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Samples of Mathematics Short Tricks Book for JEE(Main) Volume - I, Magical Conceptual Trick -1, Applications of The Concept of Identity, , Q. The value of (1  cot   cos ec)(1  tan   sec ) is equals to :, (a)  3, (b) 0, (c) 1, * (d) 2, S. Here the value of the expression is independent of θ,therefore given expression is, an identity in θ, so we can put any suitable value of θ to minimise the calculations., here let θ 450 then value of given expression is (1  1  2)(1  1  2)  (2  2)(2  2)=2., Q. Let k = 4 cos x cos 2x cos 3x  cos x  cos 2x  cos 3x then k is equals to :, (a)  1, (b) 0, *(c) 1, (d) 2, S. Again value of the expression is independent of x, so the expression is an identity in x,, so let x 00 then k= 4 cos 00 cos 00 cos 00  cos 00  cos 00  cos 00  4  1  1  1  1, Q. The value of the expression cos 2   cos 2(    )  2cos  cos  cos(    ) is ?, * (a) sin2 , (b)cos 2 , (c)sin 2, (d)cos 2, S. Here options are in terms of  so the value of the expression is independent of ., so we can put any suitable value of  to minimise the calculations., put =00 then, cos2   cos2(    )  2cos  cos  cos(    ),  cos2 0  cos2( 0   )  2cos 0 cos  cos( 0   ) 1  cos 2   2cos  cos   1  cos 2   sin2 ., , Q. If cos   cos   cos   0 and if cos 3  cos 3  cos 3  k cos  cos  cos  then k , (a) 1, (b) 8, *(c) 12, (d) 9, , S.  cos   cos   cos   0 so let   0,   1200 &   1200 satisfying given condition,  cos 3  cos 3  cos 3  k cos  cos  cos ,  cos 00  cos 3600  cos 3600  k cos 00 cos1200 cos1200,  1  1  1  k .1.(1/2).(1/2)  k  12,    , , then k is equals to :,  2 , , Q. If (cos   cos )2  (sin   sin )2  k sin2 , (a) 0, , (b) 1, , S. Let   900 ,   00 then (0  1)2  (1  0)2  k , , (c) 3, , *(d) 4, , 1, k 4, 2, , 1, (sink x  cos k x), x  R and k  1,then f4(x)  f6(x) , [JEE(Main) 2014], k, 1, 1, 1, 1, (a), * (b), (c), (d), 4, 12, 6, 3, 1, 1, 1, S.  fk (x)= (sink x  cos k x)  f4(x)  f6(x)  (sin4 x  cos 4 x)  (sin6 x  cos6 x), k, 4, 6, As value of above exp ression is independent of x so put x  0 in above exp ression, 1, 1, 1 1 1,  f4 (x)  f6 (x)  (sin4 0  cos 4 0 )  (sin6 0  cos6 0)   , ., 4, 6, 4 6 12, Q. Let fk (x)=, , The beauty of these short tricks is that many problems of this kind can be solved easily.

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Q. If tan  (1  2 x )1 and tan  (1  21 x )1 then the value of ( + ) is equals to :, (a) 300, * (b) 450, (c) 600, (d) 900, S. from options it is clear that value of the expression is independent of x ,so put x 0, 1, 1, then tan  and tan = and by usin g formula of tan(  ),, 2, 3, 1 1, , 5 /6, tan(  )  2 3 , =1     450., 1, 5 /6, 1, 6,  3, , , , , , , , Q. The value of 3 sin4 , – θ  sin6 (3 +θ)  2 sin6   θ  sin6(5 + θ )  is :, 2, ,  2, , , , , , , (a) 5, , (b) 0, , * (c) 1, , (d) 3, , S. Since the value of the expression is independent of θ, therefore put θ =00, , , ,  3 ,  , , ,  sin6(3 )  2 sin6    sin6 (5 )   4, ,  2,  2, , , , , in the expression. then, 3 sin4 , , , =3  1  0  2  1  0  =3 – 2  1, so option (c) is currect., Q. The value of the expression cos  sin(    )  cos  sin(   )  cos  sin(   )is ?, * (a) 0, (b) –1, (c) 1, (d) 2, S.  the value of the expression is independent of ,  and  .,  put =    =0 in the expression, then, we will get 0  0  0  0., 1 ,  2, ,  4, , sin3   sin3 ,     sin3 ,     is equals to :, ,  3, ,  3, , sin 3 , 4, 3, 3, 4, (a), (b), * (c) , (d) , 3, 4, 4, 3, 0, S.  