5. Q x1x 2 = 4 ...

(i)

Solutions and

⇒
⇒
or
x1
+
x2
x1 - 1 x 2 - 1
=2

2 x1x 2 - x1 - x 2 = 2 ( x1x 2 - x1 - x 2 + 1 )
8 - x1 - x 2 = 2 ( 4 - x1 - x 2 + 1 )
x1 + x 2 = 2
[from Eq. (i)]
…(ii)
1. We have, From Eqs. (i) and (ii), required equation is
2 (a - b ) x 2 - 11 (a + b + c ) x - 3 (a - b ) = 0 x 2 - ( x1 + x 2 ) x + x1x 2 = 0
∴ D = { - 11 (a + b + c )} 2 - 4 ⋅ 2 (a - b ) ⋅ ( -3) (a - b ) or x 2 - 2x + 4 = 0
2 2
= 121 (a + b + c ) + 24 (a - b ) > 0 6. Let f ( x ) = x 2 - 2ax + a 2 - 1
Therefore, the roots are real and unequal. Now, four cases arise:
2. Here, a < 0 Case I D ≥ 0
Cut-off Y -axis, x = 0
⇒ y =c < 0 [from graph]
∴ c<0
x -coordinate of vertex > 0
X
b –2 α β 2
⇒ - >0
2a
b ⇒ ( - 2a ) 2 - 4 ⋅ 1 (a 2 - 1 ) ≥ 0
⇒ <0
a ⇒ 4≥0
But a<0 ∴ a ∈R
∴ b>0 Case II f ( - 2 ) > 0
and y-coordinate of vertex < 0 ⇒ 4 + 4a + a 2 - 1 > 0
D D
⇒ - <0 ⇒ >0 ⇒ a 2 + 4a + 3 > 0
4a 4a
∴ D<0 [Q a < 0 ] ⇒ (a + 1 ) (a + 3 ) > 0
i.e. b 2 - 4ac < 0 ∴ a ∈ ( - ∞, - 3 ) ∪ ( - 1, ∞ )
c Case III f (2 ) > 0
∴ >0 [Qc < 0 , a < 0 ]
a ⇒ 4 - 4a + a 2 - 1 > 0

3. Sum of the roots = -

(a + 3 )
=I+ [let] ⇒ a 2 - 4a + 3 > 0
a ⇒ (a - 1 ) (a - 3 ) > 0
 3 
∴ a = - +  …(i) ∴ a ∈ ( - ∞, 1 ) ∪ (3, ∞ )
 I + 1 Case IV - 2 < x-coordinate of vertex < 2
a -3 ⇒ - 2 < 2a < 2
Product of the roots = αβ = =I+ +2 …(ii)
a ∴ a ∈ ( - 1, 1 )
and D = (a + 3 ) 2 - 4a (a + 3 ) Combining all cases, we get a ∈ ( - 1, 1 )
9 Hence, [a ] = - 1, 0
= + {( I + - 2 ) 2 - 12 } [from Eq. (i)]
(I + 1)2  - 4a 
7. We have, -   = -2
D must be perfect square, then I + = 6  2 (- 2)
From Eq. (ii), ⇒ a =2
Product of the roots = I +
+ 2 =6 + 2 =8 ∴ y = - 2x 2 - 8x + λ …(i)
4. Let α be one root of Since, Eq. (i) passes through points ( - 2, 7 )
2
x - 3ax + f (a ) = 0 ∴ 7 = - 2 (- 2)2 - 8 (- 2) + λ
⇒ α + 2 α = 3a ⇒ 3 α = 3a ⇒ 7 = - 8 + 16 + λ
⇒ α =a …(i) ∴ λ = -1
and α ⋅ 2 α = f (a ) 8. Since, the coefficient of n 2 = ( 4p - p 2 - 5) < 0
⇒ f (a ) = 2 α 2= 2a 2 [using Eq. (i)] Therefore, the graph is open downward.
⇒ f (x ) = 2x 2 According to the question, 1 must lie between the roots.
180 Textbook of Algebra

