Size: px
Start display at page:

Download ""

Transcription

1 THEORY OF EQUATIONS OBJECTIVE PROBLEMS. If the eqution x 6x 0 0 ) - ) 4) -. If the sum of two oots of the eqution k is -48 ) 6 ) 48 4) 4. If the poduct of two oots of 4 ) -4 ) 4) - 4. If one oot of is + + = hs one imginy oot +i, then its el oot is x 5x kx 4 0 x x 6x k 0 x 5x kx = is zeo, then vlue of + = is, then k = + = is the ecipocl of nothe, then the vlue of k 5 ) 4 ) 4) 5. If two oots of the condition is ) p q = e connected by the eltion αβ + = 0, then x px qx = ) p p q = 4) 6. If two of the oots of 7 4q 0 + = ) 7. If one of the oots of two, then k = x qx 0 + p + q + = 0 q + p + q + = = e equl, the condition is 4 7q 0 = ) x 6x 5x k 0 - ) ) 4) - 8. If α, β, γ e the oots of β + γ, γ + α, α + β is 7 4q 0 + = 4) 4 + 7q = = is equl to hlf the sum of the othe x px qx 0 + =, then the eqution whose oots e x px + ( p + q) x + ( qp) = 0 ) x + px + p q x + + qp = 0 ) x + px + ( p + q) x + ( + qp) = 0 4) x + px p + q x + qp = 0

2 9. If the poduct of the two oots of 4 x + px + qx + x + s = 0 is equl to the poduct of the othe two, then ) ps ps = ) = 4) p s = p s = 0. The condition tht the eqution mgnitude but opposite in sign is x px qx 0 + = my hve two oots equl in pq+=0 ) pq-=0 ) pq-=0 4) pq-=0. If the sum of two oots of ) ) 4). If α, β, γ e the oots of ) p q + q p ) 4). If α, β, γ e the oots of 8 ) 4 ) 4. If α, β, γ e the oots of ) q q + p p p q x px qx 0 + =, then pq = + =, then + + = α β β γ γ α x px qx 0 q + p + + =, then + + = β γ γ α α β x x x ) 4 + =, then + + = α β γ x px qx 0 ) 4) p q q + p

3 5. If the oots of,, 6 ), 6, ) 6. The oots of x 9x + 0x = 0 e in the tio :, then the oots e,, x 9x + x = 0 e 4),, 5 ± i ), ± i ), ± i 4), ± i 7. If,, α, β e the oots of α =, β = ) α =, β = ) α =, β = 4) α =, β = x x 5x The eqution whose oots e -, +i is ) x x x = ) x x x = 4) + + =, then x x + x 5 = 0 x + x + x + 5 = 0 9. The cubic eqution hving, the oots +, is ) ) 4) x + 5x + 5x + = 0 x 5x + 5x = 0 x 5x 5x = 0 x 5x 5x + = 0 0. The oots of the eqution ) p pq ) + = 4) p pq. If the oots of the eqution is ) b bc d 0 x px qx = e in A.P. then p pq = = p pq + = ) b + bc d = 0 4) x bx cx d = e in A.P., then the condition b + bc + d = 0 b bc d = 0. If the oots of the eqution, 4, 8 ),/,/4 ),,4 4),5,4 x 7x + 4x 8 = 0 e in G.P., then the oots e

4 . If the oots of the eqution,, 9 ) 4. If the oots of,, ),, 4 8x 4x + 7x = 0 e in G.P., then the oots e 4),,4 x + x + kx + 8 = 0 e in G.P., the vlue of k is ) - ) 4) - 5. The oots of the eqution x px qx 0 p = q ) p = q ) p 6. If the oots of the eqution men oot is q ) q 7. If the oots of ) q 7 9pq ) p q x px qx = e in G.P. Then x px qx 0 = q 4) p = q + = e in H.P., then the vlue of the 4) q p = e in H.P., then + = ) = 4) q 7 9q 8. The oots of the eqution b c c b ) c ) 4) b 9. If α, β, γ e the oots of eqution ) x 5x 5x = ) x 5x 5x 6 0 = 4) + = q 7 9pq = q 7 9pq x x bx c 0 + = e in H.P. then the men oot is + =, then x x 8x 6 0 x 5x + 5x 0 = = x 5x 5x 6 0 α, β, γ e the oots of the 0. If α, β, γ e the oots of x x + 5x = 0, then the eqution whose oots e βγ +, γα + αβ + α β γ is ) x + 0x + x + 64 = 0 ) x + 0x + x 64 = 0 4) x 0x + x 64 = 0 x 0x x + 64 = 0

5 . If α, β, γ e the oots of the eqution x + qx + = 0, the eqution whose oots β + γ γ + α α + β α β γ e,, is ) x + qx = 0 ) x qx 0 + = 4) x qx = 0 x + qx + = 0. If f ( x) 5x x x + 7 = 0, then ) 5x 7x 4x = ) 5x 7x 4x = 4). The eqution whose oots e the oots of diminished by 4 is ) 4 x x 4x 55x = ) 4 x x 4x 55x = 4) f x expessed in powe seies of (x-) is 5x + 7x 4x 9 = 0 5x 7x + 4x 9 = 0 4 x 5x 7x 7x = ech 4 x + x 4x + 55x + 9 = 0 4 x x 4x 55x + 9 = 0 4. The eqution whose oots e k times the oots of x x + x x + = 0 is 6 8 n eqution with integl coefficients. Then k = ) ) 5 4) 6 5. If -4,-,- e the oots of x 6x 9x 4 0 x + 6 x + 9 x + 4 = 0 e 4 5 5,, ) ) 4, 5, 5 4) =, then the oots of the eqution 0,, 4 5 5,, 6. The eqution whose oots e less by thn the oots of ) x x 0 + = ) x + x + = 0 4) 4x 8x 0 = 0 x x = 0 x 4x = is

6 7. If,, e the oots of 6x x + 9x = 0 e,, ),, x 9x + x 6 = 0, then the oots of the eqution ),, 4),, 8. The second tem of the eqution diminishing its oots by ) - ) 4) - x 6x x 0 9. By emoving the second tem in the eqution tnsfomed eqution is ) y 9y = ) y 9y = 4) 40. If y 9y + 6 = 0 y 9y 6 = = cn be emoved by x x x = the f x = x x + 4 = 0 hs epeted oot, then tht oot is ) - ) 4) - 4. The multiple oot of x x x 0 + = is ) - ) - 4) 4. If the eqution q = ) q x qx 0 + = hs two equl oots, then = ) 4. If α, β, γ e the oots of q = 4) q = =, then = α β x 4x 5x 0 4 ) ) 4 4) 44. If α, β, γ e the oots of x + bx + cx + d = 0, then the vlue of α β = c + bd ) bd c ) c bd 4) c + bd

7 45. If α, β, γ e the oots of x 9x 6x =, then the vlue of Σ( α + )( β + ) is 07 ) 08 ) 8 4) The oots of x x + 6x 6 = 0 e in A.P. ) G.P. ) A.G.P 4) H.P. 47. The numbe of el oots of eqution ) ) 6 4) If α is n imginy oot of nd ) α + α is x x 0 = ) x x 0 + = 4) 5 x = 0 x + x = 0 x + x + = 0 x + / x + x + / x = 0 is, then the eqution whose oots e α + α 4