The value of above exp ressions is independent of  so put   30 ., Q., , sin, , 3, , then value of given exp ression, , 300  sin3 1500  sin3 2700, sin 90, , 0, ,   1  1 1   3 ., 8, , 8, , 4, , Q. Which of the followings is not equals to unity :, sin2 , cos , (1  cot2 ) , (1  tan2 ), 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, (c) sin  cos   cos  sin   sin  sin   cos  cos  * (d)sin   cos 4   2sin2 , S.  the value of the above exp ressions should be 1., (a) cos 4   sin4   2sin2 , , (b), ,  put     450 ( sin 450  cos 450 and also tan 450  cot 450 )., 1, (a) cos 4   sin4   2sin2   0  2   1, 2, 2, sin , cos , 1, 1, 1 1, (b), (1  cot2 ) , (1  tan2 )  .2  .2    1, 2, 2, 4, 4, 2 2, 1 1 1 1, (c) sin2  cos2   cos2  sin2   sin2  sin2   cos2  cos2       1, 4 4 4 4, 1, (d)sin4   cos 4   2sin2   0  2   1 (not equals to unity)., 2, , The beauty of these short tricks is that many problems of this kind can be solved easily.

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Magical Conceptual Trick - 2, Magical Method of Substitutions and Balancing,    ,    , Q. If cos   2cos  then the value of tan , tan , is equals to :, ,  2 ,  2 , 1, 1, 1, *  b , c, a  3, , 3, 3, S. Magical Method of Substitution : , , d  , , 1, 3, , here by observation if we put   00 and   600 in the given condition then, 1, 2,    ,    , 1, then tan , tan ,  tan 300 tan 300   ., , ,  2 ,  2 , 3, , cos 00  2cos 600  1  2.  1  1(satisfied), , , , , , , , , , sin5  sin 2  sin , is equals to :, cos 5  2cos 3  2cos 2   cos , * (a) tan, (b) cos, (c) cot, (d) sin, S. Magical Method of Substitution and Balancing : , Q. The value of the exp ression, , sin1500  sin 600  sin 300, 1, , 0, 0, 2, 0, 0, cos150  2cos 90  2cos 30  cos 30, 3, 0, Now put   30 in options then (a) will match with L.H.S., Let   300 then LHS =, , Q. The value of tan + 2 tan2 +4tan4+8cot 8 is :, (a) tan, (b) tan2, * (c) cot, S. Magical Method of Substitution and Balancing : , , (d) cot2, , Let = 15º then LHS = tan15º+2tan30º+4 tan60º + 8cot120º,  1 , +4 3 +8 ,  = 2+ 3 = cot 15º  =cot , , 3, 3, , 1, , = 2  3+ 2 ×, , Q. If tan A  tan B  x and cot A  cot B  y then cot(A  B) is equals to:, (a), , 1 1, , x y, , * (b), , 1 1, , x y, , (c) x  y, , (d) x  y, , S. Magical Method of Substitution :  Let A  600 & B  300 then, x  3, , 1, 3, , , , 2, 3, , and y , , 1, 3, ,  3, , 2, 3, , ,so L.H.S.  cot(600  300 )  cot300  3, , now put the values of x and y in options then option (b) will gives, , Q. If tan A , (a) p2 +q 2, , 3 so it is R.H.S., , p, 1, and if =6  is acute angle  then (p cosec2  q sec2) is equals to:, q, 2, * (b) 2 p2 +q 2, , (c) 2 p2  q 2, , (d) p2  q 2, , S. Magical Method of Substitution :  Let A  450 then q  p,  2 2, , and LHS  p cosec15  p sec15  p , ,  3 1, , , , 2 2 ,   2 2p, 3  1, , now put q  p in options then  b  will give 2 2 p so it is R.H.S., , The beauty of these short tricks is that many problems of this kind can be solved easily.