Hence, f (1) > 0 Q g( x ) = ( x + 1 ) 2 - 6 ≥ - 6

2
⇒ 4p - p - 5 - 2p + 1 + 3p > 0 ∴ α 3 ≤ - 7, α 2 ≤ - 8, α 1 ≤ - 9
⇒ - p 2 + 5p - 4 > 0 ∴ a + b + c ≥ 719
⇒ p 2 - 5p + 4 < 0 ∴Minimum value of a + b + c is 719.
Q α1 + α 2 + α 3 = - a
⇒ ( p - 4) ( p - 1) < 0
⇒ - a ≤ - 24
⇒ 1<p<4
⇒ a ≥ 24
∴ p = 2, 3
α 1α 2 + α 2 α 3 + α 3α 1 = b
Hence, number of integral values of p is 2.
2 2 ⇒ b ≥ 191
9. We have, 3 2 x - 2 ⋅ 3 x + x+ 6
+ 3 2 ( x + 6) = 0 and α 1α 2 α 3 = - c
2
⇒ (3 x - 3 x + 6 ) 2 = 0 ⇒ - c ≤ - 504
2 ⇒ c ≥ 504
⇒ 3x - 3x + 6 = 0
2
∴ a + b + c ≥ 719
⇒ 3x = 3x + 6 ⇒ x 2 = x + 6 Hence, minimum value of a + b + c is 719.
⇒ x2 - x - 6 = 0 n
13. Q ∑ ( x + k - 1) (x + k ) = 10n
⇒ (x - 3) (x + 2) = 0 k =1
∴ x = { - 2, 3 } n
2 2
10. Given, b - 4ac = p - 4aq …(i) ⇒ ∑ x 2 + x (2k - 1) + (k - 1)k = 10n
k =1
and f ( x ) = g( x )
⇒ nx 2 + x (1 + 3 + 5 +… + (2n - 1 ))
⇒ ax + bx + c = ax 2 + px + q
2

⇒ (b - p ) x = q - c + ( 0 + 1 ⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 +… + (n - 1 )n ) = 10n
q -c n
∴ x= =α [given] …(ii) ⇒ nx 2 + x ⋅ (1 + 2n - 1 )
b-p 2
From Eq. (i), we get  n (n + 1 ) (2n + 1 ) n(n + 1 )
+ -  = 10n
(b + p ) (b - p ) + 4a (q - c ) = 0  6 2 
⇒ (b + p ) (b - p ) + 4aα (b - p ) = 0 [from Eq. (ii)] n (n 2 - 1 )
(b + p ) ⇒ nx 2 + n 2x + = 10n
or α=- [Qb ≠ p] 3
4a
(n 2 - 31 )
 b  p ⇒ x 2 + nx + =0 [dividing by n]
-  + -  3
 a  a 
= D
4 Q (α + 1 ) - α =
Sum of the roots of ( f (x ) = 0)  1
+ Sum of the roots of (g (x ) = 0) 1= D
= 
4 ⇒ D =1
2
= AM of the roots of f ( x ) = 0 (n - 31 )
⇒ n2 - 4 ⋅1 ⋅ =1
and g( x ) = 0 3
11. Let α and β be the roots of ax 2 + bx + c = 0. ⇒ 3n 2 - 4n 2 + 124 = 3
α+β b ⇒ n 2 = 121
∴ x1 = =-
2 2a ∴ n = 11
c 14. Since, 2 is only even prime.
2⋅
2αβ 2c
and x2 = = a =- Therefore, we have
α +β -b b
2 2 + λ ⋅ 2 + 12 = 0
a
∴The required equation is ⇒ λ =8
 b   2c   2bc ∴ x 2 + λx + µ = 0
x 2 -  -  +  -   x + =0

 2a   b  2ab ⇒ x 2 + 8x + µ = 0 …(i)
2 2
i.e. 2abx + (b + 4ac ) x + 2bc = 0 But Eq. (i) has equal roots.
12. Let α 1, α 2 and α 3 be the roots of f ( x ) = 0, such that ∴ D=0
α1 < α 2 < α 3 ⇒ 82 - 4 ⋅ 1 ⋅ µ = 0
and g( x ) can take all values from [ - 6, ∞ ). ⇒ µ = 16
 Chap 02 Theory of Equations 181

