4 cos 1 ( ) 2 sin 1 ( 4      x = 2 (1 + sin2 ) – cos22 ; y = 2 (1 – sin2) – cos22 x = 1 + 2 sin2 + sin22 ; y = 1 – 2 sin2 + sin22 x = (1 + sin2)2 ; y = (1 – sin2)2  2 y x   ] [Alternate : Or put  = 4  and verify ] Q.2 If in a triangle ABC, b cos2 A 2 + a cos2 B 2 = 3 2 c then a, b, c are : (A) inA.P. (B) in G.P. (C) in H.P. (D*) None [Hint: b 2 (1 + cos A) + a 2 (1 + cos B) = 3 2 c  a + b + c = 3c  a + b = 2c  a,c,b are in A.P. ] Q.3 If tanB = A cos n 1 A cos A sin n 2  then tan(A + B) equals (A*) A cos ) n 1 ( A sin  (B) A sin A cos ) 1 n (  (C) A cos ) 1 n ( A sin  (D) A cos ) 1 n ( A sin  [Sol. tan(A + B) = B tan A tan 1 B tan A tan   = A cos n 1 A cos A sin n · A tan 1 A cos n 1 A cos A sin n A tan 2 2     = A cos A sin n ) A cos n 1 ( A cos A cos A sin n ) A cos n 1 ( A sin 2 2 2 2     = ) A sin n A cos n 1 ( A cos 0 A sin 2 2    = A cos ) n 1 ( A sin  ] Q.4 Given a2 + 2a + cosec2  2 ( ) a x  F H G I K J= 0 then, which ofthe 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
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