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, and if tan 1 tan  2 tan  3...........................tan n  1, 2, then cos 1 cos  2 cos  3...........................cos  n , [ IITJEE 2001], , Q. If 0   1 ,  2,  3 ,................. n  , , (a) 2n/2, * (b) 2 n/2, (c) 2n/4, S. Magical Method of Substitution and Balancing : , , (d) 2 n/4, , Let 1   2   3  .............   n  450 (Satisfying both given conditions), Re quired value  cos 450  cos 450  cos 450............n terms, 1, 1, 1, 1, 1, , , , ..........................n tertms ,  n/2  2 n/2, n, 2, 2, 2, 2, ( 2), 7 , , 3 , 7 , , 3 , , , , , , , Q. If X  sin   , +sin      sin   , , Y  cos   , + cos      cos   , , , , , , , , , , , 12 , 12 , 12 , 12 , 12 , 12 , X Y,  , Y X, (a) 2sin 2, , then, , (b) 2 cos 2, , * (c) 2 tan 2, , (d)2 cot 2, , 0, , S. Magical Method of Substitution :  Let   15 ., then X  sin1200 +sin 00  sin 600 =, , 3, 3, 0,  3, 2, 2, , 1, 1, 3, 1, 2,  1   1, then L.H.S. , , , 2, 2, 1, 3, 3, 2, Now put   150 in the options then option (c) will gives, so it is R.H.S., 3, Y  cos1200 + cos 00  cos 600  , , and, , Q. If a  sin x  sin y, b  cos x  cos y, c  tan x  tan y then, (b) c 2, , * (a) c, , 8ab, , (a  b2 )2  4a 2, 2, , (d) 2c 2, , (c) 2c, , S. Magical Method of Substitution :  let x  y  450 ,therefore a  2 , b  2, c  2, 8 2 2, 2c, ( 2  2 2)2  4 2 2, , then, , Q. If, , 2, , (cos10  sin10 )(cos 20  sin 20 )(cos 30  sin 30 )..............(cos 450  sin 450 ), = a b,, 0, 0, 0, 0, 0, 0, cos1 cos 2 cos 3 cos 4 ............cos 44 cos 45, , where a and b are prime numbers and a  b then a  b , (a) 22, S., , 0, , (b ) 23, 0, , 0, , 0, , (c) 24, 0, , 0, , *(d) 25, 0, , 0, , (cos1  sin1 ) (cos 2  sin 2 ) (cos 3  sin 3 ), (cos 44  sin 44 ) (cos 440  sin 440 ), .........., cos10, cos 20, cos 30, cos 440, cos 450, (1+tan1°)(1+tan2°)......................(1+tan43°)(1+tan44°)(1+ tan45°),  1  44  2  43  3  42  .........  450, So first we find out if     450 then (1+ tan )(1+ tan)  ?, By method of substitution let   0 and   45 then(1+ tan )(1+ tan)  (1)  2  2, (1+tan1°)(1+tan44°)(1+tan2°)(1+tan43°).............(1+tan22°)(1+tan23°).  (1+1) = a b, , , 22 pairs, , 22, , 2, , k, , 23, , b, ,  2 = 2  2 = a  a  2 and b  23 and hence a  b  25, , The beauty of these short tricks is that many problems of this kind can be solved easily.

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Samples of Mathematics Short Tricks Book for JEE(Main) Volume - I, Inverse Trigonometric Functions (I.T.F.),  xz ,  xy ,  zy , Q. If x 2  y 2  z 2  k 2 then value of tan1    tan1    tan1   is equals to ?,  zk ,  xk ,  yk , , 3, * (a), (b) , (c), (d) 0, 2, 2, S. Method of Substitution :  Let x  y  z 1,then k 2  3  k  3,  1 ,  1 ,  1 , ,  tan1 ,  tan1 ,  tan1 ,  300  300  300  900 , , , ,  3,  3,  3, 2, , Q. If sin1  sin1   sin1  , , a , , b, , 0, , 3, 300, then value of 1000 (     )  2, is :, 2,   2   2, 2890, , c 1900, , * d  2900, , S. Method of Substitution : ,  sin-1  sin1   sin1  , , 3, ,, 2, , ,  ,   =  =  =sin   =1,  2, 2, 300, L.H.S.  1000 (     )  2,   2   2, 300,  1000(1+1+1) , = 3000  100  2900, 1+1+1, ,  Let sin-1  sin1   sin1  , ,  x,  y, Q If cos-1    cos-1    θ and if 9x 2 –2xy cos θ 4y 2  k 2sin2θ then k,  2,  3, , b  1, , (a)  3, , * c   6, , d   2, ,  1, S. Method of Substitution :  Let x1 and y0, θ  cos-1    cos-1(0)  60°  90° 150°,  2, ,  9x 2 – 2xy cos θ 4y 2  k 2sin2θ, 2, , k2,  1,  9 × 1 – 0 0k    9 ,  k 2 36  k  6.,  2, 4, 2, ,  x  ,  x  ,   1,   1, Q. The value of tan   cos –1     tan  – cos –1    is equals to :,  y  ,  y  ,  4 2,  4 2, , (a), , x, y, , (b), , y, x, , (c), , 2x, y, , * (d), , 2y, x, , , ,  1, S. Method of substitution : - Let x  1 and y  2  cos –1    300 ,  2, , ,  1, ,  1, ,  L.H.S.  tan   .300   tan  – .300 , 4 2, 4 2, , , ,  tan 750  tan150  cot150  tan150  2  3  2  3  4, Now put x  1 and y  2 in options then (d) will give 4 and (d) is right choice., , The beauty of these short tricks is that many problems of this kind can be solved easily.