15. We have, x + x - (1 - x ) = 1 9 ⋅ 3 2 x + 6 ⋅ 3 x + 4 (3 x + 1 ) 2 + 2 ⋅ 3 x + 1 + 4
Let y = =
9 ⋅ 3 2 x - 6 ⋅ 3 x + 4 (3 x + 1 ) 2 - 2 ⋅ 3 x + 1 + 4
⇒ x - 1 - x =1 - x
t 2 + 2t + 4
On squaring both sides, we get = , where t = 3 x + 1
t 2 - 2t + 4
x - 1 - x =1 + x -2 x
⇒ (y - 1 ) t 2 - 2 (y + 1 ) t + 4 (y - 1 ) = 0
⇒ - 1 - x =1-2 x By the given condition, for every t ∈ R,
Again, squaring on both sides, we get 1
<y <3 …(i)
3
1 - x = 1 + 4x - 4 x
But t = 3x + 1 > 0
4 x = 5x
We have, product of the roots = 4 > 0, which is true.
4 2 (y + 1 )
⇒ x = [on squaring both sides] And sum of the roots = >0
5 (y - 1 )
16 y +1
⇒ x= ⇒ >0
25 y -1
Hence, the number of real solutions is 1. ∴ y ∈ ( - ∞, - 1 ) ∪ (1, ∞ ) …(ii)
From Eqs. (i) and (ii), we get
16. Let x = 7 + 7 - 7 + 7 -… ∞
1 <y <3
⇒ x= 7+ 7-x [on squaring both sides] 19. Since α , β and γ are the roots of

⇒ x2 - 7 = 7 - x (x - a ) (x - b ) (x - c ) = d
⇒ (x - a ) (x - b ) (x - c ) - d = (x - α ) (x - β) (x - γ )
⇒ (x 2 - 7)2 = 7 - x [again, squaring on both sides] ⇒ (x - α ) (x - β) (x - γ ) + d = (x - a ) (x - b ) (x - c )
⇒ x 4 - 14 x 2 + x + 42 = 0 ⇒ a, b and c are the roots of
(x - α ) (x - β) (x - γ ) + d = 0
⇒ ( x - 3 ) ( x 3 + 3 x 2 - 5 x - 14 ) = 0
20. Since, all the coefficients of given equation are not real.
⇒ (x - 3) (x + 2) (x 2 + x - 7) = 0 Therefore, other root ≠ 3 + i.
- 1 ± 29 Let other root be α.
⇒ x = 3, - 2,
2 2 (1 + i )
∴ x =3 [Q x > 7] Then, sum of the roots =
i
17. Let y = 2 (k - x ) ( x + ( x 2 + k 2 ) 2 (1 + i )
⇒ α + 3 -i =
i
⇒ y - 2 (k - x ) x = 2 (k - x ) ( x 2 + k 2 ) ⇒ α + 3 - i = 2 - 2i
On squaring both sides, we get ∴ α = -1 -i
⇒ y 2 + 4 (k - x ) 2 x 2 - 4 xy (k - x ) = 4 (k - x ) 2 ( x 2 + k 2 ) 21. We have, |[ x ] - 2 x | = 4
⇒ y 2 - 4 xy (k - x ) = 4 (k - x ) 2k 2 ⇒ |[ x ] - 2 ([ x ] + { x })| = 4
⇒ |[ x ] + 2 { x }| = 4
⇒ 4 (k 2 - y ) x 2 - 4(2k 3 - ky ) x - y 2 + 4k 4 = 0
which is possible only when
Since, x is real.
2 { x } = 0, 1
∴ D≥0
1
⇒ 16 (2k - ky ) - 4 ⋅ 4 (k 2 - y ) ( 4k 4 - y 2 ) ≥ 0
3 2 If { x } = 0, then [ x ] = ± 4 and then x = - 4, 4 and if { x } = ,
2
[using, b 2 - 4ac ≥ 0] then
⇒ 4k 6 + k 2y 2 - 4k 4y - ( - k 2y 2 + 4k 6 + y 3 - 4yk 4 ) ≥ 0 [x ] + 1 = ± 4
⇒ [ x ] = 3, - 5
⇒ 2k 2y 2 - y 3 ≥ 0
1 1
⇒ y 2 (y - 2k 2 ) ≤ 0 ∴ x = 3 + and - 5 +
2 2
∴ y ≤ 2k 2 7 9 9 7
⇒ x = , - ⇒ x = - 4, - , , 4
1 x 2 - 2x + 4 2 2 2 2
18. We have, < < 3, ∀ x ∈ R 2 3 2
3 x 2 + 2x + 4 22. We know that, x + x + 1 is a factor of ax + bx + cx + d .
1 x 2 + 2x + 4 Hence, roots of x 2 + x + 1 = 0 are also roots of
< < 3, ∀ x ∈ R
3 x 2 - 2x + 4 ax 3 + bx 2 + cx + d = 0. Since, ω and ω 2
182 Textbook of Algebra

