# QUADRATIC EQUATIONS OBJECTIVE PROBLEMS

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1 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) re p q r p (), r p p q, q r p q (), (d), q r p q. If + y 5, y, then () {, 4} () {, } {, 4,, 4} (d) {, } 4. The roots of the eqution ( + ) ( + ) 0 re (), (),, 5. The vlue of + is () () + ± 6. The numer of rel solutions of the eqution () () (d) 4 7. The roots of the eqution re () ±, ± () ±, ± i + 0 re ±, ± i

2 8. Let one root of + + c 0 where,, c re integers e + 5, then the other root is () 5 () 5 9. The roots of the eqution + + re () 0 () 0, (d) None 0. The vlue of is () () (d). If P ( ) + + c nd Q ( ) + d + c where c 0, then P( ). Q( ) 0 hs t lest () Four rel roots Four imginry roots () Two rel roots (d)none of these. The rel roots of the eqution re (), 4 (), 4 4, 4. If the roots of the eqution ( p + q ) in () A.P. H.P. () G.P. q ( p + r) + ( + r ) 0 q e rel nd equl, then p q, r, will e 7 4. Let α nd β e the roots of the eqution The eqution whose roots re α 9, β is () 0 () (d) If the product of the roots of the eqution e 6. If () () (d) , then / / () {5} () {8} φ (d) {5, 8} + 0 α is α, then the vlue of α will

3 7. The numer of roots of the eqution is () () (d) 4 8. The eqution ( + ) ( ) (4 ) hs () No solution () One solution Two solutions (d) More thn two solutions 9. The numer of solutions of log 4( ) log ( ) () () (d) 0 0. If the roots of the given eqution ( λ ) λ e equl in mgnitude ut opposite in sign, then λ () () (d) /. If root of the eqution + p + 0 is 4, while the roots of the eqution + p + q 0 re sme, then the vlue of q will e () 4 () 4/49 49/4. The eqution e 0 hs () Only one rel root 0 () At lest two rel roots Ectly two rel roots (d) Infinitely mny rel roots. The numer of solutions for the eqution is () 4 () (d) 4. The vlues of '' nd '' for which eqution () 6, 4 () 6, 5 6, 4 (d) 6, If + + c 0, 0,,, c Q, then oth the roots of the eqution hve four rel roots c re () Rtionl Irrtionl () Non-rel (d) Zero

4 6. If + + c 0, then the roots of the eqution c 0 re () Equl Rel () Imginry 7. If the roots of the given eqution (cos p ) + (cos p) + sin p 0 re rel, then π π () p ( π,0) () p, p ( 0, π ) (d) p ( 0,π ) 8. The roots of the eqution ( ) t ( c d) t ( c d ) 0 () re equl, then dc () c d d + c 0 (d) 9. The epression + + c hs the positive vlue if () 4c > 0 () 4c < 0 c < c d (d) < c 0. If the roots of the equtions p + q + r 0 nd () q pr q + 0 e rel, then p q () q pr (d) r pq p qr. If l, m, n re rel nd l m, then the roots of the eqution ( l m) 5( l + m) ( l m) 0 re () Comple Rel nd equl () Rel nd distinct. The lest integer k which mkes the roots of the eqution k 0 imginry is () 4 () 5 6 (d) 7. The condition for the roots of the eqution, to e equl is ( c ) ( c) + ( c) 0 () 0 () 0 c 0 4. Roots of + 0 re rel nd distinct if () > 0 () < 0, > 0 (d), < 0

5 5. The epression y + + c hs lwys the sme sign s c if () 4c < () 4c > c < (d) c > k ( ) 0 is perfect squre for how mny vlues of k () () 0 (d) 7. Let p, q {,,, 4}. The numer of equtions of the form p + q + 0 hving rel roots is () 5 () 9 7 (d) 8 8. If the roots of eqution re rel, then () [,8] () [, 8] (, 8) (d) (, 8) 9. If root of the eqution + + c 0 e reciprocl of root of the eqution then + + c 0, then () ( c c ) ( c )( c ) () ( ) ( c c )( c ) ( c c ) ( + c )( + c ) 40. If α nd β re the roots of the eqution , then + α β () () 7 7 (d) If the roots of the eqution A + B + C 0 re α, β nd the roots of the eqution + p + q 0 re α, β, then vlue of p will e B () B AC A 4 AC A AC B () A 4. If α nd β re the roots of the eqution ( + ) 0 then ( α + )( β + ) () () (d)