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Q Let sin1a  sin1b  sin1c   then a (1  a 2 )  b (1  b2 )  c (1  c 2 ) , 1, *  a  2abc, c  2 abc, b abc, S. Method of Substitution : , , (d), , abc, 3, ,  sin1 a  sin1 b  sin1 c  180° ,, Let sin1 a  s in1 b  sin1 c  60°  a = b =c =, , 3, 2, , L.H.S.  a 1  a 2 + b 1  b2 +c 1  c 2, 3, 3, 3, 3, 3, 3 3 3, 1 +, 1 +, 1 =, 2, 4, 2, 4 2, 4, 4, after putting the values of a ,b,c in theoptions a  will give the R.H.S., =, , Q If cos-1x  cos-1, * (a) 4sin2, , y,   then 4x 2 – 4xycos  y 2 is equals to :, 2, (b) – 4sin2, (c) 2sin2, , (d) 4, , S. Method of substitution : - Let x0 and y1,   cos-10  cos -1, , 1,  900  60°  30°., 2, , L.H.S.  4x 2 – 4xy cos y 2  0  0  1  1, now put the value of  in options then only option (a) will match with L.H.S., Q. If 0  x  1, then, , 12, , 1  x 2 { x cos ( cot 1 x )  sin( cot 1 x ) } 2  1, , , [ IITJEE2008], , (a ), , x, 2, , 1 x, S. Method of Substitution : , , *( c ) x 1  x 2, , (b) x, ,  0  x  1 therefore let x , , 1, 3, , ( d ) 1  x2, , , , 1,  3  0.57 approx. , 12, ,  1  x 2 { x cos ( cot 1 x )  sin( cot 1 x ) }2  1, 1,  1, 3, , 2, ,  1, 0, 0 ,  3 cos ( 60 )  sin( 60 )  1, 2, , 2  1 1, 3, 2 1 3, 1, 3, ,  ,  2, , 1, ,  1 , 3  3 2 2, 3 12 4, 2 3 2, 2 1 3 1, 2 1 9  6, 2 1, 2,   , , ., , 12, 3 12 4 2, 3, 3 3 3, 1, 2, now put x , in options then (c) will give ,so it is right choice., 3, 3, , , The beauty of Concepts of identities and Methods of substitutions is that, many problems can be solved very easily and speedly by these methods., , The magical book Volume - 1 contains many such different conceptual techniques, For any queary or purchasing of books contact at 07014858096, 08769855992., , " Stock of the books is limited"

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Samples of Mathematics Short Tricks Book for JEE(Main) Volume - I, Quadratic Equations, Q. If the roots of the equation ax 2 + bx + c = 0 are real and distinct then :, b, b, (a) both roots are greater than , ., (b) both roots are less than , ., 2a, 2a, b, b, * (c) one of roots exceeds , ., (d) both roots are exceeds , ., 2a, 2a, S. Master quadratic equation :  Let   1,   2 then x 2  3x  2  0  a  1, b  3,c  2,  a =1,b = – 3,c = 2 so, , –b – (– 3) 3,  b , =, =  2  One of the roots exceeds  –,  2a , 2a 2  1 2, , Q. If ,  are the roots of x 2 –ax + b = 0 and if  n + n =Vn , then :, (a) Vn 1= aVn + bVn–1, , *(b) Vn+1= aVn –bVn–1, , (c) Vn 1=aVn + aVn–1, , (d) Vn+1= aVn–1 – bVn, , S. Method of substitution : - Let =1 &  = 2 then eq. is x 2 –3x + 2= 0  a=3, b = 2, Let n=0 then V0 =2, n=1 then V1=+=3 n=2 then V2 = 2 +2 =1+4=5, Now check the options. Put n=1 in the options then, ( a) V2 =aV1+ b V0  5 = 3  3 + 2  2  5 = 9+4 =13, (b) V2 = aV1  b V0  5 = 3  3  2  2  9 – 4=5 so (b) is the Right choice., Q. If the roots of x 2 –bx  c  0 are two consecutive integers, then b2 – 4c =, * (a) 1, (b) 2, (c) 3, S. Method of substitution : -, , (d) 4, ,  Roots are two consecutive integers so let =1 & =2., then equation is x 2 – 3x+2=0  b  3, c=2 therefore b2 –4c=(3)2 –4  2=9–8 =1, Q. If, , 1, 1, and, are the roots of ax 2  bx  c  0  a  0,a, b  R  then the equation, , , , x(x  b3 )  (a 3  3abx)  0 has roots : , (a), , 3, ,  2, , and, , 3, ,  2, , * (b), , 3, 2, , and, , 3, 2, , [JEE (Main) online 2014], (c), , 1,  2, , and, , 1,  2, , (d)  and , , S. Method of substitution : - Let   4 and   9 then the equation is 6x 2  5x  1  0, and a  6, b  5,c  