 1 3i  2 26. Since, 3789108 is an even integer. Therefore, x 4 - y 4 is also an

 where ω = - +  are two complex roots of x + x + 1 = 0.
 2 2 even integer. So, either both x and y are even integers or both
Therefore, ω and ω 2 are two complex roots of of them are odd integers.
ax 3 + bx 2 + cx + d = 0. Now, x 4 - y 4 = (x - y ) (x + y ) (x 2 + y 2 )
We know that, a cubic equation has atleast one real root. Let ⇒ x - y , x + y , x 2 + y 2 must be even integers.
real root be α. Then, Therefore, ( x - y ) ( x + y ) ( x 2 + y 2 ) must be divisible by 8. But
d d
α ⋅ ω ⋅ ω2 = - ⇒α=- 3789108 is not divisible by 8. Hence, the given equation has no
a a solution.
23. We have, (5 x 2 - 8 x + 3 ) - (5 x 2 - 9 x + 4 ) ∴ Number of solutions = 0
27. We have, x 3 + ax + 1 = 0
= (2 x 2 - 2 x ) - (2 x 2 - 3 x + 1 )
or x 4 + ax 2 + x = 0 …(i)
⇒ (5 x - 3 ) ( x - 1 ) - (5 x - 4 ) ( x - 1 ) 4 2
and x + ax + 1 = 0 …(ii)
= 2 x ( x - 1 ) - (2 x - 1 ) ( x - 1 ) From Eqs. (i) and (ii), we get
⇒ x - 1 ( 5x - 3 - 5x - 4 ) = x - 1 ( 2x - 2x - 1 ) x -1 = 0
⇒ x =1
⇒ x -1 = 0 which is a common root.
⇒ x =1 ∴ 1 +a + 1 = 0
24. We have, (a + b ) (a - b ) = a 2 - b = 1 [given] ⇒ a = -2
∴ (a + b ) x 2 - 15
+ (a - b ) = 2ax 2 - 15 28. Q(1 - a 2 ) x 2 + 2ax - 1 = 0
2 1 1 - a2 ≠ 0
⇒ (a + b ) x - 15 + 2
= 2a
(a + b ) x - 15  2a   1 
x2 +  x -   =0
2 1 - a2 1 - a2
Let y = (a + b ) x - 15
1  2a   1 
⇒ y + = 2a ⇒ y 2 - 2ay + 1 = 0 Let f (x ) = x 2 +  2
x -  
y 1 - a   (1 - a 2 )

4a 2 - 4
2a ±
⇒ y = = a ± a2 - 1
2
∴ y = a ± b = (a + b ) ± 1 [Qa 2 - b = 1]
2
- 15 X
⇒ (a + b ) x = (a + b ) ± 1 0 α β1
2
∴ x - 15 = ± 1 The following cases arise:
⇒ x 2 = 15 ± 1 ⇒ x 2 = 16, 14 Case I D ≥ 0
2
⇒ x = ± 4, ± 14  2a   -1 
 2
 - 4 ⋅1 ⋅   ≥0
25. We have, x 2 - xy + y 2 = 4 ( x + y - 4 ) 1 - a  1 - a2
⇒ x 2 - x (y + 4 ) + y 2 - 4y + 16 = 0 4a 2 4
⇒ + ≥0
Q x ∈R (1 - a 2 ) 2 (1 - a 2 )
∴ ( - (y + 4 )) 2 - 4 ⋅ 1 ⋅ (y 2 - 4y + 16 ) ≥ 0 4a 2 + 4 - 4a 2
⇒ ≥0
[using, b 2 - 4ac ≥ 0 ] (1 - a 2 ) 2
2 2
⇒ y + 8y + 16 - 4y + 16y - 64 ≥ 0 4
⇒ ≥0 [always true]
⇒ 3y 2 - 24y + 48 ≤ 0 (1 - a 2 ) 2