6 4. If the sum of the roots of the qudrtic eqution squres of their reciprocls, then / c, /, c / re in c () A.P. () G.P. H.P is equl to the sum of the 44. If the roots of the eqution + m + m m re sme, then the vlue of m will e () () 0 (d) 45. If the sum of the roots of the eqution + + c 0 e equl to the sum of their squres, then () ( + ) c () c( + c) ( + ) c (d) ( + ) c 46. If α, β re the roots of + p + 0 nd γ, δ re the roots of + q + 0,then p () ( α γ )( β γ )( α + δ )( β + δ ) () ( α + γ )( β + γ )( α δ )( β + δ ) ( α + γ )( β + γ )( α + δ )( β + δ ) q 47. If + i is root of the eqution + p + q 0, where p nd q re rel, then (, q) () ( 4,7) () ( 4, 7) (4, 7) (d) ( 4, 7) 48. If the roots of the eqution + + c 0 e α nd β, then the roots of the eqution c re () α, β () α, β p, α β 49. The qudrtic in, such tht A.M. of its roots is A nd G.M. is G, is () t At + G 0 () t At G 0 t + At + G If the sum of the roots of the eqution + p + q 0 is three times their difference, then which one of the following is true () 9 p q () q 9 p p 9q (d) 9q p

7 5. A two digit numer is four times the sum nd three times the product of its digits. The numer is () 4 () 4 (d) 5. If the product of roots of the eqution, e m m ( ) 0 is, then the vlue of m will () () (d) 5. If the roots of the eqution + + p + q r re equl in mgnitude ut opposite in sign, then the product of the roots will e () p + q p q () (d) ( p + q ) ( p q ) α β β + α If α, β re the roots of the eqution + + c 0, then + () c () (d) 55. The eqution formed y decresing ech root of + + c 0 y is , then () () c c (d) + c 56. If the rtio of the roots of + + c 0 nd + q + r 0 e the sme, then () r c q () r c q r cq (d) rc q 57. If the rtio of the roots of + + c 0 is sme s the rtio of the roots of + q + r 0 then p, () () c q pr c q pr q c pr

8 58. If the sum of the roots of the eqution + p + q 0 is equl to the sum of their squres, then () p q 0 () p + q q p p q Let α, β e the roots of + p 0 nd γ, δ e the roots of 4 + q 0. If α, β, γ, δ re in G.P., then integrl vlues of p, q re respectively (), (), 6, (d) 6, 60. If the roots of + + c 0 re α, β nd the roots of + B + C 0 is equl to () 0 () A re α k, β k, then B 4 AC 4c A (d) A 6. If α, β re the roots of , then the eqution with the roots, α β is () () (d) If p nd q re the roots of + p + q 0, then () p, q () p, q p, q 0 (d) p, q 0 6. If α, β re the roots of + + c 0 then () 0 () 0 0 c (d) c 0 nd α + β, α + β, β 64. If i is root of the eqution + 0, then () () (d) 65. If α, β re the roots of the eqution , then + is equl to () () (d) 4, α + re in G.P., where 4c α β

9 66. If nd re roots of p + q 0, then + () p () q p (d) q p 67. Product of rel roots of the eqution t () Is lwys positive Does not eist ()Is lwys negtive 68. If α, β re the roots of + + c 0, then the eqution whose roots re + α, + β is () + (4 ) c 0 () + (4 ) c 0 + ( 4) c 0 (d) + ( 4) c If one root of the eqution + p + q 0 is the squre of the other, then () p + q q( p + ) 0 () p + q + q( + p) 0 p + q + q( p ) 0 (d) p + q + q( p) Let two numers hve rithmetic men 9 nd geometric men 4. Then these numers re the roots of the qudrtic eqution () () (d) If α β ut α 5α nd β 5β, then the eqution whose roots re α / β nd β / α is () () (d) Difference etween the corresponding roots of nd is sme nd, then () () (d) If is root of + k 4 0, it is lso root of () k 0 () 5 + k 0 k (d) + k + 4 0

