1. a) 28 integers are chosen from the interval [104, 208]. Show that there exit two of them having a

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1 NMTC STAGE-II 0 JUNIOR GROUP (IX AND X) Note : Answer as many questions as possible. Elegant and noval solution will get extra credits. Diagrams and explanations should be given wherever necessary. Fill in FACE SLIP and your rough working should be in the answer book itself Maximum time allowed is THREE hour. All questions carry equal marks.. a) 8 integers are chosen from the interval [04, 08]. Show that there exit two of them having a Sol. w common prime divisor. b) AB is a line segment.c is a point on AB. ACPQ and CBRS are squares drawn on the same side AB, Prove the S is the orthocentre of the triangle APB. No. of primes between 04 to 08 9 We have so many Integers which are divisible by 06 ( 0)...06( ) divisible by 8( 6)...( 7) divisible by 0,0...0 divisible by divisible by divisible by 69( )...4( ) divisible by 7 9 (7 7)...87(7 ) So that 9 primes and one from every,,,7, and we have total 9 + 6, Such Numbers that doesn t have any common prime factor. But, if we take 8 Numbers so their will be such Numbers that will have same prime factor. Q M P (b) S 4º x 4º x R A C B Draw a line from B which passes through S and intersect AP at M. As PC is the altitude of APB which pass through S and BM is also pass through S. In order to prove S orthocentre, now we just need to prove BM AP. Let PBS = x SPB + PSB + PBS = 80º RSB = 4º [As BS is diagonal of the square] SPB = 4º x In SPM AP is the diagonal of square ACPQ MPS = 4 PSM = 80 (90 + 4) [Linear pair] = 4 PMS = 80 (4 + 4) 90 [Angle sum proper] In PAB PC and BM are altituted which intrsecs at S. So S is the orthocenter PAGE #

2 . a) a,b,c are distinct real numbers such that a = (b +c ), b = (c +a ), c = (a + b ). Find the numericla value of abc. b) a = find [a], where [a] denotes the integer part of a. Sol. (a) a b = (b a ) a + b + ab = (a + b)..(i) b c = (c b ) b + c + bc = (c + b)..(ii) a c = (c a ) a + c + ac = (a + c)..(iii) from (i) & (ii) a c + b (a c) = (a c) a + b + c =..(iv) by adding given equation a + b + c = 6 (a + b + c ) 7 (a + b + c) (a + b + c ab bc ca) + abc = 6 (a + b + c ) 7 a b c + ab + bc + ac + abc = 6(a + b + c ) 7 ab + bc + ac + abc = 9(a + b + c ) 7 = 9[(a + b + c) ab bc ca)] 7 = 9[9 ab bc ac) 7 ab + bc + ac + abc = 6 8 (ab + bc + ca) abc = 6 (ab + bc + ca) abc = 7 (ab + bc + ca) from (i), (ii),(iii) a + b + c + ab + bc + ac = = 8 (a + b + c ) + ab + bc + ca = 8 ((a + b + c) ab bc ca) + ab + bc + ca = 8 () (ab + bc + ca) = 8 ab + bc + ca = 0..(vi) from (v) & (vi) abc = 7(0) =..(v) (b) > > and so on > 0 Add all above inequations > > PAGE #

3 0 > > > so it means is a decimal value less than one [a] = 0 = [ + decimal value less than one] =. The arthemetic mean of a number of pair wise distinct prime numbers is 7. Determine the biggest prime among them. Sol. These are pairs of prime no. whose mean is 7 (47, 7) (, 4) (4, ) (9, 7) So, biggest prime no. is bugs are placed at different squares of a 9 9 square board. A bug in each moves to a horizontal or Sol. vertical adjacent square. No bug makes two horizontal or two vertical moves in succession. Show that after some moves, there will be atleast two bugs in the same square. Total bugs 6 total squares 8 (IInd step) (st step) 8 8=64 squares (64 bugs) If we take 64 bugs, then we can arrange them together into a matrix of 8 8 square, so their is a possibility that No bugs are in same square, because we can move all the bugs vertically upward in st step, then Horizontally left in nd step vertically down in the third step, and in the 4 th step horizontaly left and so on. But, If we take 6 bugs so one horizontal or vertical row of square will fill with bugs. So we can not perform the above process in this situation [due to extra 6th bug] so after some move their will be bugs in same square. PAGE #