⇒ y 2 - 8y + 16 ≤ 0 ⇒ (y - 4 ) 2 ≤ 0 Case II f ( 0 ) > 0
-1 1
∴ (y - 4 ) 2 = 0 ⇒ >0 ⇒ <0
(1 - a 2 ) 1 - a2
∴ y =4
⇒ 1 - a2 < 0
Then, x 2 - 4 x + 16 = 4( x + 4 - 4 )
∴ a ∈ ( - ∞, - 1 ) ∪ (1, ∞ )
x 2 - 8 x + 16 = 0
Case III f (1 ) > 0
(x - 4)2 = 0 2a 1
⇒ 1+ - >0
x=4 (1 - a ) (1 - a 2 )
2

Number of pairs is 1 i.e., ( 4, 4 ).

 Chap 02 Theory of Equations 183

1 - a 2 + 2a - 1 a 2 - 2a  - 1 + 5
⇒ > 0 ⇒ <0 ∴ x ∈ 0, 
(1 - a 2 ) 1 - a2  2 

+ + + Case II If x < 0, i.e., -1 ≤ x < 0

–1 – 0 1 – 2 x - (1 + x ) < 0
⇒ x < 1+ x [always true]
a (a - 2 ) x ∈ [ - 1, 0 )
⇒ >0
(a + 1 ) (a - 1 ) Combining both cases, we get
∴ a ∈ ( - ∞, - 1 ) ∪ ( 0, 1 ) ∪ (2, ∞ )  - 1 + 5
x ∈ - 1, 
Case IV 0 < x -coordinate of vertex < 1  2 
2a a
⇒ 0<- <1 ⇒ 0 < 2 <1 30. We have, (a ⋅ 2b - 2c ⋅ a ) (2c ⋅ c - b ⋅ 2b ) = (ba - ca ) 2
2 (1 - a 2 ) a -1
⇒ 2a (b - c ) ⋅ 2 (c 2 - b 2 ) = a 2(b - c ) 2
a a
⇒ 0< and 1 - 2 >0 ⇒ 4a (c - b ) (c + b ) = a 2(b - c ) [Qb ≠ c ]
(a + 1 ) (a - 1 ) a -1
2
a ⇒ 4a (c + b ) = - a
⇒ >0
(a + 1 ) (a - 1 ) ⇒ a + 4b + 4c = 0
 b c
+ + 31. 0 < a < b < c, α + β =  -  and αβ =
 a a
– –1 0 – 1
For non-real complex roots,
b 2 - 4ac < 0
⇒ a ∈ ( - 1, 0 ) ∪ (1, ∞ )
 1 + 5  1 - 5 b 2 4c
a -  a -  ⇒ - <0
 2   2  a2 a
and >0 ⇒ (α + β ) 2 - 4 αβ < 0
(a + 1 ) (a - 1 )
⇒ (α - β ) 2 < 0
+ + +
Q 0 <a <b <c
–1 – 1– √5 1 – 1+√5
2 2 ∴ Roots are conjugate, then | α | = | β |
c
But αβ =
1 - 5  1 + 5  a
and a ∈ ( - ∞, - 1 ) ∪  , 1 ∪  , ∞
 2   2  c  c 
| αβ | = >1 Q a < c, ∴ > 1
a  a 
1 - 5  1 + 5 
∴ a ∈ , 0 ∪  , ∞ ⇒ | α| | β| > 1
 2   2 
⇒ | α | 2 > 1 or | α | > 1
Combining all cases, we get
a >2 32. Given equation is
29. We have, x - 1 - |x| < 0 …(i) Ax 2 - | G | x - H = 0 …(i)
2
which is defined only when ∴ Discriminant = ( - | G | ) - 4 A ( - H )
1 - |x | ≥ 0 = G 2 + 4 AH
⇒ |x | ≤ 1 = G 2 + 4G 2 [QG 2 = AH ]
⇒ x ∈ [ - 1, 1 ] 2
= 5G > 0
Now, from Eq. (i), we get
x < 1 - |x| ∴ Roots of Eq. (i) are real and distinct.
a+b 2ab
Case I If x ≥ 0, i.e., 0 ≤ x ≤ 1 Q A= > 0, G = ab > 0, H = >0
2 a+b
x - (1 - | x | ) < 0
[Qa and b are two unequal positive integers]
⇒ x < (1 - x ) Let α and β be the roots of Eq. (i). Then,
On squaring both sides, we get | G|
α+β= >0
x2 + x - 1 < 0 A
H
-1 - 5 -1 + 5 and αβ = - <0
⇒ <x< A
2 2
D G 5
But x≥0 and α -β = = >0
A A
184 Textbook of Algebra

