# % # & & ( ) +,. /! , ), %, 0 ) 1,2 / % ) (,2 / 3 % ) 7,2 / # % & (

Transcript

1

2 ! # % #&& () +,. /!,), %,0) 1,/ # % %) (,/ & ( %456 0 ) 7,/ )!

3 SET-1, 8 9 Choose the correct option :, ; <,= If roots of are equal then which of the following is Correct? >: ;,= If then value of will be which of the following? (Integers)? (fractional No) ) (Irrational ) ( (, (non real) : 0(Α 0(#4ΒΧ Χ, (Α 4,= If the perimeter and area of circle are equal then radius of circle will be? >4( units) 4(Pie units) 4(4 units)!4(7 units) : >ΕΕ ΕΕΦ Γ,= Median of,, 4, 5, 6 is? : (Α ΧΧΗ6Ι(,,= Longest chord of circle is? 9(arc) ϑ(diameter) 9(cord),(None of these)

4 Φ: X- )# %,,=!: Co-ordinate of X-axis is? ( x, 0) (0, y ) (0,0) ( x, y ),= The value of 1 > Κ: ) ),= The sum of Probabilities of events will be? Λ > Μ: 09 ϑνο,= The standard form of one variable linear equation is? :,= If then will be? &, 8 Λ Λ Fill up the blank with correct answers : - : Ν 4 Ι#4Β) Ι)::::::::::::) Χ Χ,,=

5 The ratio of areas of two similar triangles are equal to the ratio of the of corresponding sides? >: Ε>ΕΕ ΕΓ::::::::::::,= The median of 1,,, 4, 5 is : sin 50 + cos 40 = cos... : tan(90 θ) =... : (Α % :::::::::::::::::::: Χ? %,/ The tangent of circle touches at..point. Φ: A.p(? Γ<ΕΠ<::::::::::::::::::::,,= Common defference of A.p. are +ve, -ve or!: )::::::::::::::::::::::,,/ The Graph of linear equation is.. Κ: b 4ac:::::::::::::::::::::::,Ι,/ b 4ac is clled Μ: (, Ι,,:::::::::::::::::::::::::::,,/ The equations which has no any solution is called. >: )Γ;Α1Θ#4Β::::::::::::::::::::::::::::::::,,/ The total surface area of hemisphere is >: Ρ Χ, Σ Ι? Τ0( Λ,/ > Find quadratic polynomial which roots are Τ and Λ: a b c 1 1 >>:, ) 1 Χ 0Υ )! a b c,) / > a b c 1 1 Comparing, a b and 1 the linear pair of equations and c! are consistent or inconsistent. >: ς Γ<Β>,/ Ρ Ν / > Product of two consecutive positive integers is 40. Write it in the form of quadratic Equation.

6 > : ; Χ? P(ΦΕΛΦ) / > Find the distance of point P(ΦΕΛΦ) from the origin. >: Ω (Prove that) θ tanθ 1 tan 1 = sec + tan θ >Φ: Φ) 7Η ( Ι0Ξ & ( Γ / > Find HFC of 6 and 54 from Euclids division algorithm. >!: <) 64 cm,ψβ Γ Χ, Ο Γ Α1Θ#4Β / > Two cubes of each volume 64 cm added adjoining edges to form cuboid. Find surface area of cubiod. >Κ: Χ 94PA ) PB(Α % 0Υ, & APO=0 AOB=? In the given figure PA and PB are tangents of the circle and APO=0 θ > >Μ: a ( a, ) ) (, 4) Χ9 Κ,/ > Find the value of a if distance between ( a, ) and (, 4) is 8 units. : α ) β Ρ Χ, ( ) f x = ax + bx + c f ( x) ax bx c = + +?, α + β / > If α and β is Zeroes of f ( x) = ax + bx + c. find α + β : : >ΕΕ!ΕΜΕΕΓ / > Find mean of, 5, 7, 9, 10, 15. >: Ω + 50),/ Prove that + 5 is an irrational number. : Ρ x 7x + = 0 / Find the roots of the quadratic equation x 7x + = 0 : 0A.P. a = 5, d =, a n = 50 n 0( S n /

7 Find and A.P. n and S n if a = 5, d =, a n = 50 : 7 Χ? ( 5, ),( 6, 4) ) ( 7, ) ΡΧ, %1,= Are the points ( 5, ),( 6, 4) and ( 7, ) vertices of isoceles? Φ: 4 Ι Ι)BC, CA) ABς%D, E, F Γ Χ?, ar( DEF) ar( ABC) / Σ D, E, F are mid points of the sides BC, CA and AB respectively of ABC Determine ar( DEF) ar( ABC).!: Ω (prove that) 1 secθ tanθ secθ tanθ = + Κ: tan A = cot( A 18 ) Ι,Υ A?,/ A / If tan A = cot( A 18 ) where A is an acute angle Find the value of A. Μ: ΧΓ) Γς%>Φ:Κ) >!:Μ, Χ, / If mean and median of symmetric distribution is 6.8 and 7.9 then find mode. : 0.6 Νϑ7 / Express 0.6 as rational number in simplest form : 0( ϑ<ς 10,/Σ / Sum of number and its reciprocal is 10 Find the number. >: Χ Χ ΧΧ,Σ / Calculate the mode of the given frequency distribution. ()? C.I

8 (Ζ & No. of students : Β ( Γ, (Solve graphically) :- 5 x y = 1 x + y = 8 : )Υ Γ)0Η6Ι,),) Η6Ι, Η6 % Ι[,) &0 Χ,/Η6 Χ? Ι,ΥΗ6 % Ι[,Κ,Ε Η6 Υ9 / A tree break due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 0 with ground. The distance between the Foot of the tree to the point where the top touches the ground is 8 m. find the height of the tree, : Ω 4 Ι()? Ι)(Χ Χ,,/ In an right triangle, the square of hypotenuse is equal to the sum of the squares of other two sides. Φ: 0Η6, Ρ #4Β / The lengths of minute hand of a clock is 14 cm. Find the area in 5 minutes covered by minute hand.!: AB=1 cm.9) :5 ) Χ/ Draw a line AB=1 cm. and divide it in the ratio :5.

9 8 (ANSWERS) ( cotθ Χ? 16.? ( (7 19. ) 0. #

10 >: Β = α + β = = Β = αβ = ( + 1)( 1) ( ) (1) 1 = = = Ρ Χ, x x x ( α + β) x + αβ ( ) x + x + >>: x y = 8 a1 x b1 y c = 4x 6y = 9 ax b y c 0 Comparing we get 8 = = 4 9 a1 b1 c1 = a b c 0 Ψ),/ + + = >: ς 0 x) x + 1,/ Ε ( 1) 40 x x + = x x + x = 40 + x 40 = 0 > : Χ? ( 0,0) p( 6, 6) ( 6 0) ( 6 0) = + ( D ( x ) x1 ) ( y y1) = = + = 7 = 6 4

11 >: L.H.S = tan θ 1 tanθ 1 = θ θ θ (tanθ 1) (tan 1)(tan + tan 1+ 1 ) { a b = ( a b)( a + ab + b ) = θ + + θ (tan 1 tan ) = sec θ + tanθ = R.H.S L.H.S=R.H.S >Φ: 54)6(6 7Η ( Ι ( Γ 4 6 = )54(4 54 = = )1( 1 >!: 0 Ι = a cm. v = a 64 = a a = 64 = 4 cm. Ε0 Χ,/ 9Η = 4cm + 4cm = 8 cm. Α1Θ#4 = ( lb + bh + lh) cm ( ) ( cm 16cm cm) 80cm 160cm >Κ:,), APO = 0 = + + = + + = = OA APΓ OAP = 90 AOP = 180 ( ) = = 60 AOB = 60 = 10

12 >Μ:, (, ), (, 4) A a B = 84 A) BΧ9 ( a) (4 ) = + 8 ( a) = + { D ( x x ) ( y y ) = ( a) 4 = ( a) = 60 ( a) = a 60 = 60 a ± = a = ( ± 60) : 9Υ α 0( β Ρ Χ, ax + bx + c?,/ b c α + β =, a αβ = a α + β = α + β β ( ) b c =. a a b c b ac = = a a a : b ac a α + β =,5,7,9,10,15 ;Γ ( ) x =

13 = x = 8 48 = = 8 6 >: + 50,/ 5 : + = a b 5 a b = a 5 = b,,? 50),/,Υ ( Γ,/) + 5 0,/ 0() )),,/ Ρ x x + = 7 0 a =, b = 7, c = D b ac = 4 = ( 7) 4 = 49 4 = 5 b ± D x = a ( 7) ± 5 7 ± 5 = = =, =, =, =, : A. P,), a = 5, d =, a n = 50 a = a + ( n 1) d n 50 = 5 + ( n 1) = ( n 1) 45 ( n 1) =

14 45 = ( n 1) n 1 = 15 n = = 16 n = 16 n Sn = { a + ( n 1) d} 16 { 5 + (16 1) } = 8( ) = 8 55 = 440 n = 16, S = 440. : n 4 Ι %1) AB = (5 6) + ( + 4) = + = (6 7) ( 4 ) BC = = + = AB BC = Φ: ABC ΡΧ,,/ 9Υ D) Eς BC ) AC Γ Χ?,/ DE AB DE FA:::::::::::::::::::::: (i) D) F ς BC ) ABΓ Χ?,/ DF AE ::::::::::::::::::::::::(ii) (i) ) (ii) AFDE ) BDEF 0? 9 Ι,/ DF AC EDF = A, DEF = B DEFABC 1 AB ar DEF DB = = = = ar ABC AB AB =1:4 ( ) ( ) 1 1 ( ) ( ) 4

15 !: 1 L. H. S = secθ tanθ 1 secθ + tanθ = secθ tanθ secθ + tanθ secθ + tanθ secθ + tanθ = = sec θ tan θ 1 = secθ + tan θ = R. H. S L. H. S = R. H. S. θ θ ( sec tan = 1) Κ:,), tan A = cot( A 180 )?,/ cot(90 A) = cot( A 18 ) 90 A = A 18 Μ: { tan θ = cot(90 θ ) = A + A 108 = A 108 A = = 6 A = 6,), ;Γ = 6.8 ; Γ 7.9 = Χ, = ;Γ ;Γ = = = 0.1 Χ, 0.1 = :

16 x = 0.6 x = ( i) 10x = ( ii) ( ii) ( i) : >: 9x = 6 6 x = = 9 x = x,/ ϑ<ς = 1 x % Ε 10 x + = 1 10 x + = x x( x ) 1( x ) = 0 x x x + = x x x + = ( x )(x 1) = 0 x = 0 x = ) 1 : 0( x 1 = 0 x = 1 x = 1 ()? (Ζ & f Λ> > >Λ Λ! Λ! Φ!ΛΚ Φ ΚΛ Φ,Υ( ΓΧ Χ!Ι Λ(,/) Χ,( Λ,/ l = 40, f = 7, f 1 =, f = 6, i =

17 : f f 1 0 f0 f 1 f1 7 = = = = 4 0 Χ, ( M ) = l + M = 4 0 x y 1 = y = x 1 x y Λ > 0( x y 8 + = y 8 x = x y Κ > Β,, 0Υ(, ) 0,/) x, y : Η6 Υ9 AB, P C Χ? [,/ AP PC x = = & PCB = 0 Χ,/ BC 8 m. PBC = = =

18 cos0 = 8 = x BC PC x = 8 16 x = m. 16 AP = m. Β tan 0 PC = BC 1 PB = 8 8 PB = AB = AP = PB = + = 4 8 = m. = m = 8 m. Η6 Υ9 = 8 m. : ABC 0, Ι ABC = 90 )& B,/, Ω, AC = AB + BC Χ(Λ BD AC 9/ ΛADB 0(ABC ADB = ABC = 90 A = A ADBABC AA

19 AD AB = AB AC AD AC = AB i....( ) BDC ABC CD BC = BC AC CD AC = BC ii....( ) ( i) ) ( ii) ΙΗ6 AD. AC + CD. AC = AB + BC AC( AD + CD) = AB + BC Ε AC. AC = AB + BC ( AD + CD = AC) AC = AB + BC, Ω &/ Φ: Χ,/ 1 5 ]]]] = ]]]] ]]]]5 6 = 0 ]]]] ;Χ( r) 14 cm. = Ρ # τ r θ = 60!: ( ) = cm = cm 616 = cm = 05. cm. i AB 1 cm. = 9/ ( ii) BAX = ABY 9/

20 ( iii) AB + 5 = 8Χ Χ ( A1... A8 ) ( 7 / 9 BY 8Χ Χ ( B1, B... B8 ) ( 7 / ( iv) A B 5 /, AB P Χ?,/ AP 4.5 = cm 0( PB = 7.5 cm.

21 ! # % #&& () +,. /!,), %,0) 1,/ # % %) (,/ & ( %456 0 ) 7,/ )!

22 +,% ).,/ #,// Λ Λ, 8 0 &1 :π,/ 01 0!1) 01 ) ) 01 π 01!0!1!01! 01 >: ; ), 4! 010!1!0101Μ : Ρ Χ,,) 56?,/ 47 (,) ,%8 10!10,8, 1010%8, 1010%8 1 Χ,, 5?α 8 β, ,/ α β 7 (, 5α 8 β 1 α 56 1 β 01,0!101/01 ) 4 Ι&9;& :&9 59 9, &&9& :& ! 0!1Φ 01 01Μ

23 Φ: ) Γ,,/!! ( 010!1>0101, Χ? 08, 1) 0681 ( ΗΓΛ Χ?,Λ,,, ;08, 10681! 010)8%10!10)81010)8 1010,8 1 <.0=/ /,&1Χ Χ,,/.0=/ /,&1 01.&0!1,.&01,&01& =(ΑΑ Γ0(ϑ),,/! 01π 0!1 01π 01 /&Υ9 ( )Γ,Λ! 0160!101)01< Ψ:::::::::::::::::,,,/!,>/ 5! 5:::::::::::::::::,/!,>/ 5! 5 %)ΩΛ(ΑΑ :::::::::::::::::::::::::::::,,/

24 9Χ Χ (ς:::::::::::::::::::::::,ι,/ ( ) /,. % / ::::::::::::::::::::,/ /,. % / 6(Α Ω,&.::::::::::::::::::::,,/.?!: :)) )#::::::::::::::::::,,/ Α :) < & & ) (?,.:::::::::::::::::::::,/ &!.: Μ: Χ, 4 Ι,,/:::::::::::::::::::::::::: & >: ) ) )Β:::::::::::::::::::::::::,,/! Ρ Χ,?Λ>0(Λ, Χ, Σ / > # ( Β,,) Ρ Β) Βς_ 0(_, Ρ / > _ Χ (# % 0 5 (:)) % 5(:,) Ρ,/ >

25 # 5 (:)% 5(: Ρ,6 5%:/ΑΑ / > #,6 5%:/ ) 4 Ι&9&:6 8&9:6 89:6 ::::::::::9 / &9&:6 8&9:6 89:6 9 6 Χ? 0 8, 1) 0, 8 1 ( ΗΓ Χ? / > #,, ; 0 8, 10, 8 1,), &9 +# &#4Βς6 ),/ +#:), 9Σ / > Α &9 +# 6 (+#:) 9 >Κ: 4 Ι? Ι 0%8,)180,81) 0/8, 1,/ > # 0%8,)180,810/8, 1 = &:%Χ &0( & / > &:%Χ & & %/ ) / 5 %/ / 5. 6/ / / > # ) / 5 %/ / 5. 6/ / :& %8<8%8<8,,,,,,,,,,,,,,,,,!Κ, 4 & %8<8%8<,,,,,,,,,,,,< % Ω Λ %0),/ Λ %!

26 %%Η ( ΓΚΕΜ) > / # #<8= )!(# Β % )) ς>), &,, / Ε #! /) (!)! %) 0,0,/ # %,) 5 :/ %6)Α Φ&Φ9:ΦΦ Ω &: ) 9: % Φ&Φ9:ΦΦ &:? 9: %0 4 Ι#4Β Ι 0%8180,8%1) 0<861,/ # 0%8180,8%10<861 %<.0=/ /,1 4 ) ( Γ / #.0=/ /,1!(

27 %=0Χ ΒΙ,/ [ Χ Χ Ο, / & #!! (! / Ω , 1 : 1 Sin Q % 1+ SinQ ( ΕΕΕΚΕ>>Ε>ΜΕ Ε>( &>ΜΓ> 5( /% # 5(&! /8)8<8 8 =8 8% 8( = ) Γ 0# 1 ()? /,) ),)/ )/,)) )),6/ 6/,6) 6),/ /,) Χ Χ > Κ Φ Φ > :Β ( Γ, /0.!( 1 ) 5(:6,(: 0 4 %) (0),Ι 4 (0 )Ω )Γ,/ Υ9)), ΑΑ Θ #4Β / & (! %)!( ()) (

28 :0 Ι Η6 [ ) Γ;Χ,Ι,ΙΧ?Φ,Ι,/ Υ9 /!/.Γ Φ # 6 4 Ι0 4 Ι0Χ Χ,), 4 ΙΝ,/ Ω / 8!:0 4 Ι&9Χ( Ι9: 8&9:% 4 ΙΝ 4 ΙΧ( Ι Ι0 4 Ι Ι))Χ%,/ &99: 8&9:% )Χ%

29 8 8! 0!1 0!1%010!1)016010!1<01=01/010>:( () ): n π : : :?Φ:(Α Ω!:? Κ. : : 50,1 ΗΜ:Ν >:)?Β:α 5 β :, 50,)1:,6?Β:α β :, 0,)1:/ Ρ Χ,,0α 5 β 1 5α β,0, >>:Β: 1 4 Β: 1 4 Ρ,0Β1 5Β:/, Χ:/ 4x x :/, 5:/ % 5 (5) 5! (5 :/ % 5(: 5! (5 :/

30 ,) Ρ,ΙΧ a a = b b 1 1 P = 1 P = 6 > :,6 5%:/ : 8!:,68:% :!,:0,61, %:%6, : >/( () ),/ >: 4 Ι&9 &9 :06 1:%6 9 :06 1:%6 & : 06 1 : &9 5 9 :%6 5%6 : : & & : &9 59 ) &90,/ 9:=/ / 6 0 Χ? 0 8, 1) 0, ( ΗΓ Χ?