8 . Ans () THEORY OF EQUATIONS HINTS AND SOLUTIONS Given +i is one oot -i will lso be oot. Sum of the two oots = +i+-i= s = sum of ll the oots = 0. The thid oot = 0- = -. Ans.() Let the oots be,, s α α β = α + β = = β = Since is oot of the given eqution,. Ans () 6 + k = 0 k = 48 Let α, β, γ be the oots so tht αβ =. s = αβγ = 4 γ = 4 γ = k = = 4 k = 4. Ans.( Let the oots be α,, β α s = α β = ( 4) = 4 β = 4 α 64-5(6)+4k-4=0 4k = 0 k = 5 5. Ans.(). But β is oot of the G.E. Let the oots be α, β, γ. Then αβγ =. Given αβ =. γ = γ =. But γ is oot of G.E. + p + q + = 0 + pq + q + = 0 is the condition. 6. Ans ( Let the oots be α, α, β. Then α + α + β = 0 β = α

9 α. α. β = α β = α = α = But α is oot of G.E. α qα + = 0 + q + = qα = q α = q = 8 8 Requied condition is 7. Ans () 4q + 7 = 0 4q + 7 = 0 Given one oot is equl to hlf the sum of the othe two. The oots e in A.P. Let the oots be c, + d f = 0 6 s = d d = = = = (- is oot of f(x) = k = 0 k = 8. Ans ( Given Σα = p, Σαβ = q αβγ = = β + γ = α + β + γ α = α = Let y p p x Requied eqution is α = x x = p y) p y p p y + q p y = 0 y py + p + q y + pq = 0 p y p y + py p py + p y + pq qy = = 9. Ans () x px p q x pq 0 Given Σα = p, Σαβ = q, Σαβγ =, αβγδ = s. αβ αβ = s α β = s. Also given αβ = γδ. Now αβγ + αβδ + αγδ + βγδ = αβ( γ + δ ) + γδ( α + β ) = αβ( α + β + γ + δ ) = ( γδ = αβ) αβ( p) = α β p = sp =.

10 0. Ans () Let the oots be α, α, β s = α α + β = p = p β = p x = p But β is oot of G.E. p p.p + qp = 0 pq = 0. Ans () Let the oots be α, αβ S = α α + β = p β = p But β is oot of the given eqution β pβ + qβ = 0 p p p + qp = 0 pq 0 pq. Ans () Given α +β+ γ = p, αβ +βγ + γα = q, αβγ = G.E.. Ans () γ + α + β = = α β γ ( α + β + γ) ( αβ + βγ + γα) ( αβγ) Given α + β + γ =, αβ + βγ + γα =. αβγ = 4 α + β + γ + + = β γ γ α α β α β γ = α + β + γ ( αβ + βγ + γα ) ( αβγ) = =. = 4 = = = p q β γ + γ α + α β 4. Ans () α + β + γ = p, αβ + βγ + γαq, αβγ = + + = α β γ α β γ ( βγ + γα + αβ) αβγ ( α + β + γ) ( αβγ) 5. Ans ( Let the oots be α, α, β. α + α +β = 0...( q = β = 9 4α, Fom () : α ( 9 4α ) = α. α. β =...() p

11 9α 4α = 4 4α 9α 4 = 0. By inspection : α = β = 9 8 = Hence, the oots e,6,,,6 6. Ans () By inspection x = is oot of the eqution. ( x ) is fcto of x x + x + = + = x x x x x 0 x =, x = ± i 7. Ans(4) + α +β =. α + β = αβ = 6 α nd ( α ) = 6 α α = α α + = α = 6 0 0,, β =, α =, β = 8. Ans () Given -, +i e the oots of given eqution. i is lso oot of the given eqution. Requied eqution is (x+ (x--i) (x-+i) = 0 x + x + = 0 x + x 4x + 5 = = 9. Ans () x x x 5 0 Itionl oots occu in pis is lso oot. The cubic eqution is ( x )( x + )( x = 0 x ) x = 0 x 4x + x = 0 x 5x + 5x = 0

12 0. Ans() Let the oots be d,, +d. d d = p = p = p But = p + p p + q p + = = p p pq 0 p + = pq is the equied condition.. Ans ( Given eqution is b c d =...( x x x 0 Let the oots be α β, α nd α + β. b b b sα β + α + α + β = α = α = Since α is oot of (, we hve b b b b d = b bc d + = 0 b bc + d = 0is the condition.. Ans() Let the oots be,, s = ( )( ) = ( 8) = 8 = 8 = s = + + = ( 7) = = = 0 = 0 o / The oots e,,4.. Ans() = Given =...(let Let the oots of ( be x x x 0 s =.. = = 8 = = 8,,

13 4 4 7 s = + + = = + + = = + = ( ( ) = 0 = The oots be 4. Ans (,, = = 8 = o. But is oot of the given eqution k + 8 = 0 k = 4 k = 5. Ans (4) Let the oots be,, = = = / s But is oot + p + q + = 0 / = / p q 0 q p = q p = q 6. Ans () / / Let the oots be,, d d ( d) ( d)( + d) ( + d) p = q p = q p = q / / /.. = = + d + d ( d) ( + d) + + = q + d + + d q = q = q = d d ( ) ( + ) Men oot is = q

14 7. Ans (,, Let d + d be the oots s = =, s = + + = q ( d) ( + d) ( d) ( d)( + d) ( + d) + d + + d = q ( d) ( + d) ( ) = q = q But is oot + q. + = p. + q. 0 + = q q q + + = 7 9pq q q = 7 9pq q q 0 + = + = 8. Ans (4) 7 9pq q 0 q 7 9pq Put x = /y in G.E. Then cy by y 0 + =...( The oots of () e in A.P. let α β, α, α +β be the oots of ( b b b b s = α β + α + β = α = α = α = c c c c b c The men oot of ( c. The men oot of G.E. is b. 9. Ans(4) Given α + β + γ =, αβ +βγ + γα = 8αβγ = 6 Requied eqution is x x x 0 α + β + γ + α β + γ α +β γ α β γ = = x x α + β + γ Σαβ + x Σαβ αβγσα αβγ = 0 + = x x 6 x x + 5x + 5x 6 = 0

15 0. Ans. () Given Σα =, Σαβ = 5, αβγ = Let y = αβγ βγ + = = = = α α α α x ( ) = x x = 4 y Requied eqution is + = 0 y y y + = 64 0y y 0 y 0y + y 64 = 0 o. Ans () Given Σα = 0, Σαβ = q, αβγ = β + γ α + β + γ 0 α α α α α α Let y = = = = = = x = α x y x 0x 64 0 =. Putting this vlue in G.E., we hve: + q 0 + = y y y qy = 0 o. Ans () Given x qx = 0 f x = 5x x x + 7 = 0 f x = 0 We hve to diminish the oots by Diminishing the oots of ( by (), we get The tnsfomed eqution is 5x + 7x 4x 9 = 0

16 . Ans () The tnsfomed eqution is 4. Ans (6) G.E is x x + x x + = x x + x x + = 0 0 Multiplying the oots of ( by 4 x + x + 4x + 55x 9 = 0...(. = 6, we emove the fctionl coefficients. k = 6 5. Ans () Diminishing the oots of the given eqution by -/ we get the second eqution. The oots of the second eqution e : + 4 +, +, + 0,, 6. Ans () Diminish the oots of G.E. by, we hve: Tnsfomed eqution is x + x + = 0.

