HALF YEARLY EXAMINATION Class-10 - Mathematics - Solution

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1 . Let the required roots be ad. So, k k =. Smallest prime umber = Smallest composite umber = 4 So, required HF =. Zero of the polyomial 4x 8x : 4x 8x 0 4x (x + ) = 0 x = 0 or 4. Sice, a7 a 6d 4 = a + 6 ( 4) 4 = a 4 a = 8 5. cos = 0 (because cos 90 = 0) 6. sice, B PQR x y z, ar( B) B : 9 ar( PQR) PQ 9 B xy z So, LM (,B) = S x y z SETION- (Product of roots) SETION-B Sice, a S S a 4 a 4 So, a a b c 9. for ifiite solutio : a b c or so, k = k k k k k = k k = k if k k if k 0. Let the required itegers be x ad (x + )..T.Q. : x x 54 HLF YERLY EXMINTION lass-0 - Mathematics - Solutio (commo base havig largest power) x x 4x 4 54 Half Yearly Exam - 08 lass-0 (Sol.)

2 x 4x 4 54 x x 57 x x 55 0 x 7x 5x 55 0 x (x + 7) 5 (x + 7) = 0 (x 5) (x + 7) = 0 So, x = 5 or 7 But x is a positive iteger here So, required odd itegers are : 5 ad Sice, ad are the zeros of So, + =, = x x Now, the quadratic polyomial whose zeros are ( + ) ad ( + ) = x ( )x ( )( ) = x ( 4)x 4 = = x ( 4)x 4 x 6x. Give : BD is a rhombus ad diagoal ad BD meet at O R.T.P : Proof: B B D D BD Sice, diagoals of rhombus bisect each other at right agle. Now, i right agled OB : B B O OB (Pythagoras Theorem) BD BD B 4 4 O 4B BD B B B B BD B B D D BD (ll sides of rhombus are equal) Hece, proved D. Give : B ~ DEF R.T.P. : ar B B B ar DEF DE EF DF SETION- D ostructio: Draw L B ad DM EF Proof : arb B L B L ardef EF DM EFDM B.. () L E M F Now, i s BL ad DEM : B = E (.P.S.T.) LB = DME (each 90 ) Half Yearly Exam - 08 lass-0 (Sol.)

3 So, ad BL DEM ( B BL L () DE EM DM B B (.P.S.T.) () EF DE from equatio (), () ad (): Sice, So, 4. L.H.S. ar B B B B ar DEF DE DE DE B B (.P.S.T.) DE EF DF ar B B B ar DEF DE EF cos si ta cot DF cos si si cos cos si cos si cos si si cos cos cos si cos si si Hece Proved cos si cos si cos si cos si cos si = cos + si = R.H.S. Hece, Proved 5. Here, B represet a buildig of height 7 m ad E represet a tower of height h m. B = D = 45 (. I..) B B Now, i B : ta45 B B B = B = 7 m ED ED ad, i ED : ta60 D B (D = B) ED ED 7 m 7 So, height of the tower is = ED + D = ED + B h = m (D = B) 7cm Buildig E D able Tower 6. Give equatio are: x y 0 a b ax by a b 0 Half Yearly Exam - 08 lass-0 (Sol.)

4 x y a b a b b a b a a b b a x y a b b a a b b a b a ab bx ay ab a b a b a b If, b bx x b a b ab a b a b bx = ab x = a ay ab ad, if ay = ab y = b a b a b 7. Usig, sum of zeros (a b) + a + (a + b) = ( ) a = a = Now, usig product of zeros (a b) a (a + b) = y a a b a b a b a a(a b ) ( b ) (puttig a = ) b b b So, a =, b 8. Here, a b, B = (ac + bd), c d for equal roots : D = 0 B 4 0 (ac bd) 4 a b c d 0 4 a c b d abcd 4 a c a d b c b d 0 4a c 4b d 8abcd 4a c 4a d 4b c 4b d 0 4a d 4b c 8abcd 0 a d b c abcd 0 ad bc 0 ad = bc a c Hece, Proved b d 9. Let us assume to cotrary that : 5 be a ratioal umber. So, 5 r (where r is a ratioal umber) r 5 4 Half Yearly Exam - 08 lass-0 (Sol.)

5 r 5 (squarig both sides) r 5 5 r 5 r r 5 5 r r r Sice, r represet a ratioal umber, so will also represet a ratioal. r 5 will also be a ratioal but, this cotradicts our assumptio because 5 is irratioal. Thus, 5 is a irratioal umber. 0. Give : a4 a8 4 a + d + a + 7d = 4 a + 0d + 4 a + 5d =. () ad, a6 a0 44 a + 5d + a + 9d = 44 a + 4d = 44 a + 7d =.. () Now, subtractig equatio () from (): d = 5 a 5d a 7d d 0 From equatio (): a = a + 5 = a = So, the required.p. is :, 8,,... Give: I PQR, bisectors of POQ, QOR ad POR meet the sides PQ at, QR at B ad PR at. R.T.P.: Proof : P BQ R = Q BR P Usig iteral agle bisector theorem: I POQ : I OQR : I OPR : OP P.. () OQ Q OQ QB.. () OR BR OR R.. () OP P Now, multiplyig equatios (), () ad (): OP OQ OR P BQ R OQ OR OP Q BR P Q P B O R P BQ R P BQ R = Q BR P Hece Proved Q BR P 5 Half Yearly Exam - 08 lass-0 (Sol.)