31 + ( ) =, =, = (0,0), &9 +# ( ) ( ) ar ABC BC = ar DEF EF 64cm BC = = 11cm 15.4 ( cm ) BC ( ) 64cm X 15.4 cm = 11cm ( ) 64cm X 15.4 cm BC = 11cm 8X15.4 = cm = 8X1.4cm 11 : < 4 Ι 0%8,)180,81) 0/8, 1? 0 8(1 x x + x + x 1 = ( ) y y + y + y 1 = = = 6 = = = = 1 ( )? :0 81 =& = 4.&: 1 Cos A : 1 4 : :

32 .&: 7 4 Cos A &: 4 Sin A = 7 : 7 4 %/ ) / 5 %/ /,. 6/ / : ( 1) + : 1 + = 4 4 %& %8<8%8,,,,,,,,,,,,,,,,,,,, :%8:<,%:)8 :<8: :50,1 <:%50,1 ) <,%:0,1 ) ):0,1 ) 0,1: 75 5 :),:) :)5:6 :6 % Λ : p 0, ),/ q ),Υ ( Γ,/

33 ) ) ) ),,/ ), 0),/ %%<: =:% % ):) ) #0<8=8 )1: %, :,: : : / %)x 5x + p = 0 :%8!:,)8: :!,:0,)1, % : ), 9Υ,,/ :/ ), :/ : ) : 5 1 %6,Φ&Φ9:ΦΦ

34 OA OD : OC OB &Φ Φ9 &Φ : Φ90;,1 & OA OD : OC OB 0,1 &Φ Φ9 &: 0( : 9 % 4 Ι 0%8180,8%10(0<861,/ 1 4 Ι#4Β: 1( y y ) + ( y y ) + ( y y ) x x x ( 6) + ( 4)(6 4) + 8(4 ) : [ ] 1 = [ ( ) ) ] 1 1 = [ ) ] = 9 :)( Κ: &90,/

35 9:=/ / :01 &:=/ /,.: AB AC : BC AC 01.0=/ /,1: BC AC 01 01) 01.0=/ /,1: %=ΙΧ0Χ ΒΙ, [:8 8%8,/ Ο :Φ [Ο, : 4 6 : 1 Sin q = = Cos q Cos q / L. H. S. ( sec q tan q) ( ) ( ) 1 Sin q 1 Sin q 1 Sin q = = = Cos q Cos q 1 Sin q ( ) ( ) ( ) ( ) 1 Sin q 1 Sin q 1 Sin q = = = R. H. S. 1+ Sin q 1 Sin q 1+ Cos q

36 Ε.:. Γ:) ):/Ι5)5<5 5 =5 5% 5(5 = Μ ) =:<)5 5( ),<): 5( /: 5( 5(:/ ()? 01 Χ Χ 0#1 9Χ Χ 01 /,) ),)/ % ) )/,)) < % )),6/ 6 = 6/,6) 6 ) 6),/ % < /,) %/ :%/

37 N 0 : :)8,9Χ Χ Μ)? ), Ι(Λ)? ΛΦ,/ ) Γ(ΛΦ,/ :))8:%8:68:) N c. f Γ = l + i f 15 1 = = = = = : 5(:6, (: (:6, (:, (:x- > ( Φ > / (, /

38 Β,, 0Υ0 Χ? >Ε>,/ ) : 8(: : Υ9:)

39 4 :%) Υ9:)),%) : = + = [ Υ9 1 + (.5 ) h r cm = cm = 156.5cm = 1.5cm ΑΑ Θ#=)Ω(ςΑΑ Θ# 65(ςΑ Θ# :>π 5π : π 0 51 : 7 %)0 %)5 )1: /)05 )1 : =) : ) ) &9, Ι Υ9,/

40 9 [,/9: 9 :0 5/1 &9 6/ / : AB BC : p b : h x : h x = 6 4 Ι&90( 4 Ι +#, &: 0( AB AC : DE DF

41 Ω, &9 +# Χ(Λ &9&9&: + &&: +0( & /, &0( +#/ 94 &: +0Χ(1 &: # &: 0,1 & +#0.&.1 &: +8 &+: #01, AB AC : DE DF AB AM : AC AN 0 +:&8 #:&1 ϑϑ9 : 90( : 01 : +0( : # 9: + & : # )Χ &90( +# &: 0,1 9: +8 : #0 1 &90( +#,0/

42 & +#0&&&1 <!:9Υ 4 Ι Ι0Υ% 0(,,/),,/ 0ΧΙ,/ 9 ΛΛΛΛ 019: 9/ 01 Χ? 9 Μ Χ/&9:% /& /&90 Χ/ 019 & ) 0?Χ,0 9Κ9/

43 01)0 5 ) ΧΗ6 Χ? % 89 80(9 ) 89Κ ) 99 8:9 9 8:9 9 %: 9 % 9 : 9 9 ),/ 019 % 0 Χ? 5 ) [ 0(9 ), Ι(9 %? 0 0Χ56 Η 9Γ α[,9/ 01Γ, Ι(&? 0 Χ56 Η9&&Γ α[,9/ &ΓΓ / &Γ9ΓΟ,)/ &9 &Γ9Γ

44 ! # % #&& () +,. /!,), %,0) 1,/ # % %) (,/ & ( %456 0 ) 7,/ )!

45 Λ Λ Λ, 8 9(Choose the correct answer) : Ρ,Ε Β,/ Sum of the roots of the quadratic equation will be (a) b c (b) b (c) a b c (d) a a >:,/ The roots of the equation will be a p (a) q a (b) ap (c) p aq ± (d) aq p : 0Α1Θ#4Β,,/ Total surface area of a sphere is (a) r sq unit (b) r sq unit (c) 4r sq unit (d) r sq unit : ; ),= Which of the following is a prime number? (a) 6 (b) 9 (c) 15 (d) 11 : Χβ Χ? (Α ) Γ % 0Υ9Ι,/ The maximum no. of tangents drawn from a external point to a circle is (a) (b) (c) 1 (d) 4 Φ: Χ? )A(,-4) ) B(4,-) ( ΗΓ Χ?, The mid point of the line-segment joining the points A(,-4) andb(4,-) is (a) (6,-6) (b) (-1,-) (c) (,-) (d) (-,)

46 !: ) (,, The probability of an impossible event is Κ: (a) zero (b) 1 (c) -1 (d) 1+ tan θ, 1+ tan θ is equal to (a) sec θ (b) cos ec θ (c) tan θ (d) cot θ Μ: 4cos ec θ 4cot θ Χ Χ, 4cos ec θ 4cot θ is equal to (a) 1 (b) 0 (c) (d) 4 : )χη:) Γ 0(?8),Ι, (a) (b) ()? (c) Γ (d), The difference of maximum value and minimum value of a data is called (a) Range (b) Class Inverval (c) Mid Point (d) None of these 7 &, 8 Ρ /(Fill up the blanks with correct answer) : (Χ, ::::::::::::::::::::::::::::::::::::::,,/ The polynomial of power two is called. >: (Α (, ::::::::::::::::::::::::::::::::::::::Χ Χ,/ Two circles are angruent if their.. are equal. : Χ? (-4, ) ::::::::::::::::::::::::::::::::::::::9 &%Η:/ Point (-4, ) lies in the..quadrant. : )# Χ? % ::::::::::::::::::::::::::::::::::::::,,/ The Co-ordinate of any point an y-axis is... : Χ? ::::::::::::::::::::::::::::::::::::::,/

47 Distance of P(x,-y) from the origin will be... Φ: sinθ = 1, ::::::::::::::::: sinθ = 1, then : 0 0!: sin 5 + cos 5 ::::::::::::::::,,/ 0 0 sin 5 + cos 5 is Κ: ΕΕΦΕ!ΕΜΓ::::::::::::::::,/ Μ: %8)8688= 0::::::::::::::::,/ is an number. >: 4 Ι, 4 Ι:::::::::::::::::::::::::::::::,,/ If two triangles are equiangulars then triangles will be. >: Ρ Χ, Σ Ι?Λ) Β>,/ Sum and product of zeros are - and find the quadratic polynomial. >>: x / If then find the value of x. >: 0 ΡΧ, 4 ΙABC B=90 0 Ω AC =AB In Isocels triangle ABC, B=90 0 then to prove that AC =AB > : Χ? Ψ(,) &(4,1)Χ9 Σ Ι0 Find the distance between the pair of points (,) and (4,1) >: 0Φcm 4 ( 4 Η 4 ΗΦ,/ Find the perimeter of sector whose radius is 6 cm and angle of sector is >Φ: ΕΚΕΕΚ::::::::::: Κ,/ Which term of A.P. is,8,1,18 is 8. >!: If 5 tan = 1 θ sin θ + cosθ / 5 tan θ = sin θ + cosθ 1

48 >Κ: SinA = CosB & A) B? Ω A + B = 90 0 If SinA = CosB&89 A + B = >Μ: 0)Ω(Α ΓΦcm, ϑ / If circumference of semi circle is 6 find its diameter. : 4 ΙΝ, 0 Χ?Γ / Write the condition for similarity of two triangles. : 0(Α #4Β) Γ, (Α 4 7,= If the Values of area and circumference of a circle are equal, what is the value of radius of the circle. >: Ω 0),/ % Prove that! : Ρ x x + 1 = 0 Find the roots of quadratic equation x x + 1 = 0 : 7Η ( Ι0Ξ & &>> / Using Eaclids division algorithm find H.C.F. of 15 and 5. : (ΞΧ, 4 4cm) 4 cm #4Β / The inner and outer radius of ring are cm and 4cm find its area. Φ: 4 Ι? Σ Ι 1(,-5) (-7, 4)) (10,-),/ Find centroid of triangle whose co-ordinates of vertices are (,-5) % % % % (-7, 4)(10,-). 4 x!:, + y = 8 6 Solve 4y = 5 x Κ: Φ Υ9(0 Γ(Γ [4m, 0 [8m, 9 / The shadow of 6m vertical pilar is 4m, just then the shadow of tower is 8m % % find height of tower.

49 Μ: Ω (Prove it) 1+ sinθ cosθ + = secθ cosθ 1+ sinθ : sin θ = cosθ sin θ + cosθ / If sin θ = cosθ then find sin θ + cosθ. : 0APλΕ 0( Μ, & / In an AP λ, and total terms be 9 then find first term. >: ; ΧΓ, / If mean of the following distribution is 50 find the value of ()? 01 Χ Χ (f) : Υ9) ϑ(0θχ Υ9) ϑ (0 7δ( Ι, 1Χ9Θδ1Θ#4Β / From a solid cylinder of height.4cm and diameter 1.4cm, a conical cavity of same height and same diameter is cut of. Find the surface area of the remaining solid. : Β ( Γ, Λ Solve from graphical method. x + y = x+5y = 1 : Ω Ν 4 Ι#4Β) Ι)( ) Χ Χ,,/ The ratio of area of two similar triangles is equal to the ratio of the squares of corresponding sides. Φ: )Γ 0 0(Μ & Χ? ) %?, Υ9 / The angle of elevations of the top of tower from two points an the ground at distance of 4m and 9m from the base of tower in the same straight line are complementary find height of tower. % % % % ) ) ) )

50 Answer ( 8 8 )of Set b c c d a c a a d a Ρ 4 0Υ Ρ () x + y Φ ) Ν >:?Β?Β Ρ Χ,?Β?Β >>: 9x 5 = 0 9x = 5 x 5 = 9 x = 5 5 = ± 9 >: ΡΧ, AB = BC & AC = AB + BC = AB + AB ( BC = AB) = AB = AC = AB

51 > : Χ? )Χ9 & Χ9 = ( 4 ) + ( 1 ) ( ) = = = 8 = >: 4 Η πrθ = r = = 1 cm + cm 7 = 1 cm cm = 18.5cm >Φ: a=, d=8 = 5, an =8, n=? 4 a n = a + (n - 1)d 8= + (n - 1)x5 8 - = (n - 1)x5 5 = (n - 1)x5 (n - 1) = 7 n = 8 >!: 5k tanθ = = 1 k AB BC AB = 5 K, BC = 1K AC + = AB + BC = ( 5K) (1K ) = 5K + 144K = 169K = 1K

52 >Κ: AB 5K sin θ = = = AC 1K BC 1K cos θ = = = AC 1K sin θ + cosθ = + = 1 1 sin A = cos B sin A = sin(90 B) A = 90 B A + B = 90 >Μ: )Ω(Α Γ πr + r = 6cm r( π + ) = 6cm r = 6cm = πr + r! r = 6cm 6 cm r = 6cm 7 r = r 7cm : 4 ΙΝΟ,, (i) Χ Χ, & = (ii) Ι0Υ0,),/ : (Α #4Β = πr ( (Α Γ = πr,(α #4Βε(Α Γ πr πr = r =

53 >:,, 0, r = ( s 0) s ; r ) sε) 7 1ΘΗ,/ a = b Ο,Ι,Υ a) b,),/ b = a ) b = a ( Ι,/ ( Ι /,Ι,Υ,/ ) ) 0 1ΘΗ>,/,Υ ( Γ Ο,)/) 0),/ : x x + 1 = 0 D = ( ) 4 1 = 8 8 = 0 b ± a D 0 ± ± 0 1 1, 1 4 : 7Η ( Ι

54 : (Ξ# ( R r ) = π = π 4 cm ( ) = π 16 9 cm ( ) Φ:? ( x, y),/ = 7πcm 4 Ι 1(, 5),( 7,4) ) ( 10, ),/ x1 + x + x x = y = y + y + 1 y x = y = x = y = x = y = 1? %(, 1),/ 4 x!: + y = 8 6 4y = 5 x Φ0( 4 x + 18y = y = 0 x # # 4y

55 68 y = = = 8 x 4 = 8 6 = x 4 x = 4 x = = x =, y = Κ: Υ9,/ ABC ~ DEF AB = DE BC EF 6 = h h = 6 8 h = 6 8 = θ θ θ θ 1+ sin cos Μ: # = + cos 1+ sin = ( 1+ sinθ ) + cos cosθ ( 1+ sinθ ) θ 1+ sin θ + sinθ + cos θ cosθ = ( 1+ sinθ ) 1+ sin θ + cos θ + sinθ = cosθ + sinθ = cosθ ( 1+ sinθ ) ( 1+ sinθ ) + sinθ = cosθ sinθ cosθ = ( 1+ sinθ ) ( 1+ sinθ ) 1 cos ( + sinθ ) θ ( 1+ sinθ ) =

56 = = secθ % cosθ

57 : sin θ = cosθ sin = cos θ θ AB tan θ = = = K ( let) BC AB = K BC = K AC AB + BC = ( ) = K + K = K + K = 4K = K sinθ = AB AC K K = = cosθ = BC AC K 1 = = K sin + cos θ θ :,Υ & n = s n ( a + l) 9 a 144 = ( + 8) 144 = a = + = + 8 = a a = 4

58 >: c. i f x fx fx x = N f = 96 + x fx = 70 x x = 96 + x ( 96 + x) = 70x x = 70x x 50x = x = x = = 4 0 x = 4 : ΧΑ1Θ#εΧΑ1Θ#Τ)Γ #Τ (ςα1θ# = π rh + πr + πrl = πr ( h + r + l) ( ) = cm 7 7 = cm 7 10 ( )

59 176 = 8.0 cm = cm = 17.6cm ( ) l + = h r = ( ) ( ) = = 6.5 =.5cm : y 1 x 5 = : 4 ΙABC) PQR,/, ABC ~ PQR < Ω, ar ar ( ABC) ( PQR) = AB PQ = BC QR = [ 4 Ι#4ΒΣ 0 CA RP

60 1;Χ() )α φ 1 ar ( ABC) = BC AM 1 ar ( PQR) = QR PN ar ar 1 BC AM ( ABC) BC AM = = ( PQR) 1 QR PN QR PN )Χ +(0(,) +, +.,% ( ) <Μ +(.,) AM = PN AB PQ 0( +.,% AB BC CA = = PQ QR RP ar( ABC) ar( PQR) = AB PQ AM PN ) AB AB = > PQ PQ AB = PQ Θ Ω,

61 ar( ABC) ar( PQR) = AB PQ = BC QR Φ: Υ9,/ /, & = CA RP ABC 0 tan(90 θ ) = AB BC tan(90 θ ) = 4 0 h h cotθ = 4 ABC tanθ = AB BD h tanθ = 9 cot θ tan θ = h h 4 9 h 1= h = 6 6 h = 6 m = 6m!: 4 (Α 9/ (Α? 0 Ρ Ι / Ρ Ι Χ? (? (Α 9/,,(Α 0(+ Χ?,/ Χ? ) + % 9Σ

62 ! # % #&& () +,. /!,), %,0) 1,/ # % %) (,/ & ( %456 0 ) 7,/ 15 Minutes of extra time have been allotted for Candidates to read the Questions.

63 xf.kr (lsv&4) Lke; %?kavk 15 feuv MATH (SET-4) Time : Hrs 15 Min. Full Marks : 100 lgh mùkj pqus %& Choose the correct answer :-,d f}?kkr lehdj.k dk?kkr gksxka 1 (d) ([k) 0 (x) 1 (?k) buesa ls dksbz ugha The degree of a quadratic equation will be (a) (b) 0 (c) 1 (d) None of these - ;fn sin θ = cos θ rks θ dk eku gksxka 1 (d) 0 ([k) 45 (x) 60 (?k) 90 If sin θ = cos θthen the value of θwill be (a) 0 (b) 45 (c) 60 (d) 90 - pøh; prqhkzqt ds leeq[k dks.kksa dks ;ksx gksxka 1 (d) 60 ([k) 180 (x) 90 (?k) 0 The sum of either pair of the opposite angles of a cyclic quadrilateral will be (a) 60 (b) 180 (c) 90 (d) 0 4- fcunq (µ5] µ4) fdl prqfkkza'k esa flfkr gs \ 1 (d) izfke ([k) f}rkh; (x) r`rh; (?k) prqfkz In which quadrant does the point ( 5, 4) lie? (a) 1 st (b) nd (c) rd (d) 4 th rfkk BAC = ABCesa BCdks D fcunq rd c<+k;k x;k gs ftlls ACD =110 rks ABCdk eku gksxka 1 (d) 5 ([k) 57 (x) (?k) 1 In ABC, side BC is extended up to D such that ACD =110 and BAC = 57 then the value of ABCwill be (a) 5 (b) 57 (c) (d) 1 6- ;fn 6, 8, 9, xrfkk 1 dk ekè; 10 gks rks xdk eku gksxka 1 (d) ([k) % (x) (?k) ) If the mean of 6, 8, 9, x, 1 is 10 then the value of x will be,))

64 (a) (b) % (c) (d) ) 7-,d iklk dks mnkyus ij,d vhkkt; la[;k vkus dh izkfkfedrk gksxha 1 (d) 1 ([k) 1 (x) 1 4 (?k) buesa ls dksbz ugha In a throw of dice, the probability of getting a prime number will be (a) 1 (b) 1 (c) 1 4 (d) None of these 8- A.P. 40, 15,10, 5,...dk 10oka in gasa 1 In A.P. 40, 15,10, 5,...10 th term is... (d) 184 ([k) 186 (x) 18 (?k) ABCesa AB = ACrFkk < BAC = 50 rks < ABCdk eku gksxk 1 (d) 60 ([k) 65 (x) 70 (?k) 75 In ABC, AB = AC and < BAC = 50 then the value of < ABCwill be (a) 60 (b) 65 (c) 70 (d) 75 sin dk eku gksxka 1 cos1 (d) 0 ([k) µ1 (x) 1 (?k) buesa ls dksbz ugha The value of sin 59 will be cos1 (a) 0 (b) 1 (c) 1 (d) None of these fjdr LFkkuksa dh iwfrz djs %& Fill in the blanks :- 11- f}?kkr lehj.k ax + bx + c = 0ds ewy oklrfod vksj vlkeu gsa rks foospd 'kwu; ls gksxka 1 If the roots of the quadratic equation value of discriminant will be... than zero. ax + bx + c = 0is real and unequal then the 1- cosec θ 1dk eku gksrk gsa 1 The value of cosec θ 1 is... 1-?kukHk esa fdukjs gksrs gsaa 1 The number of edges in a cuboid is x v{k ij flfkr fdlh fcunq dk y fu;ked gksrk gsa 1 The ordinate of any point situated upon x-axis is izfke ik p fo"ke la[;kvksa dk ekè; gksxka 1 The mean of first five odd numbers will be...