4 OR We have consider 6 shaded squares. Let we have a bug in the shaded Square. So in at most 4 moves, Bug will be in any shaded square again. And if we have a bug in the un-shaded square, in at most moves, bug will be in any shaded square again So, if we have total 6 bugs in these 8 squares, some of them will be in shaded square and some of them in un-shaded square. So after or 4 moves all the bugs need to be in shaded square. So their will exist atleast one move in which bugs will get into the same shaded square-. f(x) is a fifth degree polynomial. It is given that f(x) + in divisible by (x ) and f(x) is divisible by (x+). Find f(x). Sol. Let f(x) = k (x )(x ) (x ) f(x) = k (x ) (x ) (x ) k (x ) (x ) (x ) = k (x ) (x ) (x + ) k (x ( + )x + ) (x x + x ) = k (x ( + )x + ) (x + x + x + ) comparing cofficient of x k = k = k..(i) comparing cofficient of x 4 k k k = k k k + 6 = 0..(ii) comparing cofficient of x k + k + k + k = k k k + k = 0..(iii) comparing cofficient of x k k k k = k k k + k = + + = 0..(iv) comparing cofficient of x k + k + k = k k + k = 0..(v) comparing constant term : k = k k + k = k ( + ) =..(vi) (v) (iii) = = 0 _ = = 0..(vii) PAGE # 4 4

5 (ii) + (vii) + = = 0 ( + ) = 6 + = and ( + ) = ( ) put in (iv) () ( ) ( + ) = ( k ) = from (vi) 8 + k 6 = k 6 = 6 k = 8 + = 6 8 = now put + = and ( + ) = ( ) in (v) + + ( ) = 0 = = = 8 f(x) = k(x ) (x ) (x ) 8 = [x ( )x + ] (x ) 8 = x 8 9x 8 [ ] (x ) = ( 9x 7x 4)(x x 4 x ) 4 = 4 ( 9x + 7x 4 7x + 9x 7x 4 + 8x 8x + 7x 4x + 7x 7x + 4 4) = 4 [ 9x + 0x 4x] = 8 [ x + 0x x] f(x) + = k (x ) put x = f() + = 0 f() = Verification : f(x) = 8 [ x + 0x x] PAGE #

6 put x = 8 RHS = [ + 0 ] = 8 8 = ( ) f(x) + is divisible by (x - ) f(x) + = (x - ) Q (x) OR f (x) = (x - ) Q (x) + (x - ) Q (x) = (x - ) [(Q (x) + Q (x)(x )] so we can say f (x) is a multiple of (x ) f(x) is divisible by (x + ) f(x) = (x + ) Q (x) f (x) = (x + ) Q (x) + (x + ) Q (x) = (x + ) [(Q (x) + Q (x)(x +)] so we can say f (x) is a multiple of (x + ) f (x) = (x ) f (x) = (x 4 x + ) f(x) = ( x x + x) + C As f() = and f( ) = this gives C = 0 & = 8 f(x) = x + 8 x x ABC and DBC are two equilateral triangles on the same base BC.A point P is taken on the circle with centre D, radius BD. Show that PA, PB, PC are the sides of a right triangle. Sol. Let DBP PAGE # 6 6

7 r sin0º PC PB sin(60 ) sin(90 ) PB = r cos PC = r sin (60+ ) PC = r ( cos sin ) cos(0 + ) = r 4r cos AP.r. r cos r 4r cos AP sin(0 + ) = 4r cos AP = r +4r cos + 4r cossin(0 + ) = r [+4cos + 4cos (sin0cos + cos0ºsin) cos sin = r [+4cos + 4cos AP = [ + 4 cos + cos + AP = [ + 6 cos + Now, PB + PC sin cos] 4r cos + r ( cos + sin) sin cos] = r [4cos + cos + sin + sincos] = r [ + cos + 4cos + sin cos] AP = [ + 6 cos + = AP sin cos] 7. a,b,c are real numbers such that a + b + c = 0 and a + b + c =. Prove that a b c When does 4 the equality hold? Sol. (a b) 0 a + b ab a + b + ab ab...() a + b + c = 0 c = a b a + b + c = a + b + ( a b) = a + b + a + b + ab = a + b + ab =...() PAGE # 7 7

8 a + b + ab ab ab ; 6 ab ab 6 a b c = a b (a + b + ab) a b ab a b ab a b ab 4 ; a b c 4 Equality holds when a = b = and C = a b 6 C then a b c = 6 6 = 4 PAGE # 8 8

Worksheet 1- of 17. Preparatory Program - AMTI-NMTC Final Primary Level (Std V-VI)

Worksheet 1- of 17. Preparatory Program - AMTI-NMTC Final Primary Level (Std V-VI) Preparatory Program - AMTI-NMTC Final Primary Level (Std V-VI) Worksheet 1- of 17 Note - Elegant and novel solution will get extra Credits Diagrams and explanation should be given wherever necessary. Rough