10 74. If, y, z re rel nd distinct, then is lwys u + 4y + 9z 6yz z zy () Non-negtive Zero () Non-positive 75. If root of the equtions + p + q 0 nd + α + β 0 is common, then its vlue will e (where p α nd q β ) () q β α p () pβ αq q β q α p β or pβ αq q β If + e fctor of p + q, then ( p, q) () (, 4) () (4, 5) (4, ) (d) (5, 4) 77. If the two equtions c + d 0 nd + 0 hve one common root nd the second hs equl roots, then ( + d) () 0 () + c c (d) c 78. If is rel, the epression tkes ll vlue in the intervl (), (),, 79. If is rel, the function ( )( ) ( c) will ssume ll rel vlues, provided () > c > () < < c c < > (d) < c < n n 80. If the eqution n n..., 0, n n eqution n n + ( n ) n hs positive root, which is n, hs positive root α, then the () Greter thn or equl to α Greter thn α ()Equl to α (d) Smller thn α 8. If α nd β (α < β) re the roots of the eqution + + c 0, where c < 0 <, then () 0 () α < 0 < β < α < α < β α < β < 0 (d) α < 0 < α < β

11 8. If is rel, then the mimum nd minimum vlues of epression () 4, 5 () 5, 4 4, 5 (d) 4, will e 8. If h 0, h ( > 0) () () (d) 4 h hs common root, then the vlue of h is equl to 84. If the roots of the eqution re rel nd less thn, then () < () < 4 (d) > If >, then the eqution ( )( ) hs () Both roots in [, ] () Both roots in (, ) Both roots in (, + ) (d) One root in (, ) nd the other in (, + ) 86. If S is set of P () is polynomil of degree such tht ( 0) 0, () S 0 () S + ( ) (0, ) P P ( ), P '( ) > 0 (0,), then S + ( ) R (d) S + ( ) (0,) 87. The smllest vlue of + in the intervl, / ) ( is () /4 () 5 5 (d) The mimum possile numer of rel roots of eqution is () 0 () 4 (d) The solution set of the eqution ( p + q) + ( p + q) 0 pq is () q q p p, () p pq, q p pq q, (d) p + q p + q p q, (e) p q p q, p q

12 90. If is rel nd stisfies + > + 4, then () < () > 0 < < 0 (d) < < 4 9. If α, β nd γ re the roots of eqution then y α + αβγ stisfies the eqution () y + y + 0 () y y y 0 y + y y 0 (d) y + 4y + 5y If α, β nd γ re the roots of + 8 0, then the eqution whose roots re α, β nd γ is () 8 0 () (d) If > 0 for ll R, then () 5 < < () < 5 > 5 (d) < < If α, β, γ re the roots of the eqution , then ( α + β) + ( β + γ ) + ( γ + α) () () 4 (d) 5 QUADRATIC EQUATIONS HINTS AND SOLUTIONS ( 0). (d) ( ) 0,.. Given eqution is ( p q) + ( q r) + ( r p) 0 ( r q) ± ( q r) ( p q) 4( r p)( p q) ( r q) ± ( q + r p) r p, ( p q) p q

13 . + y 5 nd y ( 6)( 9) 0 6 ± 4 nd ±. 4. () Eqution ( + ) ( + ) 0 ( + ) + 0 nd 9 ( )( ) 0,. 5. () Let ± But the vlue of the given epression cnnot e negtive or less thn, therefore + is required nswer. 6. (d) Given + 0 ( )( ) 0 nd 7. () Eqution ±, ± ( 9) + ( 9) 0 ( + )( 9) 0 ± i, ±. 8. () If one root of qudrtic eqution with rtionl coefficients is irrtionl nd of the form α + β, then the other root must lso e irrtionl nd of the form α β. 9. (d) Given eqution is Squring on oth sides, we get (Irrtionl function) Thus 0 nd ( )( + ) 0,, since eqution is non-qudrtic eqution.