65 16- fdlh vo';ahkkoh ;k fuf'pr?kvuk Edh izk;f;drk P(E) =... 1 The probability P(E) of an sure event or certainly E = fdlh ckg~; fcunq ls o`r ij vf/dre [khaph tkusokyh Li'kZ js[kkvksa dh la[;k gksrh gsa 1 The number of maximum tangents from an external point is a circle is fdlh laekrj Js.kh esa nks Øfed inksa dk cjkcj gksrk gsa 1 In a arithmetic progression... between two consecutive terms are equal. (, ) 19- nks fcunqvksa ( 1, 1 ) x y vksj x y ds chp dh nwjh gksxh The distance between two points ( x1, y1 ) and ( x, y) will be ifjes; la[;k esa gj ds cjkcj dhkh ugha gksrk gsa 1 Denominator of a rational number never equal to... vfr y?kq mùkjh; iz'u%& Very Short Questions 1- f}?kkr cgqin x + 8x + 15ds 'kwu;d Kkr dhft,a Find the zeroes of quadratic polynomial x + 8x k dk eku fudkysa ftlds fy, x = 1 f}?kkr lehdj.k x + kx + = 0 dk,d ewy gksa Find the value of k for which x = 1 is a root of the quadratic equation x + kx + = 0 - ABC dh Hkqtk AB vksj AC ij D vksj E fcunq Øe'k% bl izdkj gs fd DE BC vksj AD =.6 cm, AB = 10 cm, AE = 4.5 cm rks AC dk eku fudkysaa If D and E are the points on the sides AB and AC of ABCsuch that DE BCand AD =.6 cm, AB = 10 cm, AE = 4.5 cm then find the value of AC. 4- fl¼ djsa fd o`r dh leku thok, dsuæ ij leku dks.k cukrh gsaa Prove that equal chords of a circle subtend equal angles at the centre. 5- fcunqvksa (µ4] 7) vksj (1] µ5) ds chp dh nwjh Kkr dhft,a Find out the distance between the points ( 4, 7) and (1, 5). 6- ;fn fdlh o`r ds thok dh yeckbz 4 lsehú gks vksj o`r ds dsuæ ij thok Mkyk x;ka yec dh yackbz 5 lsehú gks rks o`r dh f=kt;k fudkysaa If the length of the chord of circle is 4 cm and the perpendicular drawn from the centre is 5 cm then find the radius of the circle. 7- nks?kuksa ds vk;ruksa esa 1 % 8 dk vuqikr gs rks mudh dksjksa esa D;k vuqikr gksxk \ The ratio in volumes between two cubes is 1 : 8 then find the ratio between their edges. 8- tanθ dk eku sinθ ds :i esa O;Dr djsaa

66 Express tanθ in terms of sinθ. cos cos 59 cosec1 dk eku fudkysaa sin10 Evaluate cos80 + cos59 cosec1. sin10 0- fcunq (µ5] 4) vksj (7] µ8) dks feykusokyh js[kk[kam ds eè; fcunq dk fu;ked fudkysaa Find the co-ordinates of the midpoint of the line segment joining ( 5, 4) and (7, 8). 1- lekraj Js.kh 6] 9] 1] 15] dk 5oka in fudkysaa Find out the 5 th term of the AP 6, 9, 1, 15,... y?kq mùkjh; iz'u %& Short Questions: 4 - ( x 1) vksj ( x + x + 1) dk eúlú fudkysaa Find HCF of ( x 1) and 4 ( x + x + 1). - ;fn nks la[;kvksa dk ;ksxiqy 15 gs vksj muds O;qRØeksa dk ;ksxiqy gks rks nksuksa la[;kvksa 10 dks Kkr djsaa The sum of two numbers is 15 and the sum of their reciprocals is then find both 10 the numbers. 4- ABCds 'kh"kz A(1, 1)vkSj B(5,1) gsaa ;fn f=khkqt ds xq:rodsuæ ds fu;ked ( 5,1) gksa rks fcunq C ds fu;ked fy[ksaa Two vertices of ABCare A(1, 1)and B(5,1). If the co-ordinates of the centroid be 5 (,1) then find the co-ordinates of the vertex C ls 00 ds chp lhkh fo"ke la[;kvksa dk ;ksxiqy Kkr djsaa Find the sum of all odd numbers between 100 and ;fn fdlh ds dks.kksa dh ekiksa dk vuqikr :: 4gks rks lcls cm+s dks.k dh eki Kkr djsaa If the ratio of measurements of angles of a triangle are :: 4then find the measure of the greatest angle. 7- ABCds 'kh"kksz A(, 0), B(5, )vksj C( 8, 5)gSa rks ds xq:rodsuæ ds fu;ked fy[ksaa Find the centroid of the triangle whose vertices are A(, 0), B(5, )and C( 8, 5). 8-,d FkSys esa mtyh rfkk dkyh xksfy;k gsaa ;n`pn;k,d mtyh xksyh fudkyus dh izkf;drk fudkysaa A bag contains white and black balls. Find the probability of drawing a white ball at random.

67 + θ θ + = θ 1 cosθ 1+ cosθ 9- fl¼ djsa fd 1 cos 1 cos cosec Proove that 1+ cosθ 1 cos θ + = cosecθ. 1 cosθ 1+ cosθ 40- ;fn m = a cos θ + bsin θ vksj n = asin θ bcos θrks fl¼ djsa fd m n a b + = fueu lkj.kkh dk ekè; fudkysaa oxz varjky 0&10 10&0 0&0 0&40 40&50 ckjackjrk Find the mean of the following data. Class Interval Frequency fueu lkj.kkh dk cgqyd fudkysaa vk;q o"kksz esa 4µ6 6µ8 8µ10 10µ1 1µ14 cppksa dh la[;k Find the mode of the following data. Age Years in Frequency nh?kz mùkjh; iz'u %& Long Question : 4-,d oxz dk {ks=kiqy fudkysa ftldk?ksjk ml vk;r ds?ksjs ds cjkcj gs ftldh,d Hkqtk 48 ehvj gs rfkk mldh pksm+kbz dh rhu xquh gsa 5 Find the area of the square whose boundary is equal to the boundary of a rectangle whose one side is 48 m and three times its breadth. 44- fueufyf[kr lehdj.k dk xzkiq [khaps rfkk bldk izfrpnsn fcunq Kkr djsaa 5 y = x and x 4y = 6 Draw a graph for the following equations and find their intersecting point. y = x and x 4y = 6

68 45-,d LrEHk ds Nk;k dh yeckbz] LrEHk dh mq pkbz ls 1 xquh gsa lw;z dk muu;u dks.k Kkr djsaa 5 The shadow of a vertical pillar is the angle of elevation of the sun. 1 times of the height of the vertical pillar. Find vfkok OR,d irax ftldh Mksjh {ksfrt ry ds lkfk 45 dk dks.k cukrh gs] 75 ehvj dh mq pkbz ij mm+ jgh gsa ;fn Mksjh esa <hy u gks rks bldh yeckbz fu/kzfjr djsaa A Kite whose thread is inclined at an angle 45 with horizontal layer, is flying at the height of 75 m. Being considered that the thread is fully straight then find its length. 46- ;fn ABC dk dks.k B U;wudks.k gks vksj AD, BC vfkok BC ds c<+k, gq, Hkkx ij yac gks] rks AC = AB + BC BC BD 5 If the angles B of a AC AB BC BC BD ABCbe acute and AD is perpendicular to BC produced then = lsehú yacs js[kk[kam dks : ds vuqikr esa var% fohkkftr djsaa 5 Divide a line segment of 10 cm in the ratio : internally. vfkok OR ;fn,d js[kk fdlh f=khkqt dh nks Hkqtkvksa dks,d gh vuqikr esa fohkkftr djs] rks og rhljh Hkqtk ds lekukarj gksrh gsa fl¼ djsaa If a line divides only two sides of a triangle in the same ratio, then the line is parallel to the third side, prove it.

69 SOLUTION (1) (d) () ([k) () ([k) (4) (x) (5) (d) (6) (x) (7) (d) (8) (?k) (9) ([k) (10) (x) (11) cm+k (1) cot θ (1) 1 (14) 0 ('kwu;) (15) 5 (16) 1 (17) (18) varj (19) ( x x ) ( y y ) (0) 0 (1) f}?kkr cgqir x = + 8x + 15 = x + (5 + ) x + 15 = x + 5x + x + 15 = x( x + 5) + ( x + 5) = ( x + )( x + 5) x + 8x + 5 = 0 or, ( x + )( x + 5) = 0 either x + = 0 or, x + 5 = 0 x = x = 5 vhkh"v 'kwu;d =, 5 () f}?kkr lehdj.k x kx = mi;qzdr lehdj.k esa x = 1j[kus ij (1) + k 1 + = 0 or, 1 + k + = 0 or, k = 4Ans. () iz'ukuqlkj] DE BCvkSj AD =.6cm, AB = 10cm, AE = 4.5cm, AC =? DB = AB AD = 10.6 = 6.4cm pwfd ge tkursa gsa fd ;fn fdlh f=khkqt dh fdlh,d Hkqtk ds lekurj dksbz js[kk [khaph tk; rks vu; nks Hkqtk,,d gh vuqikr esa fohkkftr gksrh gsaa AD AE = DB EC or, = 6.4 EC or, EC = = 8cm.6 10

70 AC = AE + EC = = 1.5cm Ans. (4) eku fy;k fd vksj o`r dh nks leku thok, gsa ftudk dsuæ O gsa fl¼ djuk gs fdµ D AB = CDiz'ukuqlkj AOB COD (SSS) AOB = COD (CPCT) AOB = COD izek.k%µ AOBvkSj COD esa] OA = OC, OB = ODo`r dh lhkh f=kt;k, leku gksrh gsaa vr% o`r dh leku thok, dsuæ ij leku dks.k cukrh gsa (5) fcunqvksa (µ4] 7) vksj (1] µ5) ds chp dh nwjh = (1 + 4) + ( 5 7) = = 169 = 1bdkbZ (6) eku fy;k fd OdsUæ okyk o`r dh thok AB dh yackbz 4 lsehú gs vksj o`r dsuæ O ls thok AB ij Mkyk x;k yac OPdh yackbz 5 lsehú gsa OPBesa] OA OP i. e., AB = 4cm, OP = 5cm OAP vksj OBPesa] OPA = OPB = 90 (iz'ukuqlkj) = OB (o`r dh f=kt;k, cjkcj gksrh gs) = OP (Common) OAP OBP(RHS) 4 AP BP = = 1cm OPB = 90 (OB) = (OP) + (PB) = = = 169 OB = 169 = 1cm Ans.

71 (7) eku fy;k fd igys vksj nwljs?kuksa ds dksjksa dh yackbz Øe'k% xvksj ybdkbz gsa iz'ukuqlkj] ifz ke?ku dk vk;ru 1 = nlw js?ku dk vk;ru 8 x 1/ 1/ or, x or, x = y 8 y = = = = 8 y 8 dksjksa dk vhkh"v vuqikr = x : y = 1: Ans. (8) sin θ tan θ = = cos θ sin θ 1 sin θ Ans. (9) cos80 + cos59 cosec sin10 cos(90 10 ) = + cos(90 1 ) cosec1 sin10 sin10 = + sin 1 cosec1 sin10 = = Ans. (0) fcunq (µ5] 4) vksj (7] µ8) dks feykusokyh js[kk[kam ds eè; fcunq ds fu;ked (1) lekurj Js.kh 6] 9] 1] 15] ( 8) =, 4 8 =, ;gk a = 6, d = t t1 = 9 6 = t = a + ( n 1) d n t 5 = 6 + (5 1) = 6 + (4 ) = = 108 Ans. () x 1 ( x) (1) = 4 = 1, = (1, ) Ans. = ( x 1)( x + x + 1) 4 x + x + 1 = ( x ) + x 1 + (1) x

72 = ( x + 1) x = ( x + 1 x)( x x) = ( x x + 1)( x + x + 1) vhkh"v eúlú = x + x + 1 Ans. () eku fy;k fd izfke la[;k = x iz'ukuqlkj] iz'ukuqlkj] or, or, or, or, or, or, or, f}rh; la[;k = 15 x 1 1 x + 15 x = x + x = x(15 x) x x = 150 x 45x = 0 ( x 15x + 50) = 0 x 15x + 50 = 0 ( 0 ) x (10 + 5) x + 50 = 0 x 10 x 5x + 50 = 0 or, x( x 10) 5( x 10) = 0 or, ( x 5)( x 10) = 0 x 5 = 0 or, x 10 = 0 x = 5 x = 10 vhkh"v la[;k x = 5, 10 Ans. (4) eku fy;k fd fcunq C ds fu;ked ( x, y) f=khkqt ds xq:ro dsuæ ds fu;ked x y =, 6 + x y =, gsaa rks iz'ukuqlkj] xq:rodsuæ dk xfu;ked 5 =

73 i. e., 6 + x 5 = or, 6 + x = 5 or, x = 5 6 = 1 iqu%] xq:ro dsuæ dk y fu;ked = 1 y or, = 1 or, y = vr% fcu C dk fu;ked = ( 1, ) Ans. (5) vr% 100 ls 00 ds chp lhkh fo"ke la[;kvksa dk ;ksxiqy ;gk ] a = t n = d = t t = = a + ( n 1) d = 199 or, ( n 1) = 199 or, ( n 1) = = 98 or, 98 n 1 = = 49 n = = 50 n Sn = { a + ( n 1) d} 50 = { (50 1)} = 5{0 + 98} = 5 00 = 7500 Ans. (6) eku fy;k fd vuqikr dh jkf'k xgsa Rkks] dk izfke dks.k = x f}rh; dks.k = x r`rh; dks.k = 4x pwafd ds rhuksa dks.kksa dk ;ksxiqy 180 gksrk gsa x + x + 4x = 180 or, 9x = 180

74 180 or, x = = 0 9 lcls cm+k dks.k = 4x = 4 0 = 80 Ans. (7) p wfd ge tkurs gsa fd] ( x 1, y 1 ), ( x, y ) vksj ( x, y) ls cus 'kh"kksza okys ds xq:ro dsuæ ds fu;ked x 1 + x + x, y 1 + y + y = = , 6 =, = (,1) Ans. (8) ekuk fd S = dqy Ifj.kkeksa dk lewg = {W, W, W, B, B} ;gk W = mtyh xksyh vksj B = dkyh xksyh n(s) = 5 Ekku fy;k E mtyh xksyh fudkyus dh?kvuk dks iznf'kzr djrk gsa vfkkzr~ n(e) = i. e., E = {W, W, W} n(e) P(E) = = Ans. n(s) 5 (9) L.H.S. = 1 + cosθ 1 cosθ + 1 cosθ 1 + cosθ 1 + cosθ 1 + cosθ 1 cosθ 1 cosθ = + 1 cosθ 1 + cosθ 1 + cosθ 1 cosθ (1 + cos θ) (1 cos θ) = + 1 cos θ 1 + cos θ 1 + cosθ 1 cosθ = + sin θ sin θ 1 + cosθ 1 cosθ = + sin θ sin θ 1 + cos θ + 1 cosθ = sin θ = sin θ = cosecθ = R.H.S. Proved (40) iz'ukuqlkj] m = a cosθ + bsin θ

75 vksj n = asin θ bcos θ L.H.S. = m + n = ( a cosθ + b sin θ) + ( a sin θ b cos θ) = a cos θ + b sin θ + absin θcosθ + a sin θ + b cos θ absin θcosθ = a (cos θ + sin θ) + b (sin θ + cos θ) = a 1 + b 1 = a + b = R.H.S. Proved (41) oxz vurjky ckjeckjrk ( f ) eè; fcunq ( x ) f x 0 µ µ µ µ µ Σ f = 50 Σ fx = 1100 (4) Ekkè; vk;q o"kz esa Σfx 1100 = = = Ans. Σf 50 cppksa dh la[;k 4 µ 6 6 µ µ µ µ 14 7 cgqyd (Mode) M o = l + i fo f 1 fo f 1 f1

76 tgk ] l = Modal class vfkkzr~ vf/dre ckjeckjrk okyk oxz dh fueu lhek fo = Modal class dh ckjeckjrk f 1 = Modal class ds Bhd igys okyk oxz dh ckjeckjrk f 1 = Modal classds Bhd ckn okyk oxz dh ckjeckjrk i =oxz vurjky dh yackbz iz'ukuqlkj]] vf/dre ckjeckjrk okyk oxz vurjky (8 µ 10) ftldh ckjeckjrk 0 gsa vfkkzr~ l = 8, fo = 0, f 1 = 8, f1 = 1, i = 6 4 = 0 8 vhkh"v cgqyd = 8 + ( 0) = 8 + = = 8 + = = 9.o"kZ Ans. 10 (4) eku fy;k fd vk;r dh pksm+kbz = xehú iz'ukuqlkj] x = or, x = = 16 ehú vk;r dk?ksjk vfkkzr~ ifjfefr = (yeckbz $ pksm+kbz) = ( ) = 64 = 18 ehú iz'ukuqlkj] oxz dk?ksjk vfkkzr~ ifjfefr = vk;r dk?ksjk vfkkzr ifjfefr 4 Hkqtk = 18 or, Hkqtk = 18 = ehú 4 oxz dh {ks=kiqy = (Hkqtk) = () 104 y = x...(1) (44) and x 4y = 6...() lehú (1) y x = = oxz ehvj Ans. x 1 1 y 1 5 (x, y) (1, 1) (, ) ( 1, 5)

77 lehú () x 4y = 6 x 6 y 0 (x, y) (, 0) (6, ) (, )

78 (45) eku fy;k fd AB,d h ehvj mq pk LrEHk gsa lw;z ds izdk'k esa LrEHk dh Nk;k BC gs ftlls lw;z dk muu;u dks.k /jkry ds C fcunq ij θ dks.k cukrk gsa iz'ukuqlkj] LrEHk Nk;k BCdh yeckbz ABCesa] ABC = 90 vksj ACB = θ AB tan ACB = BC h or, tan θ = h / 1 = h h = = h = h = tan 60 θ = 60 Ans. ^vfkok* or AB sin ACB = sin45 = AC eku fy;k fd,d irax i`foh ds /jkry B ls 75 ehvj dh mq pkbz ij AfcUnq ij mm+ jgh gs ftldh Mksjh ACdh yeckbz Kkr djuh gs] {ksfrt ry C fcunq ij 45 dk dks.k cukrh gsa vfkkzr~ ACB = 45 ABCesa ABC = = AC or, AC = 75 ehvj Ans. = = ehVj Ans.

79 (46) ledks.k f=khkqt ADB esa D =,d ledks.k ikfkkxksjl izes; ls AB = AD + BD --- (1) fiqj ledks.k f=khkqt ADCesa D =,d ledks.k ikfkkxksjl izes; ls AC = AD + DC --- () fp=k ( I ) ls DC = BC BD DC = (BC BD) = BC BC BD BD fp=k ( II ) ls DC = DB CB DC = (DB CB) = DB DB CB + CB = BD BC BD + BC = BC BC BD + BD nksuksa fp=k esa] DC = BC BC BD + BD nksuksa vksj AD tksm+us ij AD + DC BC BC BD + AD BD (1) vksj () dks iz;ksx esa ykus ij AC = BC BC BD + AB AC = AB + BC BC BD Proved (47) jpuk ds pj.k %& (i ) AB = 10cm,d js[kk[kam [khapka fp=k ( I ) = + (ii ) ABds lkfk fcunq Aij dksbz U;wudks.k cukrh gqbz js[kk AD [khapka fp=k ( II ) (iii ) ADij 5 ($) cjkcj pki AL, LM, MN, NQ, QC dkvdj L, M, N, Q, Cvafdr djsaxsa (iv)cvksj Bdks feyk nsaxsa ( v ) rhljs fcunq (va'k = ) Nls NP BC[khapk tks AB dks Pij dkvrh gsaa bl izdkj PfcUnq ABdks : ds vuqikr esa var% fohkkftr djrk gsa

80 (48) eku fy;k fd ABC,d f=khkqt gsa blesa,d js[kk DEtks Øe'k% AB,AC ij flfkr gsa bl izdkj AD AE = DB EC Rkks fl¼ djuk gs fd DE BC cukov µ eku fy;k fd DE, BCds lekurj ugha gs rks BCds lekurj,d vu; js[kk DE [khapka izek.k µ cukov ds vuqlkj DE BC FkSYl izes; ls] AD AE = DB E C IjUrq fn;k gqvk gs AD AE = DB EC AE AE = EC E C nksuksa rjiq 1 tksm+us ij AE AE + 1 = +1 EC E C AE + EC AE + E C = EC E C AC AC = EC E C EC = E C ;g rhkh lahko gs tc E,oa E,d gh fcunq ij gksa vr% BC ds lekurj js[kk DE gh gs] DE ugha gsa DE BC Proved

81 ! # % #&& () +,. /!,), %,0) 1,/ # % %) (,/ & ( %456 0 ) 7,/ )!