1 so x(x  b3 )  (a 3  3abx)  0 becomes x 2  35x  216  0, 3, , 3, , 3, , 3, , and  2 and  2 i.e. 4 2 and 9 2 or 8 & 27 satisfy this., Q. If p and q are nonzero real numbers and  3  3   p,   q then the quadratic equation, whose roots are, ,  2 2, ,, is : ,  , , * (a) qx 2  px  q 2  0, , (b)qx 2  px  q 2  0, , [JEE (Main)online, 2014], (c) px 2  qx  p2  0, , (d) px 2  qx  p 2  0, , S. Method of substitution : - Let   1 and   2 then the equation is x 2  3x  2  0, and p  9,q  2 and roots of r e quired equation are, , 2 1, 2 4,  and, , , 2, , 1, , and r e quired equation is 2x 2  9x  4  0 or qx 2  px  q 2  0, , The beauty of these short tricks is that many problems of this kind can be solved easily.

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Q. If  and β (  <β ) are the roots of x 2 + bx + c=0,where c <0 <b , then, (a) 0<  < β, * (b) < 0<β<|  |, (c) < β< 0, S. Combination of methods of substitution and balancing : -, , [IITJEE2000], (d)  <0<|  |< β, , let c =  2,b = 1 ( c <0 <b ) then given equation will be x 2  x  2=0   =1,  = –2, After putting the values of  and  in options, option (b) will be satisfied., Q. If b  a , then the equation ( x  a )( x  b )  1  0 has :, [IITJEE2000], (a) both roots in (a , b), (b) both roots in (  ,a), (c) both roots in (b ,  ), * (d) one root in ( ,a) and the other in (b ,  ), S. Methods of substitution : -  b  a so let a  1, b  2,  given equation will be ( x  1)( x  2 )  1  0  x 2  3x +1= 0  x=,  x=, , 3 5 3 5, ,, 2, 2, , 3 5,  2  x  b  one root lies in ( b ,  )., 2, , 3 5,  1  x  a  one root lies in ( ,a ) so option (d) is currect ., 2, Q. If one root of the equation x 2 + px + q=0 is square of the other then:, [IITJEE2004], again x =, , *  a  p3  q  3p  1 + q 2, ,  b p3  q  3p+1 + q 2, ,  c  p3 + q 3p  1 +q 2=0 d  p3 + q 3p+1 +q 2=0, , S. Combination of methods of substitution and balancing :  One roots is square of the other roots, therefore let =1 and =1, then the equation is x 2  2x+1=0 , p  2 and q=1, Put these values of p & q in the options then (a) will satisfy so it is right option., Q. If the equation a n x n  a n 1x n 1  ...  a1x  0, a1  0, n  2, has a positive root x =  then, the equation na n x n 1  ( n  1)a n 1x n  2  ...........  a1  0 has a positive root, which is :, (a) Greater then or equal to ., , (b) Equal to . (c) Greater then ., n, , S. Method of substitution : - Let n = 2 then, a n x  a n 1x, a 2 x 2  a1x  0  x=0 and x = –, , n 1, , *(d) Smaller then ., ,  ......  a1x  0  a 2x 2  a1x  0, , a1, =  (let), a2, , na n x n 1   n  1 a n 1x n  2  ..  a1  0 will be 2a 2 x + a1=0  x = –, , a1 , =, smaller than   ., 2a 2 2, , Q. Let p and q be real numbers such that p  0 , p3  q and p3  q If  &  are non-zero, complex numbers satisfying      p and  3  3  q then a quadratic equation having, , , and, as its roots is : , [IITJEE2010], , , ,  a   p3  q  x 2   p3  2q  x   p3  q   0.,  c  p3  q  x 2  5p3  2q  x   p3  q   0., , , d p, ,  ,  q  x  p, ,  ,  2q  x   p, , ,  q   0., , *  b p3  q x 2  p3  2q x  p3  q  0., 3, , 2, , 3, , 3, , S. Combination of methods of substitution and balancing :   and  are nonzero complex numbers so let    and    2, ,  , ,         2  1   p  p  1 and  3  3   3   2, , 3, , 2q, ,  , so required equation is x 2     x  1  0  x 2  x  1  0,   , put the value of p & q in options then  b will give required result., , The beauty of these short tricks is that many problems of this kind can be solved easily.