17 7. Ans ( G.E. is x x x x 9x + x 6 = 0. Put /x in the plce of x = 0 oots e /,,. 6x x + 9x = 0 is the ecipocl eqution of G.E. whose Hence, the oots of tnsfomed eqution /,, /. 8. Ans (4) G.E. is + + = whee o =, = 6, =, = x 6x x 0 To emove the second tem, diminish the oots by h = 9. Ans ( Hee 0 = =, h = = = n0 Diminishing the oots by to emove the second tem The tnsfomed eqution is 40. Ans ( f x = x x + 4 Given y + 9y + 6 = 0 f x = x 6x = x x = 0 x = 0, 0 6 = = n f = = 0 f. But ( 0) 0 is epeted oot. 0

18 4. Ans ( Let f x = x x x + f x = x x f ( x) 0 x x 0 = = x x + = 0 x =, / But f( = 0 nd f ( / ) 0. is epeted oot. 4. Ans (4) Let the oots be α, α, β. Then α + α + β = 0 α. α. β = β = α, α ( α ) = α = 4. Ans ( α + β + γ = 4, αβ + βγ + γα = 5, αβγ = γ + α + β Σ = + + = α β α β β γ γ α α β γ = α + β + γ ( αβ + βγ + γα ) ( αβγ) 44. Ans () G.E. is x x x 0 ( ) b c d =...( b c d Σα = Σαβ =, αβγ = = = = 4 Σα β = α β + β γ + γ α = ( αβ + βγ + γα) αβγ ( α + β + γ) = 45. Ans ( = = c d b c bd Let y = α + = x + x = y. The tnsfomed eqution is y 9 y + 6 y 4 = 0 y + 7 9y + 7y 9y 8+ 54y + 6y 78 4 = 0

19 y 8y + 07y 0 = 0 Whose oots e α +, β +, γ + s = Σ α + β + = Ans () x x + 6x 6 [ ] = x 6 x x 6 = 0 x 6 x + 6x + 6 x = 0 x 6 x 5x + 6 = 0 x 6 x x = 0 x =, 6, These oots e in G.P. 47. Ans (4) Given eq. is x + + x + = 0 x x x + x + + = 0 x + + = 0 x + = 0( o) x + = 0 (O) x x x x x + + = 0 x Thee e no el oots. 48. Ans ()

QUADRATIC EQUATIONS OBJECTIVE PROBLEMS

QUADRATIC EQUATIONS OBJECTIVE PROBLEMS QUADRATIC EQUATIONS OBJECTIVE PROBLEMS +. The solution of the eqution will e (), () 0,, 5, 5. The roots of the given eqution ( p q) ( q r) ( r p) 0 + + re p q r p (), r p p q, q r p q (), (d), q r p q.

More information

Maths Extension 2 - Polynomials. Polynomials

Maths Extension 2 - Polynomials. Polynomials Maths Extension - Polynomials Polynomials! Definitions and properties of polynomials! Factors & Roots! Fields ~ Q Rational ~ R Real ~ C Complex! Finding zeros over the complex field! Factorization & Division

More information

B.Sc. MATHEMATICS I YEAR

B.Sc. MATHEMATICS I YEAR B.Sc. MATHEMATICS I YEAR DJMB : ALGEBRA AND SEQUENCES AND SERIES SYLLABUS Unit I: Theory of equation: Every equation f(x) = 0 of n th degree has n roots, Symmetric functions of the roots in terms of the

More information

Draft Version 1 Mark scheme Further Maths Core Pure (AS/Year 1) Unit Test 5: Algebra and Functions. Q Scheme Marks AOs. Notes

Draft Version 1 Mark scheme Further Maths Core Pure (AS/Year 1) Unit Test 5: Algebra and Functions. Q Scheme Marks AOs. Notes 1 b Uses α + β = to write 4p = 6 a TBC Solves to find 3 p = Uses c αβ = to write a 30 3p = k Solves to find k = 40 9 (4) (4 marks) Education Ltd 018. Copying permitted for purchasing institution only.

More information

STD: XI MATHEMATICS Total Marks: 90. I Choose the correct answer: ( 20 x 1 = 20 ) a) x = 1 b) x =2 c) x = 3 d) x = 0

STD: XI MATHEMATICS Total Marks: 90. I Choose the correct answer: ( 20 x 1 = 20 ) a) x = 1 b) x =2 c) x = 3 d) x = 0 STD: XI MATHEMATICS Totl Mks: 90 Time: ½ Hs I Choose the coect nswe: ( 0 = 0 ). The solution of is ) = b) = c) = d) = 0. Given tht the vlue of thid ode deteminnt is then the vlue of the deteminnt fomed

More information

QUADRATIC EQUATION EXERCISE - 01 CHECK YOUR GRASP

QUADRATIC EQUATION EXERCISE - 01 CHECK YOUR GRASP QUADRATIC EQUATION EXERCISE - 0 CHECK YOUR GRASP. Sine sum of oeffiients 0. Hint : It's one root is nd other root is 8 nd 5 5. tn other root 9. q 4p 0 q p q p, q 4 p,,, 4 Hene 7 vlues of (p, q) 7 equtions

More information

UNIT-I CURVE FITTING AND THEORY OF EQUATIONS

UNIT-I CURVE FITTING AND THEORY OF EQUATIONS Part-A 1. Define linear law. The relation between the variables x & y is liner. Let y = ax + b (1) If the points (x i, y i ) are plotted in the graph sheet, they should lie on a straight line. a is the

More information

Class Summary. be functions and f( D) , we define the composition of f with g, denoted g f by

Class Summary. be functions and f( D) , we define the composition of f with g, denoted g f by Clss Summy.5 Eponentil Functions.6 Invese Functions nd Logithms A function f is ule tht ssigns to ech element D ectly one element, clled f( ), in. Fo emple : function not function Given functions f, g:

More information

Roots and Coefficients Polynomials Preliminary Maths Extension 1

Roots and Coefficients Polynomials Preliminary Maths Extension 1 Preliminary Maths Extension Question If, and are the roots of x 5x x 0, find the following. (d) (e) Question If p, q and r are the roots of x x x 4 0, evaluate the following. pq r pq qr rp p q q r r p

More information

UNIT-2 POLYNOMIALS Downloaded From: [Year] UNIT-2 POLYNOMIALS

UNIT-2 POLYNOMIALS Downloaded From:   [Year] UNIT-2 POLYNOMIALS UNIT- POLYNOMIALS Downloded From: www.jsuniltutoril.weebly.com [Yer] UNIT- POLYNOMIALS It is not once nor twice but times without number tht the sme ides mke their ppernce in the world.. Find the vlue