82 Lke; %?kavk 15 feuv xf.kr (lsv&5) MATH (SET-5) Time : Hrs 15 Min. Full Marks : 100 lgh mùkj pqus %& Choose the correct answer :- pøh; prqhkzqt ds pkjks dks.kksa dks ;ksx gksxka 1 (d) 180 ([k) 540 (x) 60 (?k) lhkh Sum of four angles of quadrilateral will be? (a) 180 (b) 540 (c) 60 (d) All - cot 60 dk eku gksxka 1 (d) ([k) 1 (x) 1 (?k) 0 Value of cot 60 will be 1 (a) (b) (c) 1 (d) 0 - fuf'pr?kvuk dk izkf;drk gksrh gsa 1 (d) 1 ([k) 0 (x) 1 (?k) 1 Probability of sure event is? 1 (a) (b) 0 (c) 1 (d) 1 4- sin(90 A) fueu esa fdlds fdlds cjkcj gs \ 1 (d) cos A ([k) tan A (x) sec A (?k) coseca sin(90 A) is equal to which of the following? (a) cos A (b) tan A (c) sec A (d) coseca 5- izfke ik p le izkd`r la[;kvksa dk ekè; gksxkµ 1 (d) 8 ([k) 10 (x) 4 (?k) 6 Mean of the first five even natural number will be (a) 8 (b) 10 (c) 4 (d) 6 6- nks o`rksa dh f=kt;kvksa dk vuqikr 1 % gs rks {ks=kiqyksa dk vuqikr D;k gksxk \ 1 (d) 1 % 9 ([k) % 9 (x) % (?k) %

83 If the ratio of radius of two circles is : then the ratio of their area will be (a) 1 : 9 (b) : 9 (c) : (d) : 7- ;fn lw;z dh muufr dks.k 60 gs] rc 0 ehú yach Nk;k cukus okyh mæx ehukj dh mq pkbz gs \ 1 (d) m ([k) 0 m (x) 5 m (?k) 0 m If the altitude of the sun is 60 then what is the height of the vertical tower making the 0 metre long shadow (a) m (b) 0 m (c) 5 m (d) 0 m 8-,d v¼zxksys dk vk;ru lsehú gs] rks v¼zxksys dk oø i`"b {ks=kiqy D;k gksxk\ 1 (d) 77 lsehú ([k) 696 lsehú (x) 6 lsehú (?k) 16 lsehú What will curved surface area of Hemisphere whose volume is cm (d) 77 cm ([k) 696 cm (x) 6 cm (?k) 16 cm 9- fcunq Pls fdlh o`r dh Li'kZ js[kk dh yeckbz 4 lsehú gs vksj dsuæ ls Pdh nwjh 5 lsehú gs] rks o`r dh f=kt;k gsµ 1 (d) 1 lsehú ([k) 8 lsehú (x) 1-5 lsehú (?k) 7 lsehú If length of the tangent of a circle from point P is 4 cm and distance from centre of P is 5 cm then the radius of the circle is (a) 1cm (b) 8 cm (c) 1.5 cm (d) 7 cm 10- (µ] µ) fdl ikn esa gksxk \ 1 (d) r`rh; ikn ([k) prqfkz ikn (x) izfke ikn (?k) f}rh; ikn Which quadrant will the point (, ) lie? (a) Third quadrant (b) Fourth quadrant (c) First quadrant (d) Second quadrant fjdr LFkkuksa dh iwfrz djs %& Fill in the blanks :- 11- ax + bx + c = 0ds ewy oklrfod vksj vlkeu gsa ;fn gksa 1 ax + bx + c = 0is real and unequal if... The roots of 1- cosec(90 A) = ] 8] 9] 7] 8] 9] 9] ] 5] 4 dk cgqyd gs A What s mode of 7, 8, 9, 7, 8, 9, 9,, 5, dk n'keyo izlkj gksrk gsa 1 5 Decimal expansions of 17 is... 5

84 15- fn;s x;s fp=k esa x dk eku gksxka 1 What will be the value of x in the given figure (triangle)? 16- nks cgqhkqt le:i gksrs gsaa ;fn muds laxr dks.k cjkcj gks rfkk laxr Hkqtk, gksaa 1 Two regular polygon will be similar if their corresonding angle are equal and corresponding sides be dkrhz; ry esa flfkr fdlh fcunq (µ4] 5) esa dksfv dk eku gksxk A 1 The value of ordinate of point ( 4, 5) in cartesian plan A.P 5, 9,1,17,... dk lkozarj gksxka 1 The common difference of A.P. 5, 9, 1, 17 is sin 60 cos0 sin0 cos60 dk eku gksrk gsa 1 The value of sin 60 cos0 sin0 cos60 is D;k x x + = 0dk,d ewy gs \ Is a root in x x + = 0... vfr y?kq mùkjh; iz'u%& Very Short Questions : 1- fn;s x, A.P 16,, 8, 4,...dk igyk in vksj lkoz varj Kkr djsaa Find the first term and common difference in the given A.P 16,, 8, 4, fl¼ djsa fd cot 48 cot cot 4 cot 57 = 1 - ;fn Prove that cot 48 cot cot 4 cot 57 = 1 If cos A = 4 rks sin AvkSj tan A dk eku fudkysaa cos A = 4 then find the value of sin Aand tan A. 4- vkd`fr esa DE BC gsa rks ECKkr dhft,a In the figure DE BCthen find EC?

85 5- cxy ds fp=k esa PAvkSj PBo`r dh Li'kZ js[kk gsa rfkk APO = 0 rks AOB D;k gksxk\ In the figure PA and PB is tangent line of circle and APO = 0 then what is the value of AOB. 6- f}?kkr lehdj.k 6 x x = 0 ds izd`fr Kkr djsaa Find the nature of the quadratic equation a b c a b c 6 x x = vuqikrksa 1, 1, 1 dh rqyuk dj Kkr dhft, fd fn;s x;s jsf[kd lehdj.k dk ;qxe laxr gs ;k vlaxra a On comparing the ratios 1 b, 1 c, 1 a b c x y = 8rFkk 4 x 6y = 9 of the given linear equations x y = 8 and 4 x 6y = 9, find out whether the given pair of linear equations consistent or in consistent. 8- fn, x, ds fp=k esa ;fn PR CBvkSj PS CDgks rks fl¼ djsa fd OR = OS OB OD In the given figure if PR CBand PS CDthen prove that OR = OS OB OD. 9-,d lef}ckgq ABCesa <C = 90 rks fl¼ djsa fd AB = AC Angle (<C) = 90 in a isosceles triangle then prove that AB = AC. 0- ml xksys dk i`"b {ks=kiqy fudkysa ftldk O;kl 14 lsehú gsa Find the surface area of sphere where diameter is 14 cm. 1- ;fn fcunq P(, )dh nwjh Q(10, y)ls 10 bdkbz gks rks ydk eku Kkr djsaa If the distance from Q(10, y) to point P(, ) is 10 unit then find the value of y.

86 y?kq mùkjh; iz'u %& Short Questions: - fl¼ djsa fd 5,d vifjes; la[;k gsa Prove that 5 is irrational. - 1 lsehú O;kl] okyh 8 lsehú yach rk cs dh,d NM+ dks,d leku pksm+kbz okys 18 ehú yacs,d rkj ds :i esa [khapk tkrk gsa rkj dh eksvkbz Kkr djsa A copper rod of diameter 1 cm and lenth 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire? 4- fl¼ dhft, fd (1 sina)(seca + tana) =1 Prove that (1 sina)(seca + tana) =1 5- vuqikrksa cosa, tanavksj seca dks sinads inksa esa O;Dr dhft,a Express the ratios cosa, tanaand seca in terms of sina. 6- fdlh fhkuu ds va'k vksj gj dk ;ksx 10 gsa ;fn gj esa tksm+ fn;k tk; rks fhkuu 1 gks tkrk gsa fhkuu Kkr djsaa The sum of numerator and denominator of any fraction is 10. If is added to denominator the fraction becomes 1. Find the fraction. 7- nks vadksa okyh fdruh la[;k, ls fohkkt; gsa \ How many number of two digit is divisible by? 8-,d FkSys esa yky vksj 5 dkyh xsan gs bl FkSys ls,d xsan ;kn`pn;k fudkyh tkrh gsa bldh izkf;drk D;k gksxh fd xsan (d) yky gs] ([k) yky ugha gsa In a bag there is red and 5 black balls. One ball is taking out atrandom. What will be probability of ball in (a) is it red? (b) is it not red? cot A cos A coseca 1 = cot A + cos A coseca fl¼ dhft, fd Prove that cot A cos A coseca 1 = cot A + cos A coseca fueufyf[kr ckjackjrk forj.k dk cgqyd Kkr djsaa Find the Mode of following frequency distribution. oxz varjky 6µ10 11µ15 16µ0 1µ5 6µ0 ckjackjrk f}?kkr lehdj.k x 6x + = 0 ds ewy Kkr dhft,a Find the roots of the quadratic equation x 6x + = ;fn sina = cos(a 6 ) gks tgk ] A,d U;wudks.k gs rks Adk eku Kkr dhft,a If sina = cos(a 6 ) where A is an acute angle. Find the value of A.

87 nh?kz mùkjh; iz'u %& Long Question : 4- xzkiqh; fof/ ls lehdj.k gy djsaa 5 Solve the equation by graphically. x 4y + 14 = 0 x + y 14 = 0 44-,d lery tehu ij [km+h ehukj dh Nk;k ml flfkr esa 0 ehú vf/d yach gks tkrh gs tc fd lw;z dk muurka'k 60 ls?kvdj 0 gks tkrk gsa ehukj dh mq pkbz Kkr dhft,a 5 The shadow of a towar standing on a level ground is found to be 0m longer when the sun s altitude is 0 then what it is 60. Find the height of the towar. 45- ikuh ls iwjh Hkjh gqbz,d v¼zxksykdkj Vadh dks,d ikbi }kjk 4 7 yhvj izfr lsdsam dh nj ls [kkyh fd;k tkrk gsa ;fn Vadh dk O;kl ehú gs rks og fdrus le; esa vk/h [kkyh gks tk;sxh \ ( π = ) A hemispherical tank full of water is emptied by a pipe at the rate of ltr/sec. How 7 much time will it take to empty half of the tank, if it is m in diameter?(take π = ) ABC esa C ledks.k gsa ;fn Cls ABij [khaps x;s yac dh yackbz P gks vksj AB = c, BC = a8 CA = bvksj CD = p rks fl¼ djsa fd = P a + b 5 ABC is a right triangle in which C = 90, CD AB. If AB = c, BC = a, CA = b and CD = pthen prove that = +. P a b vfkok ^OR 46- cxy ds fp=k esa PDC PBA, BPC = 15 ] CDP = 70 rks DPC, DCP,oa PAB Kkr djsaa In side by figure PDC PBA, BPC = 15 <BPC =15 and CDP = 70 then find DPC, DCF and PAB? 47- lsehú f=kt;k okys o`r ds ckgj flfkr,d fcunq ls ml ij Li'kZ js[kkvksa dh jpuk djsaa 5 To construct the tangents to a circle of radius cm from a point outside it?

88 SOLUTION (1) (x) () ([k) () (x) (4) (d) (5) (?k) (6) (d) (7) (?k) (8) (d) (9) (?k) (10) (d) (11) b 4ac > 0 (1) sec A (1) 9 (14) 0-68 (15) 10 (16) lekuqikrh (17) 4 (18) 1 (19) 1 (0) gk (1) igyk in ( a ) = 16 lkoz varj ( d ) =nwljk in µ igyk in = 16 = 6 Ans. () L.H.S. = cot 48 cot cot 4 cot 57 = cot(90 4)cot(90 57 )cot 4 cot 57 = tan 4 tan 57 cot 4 cot = tan 4 tan 57 = 1 R.H.S. Proved tan 4 tan 57 () Given cos A = sina = 1 cos A = 1 = 1 = = sin A tana = = = = Ans. cos A 4 4 (4) fn;k x;k gs ABCesa DE BC AD AE BD = EC 1.5 = 1 EC = EC BC = = BC = cm Ans. 15 (5) APO = 0 OA APij OAP = 90 AOP =180 ( ) = = 60 Ans. (6) fn;k gqvk gsµ 6 x x = 0 ;gk a = 6, b = 1, c =

89 D = b 4ac = ( 1) 4 6 = = 49 > 0 D > 0 (7) x y = 8 4 x 6y = 9 ewy oklrfod vksj vleku gsa a 1 =, b1 =, c 1 = 8 a = 4, b = 6, c = 9 b = = = vfkkzr~ 1 b 1 c = 1 a b c a 1 b 1 c =, =, =, a 4 b 6 c vr% jsf[kd lehdj.kksa dk ;qxe laxr gsa (8) fp=kkuqlkj OBCesa] PR CB FksYl izes; ls OR OP = --- (1) OB OC iqu% OCD esa] PS CD FksYl izes; ls OP OS = --- () OC OD vr% (1) vksj () ls OR OS = lr;kfira OB OD (9) fn;k x;k gs iz'u ls ABC,d lef}ckgq f=khkqt gs ftldk dks.k C ledks.k gsa fl¼ djuk gsµ AB = AC ACBesa] Rkks AB (0) ge tkurs gsa fd iz'u ls C = 90, AC = BC(fn;k gs) AB = AC + BC (ikbfkksxksjl izes; ls) d =14cm, = AC + AC [BC = AC] = AC fl¼ gqvk d 14 r = = = 7cm

90 Xkksys dk i`"b {ks=kiqy = πr (1) PQ = ( 10) + ( y) PQ 8 ( y) = (7) = = 46cm Ans. = + + iz'u ls PQ =10 10 = y + y or, y + 6 y 7 = 0 or, y + 9 y y 7 = 0 or, y( y + 9) ( y + 9) = 0 or, ( y + 9)( y ) = 0 y = 9, Ans. () ekuk fd 5,d ifjes; la[;k gsa a 5 =, b tgk a vksj b ( b 0) lgvhkkt; la[;k, gsa a vr% 5 b = gsa blh lehdj.k esa iquo;zoflfkr djus ij gesa izkir gksrk gsa = 5 a b pw fd a vksj b iw.kkzad la[;k, gs] blfy, 5 a b la[;k gsa,d ifjes; la[;k gs vfkkzr~ ifjes; ijarq blls bl rf; dk fojks/khkkl izkir gksrk gs fd,d vifjes; la[;k gsa gesa ;g fojks/khkkl viuh xyr dyiuk ds dkj.k izkir gqvk gs fd 5,d ifjes; la[;k gsa vr% ge fu"d"kz fudkyrs gsa fd 5,d vifjes; la[;k gsa () NM+ dk vk;ru 1 = π 8 cm = π cm leku vk;ru okys rkj dh yackbz = 18m =1800cm ;fn rkj ds vuqizlfk dh f=kt;k r gs rks] rkj dk vk;ru = π r 1800 cm vr% π r 1800 = π π 1 r = = π r = 900

91 1 1 r = = cm vr% rkj ds vuqizlfk rkj dk O;kl] rkj dh pksm+kbz 1 cm 15 (4) L.H.S. = seca(1 sina)(seca + tana) 1 1 sina = (1 sina) + cos A cosa cosa (1 sina)(1+ sina) = cos A 1 sin A cos A = = 1 R.H.S Proved cos A cos A = = (5) ge tkurs gsa fd cos A + sin A = 1 So, cos A = 1 sin A vfkkzr~] cos A = 1 sin A ± blls ;g izkir gksrk gs cos A = sina vr% tan A = = cosa sina 1 sin A 1 1 vksj seca = = cos A 1 sin A (6) ekuk fd fhkuu x = y 1 sin A iz'ukuqlkj] x + y = 10 (i) x 1 = y + ;k] x = y + x y = (ii) lehú (i) rfkk (ii) dks tksm+us Ikj x + y = 10 x y = 1 x = = 4 x = 1 x = 4 lehú (i) j[kus ij 4 + y = 10 y = 6 x 4 vr% vhkh"v fhkuu = = y 6. ;kfu 0.67mm (yxhkx)

92 (7) 10,11,1,1,..., 99esa vadksa okyh mls fohkkt; la[;k, 1,15,18, 1, 4,..., 99 ;gk ls lú{ksú ds fy, a = 1, d = 15 1 =, l = 99 l = a + ( n 1) d 99 = 1 + ( n 1) = n 1 = n 1 ;k n 1= 9, n = 0 Ans. (8) FkSys esa dqy xsan = + 5 = 8 dqy lahkkfor ifj.kkeksa dh la[;k = 8 (9) (d) P (yky xsan) = 8 5 ([k) P (yky xsan ugha) = 1 = Ans. 8 8 cos A cos A cota cos A L.H.S. = = sina cota + cos A cos A + cos A sina 1 1 cos A 1 1 sina sina coseca 1 = = = = R.H.S 1 1 coseca + 1 cos A sina sina (40) ;gk cgqyd oxz 16µ0 gsa D;ksafd bldh ckjackjrk lokzf/d gsa 16µ0 esa viorhz cukus ij ;g oxz = bl izdkj] l = 15.5, f0 = 50, f 1 = 0, fi = 40, i = 5 f 0 f Mode = l + 1 ils f0 f 1 f i M 0 = = = = 18.8 (41) x 6x + = x 6 x 6x + = x( x ) ( x ) = ( x )( x ) vr% lehdj.k ds ewy x ds os eku gsa] ftuds fy, ( x )( x ) = 0 x = x = 0

93 iqu% ] vr% x = 0 x = 6 0 x x + = ds ewy, gsaa (4) ;gk fn;k gqvk gs fd sina = cos(a 6 ) --- (1) sin A = cos(90 A) blfy, ge (1) dks bl :i esa fy[k ldrs gsaa cos(90 A) = cos(a 6 ) D;ksafd 90 AvkSj A 6 nksuksa U;wudks.k gs rks 90 A = A 6 A A = 6 90 izkir gksrk gsa (4) 116 4A = 116 A= A = 9 4 x 4y + 14 = 0...(1) and x + y 14 = 0...() lehú (1) x 4y + 14 = 0 lehú () x + y 14 = 0 x 4 y = 14 x + y = 14 x + 14 = 4y x = 14 y x y = y x = 4 x 0 y 4.5 nksuksa vkys[kksa dk dvku fcunq (] 4) gsa vr% gy & x = y = 4 x 4 0 y 1 7 4

94 (44) eku fyft, fd ABehukj dh yackbz h ehvj gsa vksj BC, x ehvj gsa DC = 0ehVj DB = DC + CB = (0 + x) ehú vr% DB = (0 + x) ehú vc ;gk nks ledks.k f=khkqt ABCvkSj ABDgSA ledks.k ABCesa AB tan 60 = BC h x AB ABDesa] tan 0 = BD 1 h or, = x + 0 (1) ls gesa izkir gksrk gsa h x = ---(1) ---() = ---() bl eku dks () esa izfrlfkkfir djus ij gesa ;g izkir gksrk gs ( x ) = x + 0;kfu fd x = x + 0 x x = 0 blfy, () ls x = 15 0 x = 0 x = = 15 vr% ehukj dh mq pkbz 15 gsa (45) v¼zxksykdkj Vadh dh f=kt;k vr% Vadh dk vk;ru = m = m 7 m = 7 99 m = 14 ml ikuh dk vk;ru] ftls [kkyh fd;k tkuk gsa 1 99 m = 14

95 = 1000 = yhvj 8 8 vc 5 yhvj ikuh [kkyh gksrk gs 1 lsdsam esaaa 7 blfy, yhvj ikuh [kkyh gksxk = = = 990 second = feuv Ans. (46) ABC esa ABdks vk/kj ysrs gq, dk {ks=kiqy 1 = AB CD 1 = C p iqu% BCdks vk/kj ekurs gq, 1 1 ABCdk {ks=kiqy = BC AC= ab 1 1 C p = ab C p = a b 1 C 1 c = = p ab p a b 1 b + a = p a b ikbfkkxksjl izes; ls] ABCesa c a b b a = + a b a b = + Proved p a b DPC + BPC =180 DPC +15 =180 DPC = = 55 CPD esa] = + vfkok (jsf[kd ;qxe vfhkx`ghr) o CDP + DPC + DCP =180 (f=khkqt ds rhuksa ;ksx dk ;ksx 180 gksrk gsa)

96 DCP =180 ;k DCP =180 ( ) = DCP = 55 iqu% PDC PBA (fn;k gs) PAB = PCD PAB = 55 ( PCD = 55 ) blh izdkj DPC = 55 DCP = 55, PAB = 55 (47) jpuk ds pj.kµ (1) dksbz fcanq 0 dks dsunzekudj lsehú f=kt;k dk,d o`ùk [khpsaxsa () o`ùk ds ckgj dksbz fcanq fcanq P ysaxsa () PO dks feyk,sxs vksj bls lef}hkkftr djsaxs eku fd POdk eè; fcunq M gsa (4) Mdks dsunz eku dj rfkk MOf=kT;k ysdj,d o`ùk [khpsaxsaa ;g fn, x, o`ùk dks Q vksj R Ikj izfrpnsn djrk gsa (5) Mdks Q rfkk R ls feyk,saxs rc PQvkSj PRvHkh"V Li'kZ js[kk, feyrh gsa

97 ! # % #&& () +,. /!,), %,0) 1,/ # % %) (,/ & ( %456 0 ) 7,/ )!