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Samples of Mathematics Short Tricks Book for JEE(Main) Volume - I, Progressions, Arithmatic Progression, Master A.P. :  1, 2, 3, 4, 5, 6, ........, Q. If S1, S2 and S3 denote the sum of first n1, n2 and n3 terms of an A.P. then, S, S, S1, (n2  n3 )  2 (n3  n1 )  3 (n1  n2 ) , n1, n2, n3, * (a) 0, , (b) 1, , (c) S1S2S3, , (d) n1n2n3, , S . Method of substitution : - Let A.P. be :– 1, 2, 3, 4, 5, 6, .........(Master A.P.), Let n1  1 then S1  1, for n2  2, S2  1  2  3 and for n3  3, S3  1+ 2+3= 6, , , S, S, S1, 1, 3, 6, (n2  n3 )  2 (n3  n1 )  3 (n1  n2 ) = (  1)  (2)  (1)  –1  3  2  0., n1, n2, n3, 1, 2, 3, , Q. If Sn denotes the sum of first 'n' terms of a A.P., then, (a) 2n –1, , * (b) 2n  1, , S3n  Sn 1, =, S2n  S2n 1, , (c) 2n, , (d), , n, 2, , S. Combination of methods of substitution and balancing : Let the A.P. be 1, 2, 3, 4, 5, 6, .........(Master A.P.), S9  S2, 45  3, 42, Let n  3, therefore LHS , , , 7, S6  S5, 21  15, 6, After putting the value of n  3 in the option (b) will give 7, so it is R.H.S., Q. If a1, a 2 , a 3 , .............. a n are in A.P., then, (a), , 1, a1a n, , (b), , 1, 1, 1, +, +...................+, =, a1a 2 a 2a 3, a n–1a n, , n, a1a n, , * (c), , n–1, a1a n, , (d), , n+1, ., a1a n, , S. Combination of methods of substitution and balancing : Let A.P. be 1, 2, 3, 4, 5, 6, .........(Master A.P.), and let n  3 then a1  1,a 2  2, a 3  3, (a n  3) then, LHS , , 1 1 2,   ., 2 6 3, , Now put the values of a1=1, a n =3 in the option then (c) will give, , 2, , so it is R.H.S., 3, ,   or 0.573. is :, Q. The value of 0.573737373 ....................or 0.573., 284, 284, 558, 567, (a), (b), (c), (d), 497, 495, 990, 990, S. Short trick : - Remove all decimals i.e.573 then subtract that number which has, no recurring decimal,Then divide by as many 9's as many digits has recurring, decimal and as many zero's as many digits has no recurring decimal.,  0.573737373................ = 0.57 3, =, , 573–5 568 284, , =, ., 990, 990 495, , The beauty of these short tricks is that many problems of this kind can be solved easily.