14 But , so it is equl to.. () Let ll four roots re imginry. Then roots of oth equtions P( ) 0 nd Q( ) 0 re imginry. Thus 4c < 0; d + 4c < 0, So + d < 0, which is impossile unless 0, d 0 So, if 0 or 0 If 0, d 0 d t lest two roots must e rel., we hve the equtions. P( ) + c 0 nd Q ( ) + c 0. Or c c ; s one of c nd c must e positive, so two roots must e rel.. (d) , 4, which is not possile. Hence, the given eqution hs no rel root.. () Given eqution is ( p + q ) q( p + r) + ( q + r ) 0 Roots re rel nd equl, then 4q ( p + r) 4( p + q )( q + r ) 0 4 q ( p + r + pr) ( p q + p r + q + q r ) 0 4 q p + q r + pq r p q p r q q r 0 4 pq r p r q 0 ( q pr) 0 Hence. Thus p, q, r in G.P. q pr 4. (d) Given [ ± i ] ( + i ), ( i ) ω,ω But nd. 9 α ω ω β ω ω Hence the eqution will e sme. α + 5. () According to condition α α + α + 0 α,. 6. (d) Given tht / 7 / / / Given eqution cn e written s ( ) 7( ) / Let, then it reduces to the eqution ( 5)( ) 0 5, Putting these vlues, we hve 5 nd 8.

15 7. (d) The eqution ( 4)( ) 0 4 ± 4 ±. 8. () Given ( + ) ( ) (4 ) Squring oth sides, we get ( ) Squring gin, we get 5 / 4 which does not stisfy the given eqution. Hence eqution hs no solution. 9. () ( ) log ( ) log 4 ( ) ( 5)( ) 0 5, ut < 0 when Only solution is 5 Hence numer of solution is one. 0. () Let roots re α nd α, then sum of the roots ( λ ) α + ( α ) 0 ( λ ) λ. Put 4 in + p + 0, we get p 7 Now second eqution + p + q 0 hve equl roots. Therefore!!. () e !! 0, 0,... 0 Hence, 0 n only one rel roots.. () Given eqution i.e., nd nd ( )( ) 0 nd ( + ).( + ) 0, nd,. i.e., Four solutions of this eqution. 4. (d) Let for rel roots re α, β, γ, δ then eqution is ( α)( β)( γ )( δ ) 0 p 4q q 49 4

16 4 ( α + β + γ + δ ) + ( αβ + βγ + γδ + αδ + βδ + αγ ( αβγ + βγδ + αβδ + αγδ ) + αβγδ 0 ) 4 α. + αβ. αβγ. + αβγδ 0 4 On compring with α 4, αβ αβγ, αβγδ Solving 4 ; 6 nd () D 4c ( c) 4c ( + + c 0) ( + c) 4c ( c) 0 Hence roots re rtionl. 6. We hve c 0 Let roots re α nd β Let D B 4 AC 9 4(4 ) 9 c Given tht, ( + + c) 0 ( + c) Putting this vlue, we get 9( + c) c 9( c) + 4c. Hence roots re rel. 7. Given eqution (cos p ) + (cos p) + sin p 0 Its discriminnt D 0 since roots re rel cos p 4(cos p )sin p 0 cos p 4 cos p sin p + 4 sin p 0 (cos p sin p) 4 sin p + 4 sin p 0 (cos sin p) + 4 sin p( sin p) 0 p..(i) Now ( sin p ) 0 for ll rel p, sin p > 0 for 0 < p < π. Therefore 4 sin p ( sin p) 0 when 0 < p < π or p ( 0, π ) 8. (d) Accordingly, {( c + d)} 4( + )( c + d ) 4 c + 4 d + 8cd 4 c + 4 d + 4 c + 4 d 4 d + 4 c 8cd 0 4( d c) 0 c d c. d 9. (d) + + c ( + ) + c ( ) + is perfect squre, therefore the given epression is positive if > 0. c or < c

