MATHEMATICS SOLUTIONS OF "ADVANCED LEVEL PROBLEMS"
Target : IIT-JEE TOPIC : QUADRATIC EQUATIONS PART - I 1.
x2 " x ! c
!
; x # R and y # R. x 2 ! x ! 2c (y – 1) x2 + (y + 1)x + 2y c – c = 0 x # R
$ $ $ $
D % 0 (y + 1)2 – 4 c(y – 1) (2y – 1) % 0 y2 + 1 + 2y – 4c [2y2 – 3y + 1] % 0 (1 – 8c)y2 + (2 + 12c) y + 1 – 4c % 0
Let y =
$
.......... (1)
Now for all y # R (1) will be true if . 1 and D & 0 8 $ 4 (1 + 6c)2 – 4 (1 – 8c) (1 – 4c) & 0 $ 1 + 36c2 + 12c – 1 – 32c2 + 12c & 0 – 6 & c & 0 $ $ 4c2 + 24c & 0 But c = –6 and c = 0 will not satisfy given condition
1 – 8c > 0 $ c <
2.
'
c # ( –6, 0)
Clearly the graph of y = x4 – 4x – 1 is
' no of positive real roots = 1 Aliter y = x 4 – 4x – 1 dy = 4x 4 x3 – 4 dx d2 y dx 2
= 12x2
when
dy = 0, then x = 1 dx
d2 y dx 2
x )1
= 12 > 0
so x = 1 is a minima point so by graph, number of positive real roots = 1 3.
$
a2 + b2 + c2 = 1 (a + b + c) 2 = a2 + b2 + c2 + 2 (ab + bc + ca) % 0 1 + 2 (ab + bc + ca) % 0
$
(ab + bc + ca) % –
!
$
1 2 2 2 2 a + b + c – (ab + bc + ca) % 0 (ab + bc + ca) & 1
'
From (1) and (2) we can say that
!
........(1) ........(2) (ab + bc + ca) #
* 1 ,+" 2 , 1/.
QUADRATIC EQUATIONS - 1
4.
5.
Since the given equation has distinct roots ' D>0 $ 16 + 4 (1 – k2) > 0 $ k2 < 5, also k 0 –1 If k = –1we will get only one solution, but we want two solutions ! ' k2 < 5, k 0 –1 !
12 3 are roots of
'
1 + 3 =
!
$ $
42
' '
=
1 2 ! 32 = 4 13
30
4
4 – 32 4 + 1 = 0 41, 42 are roots of (1) 41 + 42 = 32 and 4142 = 1 2
!
5 4 "1 and 13 = 4 4
1 3 $ + = 4 3 1 (1 + 3 )2 = 6 1 3 (4 " 1)2
4x2 – ( 4 – 1) x + 5 = 0
............(1)
41 42 ( 41 ! 4 2 )2 " 24 14 2 (32)2 " 2 = 4 2 + 41 = 4 14 2 1
= 1022
6.
Dis. of x2 + px + 3q is p2 – 12q 5 D1 2 Dis. of –x + rx + q is r2 + 4q 5 D2 Dis. of –x2 + sx – 2q is s2 – 8q 5 D3 Case 1 : If q < 0, then D1 > 0, D3 > 0 and D2 may or may not be positive Case 2 : If q > 0, then D2 > 0 and D1, D3 may or may not be positive Case 3 : If q = 0, then D1 % 0, D2 % 0 and D3 % 0 from Case 1, Case 2 and Case 3 we can say that the given equation has at least two real roots.
7.
(a – 1) (x2 + x + 1)2 – (a + 1) (x4 + x2 + 1) = 0 ........(1) 4 2 2 2 x + x + 1 = (x + x + 1) (x – x + 1) ! (1) becomes ' $ (x2 + x + 1) [(x2 + x + 1) (a – 1) – (a + 1) (x 2 – x + 1)] = 0 $ (x2 + x + 1) (x2 – ax + 1) = 0 Here two roots are imaginary and for other two roots to be real D> 0 $ a2 – 4 > 0 a # ( –6, –2) 7 (2, 6) $
8.