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Harmonic Progression, Applications of master H.P. 3, 4, 6,12......., Q. If H1, H2 , H3 .............H2n 1 are H.P.,the, , 2n, , H H, , , ,  (1)i  Hi  Hi 1 , i 1, , =, , i 1, , i, , (a) 2n–1, * (b) 2n  1, (c) 2n, S. Combination of methods of substitution and balancing : -, , (d) 2n  2, ,  H1, H2 , H3 ,......are in H.P. therefore Let H.P. be master H.P. 3, 4, 6,12........,  H1  3, H2  4, H3  6, and Let n  1,  Hi  Hi 1 , , , i=1,  Hi  Hi 1 ,  H  H2  H2  H3, 3  4 4  6,   1,  , , ,  2, 3  4 4  6,  H1  H2  H2  H3, Put n  1 in the options then only (c) will match with L.H.S., , LHS , , 2n, ,  (  1)i, , Q. If a1, a 2 , a 3 ............, a n are in H.P. then a1a 2  a 2a 3  a 3a 4  ..........+a n–1a n is equals to:, * (b) (n  1)a1a n, , (a) na1a n, , (c) (n  1)a1a n, , (d) None of these, , S. Combination of methods of substitution and balancing :  a1, a 2 , a 3 ....., a n are in H.P. therefore Let H.P. be master H.P. 3, 4, 6,12........, also let n=4, then a1  3, a 2  4, a 3  6, a 4  12,, LHS  a1a 2  a 2a 3  a 3a 4  3  4  4  6  6  12 108, After putting values of a1& a n and n in the options (b) will match with LHS., Q. If a1,a 2 ,a 3 ,........................, a100 are in H. P. then, , 99, , aa, ,  a ia i 1, i 1, , is equals to :, , 1 100, , (a) 100, * (b)99, (c)101, (d) 110, S. Master Result : - From above question If a1 , a 2 , a3 .............., a n a re in H.P., then a1a 2 + a 2 a3 + a3a 4 + ....................+ a n –1a n = (n  1)a1an, 99, , a ia i 1, 1, , a1a100, i 1 a1a100, , , , 99, , 1, ,  a ia i 1  a a, i 1, , (a1a 2  a 2a 3  a 3a 4  ...............+ a 99a100 ), , 1 100, , for n  100 ,(a1a 2  a 2a 3  a 3a 4  ......+ a 99a100 )  (100  1)a 99a100  99a 99a100, 99, , a ia i 1, 1, , .99 a 99a100  99, a1a100, i 1 a1a100, , , , Q. If a1,a 2 ,a 3 , .....are in H.P. and if f (k) , , n, ,  a r  a k , then, r 1, , a1 a 2 a 3, a, ,, ,, , ........... n are in:, f(1) f(2) f(3), f(n), , (a) G.P., *(b) H.P., (c) A.P., S. Method of substitution : - Let H.P. be master H.P. 3, 4, 6,12........, , (d) A.G.P., , Let a1 ,a 2 ,a 3 ,a 4 ....... are in H.P., , , 3, , , , 4, , , , 6, , , , 12, ,  f (k)  a1  a 2  a 3  ........  a n  a k let n  4,  f (k)  a k  a1  a 2  a 3  a 4,  f(1)  22, f(2)  21, f (3)  19, a2 a3, a, a, 3 4 6,  1 ,, ,, , .............. n , ,, , , ..... .are in H.P., f(1) f(2) f(3), f(n) 22 21 19, , The beauty of these short tricks is that many problems of this kind can be solved easily.