More information

20 MATHEMATICS POLYNOMIALS

20 MATHEMATICS POLYNOMIALS 0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of

More information

Optimization. x = 22 corresponds to local maximum by second derivative test

Optimization. x = 22 corresponds to local maximum by second derivative test Optimiztion Lectue 17 discussed the exteme vlues of functions. This lectue will pply the lesson fom Lectue 17 to wod poblems. In this section, it is impotnt to emembe we e in Clculus I nd e deling one-vible

More information

OCR Maths FP1. Topic Questions from Papers. Roots of Polynomial Equations

OCR Maths FP1. Topic Questions from Papers. Roots of Polynomial Equations OCR Maths FP1 Topic Questions from Papers Roots of Polynomial Equations PhysicsAndMathsTutor.com 18 (a) The quadratic equation x 2 2x + 4 = 0hasroots and. (i) Write down the values of + and. [2] (ii) Show

More information

EECE 260 Electrical Circuits Prof. Mark Fowler

EECE 260 Electrical Circuits Prof. Mark Fowler EECE 60 Electicl Cicuits Pof. Mk Fowle Complex Numbe Review /6 Complex Numbes Complex numbes ise s oots of polynomils. Definition of imginy # nd some esulting popeties: ( ( )( ) )( ) Recll tht the solution

More information

Radial geodesics in Schwarzschild spacetime

Radial geodesics in Schwarzschild spacetime Rdil geodesics in Schwzschild spcetime Spheiclly symmetic solutions to the Einstein eqution tke the fom ds dt d dθ sin θdϕ whee is constnt. We lso hve the connection components, which now tke the fom using

More information

PROGRESSION AND SERIES

PROGRESSION AND SERIES INTRODUCTION PROGRESSION AND SERIES A gemet of umbes {,,,,, } ccodig to some well defied ule o set of ules is clled sequece Moe pecisely, we my defie sequece s fuctio whose domi is some subset of set of

More information

Downloaded from

Downloaded from POLYNOMIALS UNIT- It is not once nor twice but times without number tht the sme ides mke their ppernce in the world.. Find the vlue for K for which x 4 + 0x 3 + 5x + 5x + K exctly divisible by x + 7. Ans:

More information

Repeated Root and Common Root

Repeated Root and Common Root Repeted Root d Commo Root 1 (Method 1) Let α, β, γ e the roots of p(x) x + x + 0 (1) The α + β + γ 0, αβ + βγ + γα, αβγ - () (α - β) (α + β) - αβ (α + β) [ (βγ + γα)] + [(α + β) + γ (α + β)] +γ (α + β)

More information

PARABOLA EXERCISE 3(B)

PARABOLA EXERCISE 3(B) PARABOLA EXERCISE (B). Find eqution of the tngent nd norml to the prbol y = 6x t the positive end of the ltus rectum. Eqution of prbol y = 6x 4 = 6 = / Positive end of the Ltus rectum is(, ) =, Eqution

More information

3.0 INTRODUCTION 3.1 OBJECTIVES 3.2 SOLUTION OF QUADRATIC EQUATIONS. Structure

3.0 INTRODUCTION 3.1 OBJECTIVES 3.2 SOLUTION OF QUADRATIC EQUATIONS. Structure UNIT 3 EQUATIONS Equations Structure 3.0 Introduction 3.1 Objectives 3.2 Solution of Quadratic Equations 3.3 Quadratic Formula 3.4 Cubic and Bioquadratic Equations 3.5 Answers to Check Your Progress 3.6

More information

BINOMIAL THEOREM SOLUTION. 1. (D) n. = (C 0 + C 1 x +C 2 x C n x n ) (1+ x+ x 2 +.)

BINOMIAL THEOREM SOLUTION. 1. (D) n. = (C 0 + C 1 x +C 2 x C n x n ) (1+ x+ x 2 +.) BINOMIAL THEOREM SOLUTION. (D) ( + + +... + ) (+ + +.) The coefficiet of + + + +... + fo. Moeove coefficiet of is + + + +... + if >. So. (B)... e!!!! The equied coefficiet coefficiet of i e -.!...!. (A),

More information

of the contestants play as Falco, and 1 6

of the contestants play as Falco, and 1 6 JHMT 05 Algeba Test Solutions 4 Febuay 05. In a Supe Smash Bothes tounament, of the contestants play as Fox, 3 of the contestants play as Falco, and 6 of the contestants play as Peach. Given that thee

More information

Subject : MATHEMATICS

Subject : MATHEMATICS CCE RF 560 00 KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE 560 00 05 S. S. L. C. EXAMINATION, MARCH/APRIL, 05 : 06. 04. 05 ] MODEL ANSWERS : 8-E Date : 06. 04. 05 ] CODE NO.

More information

Tenth Maths Polynomials

Tenth Maths Polynomials Tenth Maths Polynomials Polynomials are algebraic expressions constructed using constants and variables. Coefficients operate on variables, which can be raised to various powers of non-negative integer

More information

QUADRATIC EQUATION. Contents

QUADRATIC EQUATION. Contents QUADRATIC EQUATION Contents Topi Pge No. Theory 0-04 Exerise - 05-09 Exerise - 09-3 Exerise - 3 4-5 Exerise - 4 6 Answer Key 7-8 Syllus Qudrti equtions with rel oeffiients, reltions etween roots nd oeffiients,

More information

Math 4318 : Real Analysis II Mid-Term Exam 1 14 February 2013

Math 4318 : Real Analysis II Mid-Term Exam 1 14 February 2013 Mth 4318 : Rel Anlysis II Mid-Tem Exm 1 14 Febuy 2013 Nme: Definitions: Tue/Flse: Poofs: 1. 2. 3. 4. 5. 6. Totl: Definitions nd Sttements of Theoems 1. (2 points) Fo function f(x) defined on (, b) nd fo

More information

MATHEMATICS IV 2 MARKS. 5 2 = e 3, 4

MATHEMATICS IV 2 MARKS. 5 2 = e 3, 4 MATHEMATICS IV MARKS. If + + 6 + c epesents cicle with dius 6, find the vlue of c. R 9 f c ; g, f 6 9 c 6 c c. Find the eccenticit of the hpeol Eqution of the hpeol Hee, nd + e + e 5 e 5 e. Find the distnce

More information

Random Variables and Probability Distribution Random Variable

Random Variables and Probability Distribution Random Variable Random Vaiables and Pobability Distibution Random Vaiable Random vaiable: If S is the sample space P(S) is the powe set of the sample space, P is the pobability of the function then (S, P(S), P) is called

More information

7.5-Determinants in Two Variables

7.5-Determinants in Two Variables 7.-eteminnts in Two Vibles efinition of eteminnt The deteminnt of sque mti is el numbe ssocited with the mti. Eve sque mti hs deteminnt. The deteminnt of mti is the single ent of the mti. The deteminnt

More information

MTH 505: Number Theory Spring 2017

MTH 505: Number Theory Spring 2017 MTH 505: Numer Theory Spring 207 Homework 2 Drew Armstrong The Froenius Coin Prolem. Consider the eqution x ` y c where,, c, x, y re nturl numers. We cn think of $ nd $ s two denomintions of coins nd $c

More information

THEORY OF EQUATIONS SYNOPSIS. Polyomil Fuctio: If,, re rel d is positive iteger, the f)x) = + x + x +.. + x is clled polyomil fuctio.. Degree of the Polyomil: The highest power of x for which the coefficiet

More information

1. If y 2 2x 2y + 5 = 0 is (A) a circle with centre (1, 1) (B) a parabola with vertex (1, 2) 9 (A) 0, (B) 4, (C) (4, 4) (D) a (C) c = am m.