13 = 1232 ......(1) 2 2 1 + 3 = 1 + 3 ......(2) From (1), 13 = 0 or 13 = 1 Case 1 : If 13 = 0 then we have following possibilities (i) If 1 = 0 then from (2), 3 = 0 or 3 = 1 (ii) If 3 = 0 then from (2), 1 = 0 or 1 = 1 ' from (i) and (ii) we can say that roots are 1 = 0, 3 = 0 or 1 = 0, 3 = 1 ' Required quadratic equations are x2 = 0 or x2 – x = 0 Case 2 : If 13 = 1 then from (2) we get 1 + 3 = 12 + 32 = (1 + 3)2 – 213 ! 13 = 1 ' 1 + 3 = – 1, 2 So the required quadratic equations are x2 + x + 1 = 0 and x2 – 2x + 1 = 0 $ four equations are possible.
QUADRATIC EQUATIONS - 2
9.
Given equation can be written as x " a b x " b a – – + = 0 b x "a a x "b
$
8x " a 92 " b 2 8x " b 92 " a 2 ! b8x " a 9 a8x " b 9
$
(x – a – b) ,
$
B< a:x 2 " bx " ax ! ab ! bx " b 2 ;! b:x 2 " ax " bx ! ab ! ax " a 2 ;?< (x – a – b) A > = 0 ab8x " a 98x " b 9 <@ =<
$ $
(x – a – b) (ax2 – a2x + a2b – ab2 + bx2 – b2x + ab2 – a2b) = 0 x(x – a – b ) {x(a + b) – (a2 + b2)} = 0
*x " a ! b x " b ! a! / = 0 + b8x " a9 a8x " b9 .
roots will be
Let
x1 = a+b, x2 =
!
x1 – x2 – x3 = c (given)
'
(a + b) –
' either or
11.
f(x) =
2ab =c a!b
"bD C
C2 = b2 – 4ac and 2a 2a1 + C = – b and 2a3 – C = – b 2a1 + C = – b + 2C and 2a3 – C = – b – 2C Sum of roots of required equation = – 2b And product of roots = b2 or b2 – 4C2 = – 3b2 + 16ac Required equation is x2 + 2bx + b2 = 0 x2 + 2bx – 3b2 + 16ac = 0 3
x"2
+
and
12.
$
= c
a, c, b are in H.P.
!
again
[! given x1 > x2 > x3 ]
a2 ! b2 – 0 = c a!b
a!b
Since 1, 3 =
'
a 2 ! b2 and x3 = 0 a!b
8a ! b92 " 8a 2 ! b 2 9
i.e.
either or
a2 ! b 2 x = 0, a + b, a!b
'
$
10.
=0
4 x"3
+
5 x"4
f (2 ! ) E 6 ? <
>
f (3 " ) E "6< =
!
f (3 ! ) E 6 ? <
and
f ( 4 " ) E "6< =
>
$
f(x) = 0 has exactly one root in (2, 3).
$
f(x) = 0 has exactly one root in (3, 4).
Given equation can be Expressed as
Fe (x – F) (x – F – e) + eF (x – e) (x – F – e) + (FF + ee) (x – e) (x – F) = 0 Let
$
f(x) = Fe(x – F) (x – F – e) + eF (x – e) (x – F – e) + ( FF + ee) (x – e) (x – F) f(e) = Fe (e – F) ( – F) > 0 QUADRATIC EQUATIONS - 3
and f(F) = eF (F – e) ( –e) < 0 ; hence given equation has a real root in (e, F) f(F + e) = (FF + ee) F . e > 0
again
also
!
F + e > F2
!
F – e < e
It concluds it has a real root in (F, F + e)
Hence f(x) has two real roots in (F – e, F + e)
PART - II 1.
a2x2 + (b2 + a2 – c2) x + b2 = 0 ......(1) a + b > c $ a + b – c > 0 ......(2) ! and |a – b| < c $ a – b – c < 0 .......(3) and a – b + c > 0 .......(4) 2 2 2 2 Discriminant of equation (1) i.e. D = (b + a – c ) – 4a2b2 = (b2 + a2 – c2 – 2ab) (b2 + a2 – c2 + 2ab) = {(a – b)2 – c2} {(a + b)2 – c2} = (a – b + c) (a – b – c) (a + b + c) (a + b – c) < 0 (using (2), (3), (4)) D<0 ' roots are not real.