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Geometric Progression, Q. The sum of n terms of a G.P. is S, product is P and sum of their inverse is R, then the value, of P 2 is equals to :, n, , R, S,  R, (b), (c)  ,  S, S, R, S. Combination of methods of substitution and balancing : -, ,  S, * (d)  ,  R, , (a), , n, , 1 1 7, Let the master G.P. 1, 2 , 4 , 8, 16...... Let n  3 then S  7, P  8 , R  1+ + =, 2 4 4, 2, L.H.S  P  64 ,Now put the values of n.S,R in options then(d) will match with L.H.S., Q. If m is the A.M. of two distinct real numbers l and n ( l , n  1) and G1,G2 ,G3 are three geometric, means between l and n then G14  2G24  G34 , 2, , (a) 4 l mn, , [JEE(Main) 2015], , 2, , * (b)4 lm n, , (c)4 l mn, , 2, , (d) 4 l 2m2n2, , S. Combination of methods of substitution and balancing : Let master G.P. 2, 4 , 8 ,16 ,32 (  l , n  1) and m , l G1 G2 G3, , n, , l  n 2  32, ,  17, 2, 2, , G14  2G24  G34  44  2.84  164  44  25.44  48  44 1  32  256  44.289  4 lm2n, Q. If a, a1, a 2..........a 2n , b are in arithmetic progression and a , g1, g 2 , ........g 2n , b are in gemettric, progression and ' h ' is the harmonic mean of a and b , then the value of, a  a 2n, a  a 2n 1, a  a n 1, the exp ression 1,  2,  .......................................  n, is :, g1 g 2n, g 2 g 2n 1, g n g n 1, n, (c) nh, h, S. Combination of methods of substitution and balancing : (a) 2nh, , (b), , * (d), , 2n, h, , 2ab, ab, and a , g1, g 2 , b are in G.P.  ab  g1 g 2 ., , Let n  1 then a , a1, a 2, b are A.P. and a , g1, g 2, b are in G.P. and h ,  a , a1, a 2 , b are A.P.  a  b  a1  a 2, L.H.S , , a1  a 2 a  b 2, ,  , now put n  1 in option then (d) will give R.H.S., g1 g 2, ab, h, , Q. If the sum of first n terms of an A.P. is cn2 , then sum of squares of these n terms is : , [IITJEE 2009], n(4n2  1)c 2, n(4n2  1)c 2, (b), 6, 3, S. Method of substitution : (a), , * (c), , n(4n2  1)c 2, 3, , (d), , n(4n2  1)c 2, 6, ,  Tn  Sn  Sn-1  cn2  c(n  1)2  c(2n  1)  Tn2 = c 2 (2n  1)2 ,  T12  S12  c 2, If we put n = 1 in options the result should be c 2, so (c) is right option., The beauty of applications of Master Quadratic Equations and Master Progressions, is that many problems can be solved very easily and speedly ., , The magical book Volume - 1 contains many such different conceptual techniques, For any queary or purchasing of books contact at 07014858096, 08769855992., , " Stock of the books is limited"

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Samples of Mathematics Short Tricks Book for JEE(Main) Volume - I, Determinants, 3, 1  f(1) 1  f(2), Q.If ,   0 and f(n)     and 1  f(1) 1  f(2) 1  f(3)  K(1   )2 (1   )2 (    )2, 1  f(2) 1  f(3) 1  f(4), n, , n, , then k is equals to : , (b)  1, , (a)1, , (c) , , [JEE(Main) 2014], 1, (d), , , S. Method of substitution : Let   2,   1 then f(n)  2n  (1)n,  f(1)  21  (1)1  1, f(2)  22  (1)2  5 , f(3)  23  (1)3  7, f(4)  24  (1)4  17, 3 2 6, 2 6 8  K(1  2 )2 (1  1)2 ( 2  1)2  36K, 6 8 18, 3 2 1,  2 6 0  3(24  0)  2(8  0)  1(16  36)  72  16  30  36  36K  K  1, 6 8 4, Q. For all real numbers x, y and z, the value of the det er min ant, xy  xz, , 2x, , y, , 2x  z  1 xy  xz  yz  z 2 1  y is equals :, 3x  1, 2xy  2xz, 1 y, , (JEE(Main)P2, 2017), , (a) (y  xz)(z  x), (b)(x  y)(y  z)(z  x), (c) 0, S. Combination of methods of substitution and balancing : -, , (d) (x  yz)(y  z), ,  x, y, z  R so let x  1, y  2, z  3, 2 1 2,  6 4 3  2(12  6)  1(18  12)  2(12  16)  12  6  8  2, 4 2 3, (a) (y  xz)(z  x)  (2  3)(3  1)  2, (b)(x  y)(y  z)(z  x)  (1  2)(2  3)(3  1)  2., (d) (1  6)(2  3)  5, so (b) is right choice., a2, , b2, , Q. If (a   )2 (b   )2, (a   )2 (b   )2, , c2, , a2, (c   )2  k a, 1, (c   )2, , b2 c 2, b c (   0) then k , 1 1, (JEE (Main) online,12April 2014), , 2, , (a) 4abc, * (b) 4, (c)  4 2, S. Combination of methods of substitution and balancing : -, , (d)  4abc, , Let a  1, b  2 and c  3 and   1, 1 4 9, 1 4 9,  4 9 16  k .1 1 2 3  k  4, 0 1 4, 1 1 1, put the values a  1, b  2 and c  3 and   1 in options, then only (b) is right choice., , The beauty of these short tricks is that many problems of this kind can be solved easily.