1. If y 2 2x 2y + 5 = 0 is (A) a circle with centre (1, 1) (B) a parabola with vertex (1, 2) 9 (A) 0, (B) 4, (C) (4, 4) (D) a (C) c = am m. SET I. If y x y + 5 = 0 is (A) circle with centre (, ) (B) prbol with vertex (, ) (C) prbol with directrix x = 3. The focus of the prbol x 8x + y + 7 = 0 is (D) prbol with directrix x = 9 9 (A) 0, (B)

More information

DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING FLUID MECHANICS III Solutions to Problem Sheet 3

DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING FLUID MECHANICS III Solutions to Problem Sheet 3 DEPATMENT OF CIVIL AND ENVIONMENTAL ENGINEEING FLID MECHANICS III Solutions to Poblem Sheet 3 1. An tmospheic vote is moelle s combintion of viscous coe otting s soli boy with ngul velocity Ω n n iottionl

More information

is equal to - (A) abc (B) 2abc (C) 0 (D) 4abc (sinx) + a 2 (sin 2 x) a n (A) 1 (B) 1 (C) 0 (D) 2 is equal to -

is equal to - (A) abc (B) 2abc (C) 0 (D) 4abc (sinx) + a 2 (sin 2 x) a n (A) 1 (B) 1 (C) 0 (D) 2 is equal to - J-Mthemtics XRCIS - 0 CHCK YOUR GRASP SLCT TH CORRCT ALTRNATIV (ONLY ON CORRCT ANSWR). The vlue of determinnt c c c c c c (A) c (B) c (C) 0 (D) 4c. If sin x cos x cos 4x cos x cos x sin x 4 cos x sin x

More information

The Area of a Triangle

The Area of a Triangle The e of Tingle tkhlid June 1, 015 1 Intodution In this tile we will e disussing the vious methods used fo detemining the e of tingle. Let [X] denote the e of X. Using se nd Height To stt off, the simplest

More information

Data Structures. Element Uniqueness Problem. Hash Tables. Example. Hash Tables. Dana Shapira. 19 x 1. ) h(x 4. ) h(x 2. ) h(x 3. h(x 1. x 4. x 2.

Data Structures. Element Uniqueness Problem. Hash Tables. Example. Hash Tables. Dana Shapira. 19 x 1. ) h(x 4. ) h(x 2. ) h(x 3. h(x 1. x 4. x 2. Element Uniqueness Poblem Dt Stuctues Let x,..., xn < m Detemine whethe thee exist i j such tht x i =x j Sot Algoithm Bucket Sot Dn Shpi Hsh Tbles fo (i=;i

More information

SSC [PRE+MAINS] Mock Test 131 [Answer with Solution]

SSC [PRE+MAINS] Mock Test 131 [Answer with Solution] SS [PRE+MINS] Mock Test [nswe with Solution]. () Put 0 in the given epession we get, LHS 0 0. () Given. () Putting nd b in b + bc + c 0 we get, + c 0 c /, b, c / o,, b, c. () bc b c c b 0. b b b b nd hee,

More information

A B= ( ) because from A to B is 3 right, 2 down.

A B= ( ) because from A to B is 3 right, 2 down. 8. Vectors nd vector nottion Questions re trgeted t the grdes indicted Remember: mgnitude mens size. The vector ( ) mens move left nd up. On Resource sheet 8. drw ccurtely nd lbel the following vectors.

More information

Lecture 10. Solution of Nonlinear Equations - II

Lecture 10. Solution of Nonlinear Equations - II Fied point Poblems Lectue Solution o Nonline Equtions - II Given unction g : R R, vlue such tht gis clled ied point o the unction g, since is unchnged when g is pplied to it. Whees with nonline eqution

More information

On the diagram below the displacement is represented by the directed line segment OA.

On the diagram below the displacement is represented by the directed line segment OA. Vectors Sclrs nd Vectors A vector is quntity tht hs mgnitude nd direction. One exmple of vector is velocity. The velocity of n oject is determined y the mgnitude(speed) nd direction of trvel. Other exmples

More information

General Physics II. number of field lines/area. for whole surface: for continuous surface is a whole surface

General Physics II. number of field lines/area. for whole surface: for continuous surface is a whole surface Genel Physics II Chpte 3: Guss w We now wnt to quickly discuss one of the moe useful tools fo clculting the electic field, nmely Guss lw. In ode to undestnd Guss s lw, it seems we need to know the concept

More information

3.6 Applied Optimization

3.6 Applied Optimization .6 Applied Optimization Section.6 Notes Page In this section we will be looking at wod poblems whee it asks us to maimize o minimize something. Fo all the poblems in this section you will be taking the

More information

4.5 JACOBI ITERATION FOR FINDING EIGENVALUES OF A REAL SYMMETRIC MATRIX. be a real symmetric matrix. ; (where we choose θ π for.

4.5 JACOBI ITERATION FOR FINDING EIGENVALUES OF A REAL SYMMETRIC MATRIX. be a real symmetric matrix. ; (where we choose θ π for. 4.5 JACOBI ITERATION FOR FINDING EIGENVALUES OF A REAL SYMMETRIC MATRIX Some reliminries: Let A be rel symmetric mtrix. Let Cos θ ; (where we choose θ π for Cos θ 4 purposes of convergence of the scheme)

More information

MATHEMATICS AND STATISTICS 1.2

MATHEMATICS AND STATISTICS 1.2 MATHEMATICS AND STATISTICS. Apply lgebric procedures in solving problems Eternlly ssessed 4 credits Electronic technology, such s clcultors or computers, re not permitted in the ssessment of this stndr

More information

Physics 505 Fall 2005 Midterm Solutions. This midterm is a two hour open book, open notes exam. Do all three problems.

