QUADRATIC EQUATION EXERCISE - 01 CHECK YOUR GRASP
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1 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 re possile.. 4 < 0 Now x + x D () 4( )() ( 4) Here D < 0 lso oeffiient of x is positive. urve is lwys ove x xis expression is lwys positive.. Hint : A, B 7, C 7. Let x e the ommon ftor (i) Also (ii) from (i) nd (ii) + 0 Put in eqution : 4 x 5 8. x 5x 4. D > 0 7 > smllest integer is > 0 < 9 Also 4 (x 5) 0 (x 7)(x ). ( ) lest vlue 5 x x x x 7 y x ( y) x( y) 7y 0 D > 0 6y(y ) < 0 y [0, ). Hint : x mx + m 0 (i) f ( ) > 0 (ii) f (4) > 0 (iii) D 0 (iv) < < 4 Common solution m (, ) 4 4. Let roots of x Ax + Bx C 0 re,, A, B, C & ( + ) ( + ) ( + ) 9 ( + + ) + ( + + ) A + B + C (x + 4) (x ) x + 4 ± (x ) x + x + 0 or x x 7 0 D0 (x + ) 0 or No solution x Hve only one solution. 6. log / 0 x So, / x, x 0 0 x 00 0
2 EXERCISE - 0. Hint : os 4, x + (p + q)x + pq r(x + p + q) x + (p+q r)x + pq rp rq 0 sum of roots 0 p + q r 0 p + q r produt of roots pq r (p + q) pq (p + q ) 5. D ( + ) 84 4( + ) + ( ) > 0 ( + + 0) (p q). (A) 5. K K BRAIN TEASERS Expression is (x + (K+K )x +K(K-) 0 (x K) (x K + ) 0 x K or x K greter prt < K < (B) Opposite sign 0 K (K ) < 0 K(0,) (C) K (K ) > 0 K (, 0) U (, ) (D) K > K > K (, ) ,, Also + P, put in eqn. f( ) ve one root (, ) f( ) ve (x ) ( x ( + )) < 0 x Now x min. t < < mx. t + > 0 0, Reverse + < x < + < < 0. Put x 0 we get ƒ (x) ƒ (0) > 6 grph is shown s ƒ () lest vlue. Given expression redue to < > no solution [ (p q) (x p) + (x q)] 0 [ p + q x + p + x q] 0 hene x R.. Hint : D 0 f( ) ve 4 f( ) ve 6 one root in (, 4 ) f( 5 ) one root in ( f( ) 5, 6 ) 7. log (x + 8) log ( 4x x ) 5 5 x + 8 4x x x + ( + 4) x sine D 0 ( + 4) , At, x 4 Put in log domin No solution t 4 stisfies the domin 8. x 5 (x ) + x(x ) + y when x y > 0 < x < y > 0 x y > 0 hene y > 0 9. p + p + is lwys positive ƒ () < 0 0. x 4 px + qx rx + s 0 tn (A + B+ C + D) tn ( + D) s s tnd s s 4 p r q s
3 EXERCISE - 0 Tr ue & Flse :. D os p 4sinp(osp ) os p + 4sinp( osp) > 0 Fill in the lnks : & Mth the Column :. x + ( )x D 4( ) 4 ( + 5) 4( 4) 4 ( 4) ( + ) (A) D < 0 (, 4) (B) f()< < 0, 7 (C) (D) f(). f() < ( + 4) (7 + 8) < 0, 7 f() < 0 & f() < 0 MISCELLANEOUS TYPE QUESTIONS Comprehension # : x+. x. + x+ x x+ Let x x x x ( ) + x Similrly other root n e find out. x x x x + 4 x + 0x x x x x x x x x x x x + 7x x + 0. Comprehension # :. ( x x x x ) ( ) + ( ) ( ) 4 x x + x ( ) x ( ) 4 x x ± x,. x x ± x ±, ut x ,. x ± ( + ), ± 4 solution / / Assertion & Reson : x. f(x) (x + ) (x ) (s is root) f() + f() ( ) + ( ) 0 lso (5 6 ) (5 6 ) (8 5) 0 s hene x (5 6 ) + x x x (5 6 ) is one of the root of the eqution & degree of eqution is 4 & omplex roots our in onjugte t lest rel roots. x ± x ± & x ± domin of x > 0 x > OR x < hene x only solution.