2.
Equation is mx 2 – (m + 1) (a + b)x + ab (m + 2) = 0 (a ! b)(1 ! m) = 0 m
Sum of roots = and
(m ! 2)ab < 0 m $ b = – a
product of roots =
Case 1 : If a + b = 0 (2) becomes '
.........(1) .........(2)
(m ! 2) m!2 < 0 $ > 0 m m m # ( – 6 , –2) 7 (0, 6 ) $ $ a + b = 0 , m ( – 6 , –2) 7 (0, 6 ) Case 2 : If m = – 1, then from (2) we get
$
– a2
1.ab < 0 "1 ab > 0 m = –1 ; ab > 0
$ $ $ 3.
1 x "1 Let
1
–
4 +
x"3 x"4 2 x – 5x = y
4
–
x"2
$
3 1 < – y!6 y!4 30
$
y 2 " 20 y ! 84 > 0 ( y ! 6)( y ! 4)
<
1 30
!
4 1 –3 ! G x 2 – 5 x ! 4 x 2 – 5 x ! 6 30
4
$
y # ( – 6 , – 6) 7 ( –4, 6) 7 (14, 6 ) Now (i) if y # ( –6, –6) 2 $ x – 5x < –6. $ x # (2, 3) (ii) if y # ( –4, 6) –4 < x2 – 5x < 6 $ $ x # ( –1, 1) 7 (4, 6) (iii) if y # (14, 6 ) $ x2 – 5x > 14 $ x # ( – 6, – 2) 7 (7, 6) final answer is x # ( – 6 , – 2) 7 ( –1, 1) 7 (2, 3) 7 (4, 6) 7 (7, 6 ) ' QUADRATIC EQUATIONS - 4
4.
Let the three numbers in G.P. be a, ar, ar2 a + b + c = xb ! a c + 1 + = x b = ar, c = ar2 ! b b r2 + (1 – x )r + 1 = 0 ........(1) $ r is real ! ' $ for (1) D % 0 (1 – x)2 – 4 % 0 2 $ x – 2x –3 % 0 $ x & –1 or x % 3 Note : If we put x = – 1 and x = 3 in (1) we get r = – 1 and r = 1 respectively which is not possible because in both cases the three numbers will not be distinct therefore x < – 1 or x > 3
5.
f(x) = x2 – (m – 3) x + m > 0 H x # [1, 2] Here D = (m – 3)2 – 4m = m2 – 10m + 9 = (m – 1) (m – 9) All possible graphs are Case 1 :
D<0
$
m # (1, 9)
Case 2 : (i)
(iii)
b < 1 2a D % 0
(i)
f(2) > 0
(ii)
'
(i) I (ii) I (iii),
$ 4 > 0 always true
f(1) > 0
$
–
m<5
$ m # ( –6, 1] 7 [9, 6)
we get m # ( – 6, 1]
Case 3 :
(ii) (iii)
'
"b
$
J2 $
2a D % 0
$
m < 10 m>7 m # ( – 6 , 1] 7 [9, 6)
(i) I (ii) I (iii), we get m # [9, 10)
Now final Answer is (Case 1) 7 (Case 2) 7 (Case 3) we get m # ( – 6, 10) 6.
!
f(x) = ax2 + (a – 2) x – 2 f(0) = –2
and
f( –1) = 0
Let
Since the quadratic expression is negative for exactly two integral values f(1) < 0 and f(2) % 0 $ $ a + a – 2 – 2 < 0 and 4a + 2a – 4 – 2 % 0 $ a<2 and a % 1 ' a [1, 2) QUADRATIC EQUATIONS - 5
7.
Let roots be 1, 1 and 3 20 4
'
1 + 1 + 3 = –
$
21 + 3 = – 5
'
1 . 1 + 13 + 13 = –
$
12 + 2 13 = –
.............(1) 23 4
23 4
.............(2)
6 3 = – 4 2 from equation (1) put 3 = – 5 – 21 in (2), we get
123 = –
and
12 + 2 1 ( – 5 – 2 1) = –
8.
$
12 12 + 401 – 23 = 0
'
1 = 1/2, – 1 2
If 1 =
(ii)
If 1 = –
Note :
!
'
required roots are
!