Physics 505 Fall 2005 Midterm Solutions. This midterm is a two hour open book, open notes exam. Do all three problems. Physics 55 Fll 5 Midtem Solutions This midtem is two hou open ook, open notes exm. Do ll thee polems. [35 pts] 1. A ectngul ox hs sides of lengths, nd c z x c [1] ) Fo the Diichlet polem in the inteio

More information

JEE(MAIN) 2015 TEST PAPER WITH SOLUTION (HELD ON SATURDAY 04 th APRIL, 2015) PART B MATHEMATICS

JEE(MAIN) 2015 TEST PAPER WITH SOLUTION (HELD ON SATURDAY 04 th APRIL, 2015) PART B MATHEMATICS JEE(MAIN) 05 TEST PAPER WITH SOLUTION (HELD ON SATURDAY 0 th APRIL, 05) PART B MATHEMATICS CODE-D. Let, b nd c be three non-zero vectors such tht no two of them re colliner nd, b c b c. If is the ngle

More information

π,π is the angle FROM a! TO b

π,π is the angle FROM a! TO b Mth 151: 1.2 The Dot Poduct We hve scled vectos (o, multiplied vectos y el nume clled scl) nd dded vectos (in ectngul component fom). Cn we multiply vectos togethe? The nswe is YES! In fct, thee e two

More information

PART- A 1. (C) 2. (D) 3. (D) 4. (B) 5. (D) 6. (C) 7. (D) 8. (B) 9. (D) 10. (B) 11. (B) 12. (B) 13. (A) 14. (D)

PART- A 1. (C) 2. (D) 3. (D) 4. (B) 5. (D) 6. (C) 7. (D) 8. (B) 9. (D) 10. (B) 11. (B) 12. (B) 13. (A) 14. (D) PRACTICE TEST-4 KISHORE VAIGYANIK PROTSAHAN YOJANA (KVPY) 7 STREAM (SA)_ DATE : -9-7 ANSWER KEY PART- A. (C). (D) 3. (D) 4. (B) 5. (D) 6. (C) 7. (D) 8. (B) 9. (D). (B). (B). (B) 3. (A) 4. (D) 5. (C) 6.

More information

Algebra Based Physics. Gravitational Force. PSI Honors universal gravitation presentation Update Fall 2016.notebookNovember 10, 2016

Algebra Based Physics. Gravitational Force. PSI Honors universal gravitation presentation Update Fall 2016.notebookNovember 10, 2016 Newton's Lw of Univesl Gvittion Gvittionl Foce lick on the topic to go to tht section Gvittionl Field lgeb sed Physics Newton's Lw of Univesl Gvittion Sufce Gvity Gvittionl Field in Spce Keple's Thid Lw

More information

The graphs of Rational Functions

The graphs of Rational Functions Lecture 4 5A: The its of Rtionl Functions s x nd s x + The grphs of Rtionl Functions The grphs of rtionl functions hve severl differences compred to power functions. One of the differences is the behvior

More information

Orthogonal Polynomials

Orthogonal Polynomials Mth 4401 Gussin Qudrture Pge 1 Orthogonl Polynomils Orthogonl polynomils rise from series solutions to differentil equtions, lthough they cn be rrived t in vriety of different mnners. Orthogonl polynomils

More information

Chapter 7. Kleene s Theorem. 7.1 Kleene s Theorem. The following theorem is the most important and fundamental result in the theory of FA s:

Chapter 7. Kleene s Theorem. 7.1 Kleene s Theorem. The following theorem is the most important and fundamental result in the theory of FA s: Chpte 7 Kleene s Theoem 7.1 Kleene s Theoem The following theoem is the most impotnt nd fundmentl esult in the theoy of FA s: Theoem 6 Any lnguge tht cn e defined y eithe egul expession, o finite utomt,

More information

Mark Scheme (Results) January 2008

Mark Scheme (Results) January 2008 Mk Scheme (Results) Jnuy 00 GCE GCE Mthemtics (6679/0) Edecel Limited. Registeed in Englnd nd Wles No. 4496750 Registeed Office: One90 High Holbon, London WCV 7BH Jnuy 00 6679 Mechnics M Mk Scheme Question

More information

ALL INDIA TEST SERIES

ALL INDIA TEST SERIES Fom Classoom/Integated School Pogams 7 in Top 0, in Top 00, 54 in Top 00, 06 in Top 500 All India Ranks & 4 Students fom Classoom /Integated School Pogams & 7 Students fom All Pogams have been Awaded a

More information

Polynomials and Division Theory

Polynomials and Division Theory Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the

More information

along the vector 5 a) Find the plane s coordinate after 1 hour. b) Find the plane s coordinate after 2 hours. c) Find the plane s coordinate

along the vector 5 a) Find the plane s coordinate after 1 hour. b) Find the plane s coordinate after 2 hours. c) Find the plane s coordinate L8 VECTOR EQUATIONS OF LINES HL Mth - Sntowski Vector eqution of line 1 A plne strts journey t the point (4,1) moves ech hour long the vector. ) Find the plne s coordinte fter 1 hour. b) Find the plne

More information

Each term is formed by adding a constant to the previous term. Geometric progression

Each term is formed by adding a constant to the previous term. Geometric progression Chpter 4 Mthemticl Progressions PROGRESSION AND SEQUENCE Sequence A sequence is succession of numbers ech of which is formed ccording to definite lw tht is the sme throughout the sequence. Arithmetic Progression

More information

U>, and is negative. Electric Potential Energy

U>, and is negative. Electric Potential Energy Electic Potentil Enegy Think of gvittionl potentil enegy. When the lock is moved veticlly up ginst gvity, the gvittionl foce does negtive wok (you do positive wok), nd the potentil enegy (U) inceses. When

More information

NS-IBTS indices calculation procedure

NS-IBTS indices calculation procedure ICES Dt Cente DATRAS 1.1 NS-IBTS indices 2013 DATRAS Pocedue Document NS-IBTS indices clcultion pocedue Contents Genel... 2 I Rw ge dt CA -> Age-length key by RFA fo defined ge nge ALK... 4 II Rw length

More information

Progression. CATsyllabus.com. CATsyllabus.com. Sequence & Series. Arithmetic Progression (A.P.) n th term of an A.P.

Progression. CATsyllabus.com. CATsyllabus.com. Sequence & Series. Arithmetic Progression (A.P.) n th term of an A.P. Pogessio Sequece & Seies A set of umbes whose domai is a eal umbe is called a SEQUENCE ad sum of the sequece is called a SERIES. If a, a, a, a 4,., a, is a sequece, the the expessio a + a + a + a 4 + a

More information

4 Unit Math Homework for Year 12

4 Unit Math Homework for Year 12 Yimin Math Centre 4 Unit Math Homework for Year 12 Student Name: Grade: Date: Score: Table of contents 3 Topic 3 Polynomials Part 2 1 3.2 Factorisation of polynomials and fundamental theorem of algebra...........

More information

This immediately suggests an inverse-square law for a "piece" of current along the line.