4 EXERCISE - 04 [A] CONCEPTUAL SUBJECTIVE EXERCISE. D p 4q & D r 4s Add p + r 4(q + s) D + D p + r pr (p r) 0 D +D 0 possile only if t lest one of D is 0 5. x + 8x + 45 x x 8 x 45 6 x 8 x 45 ± 4 x 8 x 45 ± 4 + 5, x + 8x , (Rejet ) x + 8x Produt of root Considering denomintor x 8x + D < 0 nd > 0 So denomintor is lwys positive x + ( + )x < 0 < 0 & 4( + ) 4 (9 + 4) < 0 4( ) < 0 4( 8 + ) < > 0 (4 ) ( + ) > 0, x x 4x ( )x 7. Hint : 0 0 x x x x s Denomintor > 0 Numertor > 0 nd x x 0 x x x ( )x 4 0 x x s Denomintor > 0 Numertor < 0 Now solve it. 9. Let x x + 0 hs root & x(x px + q) 0 hs root 0, ut x 0 is not the ommon root (sine 0) & x px + q 0 hve two equl root D 0 p 4q 0 p p stisfy eqution x x + 0 is its root. p 4 p + 0 p 4 + p (q ) p. Let x x t. (t )(t 7) + 5 < 0 t 8t + < 0 (t 6)(t ) < 0 < x x < 6 x x 6 < 0 & x x > (x )(x + ) < 0 & (x )(x + ) > 0 x (, ) (, ) (x x) 9 (x ) 0 x x (x x )(x x ) x 0 x(x ) (x x )(x )(x ) x 0 x(x ) x (, ] (0,] (,]. Hint : D 0 & ƒ (0). ƒ () > 0 & 0 < ( /) <. 4. x + (K ) x + K D > 0 4(K K 4) 0 (K 4) (K + ) 0 K (, ) (4, ) Now sutrt those ses in whih oth roots re negtive (i) f(0) > 0 K + 5 > 0 K > 5 (ii) < 0 K < 0 K > So for K > oths root re negtive hene for tlest one positive root K (, ) 5. Let y (x )(x ) + (x )(x d) then y() > 0 y() < 0 y() < 0 y(d) > 0 y hve one rel root etween & & one rel root etween & d
5 EXERCISE - 04 [B] BRAIN STORMING SUBJECTIVE EXERCISE os x. Let y y y p os x 0 y Also y p y p os x y p 4 p 5 4 p, 5 4. Let the roots re rtionl + m + n + then D (m +) 4 ( + ) (n + ) odd 4 odd odd odd even sy(p + ) (m + ) 4 ( +) (n + ) (p + ) (m + ) (p + ) 4 ( + ) (n + ) even (m p) (m + p + ) ( + ) (n + ) Cse-I : m is odd p is even LHS odd even even Not possile RHS odd Similrly for m even p odd m even m odd p even p odd hene roots n not e rtionl do not hold 6. x + x + 0 hve rel roots of opposite sign in (, ) x + x + 0 () 4 D 0 0 () f( ) > 0 4 > > 0 () f() > > > 0 (4) 4 < < > 0 (5) < < omined ondition from () & () + 4 > 0 7. z x + y + + xy + x + y + x 4x (x + y + ) + (x ) minimum vlue of z is 9. x n + px + qx + r 0 n 5 () S n S n ( n ) ( ) 0 (s oeffiient of x n, x n 0) s + + n 0... n 0 ut n i ( )n r 0 i hene ll of roots re not rel () S n + ps + qs + nr 0 ( n + p. + q. + r) + ( n + p. + q. + r) ( n n + p. n + q. n + r) S n ps qs nr nr S n nr 0. log (x 8) log ( 4x x ) 5 x + 8 4x x x + ( +4) x for only one solution Cse-I : 5 D 0 or expression is perft squre 4, t 4 x 4 stisfies oth log domin t x 4 do not stisfies the log domin 4 domin 4x x >0 x + 4x < 0 6 < x < & x + 8 > 0 Cse-II : Only one root in domin of log 6 f( 6). f() < 0 (8 6) (8 + ) < 0, 4 (4 /, ) t x 6 t 4 similrily t x 4 x do not stisfies the domin. (, 4) {4}, x still stisfy the domin 4,.