$
23 4
23 6
(i)
1=
.............(3)
then from (1), we get 3 = – 6 23 6
then from (1), we get 3 =
8 3
1 23 8 and 3 = – 6 also satisfy (3) but 1 = " and 3 = does not satisfy (3) 2 6 3 1 1 , , – 6 2 2
1 , 3 are the roots of x2 – 34x + 1 = 0 1 + 3 = 34 and 13 = 1 2
!
1 M P 1 N 1 4 " 3 4 K = 1 ! 3 – 2 (13)1/4 NO LK
!
13 = 1
2
' !
' '
1 M P 1 N 1 4 " 3 4 K = 1 ! 3 – 2 NO LK
8 1 ! 3 9 = 1 + 3 + 2 8 1 ! 3 9 = 36 2
13
2
........(1)
!
1 + 3 = 34 and 13 = 1
!
we consider the principal value
1 ! 3 = 6 put in (1), we get. 2
1 M P 1 N 1 4 " 3 4 K = 4 NO LK
'
1 4
1 4
1 " 3 =
±2
Ans.
QUADRATIC EQUATIONS - 6
9.
The equation can be rewritten as 2
P x 2 ! x ! 2 M P 2 M N K " 8a " 3 9N x ! x ! 2 K ! 8a " 4 9) 0 N x2 ! x ! 1 K N x2 ! x ! 1 K O L O L Let
x2 ! x ! 2 2
x ! x !1
=t
or
t=1+
x2 + x + 1
since
Therefore (x2 + x + 1) >
1 2
x ! x !1
=
P ! 1 M N x K O 2 L
3
2
+
$
4
3 4
P O
t # N1,
7-
/
3.
Now given equation reduces to t2 - (a - 3) t + (a - 4) = 0
P O
At least one root of the equation must lie in N1,
8a " 39 D 8a " 39 " 48a " 4 9
7-
/
3.
2
Now,
t=
P O
For one root to lie in N1, 10.
Let
$
2
7-
/ we must have 1 < a – 4 < 3.
t = a – 4, 1
7 3
$
5 < a <
19 3
x2 = t > 0, for only real solution
Again let f(t) = t2 – (a2 – 5a + 6) t – (a2 – 3a + 2)
$ a2 – 3a + 2 < 0 – b/2a > 0 $ a2 – 5a + 6 > 0 f(0) > 0
a
$
# [1, 2]
(a – 2) (a – 3) > 0
a # 8" 6,2;7 :3, 6 9 So possible 'a' from above two conditions are a = 1, 2. Now condition for D = ((a – 2) (a – 3))2 + 4(a – 1) (a – 2) > 0 is also satisfied by these two possible values of a So required value of 'a' are 1, 2 11.
12 3 are the roots of a1x2 + b1x + c1 = 0 1!3= "
$
b1 a1
and
13=
c1 a1
1 + 1 ! 3 + 1 3 =
c 1 " b1 ! a1 a1
81 ! 1 9 81 ! 39
=
a1 " b1 ! c 1 a1
....................(1)
=
a2 " b2 ! c 2 a2
....................(2)
Similarly
81 ! 39 81 ! Q 9
QUADRATIC EQUATIONS - 7
81 ! Q 9 81 ! 1 9
a 3 " b3 ! c 3 a3
=
....................(3)
Multiplying (1), (2) & (3), we get 2
2
2
81 ! 19 81 ! 39 81 ! Q 9
=
8a1 " b1 ! c1 9 8a 2 " b 2 ! c 2 9 8a3 " b 3 ! c 3 9 a1
a2
a3 1 2
$
B 3 ? < P ai " bi ! c i M< 81 ! 1 9 81 ! 39 81 ! Q 9 = A NNO ai LKK> < i)1 < @< =<
$
B 3 ? < P ai " bi ! c i M< NN KK> 1 + 81 ! 3 ! Q 9 ! 813 ! 3Q ! 1Q 9 ! 13Q = A a i L< < i)1 O <@ <=
$
B 3 ? 2 ai " bi ! c i M< < P N KK> – 1 A N 81 ! 3 ! Q 9 ! 813 ! 3Q ! 1Q 9 ! 13Q = a i L< < i)1 O <@ <=
R
1 2
R
1
12.