|G | + G 5 The given equation will have four real roots, i.e. Eq. (i) has two
∴ α= >0 non-negative roots.
2A
b
|G | - G 5 Then, - ≥0
and β= <0 a
2A
af ( 0 ) ≥ 0
Exactly one positive root and atleast one root which is
negative fraction. and b 2 - 4ac ≥ 0 [given]
33. It is clear from graph that the equation y = ax 2 + bx + c = 0 ⇒
b
≤0
has two real and distinct roots. Therefore, a
b 2 - 4ac > 0 …(i) ac ≥ 0
Q Parabola open downwards. ⇒ a > 0, b < 0, c > 0
∴ a<0 or a < 0, b > 0, c < 0
a
andy = ax 2 + bx + c cuts-off Y -axis at, x = 0. 36. Let the roots be , a and ar , where a > 0, r > 1
r
∴ y =c < 0 ∴ Product of the roots = 1
⇒ c<0 a
and x-coordinate of vertex > 0 ⇒ ⋅ a ⋅ ar = 1
r
b b
⇒ - >0 ⇒ <0 ⇒ a3 = 1
2a a
∴ a =1 [one root is 1]
⇒ b>0 [Qa < 0]
1
It is clear that a and b are of opposite signs. Now, roots are , 1 and r . Then,
r
34. Let y = ax 2 + bx + c 1
+ 1 + r = -b
r
1
a>0 ⇒ + r = -b -1 …(i)
r
–2 2
1
Q r + >2
α β r
⇒ -b -1 >2
Consider the following cases: ⇒ b < -3 [from Eq. (i)]
Case I D > 0 or b ∈ ( - ∞, - 3 )
⇒ b 2 - 4ac > 0 1 1
Also, ⋅1 + 1 ⋅r + r ⋅ = c
Case II af ( - 2 ) < 0 r r
⇒ a ( 4a - 2b + c ) < 0 1
⇒ + r + 1 =c = -b [from Eq. (i)]
⇒ 4a - 2b + c < 0 r
Case III af (2 ) > 0 ∴ b +c = 0
⇒ a ( 4a + 2b + c ) > 0 1
⇒ 4a + 2 b + c > 0 Now, first root = < 1 [Q one root is smaller than one]
r
Combining Case II and Case III, we get
Second root = 1
4a + 2| b| + c < 0
Third root = r > 1 [Q one root is greater than one]
Also, at x = 0, y <0 ⇒c<0 2
37. We have, f ( x ) = ax + bx + c
Also, since for - 2 < x < 2,
a, b , c ∈ R [Q a ≠ 0 ]
y <0 1
On putting x = 0, 1, , we get
⇒ ax 2 + bx + c < 0 2
For x = 1, a+b+c<0 …(i) |c | ≤ 1
and for x = - 1, a -b + c < 0 …(ii) |a + b + c | ≤ 1
1 1
Combining Eqs. (i) and (ii), we get and a + b + c ≤1
4 2
a + | b| + c < 0
⇒ -1 ≤ c ≤ 1,
35. Put x 2 = y .
-1 ≤a + b + c ≤1
Then, the given equation can be written as
and - 4 ≤ a + 2b + 4c ≤ 4
f (y ) = ay 2 + by + c = 0 …(i) ⇒ - 4 ≤ 4a + 4b + 4c ≤ 4
and - 4 ≤ - a - 2b - 4c ≤ 4
 Chap 02 Theory of Equations 185