98 Lke; %?kavk 15 feuv xf.kr (lsv&6) MATH (SET-6) Time : Hrs 15 Min. Full Marks : 100 lgh mùkj pqus %& Choose the correct answer :- fueu esa ls dksu f}?kkr lehdj.k x + x 6 = 0 dk,d gy gs \ 1 (d) x = ([k) x = (x) x = (?k) x = Which of the following is a solution of the quadratic equation x + x 6 = 0? (a) x = (b) x = (c) x = (d) x = - nksa la[;kvksa dk e-l- 5 rfkk y-l- 50 gsa rks budk xq.kuiqy fueu esa ls dksu gksxk \ 1 (d) 1050 ([k) 1150 (x) 150 (?k) 150 H.C.F. and L.C.M. of two numbers are 5 and 50 respectively. Which of the following is the product of the numbers? (a) 1050 (b) 1150 (c) 150 (d) fueu esa ls dksu ifjes; la[;k gs \ 1 (d) 1 ([k) 16 Which of the following is a rational number. (a) 1 (b) 16 (x) (c) (?k) fdlh ABC esa A = 90, BC =1cm, AB =1cm rks fueu esa ls dk eku dksu gs \ 1 (d) (d) lsehú ([k) 4 lsehú (x) 5 lsehú (?k) 6 lsehú In AC? ABC 8 A = 90, BC =1cm, AB =1cm 8 which of the following is the value of (a) cm (b) 4 cm (c) 5 cm (d) 6 cm 5- lekarj Js.kh 11, 8, 5,...dk izfke /ukred in fueu esa ls dksu gksxk \ 1 (d) 1 ([k) µ (x) (?k) Which of the following is the first positive term of the given A.P. 11, 8, 5,...? (a) 1 (b) (c) (d)

99 6- fdlh fcunq P( x, y) dk ewy fcunq ls nwjh fueu esa ls dksu gksxk \ 1 (d) x + y ([k) x y (x) x + y (?k) x y Which of the following is the distance of a point P( x, y ) from the origin? (a) x + y (b) x y (c) x + y (d) x y 7- f=khkqt dk xq:ro dsuæ mldh ekfè;dk dks fueu esa ls fdl vuqikr esa fohkkftr djrk gs \ 1 (d) 1 % ([k) % 1 (x) 1 % (?k) % 1 Centroid of a triangle divides its median in which of the following ratio? (a) 1 : (b) : 1 (c) 1 : (d) : 1 8- cosec60 dk eku fueu esa dksu gs \ 1 (d) ([k) (x) Which of the following is the value of cosec60? (a) (b) 9- cosec cot (c) (d) (?k) θ θdk eku fueu esa dksu gs \ 1 (d) 1 ([k) (x) (?k) 1 Which of the following is the value of cosec θ cot θ (a) 1 (b) (c) (d) o`r dk {ks=kiqy dk lw=k fueu esa dksu gs \ 1 (d) π r ([k) π r (x) π r (?k) π r Which of the following is the formula for area of a circle? (a) π r fjdr LFkkuksa dh iwfrz djs %& Fill in the blanks : (b) π r (c) π r (d) πr dk n'keyo izlkj gksxka 1 Decimal expansion of 5 is f}?kkr lehdj.k ax + bx + c = 0 ds ewyksa dk xq.kuiqy gksrk gsa 1 Product of the roots of a quadratic equation ax + bx + c = 0 is...

100 1- ABC,d ledks.k f=khkqt gs ftldk B = 90, C = 0 rks A = ABC is a right angle triangle in which B = 90, C = 0 then A = dksbz ljy js[kk fdlh f=khkqt dh nks Hkqtkvksa dks leku vuqikr esa dkvrh gs rks og js[kk rhljh Hkqtk ds gksrh gsa 1 If a line divides any two sides of a triangle in the same ratio, the line must be... to the third side. 15- ;fn lekraj Js.kh dk izfke in a rfkk bldk lokzarj dgks rks n oka in = a + (...) d 1 If the first term of an A.P. is a and its common difference is d then n th term = a + (...) d 16- o`r dh Li'kZ js[kk o`r dks ftl fcunq ij Li'kZ djrk gs mls dgk tkrk gsa 1 The tangent line touches the circle on a point is called ;fn lekarj Js.kh dk n th If th n in t n }kjk lwfpr gks rks tn tn 1 =... lwfpr gksrk gsa 1 term of an A.P. is denoted by t n then tn tn 1 is denoted by cos θ cjkcj gksrk gsa 1 1 cos θ s equal to tan0 dk eku gksrk gsa 1 The value of tan0 = ekè; µ cgqyd =... (ekè; µ ekfè;dk) 1 Mean Mode =... (Mean Median) vfr y?kq mùkjh; iz'u%& Very Short Questions : 1- fn[kk, fd fn;k x;k lehdj.k ;qxe vfojks/h] fojks/h ;k vkfjr gsa x + y = 7vkSj 6x + 5y = 11 Show that the given simultaneous equations are consistent, inconsistent or dependent. x + y = 7and 6x + 5y = 11-1 x ds fy, gy djsaa 10 x = x 1 Solve for x, 10 x = x - ABC,d ledks.k f=khkqt gs ftldk A = 90 ] AM BC. ABM rfkk CAM esa laca/ crk, A ABC is a right triangle in which A = 90, AM BC. Show the relation between ABMand CAM.

101 4- fn;s x;s fp=k esa Oo`r dk dsuæ gsa APvkSj AQckg~; fcunq Als nks Li'kZ js[kk;sa gsaa ;fn OPQ = 0 rks dk PAQ eku fudkysa In the adjoin figure, O is the centre of a circle. From external point A, two tangents AP and AQ drawn on the circle. If OPQ = 0 then find measure of PAQ. 5- fcunq (] ) rfkk (4] 1) ds chp dh nwjh Kkr dhft,a Find the distance between the points (, ) and (4, 1). 6-7 lsehú f=kt;k okys o`r ds f=kt;[kam dh ifjfefr Kkr djsa ftldk dsuæh; dks.k 90 gsa Find the perimeter of a sector of a circle whose central angle is 90 and radius 7 cm. 7-8 lsehú Hkqtk okys oxz ds vanj cus egùke oxz dk {ks=kiqy Kkr djsaa Find the area of the circle that can be inscribed in a square of side 8 cm. 8- fn[kk, fd fcunq A(0,1), B(, ), C(, 4)lajs[k gsaa Show that the points A(0,1), B(, ), C(, 4)are collinear. 9- ;fn 5sin θ = 4rks cosθvksj cotθdk eku fudkysaa If 5sin θ = 4then find cosθand cotθ. 0- fl¼ djsa fd sin 40 + sin 75 = cos15 + cos50 Prove that sin 40 + sin 75 = cos15 + cos fn;s x;s lekarj Js.kh ] 8] 1] 18] ds 15oka in dk eku fy[ksaa Write a 15 th term of the given A.P., 8, 1, 18,... y?kq mùkjh; iz'u %& Short Questions: - fl¼ djsa fd,d vifjes; la[;k gsa Prove that is an irrational number. - nksa la[;k, 96,oa 404 dk eúlú rfkk yúlú vhkkt; xq.ku[kam fof/ ls Kkr djsaa Find the HCF and LCM of 96 and 404 by prime factorization method.

102 4- gy djsa%& 6x + y 6xy = 0vkSj x + 4 y 5xy = 0 Solve the equation : 6x + y 6xy = 0and x + 4 y 5xy = 0 5- f}?kkr lehdj.k x 5x + 1 = 0ds ewyksa dh izd`fr Kkr djsaa Find the nature of roots for the quadratic equation 6- fdlh ABC esa AD BC] ;fn In any triangle triangle. ABC, AD BC. If x 5x + 1 = 0. BD DA = rks fl¼ djsa fd ABC ledks.k f=khkqt gsaa DA DC BD DA = then prove that ABC is right DA DC 7- fcunq A(5,1) dh y &v{k ls yacor nwjh Kkr djsaa Find the perpendicular distance of point A(5, 1) from y-axis. 8-,d csx esa 6 yky,oa 5 uhyk xsan gsa ;fn csx ls,d xsan dks ;n`pn;k fudkyh tk, rks blds yky ugha gksus dh izkf;drk Kkr djsaa A bag contains 6 red and 5 blue balls. If a ball is drawn at random. Find the probability the ball is not red fl¼ djsa fd cos A sin A = cos A 1 Prove that 4 4 cos A sin A = cos A tan 45 + cos 0 sin 60 dk eku fudkysaa Find the value of tan 45 + cos 0 sin fdlh Ldwy dh d{kk IXdh 51 ym+ds dh mq pkbz;ksa (lsehú esa) dk,d losz{k.k fd;k x;k vksj fueufyf[kr vkadm+s izkir fd, x,a ekè;d mq pkbz Kkr dhft,a mq pkbz (lsehú esa) ym+dksa dh la[;k 140 ls de ls de ls de ls de ls de ls de 51

103 A survey regarding the heights (in cm) of 51 boys of class IX of a school was conducted and the following data was obtained. Find the median height. Height (in cm.) Numbers of Boys less than less than less than less than less than less than fueufyf[kr vkadm+ksa dk cgqyd Kkr dhft,a izkirkad 10 ls de 0 ls de 0 ls de 40 ls de 50 ls de 60 ls de 70 ls de 80 ls de ckjackjrk Find the mode of the following data. Marks less less less less less less less less than than than than than than than than Frequency nh?kz mùkjh; iz'u %& Long Question : 4-,d vk;rkdkj dkxt dk VqdM+k ftldk eki 44cm 18cmgSA bls yackbz ds vuq:i eksm+dj csyukdkj cuk fn;k tkrk gsa csyu dk vk;ru fudkysaa 5 A rectangular sheet of paper of dimension 44cm 18cm is rolled along its length and a cylinder is formed. Find volume of cylinder.

104 44-,d VkWoj ds ikn ls,d edku ds 'kh"kz dk muu;u dks.k 0 rfkk edku ds ikn ls VkWoj ds 'kh"kz dk muu;u dks.k 45 gsa VkWoj dh mq pkbz 0 ehvj gs rks edku dh mq pkbz Kkr djsaa 5 The angle of elevation of the top of a building from foot of the tower is 0 and the angle of elevation of top of the tower from the foot of the building is 45. If the tower is 0m high, find the height of the building. vfkok OR lw;z ds muu;.k dks.k 0 ds fy, VkWoj dh Nk;k dh yackbz 0 ehvj gs rks lw;z ds 60 ds muu;u dks.k ds fy, Nk;k dh yackbz Kkr djsaa If the shadow of a tower is 0m long when sun s elevation is 0. What is the length of the shadow when sun s elevation is 60? 45- xzkiqh; fof/ ls lehdj.k ;qxe dk gy djsaa 5 x y + 1 = 0 x + y 1 = 0 Graphically solve the simultaneous equation. x y + 1 = 0 x + y 1 = fl¼ djsa fd o`r ds O;kl ds Nksj fcunqvksa ij [khaph x;h Li'kZ js[kk, lekarj gksrh gsaa 5 Prove that the tangents drawn at the ends of a diameter of a circle are parallel. vfkok OR fl¼ djsa fd ledks.k f=khkqt ds d.kz dk oxz vu; nks Hkqtkvksa ds oxksz ds ;ksx ds cjkcj gksrk gsa Prove that in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. 47-,d f=khkqt ABCdh jpuk djsa tcfd AB = 4cm, BC = 5cm, AC = 6cm A blesa ml nwljs f=khkqt dh jpuk djsa ftudh Øekuqlkj Hkqtk, ABC dh Hkqtkvksa dk xq.kk gksa 5 Construct a triangle ABC in which AB = 4cm, BC = 5cm, AC = 6cm. Then construct another triangle whose sides are times the corresponding sides of ABC. vfkok OR 7 lsehú dh,d ljy js[kk [khapdj mls % 5 ds vuqikr esa var% fohkkftr djsaa Draw a line segment of length 7cm. Find a point on it which divides internally it in the ratio :5.

105 SOLUTION (1) ([k) () (x) () (?k) (4) (x) (5) (d) (6) (x) (7) ([k) (8) (?k) (9) (d) (10) (x) (11) lkr (1) c a (1) 60 (14) lekarj (15) n 1 (16) Li'kZ fcunq (17) lkoz&varj ( d ) (18) sinθ (19) (0) 1 (1) fn;s x;s lehdj.k x + y = 7, 6x + 5y = 11esa a 1 b1 =, = a 6 b 5 a 1 b 1 a b () lehdj.k ;qxe vfojks/h gsa 1 10 x = x x x 10 1 = 10 x x 1 = 0 10 x 5x + x 1 = 0 ( x 1)(5 x + 1) = x =, 5 () fp=k esa] BAC = 90 vksj AM BC AMB = AMC = 90 <ABM CAM nksuksa f=khkqt le:i gksaxsa

106 (4) fn;s x;s fp=k esa OPQ = 0 PAQ = OPQ = 0 = 40 (5) fcunq (] ) vksj (4] 1) ds chp dh nwjh (6) f=kt;[kam dh ifjfefr (Perimeter of sector) = ( x x ) ( y y ) = ( 4) + ( 1) = 8 = bdkbz πrθ = r + r = = = 5cm (7) oxz dh Hkqtk (side of square) = 8cm o`r dh f=kt;k = 4cm o`r dk {ks=kiqy = πr 5 = 16 = cm 7 7 (8) rhu fcunqvksa A(0] 1)] B(] ) vksj C(] 4) ls cus ABC dk {ks=kiqy 1 = [0( 4) + (4 1) + (1 )] 1 = [ ] = 0

107 pwafd ABC dk {ks=kiqy = 0 vr% rhuksa fcunq lajs[k gsaa (9) 4 5sin θ = p 4 sin θ = 5 = h ;gk ] vr% p = 4, h = 5, b =? b = h p = (5) (4) = 5 16 = 9 = cos θ = rfkk 5 cot θ = 4 (0) sin 40 + sin 75 = sin(90 50) + sin(90 15) = cos50 + cos15 = cos15 + cos50 (1) lekurj Js.kh ] 8] 1] 18] ;gk a =, d = t t1 = 8 = 5, n = 15 t = a + ( n 1) d n t 15 = + (15 1) 5 = + 70 = 7 () ekuk fd,d ifjes; la[;k gsa rc = p, p,oa q/u iw.kkzad rfkk mhk;fu"b xq.ku[kam ugha gsa q oxz djus ij p = or p = q q (le la[;k) --- (1) phkh le la[;k gsa --- () iqu% ekuk fd p = m, miw.kkzad gsa p = 4m or, = m or, q (lehú1 ls) q = m qhkh le la[;k gsa --- () () vksj () ls p, q le la[;k gs blfy, p vksj q esa mhk;fu"b xq.ku[kam gksxka tks dyiuk ds foijhr gsa vr% ifjes; la[;k] ;g ekuuk xyr gsa vfkkzr~ vifjes; la[;k gsa

108 () 96 = 404 = 101 eúlú = = 4 yúlú = 101 = 9696 (4) 6x + y 6xy = 0 or, 6x + y = 6xy xyls Hkkx nsus ij 6 + = 6 y x iqu% x + 4y = 5xy xyls Hkkx nsus ij eqn.() 4 + = 5 y x eqn.(1) --- (1) --- () = 15 y x 6 + = 6 y x 9 = 9 x x = 1 lehú (1) esa x = 1 j[kus ij] 6 6 y = = y = 6 y = gy % x = 1, y = (5) x 5x + 1 = 0 a = b = c = ;gk, 5, 1 D = b 4 ac = ( 5) 4 1 = = 5 8 = < 0 vr% ewy oklrfod ugha gksxka (6) BDA rfkk CDA esa] BDA = CDA = 90 iz'u ls] BD DA = DA DC BDA CDA

109 BAD = ACD o 90 ABD = ACD vfkkzr~ ABD + ACD = 90 BAC = 90 vr% ABC,d ledks.k f=khkqt gsa (7) y- v{k ij fcunq dk fu;ked (0] 1) gsa (5] 1) ls yecor nwjh (0] 1) dk = (0 5) + (1 1) = = 5 (8) csx esa dqy xsanks dh la[;k = =11 lahko dqy ifj.kkeksa dh la[;k =11 yky xsnkas dh lúa 6 P(yky xsan) = = lha ko dqy ifj.kkekas dh lúa 11 bdkbz P (yky xsan ugha) = = 1 = = (9) cos A sin A = (cos A) (sin A) = (cos A + sin A)(cos A sin A) = cos A sin A cos A + sin A = 1 = cos A (1 cos A) = cos A 1 (40) tan 45 + cos 0 sin 60 = (1) + = 1 + = 4 4

110 (41) oxz vurjky ( c. i.) ckjeckjrk ( f ) lap;h ckjackjrk ( c f ) 140 ls de µ µ µ µ µ vc N = 51gSA N = 51 rks N = 51 = 5.5gSA N cf lw=k ls] ekè;d = l + h f varjky 145 µ 150 ekè;d oxz varjky esa vkrk gsa l (fueu lhek) = 145 ekè;d oxz 145 µ 150 tks Bhd igys oxz dh lap;h ckjackjrk ( cf ) = 11 eè;e oxz 145 µ 150 dh ckjackjrk f = 18rFkk oxz eki ( h ) = 5 gsa (4) ekè;d = = = rks ym+dksa dh ekè;d mq pkbz = 149.0cm gsa izkirkad lap;h ckajckjrk ckjackjrk 10 ls de µ (¾40 µ 5) 0 µ (¾60 µ 40) 0 µ (¾75 µ 60) 40 µ (¾95 µ 75) 50 µ (¾10 µ 95) 60 µ (¾195 µ 10) 70 µ (¾45µ 195)

111 cgqyd oxz 60 µ 70 gs D;ksafd bldh ckjackjrk lokzf/d gsa fo f (Mode) M 1 o = l + i fo f f tgk ] 1 1 l = Modal class vfkkzr~ vf/dre ckjeckjrk okyk oxz dh fueu lhek fo = Modal class dh ckjeckjrk f 1 = Modal class ds Bhd igys okyk oxz dh ckjeckjrk f 1 = Modal classds Bhd ckn okyk oxz dh ckjeckjrk i =oxz vurjky dh yackbz ;gk ] l = 60, fo = 65, f 1 = 5, f1 = 50, i = 10 vhkh"v cgqyd 65 5 = ( 65) = 60 + = = 60 + = = 66.7Ans. (4) vk;rkdkj dkxt dks yackbz ls eksm+k x;k gsa csyu dh ifjf/ = π r = 44cm or, 44 7 r = = 7cm csyu dh mq pkbz ( h) = 18cm csyu dh vk;ru = πr h 7 = 7 18 = 77cm Ans.