This immediately suggests an inverse-square law for a piece of current along the line. Electomgnetic Theoy (EMT) Pof Rui, UNC Asheville, doctophys on YouTube Chpte T Notes The iot-svt Lw T nvese-sque Lw fo Mgnetism Compe the mgnitude of the electic field t distnce wy fom n infinite line

More information

f h = u, h g = v, we have u + v = f g. So, we wish

f h = u, h g = v, we have u + v = f g. So, we wish Answes to Homewok 4, Math 4111 (1) Pove that the following examples fom class ae indeed metic spaces. You only need to veify the tiangle inequality. (a) Let C be the set of continuous functions fom [0,

More information

Prerna Tower, Road No 2, Contractors Area, Bistupur, Jamshedpur , Tel (0657) ,

Prerna Tower, Road No 2, Contractors Area, Bistupur, Jamshedpur , Tel (0657) , R Pena Towe, Road No, Contactos Aea, Bistupu, Jamshedpu 8, Tel (657)89, www.penaclasses.com IIT JEE Mathematics Pape II PART III MATHEMATICS SECTION I Single Coect Answe Type This section contains 8 multiple

More information

Chapter 1: Logarithmic functions and indices

Chapter 1: Logarithmic functions and indices Chpter : Logrithmic functions nd indices. You cn simplify epressions y using rules of indices m n m n m n m n ( m ) n mn m m m m n m m n Emple Simplify these epressions: 5 r r c 4 4 d 6 5 e ( ) f ( ) 4

More information

Week 8. Topic 2 Properties of Logarithms

Week 8. Topic 2 Properties of Logarithms Week 8 Topic 2 Popeties of Logithms 1 Week 8 Topic 2 Popeties of Logithms Intoduction Since the esult of ithm is n eponent, we hve mny popeties of ithms tht e elted to the popeties of eponents. They e

More information

10 m, so the distance from the Sun to the Moon during a solar eclipse is. The mass of the Sun, Earth, and Moon are = =

10 m, so the distance from the Sun to the Moon during a solar eclipse is. The mass of the Sun, Earth, and Moon are = = Chpte 1 nivesl Gvittion 11 *P1. () The un-th distnce is 1.4 nd the th-moon 8 distnce is.84, so the distnce fom the un to the Moon duing sol eclipse is 11 8 11 1.4.84 = 1.4 The mss of the un, th, nd Moon

More information

Objective Mathematics

Objective Mathematics Multiple choice questions with ONE correct answer : ( Questions No. 1-5 ) 1. If the equation x n = (x + ) is having exactly three distinct real solutions, then exhaustive set of values of 'n' is given

More information

, MATHS H.O.D.: SUHAG R.KARIYA, BHOPAL, CONIC SECTION PART 8 OF

, MATHS H.O.D.: SUHAG R.KARIYA, BHOPAL, CONIC SECTION PART 8 OF DOWNLOAD FREE FROM www.tekoclsses.com, PH.: 0 903 903 7779, 98930 5888 Some questions (Assertion Reson tpe) re given elow. Ech question contins Sttement (Assertion) nd Sttement (Reson). Ech question hs

More information

Section 35 SHM and Circular Motion

Section 35 SHM and Circular Motion Section 35 SHM nd Cicul Motion Phsics 204A Clss Notes Wht do objects do? nd Wh do the do it? Objects sometimes oscillte in simple hmonic motion. In the lst section we looed t mss ibting t the end of sping.

More information

MEI STRUCTURED MATHEMATICS FURTHER CONCEPTS FOR ADVANCED MATHEMATICS, FP1. Practice Paper FP1-A

MEI STRUCTURED MATHEMATICS FURTHER CONCEPTS FOR ADVANCED MATHEMATICS, FP1. Practice Paper FP1-A MEI Mathematics in Education and Industry MEI STRUCTURED MATHEMATICS FURTHER CONCEPTS FOR ADVANCED MATHEMATICS, FP Practice Paper FP-A Additional materials: Answer booklet/paper Graph paper MEI Examination

More information

(n 1)n(n + 1)(n + 2) + 1 = (n 1)(n + 2)n(n + 1) + 1 = ( (n 2 + n 1) 1 )( (n 2 + n 1) + 1 ) + 1 = (n 2 + n 1) 2.

(n 1)n(n + 1)(n + 2) + 1 = (n 1)(n + 2)n(n + 1) + 1 = ( (n 2 + n 1) 1 )( (n 2 + n 1) + 1 ) + 1 = (n 2 + n 1) 2. Paabola Volume 5, Issue (017) Solutions 151 1540 Q151 Take any fou consecutive whole numbes, multiply them togethe and add 1. Make a conjectue and pove it! The esulting numbe can, fo instance, be expessed

More information

DISCUSSION CLASS OF DAX IS ON 22ND MARCH, TIME : 9-12 BRING ALL YOUR DOUBTS [STRAIGHT OBJECTIVE TYPE]

DISCUSSION CLASS OF DAX IS ON 22ND MARCH, TIME : 9-12 BRING ALL YOUR DOUBTS [STRAIGHT OBJECTIVE TYPE] DISCUSSION CLASS OF DAX IS ON ND MARCH, TIME : 9- BRING ALL YOUR DOUBTS [STRAIGHT OBJECTIVE TYPE] Q. Let y = cos x (cos x cos x). Then y is (A) 0 only when x 0 (B) 0 for all real x (C) 0 for all real x

More information

Bridging the gap: GCSE AS Level

Bridging the gap: GCSE AS Level Bridging the gp: GCSE AS Level CONTENTS Chpter Removing rckets pge Chpter Liner equtions Chpter Simultneous equtions 8 Chpter Fctors 0 Chpter Chnge the suject of the formul Chpter 6 Solving qudrtic equtions

More information

(e) if x = y + z and a divides any two of the integers x, y, or z, then a divides the remaining integer

(e) if x = y + z and a divides any two of the integers x, y, or z, then a divides the remaining integer Divisibility In this note we introduce the notion of divisibility for two integers nd b then we discuss the division lgorithm. First we give forml definition nd note some properties of the division opertion.

More information

Downloaded From:

Downloaded From: Downloded From: www.jsuniltutoril.weel.com UNIT-3 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES Like the crest of pecock so is mthemtics t the hed of ll knowledge.. At certin time in deer prk, the numer of

More information

Chapter 1: Fundamentals

Chapter 1: Fundamentals Chpter 1: Fundmentls 1.1 Rel Numbers Types of Rel Numbers: Nturl Numbers: {1, 2, 3,...}; These re the counting numbers. Integers: {... 3, 2, 1, 0, 1, 2, 3,...}; These re ll the nturl numbers, their negtives,

More information

SUMMER KNOWHOW STUDY AND LEARNING CENTRE

SUMMER KNOWHOW STUDY AND LEARNING CENTRE SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18

More information

defined on a domain can be expanded into the Taylor series around a point a except a singular point. Also, f( z)

defined on a domain can be expanded into the Taylor series around a point a except a singular point. Also, f( z) 08 Tylo eie nd Mcluin eie A holomophic function f( z) defined on domin cn be expnded into the Tylo eie ound point except ingul point. Alo, f( z) cn be expnded into the Mcluin eie in the open dik with diu

More information

St Andrew s Academy Mathematics Department Higher Mathematics VECTORS

St Andrew s Academy Mathematics Department Higher Mathematics VECTORS St ndew s cdemy Mthemtics etment Highe Mthemtics VETORS St ndew's cdemy Mths et 0117 1 Vectos sics 1. = nd = () Sketch the vectos nd. () Sketch the vectos nd. (c) Given u = +, sketch the vecto u. (d) Given

More information

Homework 3 MAE 118C Problems 2, 5, 7, 10, 14, 15, 18, 23, 30, 31 from Chapter 5, Lamarsh & Baratta. The flux for a point source is:

Homework 3 MAE 118C Problems 2, 5, 7, 10, 14, 15, 18, 23, 30, 31 from Chapter 5, Lamarsh & Baratta. The flux for a point source is: . Homewok 3 MAE 8C Poblems, 5, 7, 0, 4, 5, 8, 3, 30, 3 fom Chpte 5, msh & Btt Point souces emit nuetons/sec t points,,, n 3 fin the flux cuent hlf wy between one sie of the tingle (blck ot). The flux fo

More information

A 12. Solve for r: A = 2π. Chapter 1 Section 2: Formulas. Answers to Problems. Solve the following equations for the variable indicated.