17 0. () Equtions p + q + r 0 nd q ( pr ) + q 0 hve rel roots, then from first 4 q 4 pr 0 q pr q pr And from second 4( pr ) 4q 0 (for rel root) pr q...(ii) 0...(i) From (i) nd (ii), we get result. q pr. () Given eqution is ( l m) 5( l + m) ( l m) 0 Its discriminnt 5 + D ( l + m) 8 ( l m) Which is positive, since l m, n l., re rel nd m Hence roots re rel nd distinct.. (d) Roots re non rel if discriminnt < 0 i.e. if 5 4. k < 0 i.e. if 4 k > 5 i.e. if 5 k > 4 Hence, the required lest integer k is 7.. () According to question, 4( c) 4( c )( ( + + c c) 0 c) 0 0 or + + c c 4. () 4 AC > 0 B 0 4 > 0 < () Let f ( ) + + c. Then f ( 0) c. Thus the grph of y f() meets y-is t (0, c). If > 0 c, then y hypothesis f ( ) > 0 This mens tht the curve y f() does not meet -is. If < 0 c, then y hypothesis f ( ) < 0, which mens tht the curve y f() is lwys elow -is nd so it does not intersect with -is. Thus in oth cses y f() does not intersect with -is i.e. f( ) 0 for ny rel. Hence f( ) 0 i.e. + + c 0 hs imginry roots nd so < 4c. 6. () Given eqution ( + k ) + ( k) + ( k) 0 If eqution is perfect squre then roots re equl i.e., ( k ) 4( + k)( k) 0 i.e.,, 0 k. Hence totl numer of vlues. 7. For rel roots, discriminnt 0 q 4 p 0 q 4 p

18 For p, q 4 q,, 4 p, q 8 q,4 p, q q 4 p 4, q 6 q 4 Totl seven solutions re possile. 8. () Since the roots re rel. 64 4( 6) 0 Or [,8] 9. () Let α e root of first eqution, nd then α e root of second eqution. Therefore α + α + c 0 nd ' + + c 0 α α or c α + α + 0 Hence α α c cc c ( cc' ' ) ( ' c' )( ' c' ). 40. () Given eqution , therefore 4 α + β nd 7 αβ 4 Now α + β + α β αβ / 4 7 / () α, β re the roots of + B + C 0 So, B α + β nd A A. C αβ A Agin α, β re the roots of + p + q 0 then + α β p nd (αβ ) q Now α + β ( α + β) αβ α + β B A C A B AC AC p p 4. Given eqution ( + ) 0 A 0 α + β, αβ ( + ) Now ( α + )( β + ) αβ + α + β + A B ( + ) + +

19 4. As given, if α, β e the roots of the qudrtic eqution, then α + β + α β ( α + β) α β αβ ( / ) (c / ) ( c / ) c c ( + + c c c c ) c + c + c c c, c, re in A.P. c c,, re in H.P. 44. () Let roots re α ndα, then α + α m α m nd. α m m + 6 α m m m + 6 m. 45. Let α nd β e two roots of + + c 0 Then α + β nd c αβ α + β ( α + β) αβ c So under condition α + β + β c ( + ) c. 46. () As given, α + β p, αβ, γ + δ q nd γδ Now, ( α γ )( β γ )( α + δ )( β + δ ) { αβ γ ( α + β) + γ }{ αβ + δ ( α + β) + δ } ( + p γ + γ )( pδ + δ ) ( pγ qγ )( pδ qδ ) ) (Since γ is root of + q + 0 nd similrly δ + q δ ( p q)( p + q) q p γ + qγ + 0 γ + qγ γδ. 47. () Since + i is root, therefore i will e other root. Now sum of the roots 4 p nd product of roots 48. α, β re roots of + + c 0 α + β nd 7 q. Hence ( p, q) ( 4,7). c αβ Let the roots of c e α, β, then α + β c nd α β c ut α + β / αβ c / c + α + β α β

20 Hence α α nd β. β 49. () If α, β re the roots, then α + β A + β A G αβ αβ G α nd The required eqution is At + G Let α, β re roots of + p + q 0 So α + β p nd αβ q t. Given tht ( α + β) ( α β) p p α β Now ( α β) ( α + β) 4αβ p 9 p 4q or p 9q. 5. () It is oviously According to condition m m m m 5. () Given eqution cn e written s + ( p + q r) + pq pr qr 0...(i) Whose roots re α nd α, then the product of roots α pq pr qr pq r( p + q)...(ii) And sum p + q 0 p + q r r...(iii) 54. (d) From (ii) nd (iii), we get {( p + q) pq} p + q α pq ( p + q) α + β, αβ ( P + q ). c ( c) nd α + β Now α β + β + α + α( α + ) + β( β + ) ( β + )( α + )