6 . L.H.S. : si cosec cos sec = = si cosec sicosec cos sec cos sec si cos si cosec cos sec sec cos ec (fter rearragemet) = = 7 + ta cot = R.H.S. Hece Proved ta + + cot (usig si cos, si cos ec cos sec sec ta cos ec cot ) SETION -D. Let the speed of the boat be x km/h ad speed of the stream be y km/h So, speed i upstream = (x y) km/h Speed i dowstream = (x + y) km/h.t.q. : 6 (x y) x y 7. () x y x y 9. () Let x y p ad x y q p + 6q = 7.. () 40p + 48q = 9.. (4) Multiplyig equatio () by 5 ad equatio (4) by 4 60p 80q 5 60p 9q 6 q (subtractig equatio (4) from equatio ()) q Now, from equatio () : p 6 7 p + = 7 p = 4 4 p 8 So, ad, x y 8 x y x y = 8 (5) x + y = (6) solvig equatios (5) ad (6) : x = 0 x = y = 8 y = Thus, the speed of boat is 0 km/h ad the speed of stream is km/h. Half Yearly Exam - 08 lass-0 (Sol.)

7 4. Usig T si cos L.H.S. T 5 si cos si 5 cos 5 T T si cos R.H.S T5 T7 si cos ) (si cos ) T (si cos ) si si cos cos si si cos cos si cos (si cos ) si si cos cos si cos si cos cos si si cos si cos si cos si cos si.cos si cos Sice, L.H.S. = R.H.S. Hece, Proved 5. Give : B, right agled at. BL ad M are medias draw o sides ad B respectively. 4 BL M 5 B R.T.P : 5 5 si ( si ) cos ( cos ) (si cos ) 5 5 si cos cos si (si cos ) si cos (si cos ) (si cos ) Proof : I BL, BL B L B B 4 B B ad, i M, M M 4 B Now, BL M B 4 4 B B 4 L M B 5 5 B B 4 4 4(BL M ) 5B Hece, Proved 6. If there are ( + ) term i a.p., the odd term of the.p.: a, a, a 5,..., a (up to ( + ) terms) Eve terms of the.p: a, a 4, a 6,..., a (up to terms) Now, sum of odd terms = a a a 5... a S a a. () odd ad, sum of eve terms = a a4 a 6... a Seve a a... () Now, a a S odd Seve a a 7 Half Yearly Exam - 08 lass-0 (Sol.)

8 a a { }d a d a d a d a d Hece, Proved 7. Here, B represet a poit o the plae ad represet a poit 0 m above it. E represet a tower of height h. Now, i DE, E DE = (h 0)m DE h0 ta45 = h 0 D D D h h ad, i EB, ta60 B D D = (h 0) m 0m 45 D h h h0 B 60 h 0 h h h 0 h 0 h 0 h 5 0 h So, h =.65 m 8. Solutio table for : x y + = 0 (i) for x = 0 0 y + = 0 y = y = (ii) for x = y + = 0 y = 0 (iii) for x y + = 0 y + = 0 y = x 0 - y 0 Solutio Table for : x y 0 (i) for x = + y = 0 y = 5 (ii) for x = y = 0 y = 6 (iii) for x = + y = 0 y = 9 5 y y Half Yearly Exam - 08 lass-0 (Sol.)

9 x 0 y Graphical Represetatio : 9. Sice, am a (m )d a m a ( )d..() m.() (m )d ( )d m md d d d (m )d From (): 9 m m (m ) m a (m ) [Subtractig eq. () from ()] d m m (m ) m m a m m a m Half Yearly Exam - 08 lass-0 (Sol.)

10 Now, S a (m )d m m m (m ) m m m m m S m (m ) 0. x a x b a b x b x a b a x a x b Let y x b x a y So, a b y y ab y a b y ab aby ab a y b y aby a y b y ab 0 ay(by a) b(by a) 0 (ay b)(by a) 0 Now, b y a or a b x a b ax a bx b x b a ax bx a b (a b)x (a b)(a b) x a b d, if x a a bx ab ax ab x b b bx ab ax ab (b a)x ab ab 0 x 0 0 Half Yearly Exam - 08 lass-0 (Sol.)

[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms.

[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms. [ 11 ] 1 1.1 Polyomial Fuctios 1 Algebra Ay fuctio f ( x) ax a1x... a1x a0 is a polyomial fuctio if ai ( i 0,1,,,..., ) is a costat which belogs to the set of real umbers ad the idices,, 1,...,1 are atural