6 EXERCISE - 05 [A]. (x ) (x ) (x ) (x ) (x ) (x ) + (x ) (x ) so (x ) (x ) + 0 hve roots, 6 4. Let roots, ( 5 ) ( ) 9 ( 5 ) ( ) ( + ) + ( ) given ( ) + so +,,,, 7. x + x... A.P.... H.P. JEE-[MAIN] : PREVIOUS YEAR QUESTIONS 9. x + px + ( p) 0 ( p) + p( p) + ( p) 0 ( p) ( p + p + ) 0 p x + x 0 x 0, 0. x + px p + 0 euse 4 is root p 7 x + px + q 0 p 4q q q 4. x ( )x 0 + ( + ) ( ) ( ) + 5 is min. t hs equl root. For onseutive integers roots 4 4. f(x) n x n + n x n x 0 f(0) 0 f() so ording to Roll's theorem f'(x) 0 hve t lest one root (0, ) so root of eqution n n x n + (n ) n x n hs roots less thn. 5. x mx + m 0 7. y (x m) x m ± 4 (m ) (m+) m + < 4 nd m > m < nd m > so m (, ) x 9x 7 0 x 9x 7 0 y + x 9x 7 x + x AM GM is min t x y + 0 p p is min then y mx p x + 9x + 7
7 p min p min 4 D 4 y mx + (8 7) 0 / x 6x + 0, x x + 6 0, let ommon root nd integer root of () eqution x 6x x, 4 n't e equl to Aliter : P(x) ( )x + ( )x + ( ) 0 only x P(x) 0 So roots re equl. it mens D 0 ( ) 4( ) ( ) q 4pr Let p q r P( ) 0 p q + r 0... () 4p q + r... () 4p + q + r? From () q p + r (p + r) 4pr 0 (p r) 0 p r from eq. () q r So from eq. () 4r 4r + r r So 4p + q + r 4r + 4r + r 9r 8 4. Given e sinx e sinx 4 let e sinx y y y 4 y 4y 0 y 5 euse t whih is not integer so 4 nd ommon root. P(x) k(x + ) P( ) k( ) k P(x) (x + ) P() 8 sinx e + 5 ut we know tht e e sinx e sinx e 5 so e sinx + 5 nd 5 so No rel solution of given eqution.
8 EXERCISE - 05 [B] JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS. x x x 0 Cse-I : x + 0 x x + x > 0 p (q p) 4p q x > 0 x + +,, Cse-II : x + < 0 x + x + + x > 0 x + x + > 0 x < is solution,, 6. () x 0 x d 0 x 0x d...(i) & + d 0...(ii) dd (i) & (ii) d 0( + ) surt (i) & (ii) ( ) + ( d) 0( ) d ( )...(iii) lso 0 d 0...(iv) from (iv) & (v) 0 0 (d ) ( ) ( + ) (d ) ( + ) (from (iii)) nd d 0 ( + ) () x px + r 0 + p, r r (q p) x qx + r...(v) / + q + 4 q q p nd r 9 8. x + px + q 0 then (p q) (q p) + p & q nd x + x + 0 & q is rel is rel so (p q) ( ) 0 Let hene S(I) is true p nd q x + x 0 x + px + q 0 ± (not possile) Hene S(II) is True ut S(II) is not the orret explintion of S(I) 0. q ( ) ( ) q p q p sum of the roots p q p p p q p q p q p Produt of the roots. Required eqution is (p + q)x (p q)x + (p + q) 0., re roots of x 6x & ( ) 9 9 ( ) ( ) ( ) 9 9 ( ) ( )
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