21 ( α + β ) + ( α + β) αβ + ( α + β) + ( c) + c + + c c. c + c 55. () α, β e the roots of + + c 0 α + β /, αβ c / Roots re α, β Sum, α + β ( / ) 8 / 4 Product, ( )( β ) αβ ( α + β) + α c / + / + New eqution is / i.e., lso c c Let α, β e the roots of + + c 0 nd α ', β ' e the roots of + q + r 0. Then α + β, αβ c, α' + β ' q, α' β ' r It is given tht α α' α + β α' + β ' β β ' α β α' β' ( α + β) ( α β) ( α' + β' ) ( α' β' ) 4c q q 4r r q c 57. () If the roots of eqution + + c 0 re in the rtio m : n, Then we hve mn( ) ( m + n) c...(i) Also if the roots of the eqution p + q + r 0 re lso in the sme rtio m : n, then mn( q) ( m + n) pr...(ii) Dividing (i) nd (ii), we get q ( c) ( pr) or c q. pr 58. Let the roots e α nd β + β p Given, α + β α + β But α, αβ q α + β ( α + β) αβ p ( p) q + p q p q p p.

22 59. () Let r e the common rtio of the G.P. α, β, γ, δ then α r, α + β α + α r ( + r) α..(i) αβ p α ( αr) p α r p..(ii) α..(iii) γ + δ 4 αr + αr 4 r ( + r) 4 β αr δ αr γ nd nd q r 5 γδ α.αr q r q α..(iv) (p, q) (, ). 60. ( α β) ( α + β) 4αβ ( 4c) /...(i) Also {( α k) ( β k) } {( α k) + ( β k)} 4( α k)( β k) ( B / A) 4( C / A) ( B 4 AC) / A...(ii) From (i) nd (ii), ( 4c)/ ( B 4 AC) / A 4 AC 4c A 6. Given eqution is β 9 αβ α nd / 9 α β ( α β) 4 + 4αβ α, β Eqution ( α + β) + αβ () p + q p nd pq q p nd q. 6. (d) ( α + β ) ( α + β )( α + β ) c + c 4 c c c ( 4c) 0 As 0 c (d) i is root of the eqution so i ( ) i ( ) ( i) + 0

23 By comprison,,. 65. (d) Here, α + β nd αβ 4 α + β α + β ( αβ) ( α + β) αβ ( α + β) ( αβ) ( ) ( )(4) 6. (4) (d) Roots of given eqution p + q 0 is nd i.e., Then + p (i) nd q + p +. q (ii) 67. Note tht for t R, t nd hence the given eqution cnnot hve rel roots. 68. (d) We hve α + β nd c αβ Now sum of the roots + α + + β 4 And product of the roots ( + α )( + β ) c 4 + c 4 + Hence the required eqution is 4 + c (4 ) c 0 + ( 4) c (d) Let root of the given eqution + p + q 0 re α nd Now,. α α q, α α α p + Cuing oth sides, α + ( α ) + α. α ( α + α ) p α. q + q + q( p) p p + q + q( p) () Let the two numer is nd + 9 nd nd 6 Eqution (Sum of roots) + Product of roots 0 Required eqution

24 7. (d) α 5α + 0..(i) β 5β + 0..(ii) From (i) (ii), ( α β ) 5α + 5β 0 α β 5( α β) α + β 5 From (i) + (ii), ( α + β ) 5( α + β) ( α + β ) α + β 9 Then ( α + β) α + β + αβ αβ αβ Then the eqution, whose roots re α nd β, is β α α β α β β α β α α + β + 0 αβ () Let α, β re the roots of the eq n ± α, β 4 And α, β re the roots of the eqution So, α, + β 4 Now α β 4 ; α β 4 Given, α β α β ( ) Eqution + k 4 0 hs one root is. k 4 0 k 5 Put nd k 5 in options, only gives the correct nswer.