112 (44) ekuk fd edku dh mq ú h m gs rfkk BD = x cm ABD esa] 0 tan 45 = x or, x = 0 m CBD esa] tan 0 = h 0 1 or, = h 0 0 or, h = 0 or, h = = 10 m = m edku dh mq ú = 17. m ABC esa] vfkok OR tan 0 = h 0 1 or, = h 0 or, h = 10 iqu% ABD esa] tan 60 = h x 10 or, = x or, x = 10m

113 Nk;k dh yackbz = 10m (45) lehdj.k x y + 1= 0ls y = x (1) lehdj.k x + y 1 = 0ls lehú(1) dk eku lkj.kh 1 x y = --- () x 0 1 y 1 0 lehú() dk eku lkj.kh x 0 4 y 6 0 lehdj.kksa ds vkys[k (] ) ij izfrpnsn djrs gsaa vr% gy x =, y = (46) ekuk fd,d o`r ftldk dsuæ OgSA ABo`r dk O;kl gsa CDrFkk EF nks Li'kZ js[kk;sa gsaa AB CDrFkk AB EF CAB = 90 rfkk ABF = 90 CAB = ABF rfkk ;s,dkarj dks.k gsaa CD EF Proved ekuk fd ABC,d ledks.k f=khkqt gsa ftldk B = 90 ACd.kZ gsa BD ACMkyk ADB rfkk ABC esa vfkok OR

114 BAD = BAC (mhk;fu"v dks.k) ADB = ABC = 90 ADB ABC AB AD or AB AC AD AC = AB = --- (1) blh izdkj BDC rfkk ABC Hkh le:i gsaa BC DC or BC AC DC AC = BC = --- () (1)$() djus ij AB + BC = AC(AD + DC) = AC AC = AC or, AC = AB + BC Proved. (47) cukov ds pj.k%µ BC = 5cm BdsUæ ls AB = 4cm pki [khapk C dsuæ ls AC = 6cm pki [khapka dvku fcanq feyk;ka ABC BX ij BB 1 = B1B = BBfpUg yxk;saa cuk B,oa Cdks feyk;saa B ls B C B C[khapsaA C ls C A CA[khapsaA vhkh"v f=khkqt A BC cuka

115 cukov ds pj.k%µ vfkok OR (i) AB = 7cm[khapkA (ii) A fcunq ls U;wudks.k ysdj AX [khapka A X ij AA 1 = A1A = AA,... = A7A 8, 5 + = 8 cjkcj pki [khapka (iii) BA8dks feyk;saa A P BA [khapka fcunq P, ABdks :5esa ck Vrh gsa (iv) 8 AP : PB = :5 bl izdkj P fcunq ABdks %5 ds vuqikr esa var% fohkkftr djrk gsa

116 ! # % #&& () +,. /!,), %,0) 1,/ # % %) (,/ & ( %456 0 ) 7,/ )!

117 Lke; %?kavk 15 feuv xf.kr (lsv&7) MATH (SET-7) Time : Hrs 15 Min. Full Marks : 100 lgh mùkj pqus %& Choose the correct answer :- fueufyf[kr esa dksu vifjes; gs \ 1 (d) 64 6 ([k) 81 (x) 15 (?k) 49 9 Which of the following number is irrational? (a) 64 6 (b) 81 (c) 15 (d) - fueukafdr esa dksu cgqin ugha gs \ 1 (d) 7x ([k) y + (x) x + x + 7(?k) Which of the following is not polynomials. (a) 7x (b) - ;fn y + (c) x + x + 7 (d) x x x x + 7 sin θ = gks rks cosθeku gksxkµ 1 5 (d) 4 5 ([k) 4 (x) 5 6 (?k) 5 If sin θ = then value of cosθ 5 (a) 4 5 (b) 4 (c) 5 6 (d) 5 4- ;fn cosθ = gks rks θdk eku gsµ 1 (d) 45 o ([k) 60 o (x) 0 o (?k) 90 o If cosθ = then the value of θ. (a) 45 o (b) 60 o (c) 0 o (d) 90 o 5- fueufyf[kr esa dksu&lh?kvuk dh izkf;drk ugha gks ldrk gsa 1 (d) ([k) 15 (x) 15% (?k) 0.7% In the following which is not the probability of an event? (a) (b) 15 (c) 15% (d) 0.7%

118 6- ;fn fdlh o`r dh f=kt;k nqxquh dj nh tk; rks iqjkus,ao u;s o`rksa ds {ks=kiqyksa dk vuqikr D;k gksxka 1 (d) ([k) (x) (?k) If the radius of a circle becomes twice then the ratio of any of the old and new circle will be (a) (b) (c) (d) 7- fcunqvksa (µ5] 7) vksj (µ1] ) ds chp dh nwjh gsµ 1 (d) ([k) (x) 4 (?k) 5 Find the distance between the point ( 5, 7) and ( 1, ) (a) (b) (c) 4 (d) 5 8- ckg~; Li'kZ djus okys nks o`rksa ds mhk;fu"v Li'kZ js[kkvksa dh la[;k fdruh gsa 1 (d) 1 ([k) (x) (?k) 4 How many are common tangents if two circle are touched externally each other? (a) 1 (b) (c) (d) 4 9- f}?kkr cgqin ds 'kwu;kad dh la[;k fdruh gksrh gs \ 1 (d) ([k) (x) 1 (?k) 4 How many are numbers of zeroes in quadratic polynomial? (a) (b) (c) 1 (d) ;fn cgqin P( x) = x x + 5esa 'kwu;kad α, β gks rks α + βdk eku gksxkµ 1 (d) 5 ([k) µ5 (x) (?k) µ If α, β are the zeroes of the polynomial P( ) = + 5 x x x then value of α + βwill be (a) 5 (b) 5 (c) (d) fjdr LFkkuksa dh iwfrz djs %& Fill in the blanks :- 11- forj.k esa vf/dre ckjackjrk okyk oxz oxz dgykrk gsa 1 In a frequency distribution the class having maximum frequency is said to be tc fdlh vuqøe ds in fdlh fu;e ds rgr fy[ks tkrs gsa mls dgrs gsa 1 When the terms of a sequence has written in a difinite pattern than it is called π la[;k gsaa 1 π is an... number. 14- rhu?kkr okys cgqin dgykrs gsa 1 The polynomial having three degree is called...

119 15- fn;s x;s fp=k esa ABC esa A = 90 vksj AD BC A ;fn BD = 9 lsehú rfkk CD = 4 lsehú] rks AD = In the given figure A = 90 and AD BCif BD = 9cm and CD = 4cm then what will be of AD = ewy fcunq ls P( x, y) dh nwjh gksxha 1 The distance of a point P( x, y) from the origin is dkrhz; ry esa flfkr fdlh fcunq (µ4] 0) ds Hkqt dk eku gksxka 1 The value of abscissa of point ( 4, 0) situated at Cartesian plane sec θ 7 tan θ = ;fn θ = 90, rks tan θ =... 1 If θ = 90, then the value of tan θ is.. 0- ads eku ds fy, d m gksxka 1 d mfor the value of a... vfr y?kq mùkjh; iz'u%& Very Short Questions : 1- f}?kkr cgqin Kkr djsa] ftuds 'kwu;dksa dk ;ksx,oa xq.kuiqy Øe'k% µ vksj gsa Find the quadratic polynomial. If sum and product of zeroes are and respectively. - ewyksa dh izd`fr crkb, ;fn f}?kkr lehdj.k x 6x + = 0gksA If quadratic equation is 6 + = 0 x x then find the nature of its roots. - ABC,d leckgq f=khkqt gs] ftldh izr;sd Hkqtk a lsehú gks rks fl¼ dhft, fd f=khkqt dk 'kh"kzyec alsehú gksxka ABC is an equilateral triangle whose each sides are a cm then prove that altitude of the triangle will be a?

120 4- fdlh o`r ds dsuæ ls 5 lsehú nwj flfkr fcunq A ls [khaph xbz Li'kZ js[kk dh yackbz 4 lsehú gs] rks o`r dh f=kt;k fudkysaa From a point A which is at the distance 5 cm from the centre of the circle. The length of the tangent drawn on the circle is 4 cm. Find its radius. θ 5- fcunqvksa ( a cos, 0) rfkk ( 0, a sin θ ) ds chp dh nwjh fudkysaa Find out the distance between the points ( a cos θ, 0 ) and ( 0, a sin θ ). 6- ;fn,d o`r ds {ks=kiqy vksj ifjf/ dk la[;kred eku leku gs] rks o`r dh f=kt;k D;k gksxh \ What will be the radius of the circle if the area of a circle and circumferences are numerically same? 7- 'kadq dk vk;ru fudkysa] ftldh mq pkbz 8 lsehú vksj vk/kj dh f=kt;k 1 lsehú gsa Find the volume of the cone whose height 8 cm and base of radius is 1cm. k rfkk B( k, 5) 8- ;fn fcunq A(, ) ls fcunq P(0, )dh nwfj;k leku gks rks k dk eku Kkr djsaa If the point P(0, ) is equidistant from the points A(, k) and B(k, 5). Find the value of k. θ = rks fl¼ djsa fd 4 cos θ cosθ = 0 9- ;fn 0 If θ = 0 then prove that 4 cos θ cosθ = vhkkt; xq.ku[kam fof/ }kjk 7] 11] 19 dk egùke lekorzd vksj y?kqre lekorzd Kkr dhft,a Find the HCF and LCM of 7, 11, 19 by prime factorisation method. 1- laekrj Js.kh 0-6] 1-7] -8] -9] ds izfke 100 inksa dk ;ksx Kkr djsaa Find the sum of first 100 terms of the A.P. 0.6, 1.7,.8,.9,... y?kq mùkjh; iz'u %& Short Questions: - fl¼ djsa fd 7,d vifjes; la[;k gsa Prove that 7 is an irrational number. - fueufyf[kr ckjackjrk cavu fdlh eksgyys ds 68 mihkksdrkvksa dh fctyh dh ekfld [kir n'kkzrk gsa budk ekè; Kkr fdft,a Following are the frequency distribution of a colony where monthly electric consumption of 68 consumers shows then, what will be their mean? ekfld [kir (;wfuv esa) Monthly Consumption (Unit) mihkksdrkvksa dh la[;k No. of Consumers 5µ5 5µ45 45µ65 65µ85 85µ µ15 15µ

121 4- gy djsa%µ = rfkk 6 x 1 y 5 1 Solve it:- x 1 + y = and 1 x 1 y = 6 1 x 1 y =. 5- f}?kkr lehdj.k x x + 4 = 0 dk fofodrdj vksj ewyksa dh izd`fr Kkr dhft,a Find the discriminant and nature of the roots of the quadratic equation x x + 4 = ;fn ABCesa AD BCgS rks fl¼ djsa fd AB AC BD CD If in ABC8 AD BCprove that AB AC = BD CD. = 7- k dk eku Kkr dhft, ;fn rhuksa fcunq, (16, )] ( k, 8) rfkk (4, 10) lajs[kh gksa Find the value of k if three points (16, ), (k, 8) and (4, 10) are collinear. 8- fdlh dkj.k ls 10 [kjkc isu 10 vpns isuksa ls fey x;s gsaa dsoy ns[kdj ;g ugha crk;k tk ldrk gs fd dksu isu [kjkc isu gs ;k vpnk gsa bl fej.k esa ls,d isu ;kn`pn;k fudkyk tkrk gsa fudkys x;s isu dh vpnh gksus dh izkf;drk Kkr dhft,a Due to some reason 10 damage pen and 10 good pens has mix. It can not be seen only by searching which is damage or which is good. From this mix. the pens is taken out at random. Find the probability of pen which is good after taking out. 9- fl¼ dhft, fd 1 cot A = cot A 1 tan A 1 cot A Prove that = cot A 1 tan A. 40- fl¼ dhft, fd 1 cos A 1 cos A = coseca cot A, Prove that = coseca cot A 1 + cos A 1 + cos A 41- Hkwfe ds,d fcunq ls tks ehukj ds ikn fcunq ls 60 ehvj dh nwjh ij gsa ehukj ds f'k[kj dk muu;u dks.k 60 gsa ehukj dh mq pkbz Kkr dhft,a A point from a land where base of tower point s distance is 60 meter. Tower s angles of elevation is 60 then find the height of tower dk f=kdks.kferh; vuqikr fudkysaa Find the trigonometrical ratios of 45.

122 nh?kz mùkjh; iz'u %& Long Question : 4-,d f[kyksus dh f=kt;k 6 lsehú okys,d 'kadq ds vkdkj dk gs] tks mlh f=kt;k okys,d v¼zxksys ij vè;kjksfir gsa bl f[kyksus dh laiw.kz mq pkbz 11 lsehú gsa bl f[kyksus dk laiw.kz i`"bh; {ks=kiqy Kkr dhft,a 5 A toy is in the form of a cone of radius 6 cm. mounted on a hemisphere of same radius. The total height of the toy is 11 cm. Find the total surface area of the toy. 44- xzkiqh; fof/ ls gy djsa%µ 5 x + y = 6rFkk x 4y = 1 Solve graphically: x + y = 6and x 4y = ;fn uhps fn;s gq, cavu dk ekè;d 8-5 gks rks xvksj ydk eku Kkr dhft,a oxz varjky 0&10 10&0 0&0 0&40 40&50 50µ60 ;ksx ckjackjrk 4 x 1 15 y 5 60 If following are the frequency distribution of median then find the value of x and y? Class Interval Total Frequency 4 x 1 1 y ;fn OdsUæ okys o`r ij okg~; fcunq T ls nks Li'kZ js[kk, TP rfkk TQ [khaph xbz gs] rks fl¼ dhft, fd PTQ = OPQ 5 If two tangents TP and TQ are drawn to a circle with centre O from a external point T then prove that PTQ = OPQ lsehú yeck,d js[kk[kam [khafp, vksj bls % ds vuqikr esa fohkkftr dhft,a izr;sd Hkkxksa dks ekfi,a 5 Draw a line segment which is 5.6 cm long divided into : then measure each parts? OR -5 lsehú f=kt;k dk o`r [khapsa blds ifjxr,d leckgq f=khkqt cuk;saa Draw a circle whose radius is.5 cm and form circumscribe equilateral triangle.

123 SOLUTION (1) (x) () (x) () (d) (4) (x) (5) ([k) (6) (x) (7) (x) (8) (x) (9) (d) (10) (x) (11) cgqyd (1) Js<+h (1) vifjes; (14) f=k?kkrh (15) 6 lsehú (16) x + y (17) µ4 (18) 7 (19) 1 (0) 0 (1) α =, β = f}?kkr cgqin x ( ) x. = α + β + α β = x ( + ) x + ( ) = x + x 6 () f}?kkr lehdj.k x 6x + = 0 ax + bx + c = 0 ls rqyuk djus ij a =, b = 6, c = D = b 4 ac = ( 6) 4 = 6 4 = 1 1 > 0 f}?kkr lehdj.k ds oklrfod vksj fhkuu ewy gksaxsaa () ABCesa AB = BC = CA = arfkk AD BC ADB esa] a a AD = ( a) = a 4 = f=khkqt ds 'kh"kzyac a = alsehú 4 = a lsehú (lr;kfir) (4),d o`r ftldk dsuæ lsehú O gsa ckg~; fcunq A ls Li'kZ js[kk PA dh yackbz 4 lsehú gs rfkk A dh dsuæ O ls nwjh 5 lsehú gsa <APO = 90 vc ledks.k OPAesa] ;k] ;k] OA = PA + OP (5) = (4) + OP 5 16 = OP ;k] 9 = OP OP = cm

124 (5) fcunqvksa ( a cos θ, 0) rfkk (0, a sin θ ) ds chp dh nwjh (6) o`r dh ifjf/ ¾ o`r dk {ks=kiqy π R = πr R = R R = ek=kd 1 (7) 'kadq dk vk;ru ¾ πr hesa] 1 = = 1 8 cm = 196 cm (8) iz'ukuqlkj] fn;k gs%µ PA = PB ( a ) ( a ) = ( cos θ ) (0) + (0) ( sin θ) = a cos θ + a sin θ ;kfu fd] PA = PB... (1) PA = (0 ) + ( k) vc] lehdj.k (1) ls PB = (0 k ) + ( 5) (0 ) + ( k) = (0 ) + ( k ) ;k] k k k = + 9 ;k] 4k = 4 k = 1 (9) LHS = 4 cos θ cos θ = 4cos 0 cos0 = 4 4 = 8 = 0 RHS Proved = = a (cos θ + sin θ ) = a 1 = a (0) lozizfke gesa nh gqbz la[;kvksa ds vhkkt; xq.ku[kam fy[krs gsaa 7 = 7 1, 11 = 11 1vkSj 19 = 19 1 yúlú = = 146, eúlú = 1 Ans. (1) ekuk fd A.Pdk igyk in arfkk lkoz&varj d gs rks

125 a = 0.6, d = = 1.1 gesa A.P.ds 100 inksa dk ;ksx Kkr djuk gsa n Sn = [ a + ( n 1) d] esa a = 0.6, d = 1.1, n = 100j[kus ij 100 S 100 = [ (100 1) 1.1] = 50( ) = 50( ) = = 5505 Ans. () ekuk fd 7,d ifjes; la[;k gsa vfkkzr ge nks iw.kkzad a vksj b( b 0) izkir dj ldrs gsa a fd 7 = gsa b ;fn avksj b esa 1 ds vfrfjdr dksbz mhk;fu"v xq.ku[kam gksa rks ge ml mhk;fu"v xq.ku[kam ls Hkkx ysdj avksj b dks lgvhkkt; cuk ldrs gsaa vr% b 7 = agsa () nksuksa i{kksa dks oxz djus ij gesa 7b a = --- (1) izkir gksrk gsa vr% a, 7ls fohkkftr gsa blfy, 7, a dks Hkh fohkkftr djsxka ekuk fd a = 7 c, tgk,d iw.kkzad gsa adk eku lehdj.k (1) esa j[kus ij] 7 b = (7 c), 7b = 49 c, b = 7c b, 7ls fohkkftr gsa blfy, 7, bdks fohkkftr djsxka vr% avksj b esa de ls de,d mhk;fu"v xq.k[kam 7 gsa ijurq blls rf; dk fojks/khkkl izkir gksrk gs fd a vksj b lgvhkkt; gsa gesa ;g fojks/khkkl viuh =kqfviw.kz dyiuk ds dkj.k izkir gqvk fd 7,d ifjes; la[;k gsa vr% fu"d"kz izkir gqvk fd 7,d vifjes; la[;k gsa ekfld [kir oxz fpug `ckjackjrk ( x ) ( f ) f x 5µ µ µ µ µ µ µ f = 68 fx = 5400

126 Ekkè; fx 5400 = = = Ans. f 68 (4) = --- (1) x 1 y 6 1 x 1 y = --- () lehú(1) dks ls xq.kk dj lehú() ds lkfk tksm+us ij] x 1 + y + x 1 y = + or, 1 = 7 or, = 1 x 1 x 1 or, x 1 = or, x = + 1 = 4 x = 4 lehú(1) esa x dk eku gksus ij y = + y = = = = y = y 1 y = + = 5 x = 4, y = 5 Ans. (5) x x + 4 = 0 a =, b =, c = 4 D = b 4 ac = ( ) 4 4 = 9 = D < 0, dksbz oklrfod ewy ugha gsa (6) ledks.k ABD esa BD = AB AD...(1) iqu% ledks.k ADCesa CD = AC AD...() lehú(1) esa () dks j[kus ij BD CD = (AB AD ) (AC AD ) = AB AD AC + AD = AB AC Proved

127 (7) eku fy;k fd A( x1, y1 ) = (16, ) B( x, y) = ( k, 8) vksj C( x, y) = (4, 10) ;fn ABCdk {ksú ¾ 0 rks fcanq A,B, C lajs[kh gksxha 1 [ ( ) + ( ) + ( )] = x y y x y y x y y 1 16( ) + k( 10 ) + 4( + 8) = 0 [ 0] 16 + k = 0 1k + 40 = 0 1k + 7 = k = 7 k = = 6 1 vr% fn;s x, fcunq k = 6 ds lajs[kh gsa (8) isuksa dh dqy la[;k ¾ 10 $ 10 ¾ 10 P (,d vpnk isu) (9) L.H.S. = 10 1 = = Ans cos A 1 1 cot A sin A = 1 tan A sin A 1 cos A sin A cos A sin A sin A cos A cos A = cos A sin A = sin A cos A sin A cos A sin A cos A cos A cos A = = = cot A = R.H.S Proved sin A (sin A cos A) sina (40) L.H.S. = 1 cos A 1 cos A 1 cos A = 1 + cos A 1 + cos A 1 cos A (1 cos A) (1 cos A) 1 cos A = = = 1 cos A sin A sin A sin A = (coseca cota) = coseca cot A = R.H.S Proved

