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 +.

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