25 74. (), y, z R nd distinct. Now, u + 4y + 9z 6yz z y ( + 8y + 8 z yz 6z 4 y) { 4 y + 4y ) + ( 6z + 9z ) + (4 y yz + 9z )} {( y) + ( z) + (y z) } Since it is sum of squres. So u is lwys non- negtive 75. Let the common root e y. Then y + py + q 0 nd y + α y + β 0 On solving y cross multipliction, we hve y y pβ qα q β α p y q α β nd p y y y pβ qα q β (d) + e fctor of p + q 0 Hence ( + ) 0 ( )( ) 0,, Putting these vlues in given eqution So 4 p q (i) And p q 0...(ii) Solving (i) nd (ii), we get (p, q)(5, 4) 77. Let roots of c + d 0 e α, β then roots of + 0 e α, α α + β c, αβ d, α + α, α Hence ( + d) ( α + αβ ) α( α + β) c 78. () If the given epression e y, then y y + (y ) + (6y ) 0 If 0 y then 0 for rel i.e. B 4 AC 0 Or 9y + 0y + 0 or ( y + )(y ) 0 / y / If 0 y then which is rel nd this vlue of y is included in the ove rnge 79. (d) Let Or ( )( ) y ( c) y ( c) ( + ) + Or ( + + y) + + cy 0

26 ( + + y) 4( + cy) y + y( + c) + ( ) Since is rel nd y ssumes ll rel vlues, we must hve 0 for ll rel vlues of y. The sign of qudrtic in y is sme s of first term provided its discriminnt B 4 AC < 0 This will e so if 4( + c) 4( ) < 0 Or 4 ( + c + )( + c + ) < 0 Or 6 ( c)( c) < 0 or 6( c )( c ) ve c lies etween nd i.e., c < <...(i) Where <, ut if < then the ove condition will e c < Hence from (i) nd (ii) we oserve tht (d) is correct nswer. n n 80. (d) Let f( ) n n... ; f ( 0) 0; f ( α) 0 ( ) 0 f, hs tlest one root etween ( 0, α ) n n i.e., eqution n + n ) n ( n Hs positive root smller thn α. < or c > 8. () Here D 4c > 0 ecuse c < 0 <. So roots re rel nd unequl. Now, α + β < 0 nd αβ c < 0 >...(ii) One root is positive nd the other negtive, the negtive root eing numericlly igger. As < is the negtive root while β is the positive root. So, α > β nd α < 0 < β. α β, α 8. () Let y y ( + + ) ( y ) + (y 4) + y 9 0 For rel, its discriminnt 0 i.e. 4( y 7) 4( y )( y ) 0 y + y 0 0 or ( y 4)( y + 5) 0 Now, the product of two fctors is negtive if these re of opposite signs. So following two cses rise: Cse I: 4 0 y or 4 This is not possile. Cse II: 4 0 y or 4 y nd y or y 5 y nd y or y 5 Both of these re stisfied if 5 y 4

27 8. (d) Sutrcting, we get 56 Putting in ny, h 4( h > 0) h h or h () Given eqution is If roots re rel, then D 0 4 4( + ) As roots re less thn, hence f ( ) > > > 0 either < ( )( ) > 0 or > Hence < stisfy ll. 85. (d) The eqution is ( + ) + 0 Both roots re rel. Let them e discriminnt ( + ) 4( ) ( ) + 4 > 0 α, β where + ( + ) ( ) 4 α, ( + ) + β ( ) + 4 Clerly, ( + ) ( ) α < ( + ) ( ) ( > ) And ( + ) + ( ) β > + + Hence, one root α is less thn nd the other root β is greter thn. 86. (d) Let P ( ) + + c As P ( 0) 0 c 0 As P ( ) + P( ) + ( ) Now P ( ) + ( ) As P ( ) > 0 for (0,)