128 (41) ekuk fd ABehukj gs ftldh mq pkbz h ehú gsa Hkwfe ds,d fcunq C tks ehukj ds ikn Bls 60 ehú dh nwjh ij gs ;kfu BC =60 ehú ledks.k ABCesa AB h tan 60 = = BC 60 h = 60 h = 60 = = ehú 10.9 (4) ekuk fd ABC,d ledks.k lef}ckgq gsa ftldh izr;sd Hkqtk x gsa A = C = 45, B = 90, Ekuk fd AB = BC = x ikbfkksxksjl izes; ls] ledks.k ABCesa] d.kz ¾ yac $ vk/kj AC = AB + BC = x + x = x AC = x = x AB x 1 BC x 1 vr% sin 45 = = =, cos 45 = = = AC x AC x AB BC tan 45 = x x 1, cot 45 1 BC = x = = AB = x = AC AC sec 45 = x x, co sec45 BC = x = = AB = x = (4) 'kadq dh f=kt;k ¾ v¼zxksys dh f=kt;k R = 6 cm ehú f[kyksus dh dqy mq pkbz ¾ 11 lsehú 'kadq dh mq pkbz ¾ (11 µ 6) ¾ 5 lsehú 'kadq dh fr;zd mq pkbz R h = + = = = 65 = 7.81lsehú crzu dh dqy i`"bh; {ks=kiqy ¾ 'kadq dk i`"bh; {ks=kiqy $ v¼zxksys dk i`"bh; {ks=kiqy = π rl + π R = π R( l + R) = 6 ( ) = = 7.56 Ans. 7 7

129 (44) x + y = 6...(1) vksj x 4y = 1...() lehú(1) ls x + y = 6 or, x = 6 6y x 6 0 y 0 1 (x, y) (6, 0) (, 1) (0, ) lehú() ls x 4 y = 1 or, x = 1 + 4y or, 1 + 4y x = x y 0 1 (x, y) (6, 0) (8, 1) (0, ) nksuksa vkys[kksa dk dvku fcunq (6] 0) gsa vr% gy x = 6, y = 0gksxkA

130 (45) 45 + x + y = 60 oxz varjky ckjackjrk ( f ) lap;hckjackjrk ( c f ) 0µ µ0 x 4 + x 0µ x 0µ x 40µ50 y 40 + x + y 50µ x + y dqy N = 60 x + y = = 15 y = 15 x...(1) Ekkè;d ¾ 8-5 (fn;k gqvk gs) N = = 0 ekè;d oxz ¾ 0 µ 0 ;gk ] l = 0, h = 10, N = 60, C = 4 + x, f = 1 Ekkè;d N l + C = h; f 60 (4 + x ) 8.5 = (6 x) 8.5 = = x 81.5 = 10x 81.5 x = = lehú (1) ls y = = 6.85 vr% x = 8.15, y = 6.85 (46) fn;k x;k gs fd %µo dsuæ okys o`r ds okg~; fcunq Tls TPvkSj TQLi'kZ js[kk, [khaph xbz gsa tks o`r dks Øe'k% PvkSj QfcUnqvksa ij Li'kZ djrh gsaa fl¼ djuk gs fd PTQ = OPQ izek.k%µ ekuk fd PTQ = x rc TQP + TPQ + PTQ = 180

131 ds rhuksa dks.kksa dk ;ksx ¾ 180 TQP + TPQ = 180 x --- (1) pwafd o`r ds ckg~; fcunq ls [khaph xbz Li'kZ js[kkvksa dh yackbz leku gksrh gsa TP = TQ TQP = TPQ fdurq TQP + TPQ = 180 x TQP + TQP = 180 x vr% TPQ,d lef}ckgq gsa 180 x x TQP = = 90 x x = vc OPQ = OPT TPQ = QPQ = PTQ PTQ = OPQ Proved (47) jpuk ds pj.k%µ (d) AB = 5.6 cmdk,d js[kk[kam [khapka ([k) A ls mqij dh vksj U;wudks.k BAX = 60 ABY = 60 cuk;ka (x) 5 fcunqvksa dks AXvkSj BYij leku eki dk vafdr fd;ka cuk;k vksj B i uhps dh vksj leku (?k) AvkSj B 5 ] A vksj B vksj A 5, Bdks feyk;ka (M+) A, B js[kk[kam ABdks Pij % ds vuqikr esa fohkdr djrk gsa PA vksj PBdks eki dj ge ikrs gsa fd PA =.4cmvkSj PB =.cm

132 vfkok (d).5cmdh f=kt;k ls,d o`r [khapka ([k) o`r dks dsuæ Odks o`r ds fdlh fcunq Als feyk;ka (x) Oij AOB = 10 vafdr fd;k] tgk Bo`r ij gsa (?k) fiqj Oij BOC = 10 vafdr fd;k tgk ij C o`r ij gsa (M+) Aij <OAR = 90 cukdj Aij Li'khZ AQ [khapka (p) blh izdkj o`r ds fcunq B vksj Cij Hkh Li'khZ [khapka (N) rhuksa Li'kZ js[kkvksa dks c<+kdj PQR gqvka vafdr fd;ka bl izdkj PQR vhkh"v f=khkqt

133 ! # % #&& () +,. /!,), %,0) 1,/ # % %) (,/ & ( %456 0 ) 7,/ )!

134 xf.kr (lsv& 8) Maths (Set- 8) le; %?kavk 15 feuv lgh mùkj pqus%& Choose the correct answer :- 1- sin 45º dk eku gksxk A 1 1 (d) ([k) ½ (x) (?k) ¼ The value of sin 45º will be (a) (b) ½ (c) 1 (d) ¼ - fueufyf[kr esa dksu ifjes; la[;k gs \ 1 49 (d) 101 ([k) (x) (?k) 18 Which of the following is rational Number? (a) (b) 64 (c) (d) sin θ + cos θ cjkcj gs& 1 (d) ([k)&1 (x) 1 (?k) sin θ + cos θ is equal to (a) (b) -1 (c) 1 (d) 4- founq (&] ) fdl ikn esa gksxk A 1 (d) izfke ([k) f}rh; (x) r`rh; (?k) prqfkz Point (-, ) lies in which quadrant (a) First (b) Second (c) Third (d) Fourth

135 5- csyu dk oø i`"b dk {ks=kiqy gksxk 1 (d) rh ([k) πrh (x) πrh (?k) πrh The Curved Surface area of cylinder is (a) rh (b) πrh (c) πrh (d) πrh 6- fueufyf[kr esa dksu f}?kkr lehdj.k ugha gs \ 1 (d) (x + ) = x(x - 1) ([k) (x + 1) = (x - ) (x) (x - ) (x + ) = 6 (?k) x 1 + x = Which of the following is not quadratic equation (a) (x + ) = x(x - 1) (b) (x + 1) = (x - ) (c) (x - ) (x + ) = 6 (d) x 1 + x = 7- ;fn ABC esa AC = AB + BC rks B dh eki gs 1 (d) 60º ([k) 75º (x) 90º (?k) 45º What is measurement of B in triangle ABC if AC = AB + BC (a) 60º (b) 75º (c) 90º (d) 45º 8-7] ] 5] 8] 6] 10] 9] 1 dh ekfè;dk D;k gs \ 1 (d) 8 ([k) 7 (x) 6 (?k) 7-5 What is the median of 7,, 5, 8, 6, 10, 9, 1 (a) 8 (b) 7 (c) 6 (d) nks iklksa dks,d lkfk iqsadk tkrk gs A nksuksa ij 6 vkus dh izkf;drk gsa 1 (d) 6 1 ([k) (x) 6 (?k) 4 1 The probability of six on both if two dice will throw at once. (a) 1 6 (b) 1 6 (c) 5 6 (d) 1 4

136 10-4 ds fdrus xq.kt 10 vksj 50 ds chp im+rs gsa \ 1 (d) 60 ([k) 50 (x) 55 (?k) 5 How many multiples between 10 and 50. (a) 60 (b) 50 (c) 55 (d) 5 fjdr LFkkuksa dh iwfrz djsa %& Fill in the blanks : , -6, -,,..., 4 esa inksa dh la[;k gs A 1 The number of term in -10, -6, -,,..., 4 is ;fn nks la[;kvksa dk e- l-¾5 vksj y- l-¾50 rks la[;kvksa dk xq.kuiqy gksxk If H.C.F. of two numbers are 5 and L.C.M. is 50 then what will the multiplication oxz vurjky ds eè;&founq dks dgrs gsa A (Mid point of class Interval is... ) 14- f=khkqt dk {ks=kiqy = ½ x... x... 1 Area of triangle = ½ x... x ;fn tan 5º cot 5º = sina rks A dk eku gs A 1 If tan 5º cot 5º = sin A then value of A ,d la[;k gs A is a... number. 17- og lehdj.k ;qxe ftldk dksbz gy u gks lehdj.k dgykrk gs A 1 Such equation which has no solution is called... equation. 18- ;fn f}?kkr lehdj.k 7x - 5x + 1 = 0 ds 'kwu;d α] β gks] rks α + β = 1 If α] β be zeros of quadratic equation 7x - 5x + 1 = 0 then α + β =

137 19- nks cgqhkqt le:i gksrs gsa ;fn muds laxr dks.k cjkcj gks rfkk laxr Hkqtk, gksa Two polygon will be similar if their corresponding angle are equal and corresponding sides are lehdj.k px + qx + r = 0 ds fy, q - 4pr dks dgk tkrk gs A 1 For the equation px + qx + r = 0, q - 4pr is said to be fn[kk, fd x = - lehdj.k x + 5x + = 0 dk gy gs A Show that x = - is a solution of equation x + 5x + = ifjes; la[;k 4 5 ds 1 Find the decimal expansion of rational number is ter rminating or non terminating reccuring. n'keyo izlkj lkr ;k vlkar vkorhz gs] Kkr djsa A - nh xbz vkd`fr esa ;fn PQ RS gs rks fl¼ dhft, fd POQ SOR gs A 4 5 In the figures if PQ RS then prove that POQ SOR 4- ;fn fdlh vuqøe dk n ok in 4n + 5 gks rks blds izfke rhu in If nth term of sequence is 4n + 5 find first three term? 5- tk p dhft, fd fueu lehdj.k f}?kkr lehdj.k gs ;k ugha \ Kkr djsa A x x + = x, x 0 Examine following equation is quadric equation or not x ls0 eh0 f=kt;k okys fdlh o`r dh ifjf/ crkosa A Find the circumference of circle whose radius is 4.9 cm. x = x, x 0

138 7- eku Kkr djsa& Find the value of cos 0º (1 + tan 0º) 8- fdlh o`r dh ifjf/ ls0 eh0 gs A mlds prqfkkz'k dk {ks=kiqy fudkysa A Find the area of quadrant of circle if circumference of circle is cm. 9- ;fn?ku dk dqy i`"b {ks=kiqy 150 eh0 gks rks mldk vk;ru D;k gksxk A What will be volum if total surfaces Area of cube is 150 m 0- ewy foanq ls founq (4] &) dh nwjh Kkr dhft, A Find distance from origin and point (4, -). 1- ;fn cgqin x - x - (k - ) dk,d 'kwu;d &4 gks rks K dk eku fudkysa \ If one zero of the polynomial x - x - (k - ) is -4 then find the value of K. 1 - fl¼ dhft, fd,d vifjes; la[;k gs A Prove that 1 is an irrational number. - nks yxkrkj?kukred iw.kkzd Kkr djs ftuds oxks± dk ;ksx 65 gks A Find two consecutive positives no whose sum of squares is ;fn A = 60º vksj B = 0º rks fl¼ djsa If A = 60º and B = 0º then prove that tan (A - B) = tan A tan B 1 + tan A.tan B tan A tan B 1 + tan A.tan B 5- fueu lkj.kh 400 fu;kwu ysiksa dk thou dky n'kkzrk gs A thou dky (?kavks esa) ysiksa dh la[;k 1500& & & &500 86

139 500& & & ,d ysai dk ekè;d thou dky Kkr dhft,a In the following table represent the time period of 400 neon Lamp.! 1500& & & & & & & Find the median life time of a lamp. 6- fl¼ djsa 1 + cot θ = cosecθ 1+ cosecθ Prove that 1 + cot θ = cosecθ 1+ cosecθ 7- fueufyf[kr okjackjrk cavu (lkj.kh) ds fy, ekè; ifjdfyr djsa A oxz vurjky 0&8 8&16 16&4 4& &40 40&48 ckjackjrk The following frequency distribution (table) is # % 0&8 8&16 16&4 4& &40 40&48 & (

140 Find the mean. 8- f}?kkr cnqin x - 8 ds 'kwu;d Kkr dhft, A Find the zeros of quadratic polynomial x gy djsa & (x + 4) (x - 4) = 0 Solve - (x + 4) (x - 4) = jke vksj ';ke nkslr gsa A D;k izk;fdrk gs fd mu nksuksa dk tue fnu (1) fhkuu gks (),d gh gks \ (vf/o"kz dh mis{kk djsa) \ Ram and Shyam is a friend. What will be probability of both birthday is (i) different (ii) same (ignore leep year) 41- ;fn ledks.k f=khkqt ABC esa B = 90º, AB = 1 ls0 eh0] BC = 16 ls0 eh0 rks AC dk eku fudkfy, A If right angle in ABC B = 90º, AB = 1 cm, BC = 16 cm then find the value of AC? 4- ;fn fcanq (1] )] (4, y), (x, 6) vksj (] 5) blh Øe esa ysus ij,d lekarj prqhkqzt ds 'kh"kz gks rks x vksj y Kkr dhft, A If point (1, ), (4, y), (x, 6) and (,5) are the vertices of a parallelogram taken in order find x and y. 4- xzkiqhd fof/ ls gy dhft, A 5 Solve graphically. 5x + 7y = 50 7x + 5y = eh0 Å ps,d Hkou ds 'kh"kz ls fdlh ehukj ds 'kh"kz ij muu;u dks.k 60º gs vksj ehukj ds pj.k dk voufr dks.k 0º gs A ehukj dh Å pkbz Kkr djsa A 5

141 The angle of elevation on the top to 7m height a building to the top of the tower is 60º and fort of the tower angle of depression is 0º. Then find the height of the tower. vfkok (or) tehu ij flfkr fdlh founq ls 0 eh0 Å ps,d Hkou ij [km+k,d >ams ds tm+ vksj 'kh"kz ds muu;u dks.k Øe'k% 45º vksj 60º gs A >ams dh Å pkbz Kkr dhft, A The angle of elevation from a point at the ground to 0 meter height building on which a flag top and root is respectively 45º and 60º then find the height of a flag ls0 eh0 f=kt;k dk,d /krq dk xksyk fi?kykdj 6 ls0 eh0 f=kt;k okys csyu ds :i esa <kyk tkrk gs A csyu dh Å pkbz fudkysa A 5 A metal sphere of 4. cm radius is melted and form cylinder with radius 6 cm. Find the height of the cylinder? 46- ;fn,d f=khkqt ds nks dks.k nwljs f=khkqt ds laxr ds nks dks.kksa ds cjkcj gks rks os nks f=khkqt le:i gksxsa A 5 If two angles of a triangle is equal to another corresponding angles are similar then prove it? 47-5 ls0eh0 f=kt;k dk,d o`r [khaps A o`r dh nks Li'kZ js[kk,a [khaps tks vkils esa,d&nwljs ls 60º dks.k ij >qds gsa A 5 Draw a circle of radius 5 cm. draw a pair of tangents to this circle which are inclined to each other at an angle of 60º. vfkok (or) 8 ls0 eh0 yeck,d js[kk[kam AB [khfp, vksj bls fcunq P bl izdkj var% fohkkftr djsa fd AP PB = gks A

142 Draw a line segment AB of length 8cm and divide its internally at P such that AP = PB

143 ! [k )!?k! SOLUTION x 7! [k! x 1!?k /! x!?k +! d!! d! ) )! )! oxz&fpug 14-, vk0 Å º 16. vifjes; 17- fojks/h 18-./ fn;k x;k lehdj.k gs A lekuqikrh 0- foospd ) ) 0 0 leh0 esa ck, i{k esa j[kus ij ) ) 0 0 ) 0 0 ) vr% fn;s x;s lehdj.k dh gy ugha gs A - 1 x 5 1 x x 5 = = 4 65 (10) 65 = = nh xbz ifjes; la[;k lkar vkoùkhz gs A - nh xbz vkd`fr esa ;fn 54 gs] rks fl¼ dhft, fd gs A iz'u ds vuqlkj fd 54 4 (,dkarj dks.k) 5 (,dkarj dks.k) rfkk ('kh"kzfhkeq[k dks.k) vr% le:irk ds vuqlkj ;gk

144 ) 7 ) / izfke rhu in Øe'k% 9] 1] 17 gs A 5- fn;k x;k lehdj.k 0 ) ;k] = x x x + x ;k] ) izkir lehdj.k esa leh0 dk?kkr gs vr% ;g f}?kkr lehdj.k ugha gs A 6- o`r dh ifjf/ )π ) 7!+ 7 ) ))!/ 77!/!! 7- ) º 0 ) º ) º ) º ) 1 º cos 0º 8- ifjf/ )π ) )) ifjf/ )) 7 ;k] ) )) / )) / )). ) )) /.) 1 prqfkkz'k dk {ks=kiqy π ) !1) ) ?ku dk dqy i`"b {ks0¾ 1 (fdukjk) ;k] 1 )

145 ;k].1 ) ;k] ) ) 5 (dsoy?kukred eku)?ku dk vk;ru ) eh! nwjh 6 ( 4 0) + ( 0) ( ) vr% vhkh"v nwjh 6 bdkb;k (0] 0) 1- fn;k x;k cgqin ) ) 0 ) cgqin dk,d 'wku;d &4 gs A 7 7 ) 7 ) 0 ) ) ) ) + vr% dk eku + gksxk A! - ekuk fd 1,d ifjes; la[;k gs ifjes; la[;k dh ifjhkk"kk ds vuqlkj 1 p eku fy;k fd!!!!!!!!!!!!!!!!!!!!!!!!!!!!! q tgk rfkk & iw.kkzd gs rfkk & 8 vksj rfkk & esa 1 ds vfrfjdr dksbz xq.ku [kam mhk;fu"b ugha gs A ;kfu fd rfkk & lg vhkkt; la[;k, gs A

146 1 p vc (1) ls q & ) ) & ) (nksuksa rjiq oxz djus ij) () )9 & ) dks fohkkftr djrk gs A )9 & dks fohkkftr djrk gs (izes; ls) () vr% eku fy;k fd & ) vc ) ls & ) ) ) ) ) ) ) 7 ) ) ) ) ) ) )9 ) dks fohkkftr djrk gs A )9 dks fohkkftr djrk gs (izes; ls) (4) bl izdkj () rfkk (4) ls ;g fu"d"kz fudyrk (4) gs fd,oa & esa A vfrfjdr vu; mhk;fu"b xq.ku[kam gs A bl izdkj gesa,d fojks/khkkl feyrk gs A 1 vr% gekjk ;g ekuuk fd ifjes; la[;k gs xyr gs A 1 vr%,d vifjes; la[;k gs A - ekuk fd nks yxkrkj /u iw.kkzd vksj 0 gs A iz'ukuqlkj ) 0 0 ) 1 ;k] ) 0 ) 0 ) 0 1 ;k] ) ) 0 ) 17 ;k] ) 0 )

147 ;k] ) 0 7 ) ;k] ;k] (vlahko) ;k 0 7 pw fd /ukred iw.kkzd gs vr% vhkh"v la[;k, 1 vksj 14 gs A 4-!!4! 60º - 0º) = tan 0º = 1 tan A tan B R.H.S. = = 1+ tan A.tan B = tan 60º tan 0º 1+ tan 60º.tan 0º =. =. : 1+ 1 L.H.S. = R.H.S. Proved 5- ;gk ge fueu lkj.kh cukrs gsa A thou dky (?kavs esa) okjaokjrk lap;h okjaokjrk 1500& & & & & & & N 400 ;gk = = 00 N = f = 400 blds Bhd cm+h lap;h ckjeckjrk 16 okyk oxz