A 12. Solve for r: A = 2π. Chapter 1 Section 2: Formulas. Answers to Problems. Solve the following equations for the variable indicated. Chapte 1 Section : Fomulas nswes to Poblems Solve the following equations fo the vaiable indicated. 1.* Solve fo x: 5x 10. Solve fo e: 7ae 3u x 3u e 7a 3.* Solve fo a: is at 4. Solve fo v: 4v is a t v

More information

The Formulas of Vector Calculus John Cullinan

The Formulas of Vector Calculus John Cullinan The Fomuls of Vecto lculus John ullinn Anlytic Geomety A vecto v is n n-tuple of el numbes: v = (v 1,..., v n ). Given two vectos v, w n, ddition nd multipliction with scl t e defined by Hee is bief list

More information

10 Statistical Distributions Solutions

10 Statistical Distributions Solutions Communictions Engineeing MSc - Peliminy Reding 1 Sttisticl Distiutions Solutions 1) Pove tht the vince of unifom distiution with minimum vlue nd mximum vlue ( is ) 1. The vince is the men of the sques

More information

Fundamental Theorem of Calculus

Fundamental Theorem of Calculus Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under

More information

Read section 3.3, 3.4 Announcements:

Read section 3.3, 3.4 Announcements: Dte: 3/1/13 Objective: SWBAT pply properties of exponentil functions nd will pply properties of rithms. Bell Ringer: 1. f x = 3x 6, find the inverse, f 1 x., Using your grphing clcultor, Grph 1. f x,f

More information

Mathematics Extension 2

Mathematics Extension 2 00 HIGHER SCHOOL CERTIFICATE EXAMINATION Mthemtics Etension Generl Instructions Reding time 5 minutes Working time hours Write using blck or blue pen Bord-pproved clcultors my be used A tble of stndrd

More information

If deg(num) deg(denom), then we should use long-division of polynomials to rewrite: p(x) = s(x) + r(x) q(x), q(x)

If deg(num) deg(denom), then we should use long-division of polynomials to rewrite: p(x) = s(x) + r(x) q(x), q(x) Mth 50 The method of prtil frction decomposition (PFD is used to integrte some rtionl functions of the form p(x, where p/q is in lowest terms nd deg(num < deg(denom. q(x If deg(num deg(denom, then we should

More information

PLANCESS RANK ACCELERATOR

PLANCESS RANK ACCELERATOR PLANCESS RANK ACCELERATOR MATHEMATICS FOR JEE MAIN & ADVANCED Sequeces d Seies 000questios with topic wise execises 000 polems of IIT-JEE & AIEEE exms of lst yes Levels of Execises ctegoized ito JEE Mi

More information

Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : , MT TRIGONOMETRIC FUNCTIONS AND TRIGONOMETRIC EQUATIONS C Trigonometric Functions : Bsic Trigonometric Identities : + cos = ; ; cos R sec tn = ; sec R (n ),n cosec cot = ; cosec R {n, n I} Circulr Definition

More information

4 7x =250; 5 3x =500; Read section 3.3, 3.4 Announcements: Bell Ringer: Use your calculator to solve

4 7x =250; 5 3x =500; Read section 3.3, 3.4 Announcements: Bell Ringer: Use your calculator to solve Dte: 3/14/13 Objective: SWBAT pply properties of exponentil functions nd will pply properties of rithms. Bell Ringer: Use your clcultor to solve 4 7x =250; 5 3x =500; HW Requests: Properties of Log Equtions

More information

15 - TRIGONOMETRY Page 1 ( Answers at the end of all questions )

15 - TRIGONOMETRY Page 1 ( Answers at the end of all questions ) - TRIGONOMETRY Pge P ( ) In tringle PQR, R =. If tn b c = 0, 0, then Q nd tn re the roots of the eqution = b c c = b b = c b = c [ AIEEE 00 ] ( ) In tringle ABC, let C =. If r is the inrdius nd R is the

More information

SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014

SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014 SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 014 Mrk Scheme: Ech prt of Question 1 is worth four mrks which re wrded solely for the correct nswer.

More information

2.4 Linear Inequalities and Interval Notation

2.4 Linear Inequalities and Interval Notation .4 Liner Inequlities nd Intervl Nottion We wnt to solve equtions tht hve n inequlity symol insted of n equl sign. There re four inequlity symols tht we will look t: Less thn , Less thn or

More information

6. Numbers. The line of numbers: Important subsets of IR:

6. Numbers. The line of numbers: Important subsets of IR: 6. Nubes We do not give n xiotic definition of the el nubes hee. Intuitive ening: Ech point on the (infinite) line of nubes coesponds to el nube, i.e., n eleent of IR. The line of nubes: Ipotnt subsets

More information

A-Level Mathematics Transition Task (compulsory for all maths students and all further maths student)

A-Level Mathematics Transition Task (compulsory for all maths students and all further maths student) A-Level Mthemtics Trnsition Tsk (compulsory for ll mths students nd ll further mths student) Due: st Lesson of the yer. Length: - hours work (depending on prior knowledge) This trnsition tsk provides revision

More information

RELATIVE KINEMATICS. q 2 R 12. u 1 O 2 S 2 S 1. r 1 O 1. Figure 1

RELATIVE KINEMATICS. q 2 R 12. u 1 O 2 S 2 S 1. r 1 O 1. Figure 1 RELAIVE KINEMAICS he equtions of motion fo point P will be nlyzed in two diffeent efeence systems. One efeence system is inetil, fixed to the gound, the second system is moving in the physicl spce nd the

More information

Prerna Tower, Road No 2, Contractors Area, Bistupur, Jamshedpur , Tel (0657) ,

Prerna Tower, Road No 2, Contractors Area, Bistupur, Jamshedpur , Tel (0657) , R Pen Towe Rod No Conttos Ae Bistupu Jmshedpu 8 Tel (67)89 www.penlsses.om IIT JEE themtis Ppe II PART III ATHEATICS SECTION I (Totl ks : ) (Single Coet Answe Type) This setion ontins 8 multiple hoie questions.

More information

USA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year

USA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year 1/1/21. Fill in the circles in the picture t right with the digits 1-8, one digit in ech circle with no digit repeted, so tht no two circles tht re connected by line segment contin consecutive digits.

More information

1. Twelve less than five times a number is thirty three. What is the number

1. Twelve less than five times a number is thirty three. What is the number Alger 00 Midterm Review Nme: Dte: Directions: For the following prolems, on SEPARATE PIECE OF PAPER; Define the unknown vrile Set up n eqution (Include sketch/chrt if necessr) Solve nd show work Answer

More information