28 Only option (d) stisfies ove condition 87. () Therefore, smllest vlue is 4, which lie in, () Let f ( ) Then the numer of chnge of sign in f () is therefore f () cn hve t most two positive rel roots. 5 4 Now, f ( ) Then the numer of chnge of sign is. Hence f() cn hve t most one negtive rel root. So tht totl possile numer of rel roots is. 89. (d) Given eqution ( pq ) ( p + q) + ( p + q) 0 Let solution set is p + q p + q, p q Sum of roots ( p + q) pq p + q p + q + p q ( p + q) pq Similrly, product of roots ( p + q) pq p + q p + q ( p + q) p q pq. 90. () Given, + > + 4 ( + ) > ( + 4) > > 0 < or > 0 > 0 ( + ) > 0 9. () Given eqution Then + β + γ.. α, αβ + βγ + γα, αβγ 5 y Σα + αβγ ( α + β + γ ) ( αβ + βγ + γα) + αβγ 9 5 y It stisfies the eqution y y 0 9. (d) Let y. Then y y y / y 64 y 64 0

29 Thus the eqution hving roots α, β nd γ is () According to given condition, 4 4(0 ) < < 0 ( + 5)( ) < 0 5 < <. 94. If α, β, γ re the roots of the eqution. p + q r 0. ( α + β) + ( β + γ ) + ( γ + α) p + q pq r Given, p 0, q 4, r p + q pq r 0 +. 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

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### 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

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### AP Calculus Multiple Choice: BC Edition Solutions AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this

### Geometric Sequences. Geometric Sequence a sequence whose consecutive terms have a common ratio. Geometric Sequences Geometric Sequence sequence whose consecutive terms hve common rtio. Geometric Sequence A sequence is geometric if the rtios of consecutive terms re the sme. 2 3 4... 2 3 The number

### CONIC SECTIONS. Chapter 11 CONIC SECTIONS Chpter. Overview.. Sections of cone Let l e fied verticl line nd m e nother line intersecting it t fied point V nd inclined to it t n ngle α (Fig..). Fig.. Suppose we rotte the line m round

### QUADRATIC RESIDUES MATH 372. FALL INSTRUCTOR: PROFESSOR AITKEN QUADRATIC RESIDUES MATH 37 FALL 005 INSTRUCTOR: PROFESSOR AITKEN When is n integer sure modulo? When does udrtic eution hve roots modulo? These re the uestions tht will concern us in this hndout 1 The

### FORM FIVE ADDITIONAL MATHEMATIC NOTE. ar 3 = (1) ar 5 = = (2) (2) (1) a = T 8 = 81 FORM FIVE ADDITIONAL MATHEMATIC NOTE CHAPTER : PROGRESSION Arithmetic Progression T n = + (n ) d S n = n [ + (n )d] = n [ + Tn ] S = T = T = S S Emple : The th term of n A.P. is 86 nd the sum of the first

### Precalculus Due Tuesday/Wednesday, Sept. 12/13th Mr. Zawolo with questions. Preclculus Due Tuesd/Wednesd, Sept. /th Emil Mr. Zwolo (isc.zwolo@psv.us) with questions. 6 Sketch the grph of f : 7! nd its inverse function f (). FUNCTIONS (Chpter ) 6 7 Show tht f : 7! hs n inverse

### MTH 4-16a Trigonometry MTH 4-16 Trigonometry Level 4 [UNIT 5 REVISION SECTION ] I cn identify the opposite, djcent nd hypotenuse sides on right-ngled tringle. Identify the opposite, djcent nd hypotenuse in the following right-ngled

### 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

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### MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.) MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give

### Section 6.1 Definite Integral Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined

### Lecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar) Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of

### Suppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = -2. Mth 43 Section 6. Section 6.: Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot

### ( ) Same as above but m = f x = f x - symmetric to y-axis. find where f ( x) Relative: Find where f ( x) x a + lim exists ( lim f exists. AP Clculus Finl Review Sheet solutions When you see the words This is wht you think of doing Find the zeros Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor Find

### 8. Complex Numbers. We can combine the real numbers with this new imaginary number to form the complex numbers. 8. Complex Numers The rel numer system is dequte for solving mny mthemticl prolems. But it is necessry to extend the rel numer system to solve numer of importnt prolems. Complex numers do not chnge the 1 PYTHAGORAS THEOREM 1 1 Pythgors Theorem In this setion we will present geometri proof of the fmous theorem of Pythgors. Given right ngled tringle, the squre of the hypotenuse is equl to the sum of the PART MULTIPLE CHOICE Circle the pproprite response to ech of the questions below. Ech question hs vlue of point.. If in sequence the second level difference is constnt, thn the sequence is:. rithmetic