148 000&500 gs A rks ekè;d oxz 000&500 gs A ekè;d M = l + N f F x i N = 00 l = 00, f = 86, F = 10, i = = x = x = = vr% ekè;d thou dky ¾ ?kaVs A 6- L.H.S. = 1 + cot θ 1+ cosec θ = 1 + cosec θ 1 1+ cosecθ (cot θ = cosec θ - 1) = 1 + (cos ecθ + 1)(cos ecθ 1) cos ecθ ekè; dh x.kuk fueu lkj.kh cukdj djrs gsa A = 1 + cosecθ - 1 = cosecθ = R.H.S. Proved oxz vurjky #!! oxz fpug okjackjrk (f) ckjackjrk x oxz fpug! ) ) )7 1)7 ) 7 ) )7) ) 1 77 ) )) +1 f = 100, fx = 64

149 64 ekè; x = = 6.4 Ans ekuk fd fn;k x;k cgqin p(x) = x - 8 vc p(x) = 0 x - 8 = 0 x - ( ) = 0 (x + ) (x - ) = 0 x + = 0 ;k] x - = 0 x = - ;k] x = vr% fn, x, cgqin ds 'kwu;d gs - rfkk 9- (x + 4) (x - 4) = 0 ;k] x - 4 = 0 ;k] x - 16 = 0 ;k] x = 6 x = + 6 = + 6 x = 6, jke vksj ';ke nkslr gs A lahko ifj.kkeksa dh dqy la[;k ¾ 65 jke ds tue fnu dks vyx djrs gq, fnuksa dh la[;k 65&1 ¾ 64 bu 64 fnuksa esa fdlh fnu ;fn ';ke dk tue fnu gks rks nksuksa ds tue fnu fhkuu gksxsa A (1) fhkuu tue fnu gksus dh?kvuk E ds vuqdqy ifjek.kksa dh la[;kk ¾ P(E) = 65 () tue fnu,d gh gksus dh?kvuk dh izkf;drk 1 - P(E) = 1-64 =

150 41- ledks.k ABC esa B = 90º, AB = 1 cm, BC = 16 cm, AC =? AC = AB + BC = 1) (16) ( + A = = 400 = 0 cm 1cm B 90º 16 cm 4- ekuk fd lekarj prqhkqzt ds 'kh"kz A (1,), B (4, y), C (x, 6) rfkk C D (, 5) gs A ge tkurs gs fd lekarj prqhkqzt ds fod.kz ijlij lef}hkkftr djrs gsa A vr% fod.kz AC ds eè; fcanq ds funs'kkad ¾ fod.kz BD ds eè; fcanq ds funsz'kkad 1+ x y + 5, =, ;k] 1+ x, 4 ¾ 7 + 5, y 1+ x = (i) y + 5 rfkk 4 ¾ (ii) leh0 (i) ls + x = x 7 = 14 ;k] x = 14 - = 1 1 x = = 6 lehdj.k (ii) ls] y + 5 = 4 x = 8 ;k] y = 8-5 =

151 vr% x = 6 vksj y = 4-5x + 7y = (i) 7x + 5y = 46...(ii) 5x = 50-7y 50 7y x = 5 x y x = 46-5y 46 5y x = 7 7. x y 0 5 fcanqvksa(10]0)](]5)]( (4-4]4)](7-]) fcanqvksa(6-57]0)](]5)] (5-14])(7-]) dks xzkiq isij ij n'kkzdj feykus ij dks xzkiq isij ij n'kkzdj feykus ij leh0 (i) dk vkys[k feysxk A leh0 (ii) dk vkys[k feysxk A nksuksa vkys[kksa dk dvku fcanq (] 5) gs vr% gy x =, y = 5

152 44- ekuk fd ehukj dh Å pkbz AB gsa PR AB [khapk iz'u ls APR = 60º, RPB = 0º = PBQ edku dk Å pkbz PQ = 7 eh0 vc ledks.k PQB ls tan 0º = PQ 7 m = QB QB 1 = 7m QB QB = 7 vc ledks.k ARP ls AR tan 60º = PR = AR PQ = QB =7 7 AR = 7 x m = 1 m vr% ehukj dh Å pkbz = AB = AR + RB = AR + PQ (RB = PQ) = 1 m + 7 m = 8 m vfkok ekuk fd >ams dh Å pkbz AC = x eh0 gs A BC ledks.k CBP ls tan 45º = BP 0 1 = or BP = BP 0 m ledks.k ABP ls tan60º = BP AB BC + = BPAC ;gk BC= 0 eh0 AC ¾ x eh0] BP ¾ 0 eh0 = ( 0+ x) 0 or 0 = 0 + x

153 x = 0-1 = 0 x (1.7-1) = vr% vhkh"v >ams dh Å pkbz ¾ eh xksys dk vk;ru = 4 π r = x π x (4.) cm ;fn csyu dh Å pkbz h cm gks rks csyu dk vk;ru = πr h = π x (6) x h cm iz'ukuqlkj] csyu dk vk;ru ¾ xksys dk vk;ru 4 π x 6 x h = π x 4. x 4. x 4. 4 π x 4. x 4. x4. 4. x4. x4. h = x = =.74 cm π x fo'ks"k izfrkk& eku ysa fd ABC vksj DEF esa B = E rfkk C = F fl¼ djuk gs ABCC DEF mrifr ABC DEF ge tkurs gs fd A + B + C = 180º rfkk D + E + F = 180º A + B + C = D + E + F ;k A + B + CC = D + E + F (fn;s x, laca/ksa ls) ;k A = D bl izdkj nksuksa ABCC vksj DEF ledksf.kd gq, vfkkzr~ A = D, B = E, C = F vr% dks.k&dks.k&dks.k le:irk ls DEF DEF 47- jpuk ds pj.k%& pj.k&1 fcunq O dks dsuæ ysdj 5 ls0 eh0 f=kt;k dk,d o`r [khpsaxs A

154 pj.k& o`r dk dksbz O;kl POQ [khapsxs A pj.k& QOC = 60º bl izdkj cuk;sxsa fd OC o`r ls founq C ij feys A pj.k& 4 PA PQ vksj CB OC [khpsaxsa A bl izdkj PA vksj CB,d nwljs dks fcunq D ij izfrpnsn djsaxs A rks DP vksj DC o`r dh vhkh"v Li'kZ js[kk, gs tks 60º ds dks.kk ij >qds gs A

155 vfkok jpuk ds pj.k&1 AB = 8 cm [khpsaxs A pj.k& fcunq ij U;wu dks.k cukrs gq, js[kk : [khapsxs A pj.k& : ij 0 ) ) cjkcj pk; ysrs gq, # dkvsxsa rfkk # fy[ksaxs A pj.k& 4 # dks feyk;saxs A pj.k& 5 rhljs fcanq ls # dks [khpsxsa tks AB dks P ij dkvrh gs A bl izdkj fcunq dks ) ds vuqikr esa vr% fohkkftr djrk gs A

156 ! # % #&& () +,. /!,), %,0) 1,/ # % %) (,/ & ( %456 0 ) 7,/ )!

157 Set - 9 Λ Λ Λ,. 90 &1 % %(,/ & % & !1) 0, 01)( ,0 1 () ΡΓ Χ,?, + ),,/ ()! ( + ), 01, 0!1 01= 01/ % ΡΓ Χ,?,/ Β ( 010,8, 1 0!10 8, 1 010, , 1,.( 08!1 Γ.( 08!1Χ Χ,/ #08!1ΚΕ08! ! / 01! ) 6 % &: &,/ % &: Λ.&! ! Ν 4 Ι Ι0Υ>),/#4Β),/ % 01= 0!1% 01= 01 % Χ? &0%810(9061Χ9,/! &0% !1! < ΕΕ>ΕΕΕ>ΕΧ,,/ 8 8)8 8/8%8 8% 01 0!1% 01 01) = (Α ΓΥ>), 4 )),/

158 % 01= 0!1% 01 % 01% / Χ,Ο 0()?,9,0/ ) ΧΗ6, (, 01+ 0!1Μ 01Ε 01 % ) 6 < = / Χ?, 4 Ι#4Β:::::::::::::::::::::::::,/ ΝΝΝΝΝΝ 0::::::::::::::::::::::::::::,/ ΝΝΝΝΝΝΝΝ! > Λ :/::::::::::::::::::::::::::,/ > Λ :/./ / :0ΝΝΝΝΝΝ1 (:::::::::::::::::::::::::,,/ &.ΝΝΝΝΝΝΝΝ 678 8::::::::::::::::::::::::::::::::::::: Ι Ι0Υ, 4 Ι:::::::::::::::::::::::::::::::::::,/ ΝΝΝΝΝΝ Γ;Χ89Η6! & 9, 0 (;Χ:::::::::::::,/ 8! (!8! (!ΝΝ 0+1://) 80+ 1:ΝΝΝΝΝΝΝ 0+1://) 80+ 1:ΝΝΝΝΝΝΝ (Α Γ, (Α #4Β:::::::::::::::::::::,/ ΝΝΝΝΝΝΝ 0 ΡΓ Χ, Σ Ι? &Β,/ # ( Β Β

159 % ) 6 < = %/ % ΡΓ Ο%,/:/Σ / # Ο%,/:/ Ι 0 Χ? 0 8,%1) 0,68 1Χ9 4,/ # (! 0 8,%10,68(1/ (Α 4 ),) Γ)? ΚΚ,/ 4 0Υ / / << # & :%8:), (Υ / & :%8:)8 / Λ6/ / %/ / 56/ / %/ / # 6/ / %/ / 56/ / %/ / Χ? Ψ08!1 &0,8,!1Χ9 / #! 08!10,8,!1 ΕΕ>ΕΕ!Ε ΕΜ)ΥΗ6 Γ / # 8%8 8)888= 00Χ ΒΙ,/0) Ο Σ / & #!! (! 8: &.80(8 / 8: & Ι Ι0Υ% 8 0(),/ Ω Ι0,0 4 Ι,/. % 8 ) Ω Ι00),/ %! %% % ΡΓ 5 5%:/ΠΣ ΕΙΧ Χ Χ,/ # Π 5 5%:/ << / % / / 6 / :

160 %) %6 % %< %= / % Ω Ι00 1 &9:; : = <:; &>? = & ) ( ς% ) ΛΚ,?,/ % = &,< (4 & Β Κ,)# (, Χ? Σ Ι0Ι>ΕΛ) Λ>ΕΜ &,/ #, 0 8,)10, 8=1 0Χ, 4 Ι Ι,/< 1;Χ;ΧΣ / &9 # 0 Χ ΒΙ,/7, 0Χ )0/ &!! ( ) (Α Γ) 4 )?!, (Α #4Β /! % # 8 ( Α) Α Γ>, / # Λ 8 ( ΑΑ Χ, Λ0#1 ()? F /, / /,/ /,6/ 6/,</ </,// //, / / %) ) 6/ %) %/ Β ( Γ, Λ0. ( 1 : : ) 4 0(Α 0 % 0Υ 90ΕΙ Φ γ,/ % 6/ / Ω Ι0 (Α? 9 Ι9 Ι,,/!!

161 6! χ9 ( % 0(( %?Φ,) )(,/χ( χ9σ Ι0/ #!! 6/ / ) /! Φ9Η6) :, 0, / Χ9Χ,,,/ (,, #4Β 9 0ΙΧ 9 0<, )(%,,/ 4 6 ) /Π Χ %/ <

162 .. %& %01 01 )0! <01 =01 /01?0/1 :( () ) :> :Ν Φ:!:Ν Κ:Β Χ =0/=)1 /0%<) 1?Β: ):?Β: ): Ρ Χ,,0 ), ),0,,,% :8!:,%8 :,/8 : Ο:0,%1 Ο 0,/1:=5/:= >.Ε : :>+>,0Φ Γ& : 9 8> :&Η 8>0 :Α( Ι Α( % ϑ+(, 0,68(1 Χ9 1+, +, Κ 1+!, +, Κ:0,<1 50(5%1 Κ:650(5%1 Κ,6:0(5%1 Κ6:0(5%1 Κ(5%:!:6 (5%:6 Κ(5%:6 (5%:6 (:6,%:% (:,6,%:,= Ι(:%8,=,(Α 4Ι5 & (Α 4 5,/,5 & 5 :/,(Α Γ: Λ5 (Α Γ: Λ5 % Λ5, Λ5 :<< Κ Λ05 & 5 1:<<

163 ) 5 & 5 :?Μ Λ : Ν Γ 5 & 5 :0 1 : 5 & 5 : Ο01 Ι5 5 :/ 5 & 5 : 5 & :) ΙΗ6 Ι 5 & : Α : Ι5 :/, :6% 5 & : 85 :6% & :%8!:):/Ο &Η : :50Ο1 Ο &Η :%50/Ο1 ):%5= ):%5):< (χ :< 6 6/ / %/ / 56/ / %/ / : 5 : : : &08!10(90,8,!1Χ9 Π+, +, :1+Ο Ο, +Θ Θ, : Ο Θ :1+Ο Θ,:1Ο Θ < )χη6εε>εε!ε ΕΜΧ56 ςι Ε>ΕΕ ΕΕ!ΕΜ9Υ :8 ;Γ:0 9& 1Χχ :0 1(χ :(χ : ;Γ: = 00Χ ΒΙ,/ Ε>ΕΕ ΕΕΦ ) >ΕΕΕ Ι) Ο : % :& %/ 8: & :;Ρ &9: Π 9:)Π &:1+Υ, +ΑΥ, :Υ ΑΥ ΣΤ & :! Υ :%Π9

164 678: ;Ρ ;ς :&Ω % &Ω :& & Ξ8: Ρς Ω : : ;ς &Ω & 4 Ι Ι0Υ,/ &:% 89: 8&9:) & :0% 1 :6= &9 :0) 1 : ) 9 :0 1 : &9 59 : ) 5 :6= :& Ι&9 59 :& % ) Ψ0 Ι()? Ι)(Χ Χ,/ Ι,0Ψ,/ 0,/ Ι: ; Ρ Κ : Ο Θ Κ :!,,,,,,,,,,01 )& Η>,),/ ΚΟ,/::::::::::::::::::01 Ι01 :! Κ! : Ι!,Κ Θ01) 01)!,/) 1Θ Η>,/,,? ( Γ,/Ι0),/ %% ΡΓ 5 5%:/,Υ&: 8!: 8:% :,:0Π1 Ο %:Π, 9Υ, Χ Χ,/ Ι :/ ΚΠ : Π: Π:!

165 % Ε.: << / % / / 6 / : << / % / 0=/ /,<< / 10=/ /, % / 1 : << / % / << / % / 0 0=/ /,1: 1 :0 << / << / 10 % / % / 1 :Κ:0 8ΞΖ8:1 ::. ΙΕ.:. %) : &9 [ Ε.: &9:; :; ] _ [ _ : &9 [ _ [ _ 0&: _α[ : _ [ _ & 1 ; : &9; ; ; & :& : &9; &>; & χδε ; ; &>; :&>βχ &>; : &>; :. ΙΕ.:. %6 & % :8 = :,< % :5 :,010 :50,1 = :5<:,<Ο01 5 : > φγ> >%φγ& 01:5:< Ι: & :, >% (χ?,/ Ι :/ & :50,1 Κ/:<50Ο10, 1 Κ/,<:0Ο10, 1 Κ,<:0Ο10, 1 Κ0Ο1: > :5:) > : Ι:)χ9(?,/

166 % )# Χ? 081Ι&0 8,)1) 0, 8=1 &,/ 0 818&0 8,)1 & : & 1+, +Α, :1+, Α Ι & :0,1 5 ) 08/1890, 8=1 9 : 9:1+, +, 1+,, & 9 Ι & :ϑη Κ0,1 5 ):05 1 5< ΚΟ5 5 ): 555< Κ,5 =:55<) Κ,Ο:<)Ο = Κ,<:)6 Κ: % :, > Ι:, Ι,)# Χ? 0,8/1,/ %< &90Χ,Ψ, Ι< Ι,/ 1&9 & 9Η/9Υ & 89 Ρ Ι,/ Ι9 : :( Ψ&9 & 1+Ο, ηι & 1+Ο, Ο :Ο Ο :Ο :Ο Ι& : 9 ) < 1;Χ Ψ,/ %= 0 Χ ΒΙ,), ; Ο,,/ ΝΝΝΝΝΝΝΝΝΝΝΝΝ0861 / ΝΝΝΝΝΝΝΝΝΝΝΝΝ %8180%8, 1ΝΝΝΝΝΝΝΝΝΝΝΝΝ0% ΝΝΝΝΝΝΝΝΝΝΝΝΝ0861 0)8180)8 1ΝΝΝΝΝΝΝΝΝΝΝΝΝ0) ΝΝΝΝΝΝΝΝΝΝΝΝΝ06861 Β ( :Φ( Ι0Χ ),/ ) 0Χ ) : && 9Υ, (Α Γ0( 4 )? %,/ Ι #55:% Κ5+#,:% %

167 Κ5+,:% Κ5+,:% Κ5+,:% Κ ϕ :% Κ%5:% 5: κ : 5: Ι(Α #:#5: 0 1 : :) (Α #:) 5 8 5%85)8)5 ;Γ: >: +9,9+9,9+09,9+9, 0 Κ>: &9& 0 : &+>&, :%051:%5% 0 Κ%5%: : %:,%Κ%:,%Κ: :%:% ()? /, / /,/ /,6/ Χ Χ / %) ) 6/,</ 6/ </,// //, / %) %/,ΥΧ Χ Χ Φ,ΙΦ6ΛΚ()? ),/) Χ,(ΦΛΚ,/ Ιλ:6/8µ Η :6/8µ, :) 8µ :%)8δ: / / :5 µ >µ µ µ µ Οδ %Η> :6/5 ν%η>> / :6/5 &Η> /:6/5 &Η> /:6/5&%Η :6/5<:6< % Κ Κ

168 / = / ) / =, / ) 0/81 / < 6 0=8)1 08/1 08/1,, Φ 6</ Κ,, 0/81 Β,, 0Υ Χ? ΜΕ,/), ( Α 9 Λ & 01? 4 0(Α 9/ + 01(Α ϑ&φ9 / Φ 01Φ οπθρ:6/ / Χ/,0Φ(Α,/ 9 01 Χ? & οθστ:=/ / Χ & οθρυ:=/ / Χ/στ ϖ θσ &ρυϖφ 01 & ) + α[,/ &0( (Α %,Ι )Φ γ,/οσωρ:6/ / ) 0(Α Ι? Φ,/(Α? 9 Ι&9 9 Ι)ς% 88 0(. Χ? ) %,/9Υ Χ, Χ? (Α % 0ΥΧ Χ,,/ Ι& : &. 9 : 9 +<4 +Β 7 + :. Κ σω πω ρξ τξ σψ πζ ρζ τψ Κ +σψ τψ, +πζ ρζ, +σω πω, +ρξ τξ,. Κ στ πρ σπ ρτ Κ στ σπ+{στ πρ, Ισπ πρ+σπ ρτ, & 9

169 6 Ισπ πρ ρτ τσ {σπρτ09 Ι,/ &9 ( Υ9!,/ (( Υ9,/ οωσξ:6/ / οξσω:) / Ιοσζπ:) / 09&1 &9:: 6/ / &9 ) / : ;Ρ Ρ : Ρ )/ & Κ πζ Ν}πζ σξ Ν} Ψ & 6/ / : ; : Κ ϑ Κ : Ν} 9 ( Υ9: :5 :Ν}5 Ν} Ν} Ν}, 9:Φ, ::9:/ Χ/ ://// {Φ,/ Ι Θ Θ Ι%/ Θ Θ &ΗΗΗΗ %Η &ΗΗΗΗ %/:)/// %Η Ι, Ρ Χ,) :Κ9Κ, :ΚΦΚ: : 9 0<, )(%,/ ) Ι, Ρ 9 #:, : 0ΗΗΗ [ : 0ΗΗΗν&ΗΗ :)6 )//

170

171

172

173

174

175

176

177

178

179

180

181

182

183

184

185

186

187

188