R
On putting the value of p, q and r in given equation we have 2x3 - (a1 + a2 ........+ a6)x2 + (a1a3 + a3a5 .......+ a6a2)x - (a1a3a5 + a2a4 a6) = 0
Sx
3
" 8a1 ! a 3 ! a 5 9x 2 ! 8a1a 3 ! a3 a5 ! a5a1 9x " a1a3a 5 T +
Sx
3
" 8a 2 ! a 4 ! a 6 9x 2 ! 8a 2a 4 ! a 4a 6 ! a 6a 2 9x " a2a 4a 6 T = 0
$ 8 x " a1 98x " a3 98x " a5 9 ! 8x " a 2 98x " a 4 98x " a 6 9 ) 0 f(x) = 8x " a1 98x " a 3 98x " a 5 9 ! 8x " a 2 98x " a 4 98x " a 6 9
Let Now
f(a1) = 8a1 " a 2 98a1 " a 4 98a1 " a 6 9 > 0 f(a2) = 8a 2 " a1 98a 2 " a 3 98a 2 " a 5 9 < 0 f(a3) = 8a 3 " a 2 98a 3 " a 4 98a 3 " a 6 9 < 0 f(a4) = 8a 4 " a1 98a 4 " a 3 98a 4 " a 5 9 > 0 f(a5) = 8a 5 " a 2 98a 5 " a 4 98a 5 " a 6 9 > 0 f(a6) = 8a 6 " a1 98a 6 " a 3 98a 6 " a 5 9 < 0
From above results it is clear that there are three real roots lying in the intervals (a1, a2), (a3,a4) and (a5, a6) 13.
Let A1 , A2 are the roots of ax2 + bx + c = 0, 2
then (A1 – A2)
2
= (A1 + A2) – 4A1A2 =
b2 a2
"
4c b 2 " 4ac = a a2
Using same result for x2 + 2bx + c = 0
$
S83 ! cos 1 9 " 83 ! sin 1 9T = 4b – 4c 8cos 1 " sin 19 = 4b – 4c
$
cos 2 21 = 4(b 2
2
2
2
2
2
2
" c)
2
2
.....................(i) QUADRATIC EQUATIONS - 8
X2 + 2BX + C = 0
Similarly for
$ $ '
14.
4B 2 " 4C = 4B2 – 4C 1
S8Q ! cos 19 " 8Q ! sin 1 9T S8cos 1 " sin 198cos 1 ! sin 19T = 4(B – C) 4
4
2
2
cos 2 21 from (i) & (ii) B2 – C = B2 – b2 =
2
=
2
2
b2 – c C – c
H x # R
* 1 M , 6K + 4 L
Now
f(t) = t2 + at + 4 = 0
(i)
all four real and distinct roots (A) D>0 (B) f( – 1/4) > 0
(A)
b 1 > – 2a 4 D>0 $
(B)
f( –1/4) =
(C)
–
(C)
.....................(ii)
Hence Proved.
t # ,"
$
2
4(B2 – C)
=
(x2 + x)2 + a(x2 + x) + 4 = 0 Let x2 + x = t then x2 + x – t = 0 D % 0 1 + 4t % 0
2
....(1)
–
a2 – 16 > 0
1 a – + 4 > 0 16 4
$
|a| > 4
$
a < 65/4
b a 1 = – > – 2a 2 4
1 2 a # ( – 6, –4) Two real roots which are distinct a<
(ii)
$ f( – 1/4) < 0 a # (65/4, 6) (iii) all four roots are imaginary Case-(i)(A) D % 0 $ |a| % 4
? < 65 << (B) f( –1/4) > 0 $ a < 4 > < b 1 1 < $ a > (C) – < – = 2a 4 2 <
a > 65/4
* 65 M $ a # ,4, 4 K + L
Case-(ii)
D<0 ...(i) $ a # ( –4, 4) taking union of both conditions mentioned by graph of case-(i) and case-(ii)
P O
a # N " 4, (iv)
65 M K 4 L
four real roots in which two are equal (A)
D>0
(B)
(A)
|a| > 4
(B) a = 65/4
No common solution
'
f( –1/4) = 0
(C)
–
(C) a <
1 2
b 1 > – 2a 4
a#U QUADRATIC EQUATIONS - 9