On adding, we get b
Option (d) a < 0, c < 0, - <0
- 8 ≤ 3a + 2b ≤ 8 2a
Also, - 8 ≤ a + 2b ≤ 8 or a < 0, c < 0, b < 0
∴ - 16 ≤ 2a ≤ 16 ∴ abc < 0
⇒ | a| ≤ 8 41. Here, D ≤ 0
Q -1 ≤ - c ≤ 1, - 8 ≤ - a ≤ 8 and f ( x ) ≥ 0, ∀ x ∈ R
We get, - 16 ≤ 2b ≤ 16 ∴ f (3 ) ≥ 0
⇒ |b | ≤ 8 ⇒ 9a + 3b + 6 ≥ 0
or 3a + b ≥ - 2
∴ | a| + | b| + | c | ≤ 17
⇒ Minimum value of 3a + b is - 2.
- 5 ± 25 + 1200 -5 ± 35 30 - 40
38. Q x = = = , and f (6 ) ≥ 0
50 50 50 50 ⇒ 36a + 6b + 6 ≥ 0
3 -4 ⇒ 6a + b ≥ - 1
or cos α = ,
5 5 ⇒ Minimum value of 6a + b is -1.
But -1 < x < 0 42. Since, f ( x ) = x 3 + 3 x 2 - 9 x + λ = ( x - α ) 2(x - β)
4 ∴ α is a double root.
∴ cos α = - [ lies in II and III quadrants]
5 ∴ f ′( x ) = 0 has also one root α.
3 i.e. 3 x 2 + 6 x - 9 = 0 has one root α.
∴ sin α = [ lies in II quadrant]
5
∴ x 2 + 2 x - 3 = 0 or ( x + 3 ) ( x - 1 ) = 0
3
∴ sin α = - [lies in III quadrant] has the root α which can either -3 or 1.
5
If α = 1, then f (1 ) = 0 gives λ - 5 = 0 ⇒ λ = 5.
24
∴ sin 2α = 2 ⋅ sin α ⋅ cos α = - If α = - 3, then f ( - 3 ) = 0 gives
25
- 27 + 27 + 27 + λ = 0
[ lies in II quadrant] ⇒ λ = - 27
24 2
43. We have, D = (b - c ) - 4a (a - b - c ) > 0
∴ sin 2α = 2 ⋅ sin α ⋅ cosα = [lies in III quadrant]
25
⇒ b 2 + c 2 - 2bc - 4a 2 + 4ab + 4ac > 0
39. Qa + 2b + 4c = 0
⇒ c 2 + ( 4a - 2b ) c - 4a 2 + 4ab + b 2 > 0, ∀c ∈ R
2
 1  1 Since, c ∈ R, so we have
∴ a   + b  + c = 0
 2  2
( 4a - 2b ) 2 - 4 ( - 4a 2 + 4ab + b 2 ) < 0
1
It is clear that one root is . ⇒ 4a 2 - 4ab + b 2 + 4a 2 - 4ab - b 2 < 0
2
1 b ⇒ a (a - b ) < 0
Let other root be α. Then, α + =- If a > 0, then a - b < 0
2 a
1 b i.e. 0 <a <b
⇒ α=- - or b >a > 0
2 a
which depends upon a and b. If a < 0, then a - b > 0
i.e. 0 >a >b
40. Q Cut-off Y -axis, put x = 0, i.e. f ( 0) = c
or b <a < 0
b 44. We have, x 3 - ax 2 + bx - 1 = 0 …(i)
Option (a) a < 0, c < 0, - <0
2a 2 2 2 2
Then, α + β + γ = (α + β + γ ) - 2 (αβ + βγ + γα )
or a < 0, c < 0, b < 0
∴ abc < 0 = a 2 - 2b
b α 2 β 2 + β 2 γ 2 + γ 2 α 2 = (αβ + βγ + γα ) 2
Option (b) a < 0, c > 0, - >0
2a - 2 αβγ (α + β + γ ) = b 2 - 2a
or a < 0, c > 0, b > 0 and α 2 β2 γ 2 = 1
∴ abc < 0
Therefore, the equation whose roots are α 2, β 2 and γ 2, is
b
Option (c) a > 0, c > 0, - >0 x 3 - (a 2 - 2b ) x 2 + (b 2 - 2a ) x - 1 = 0 …(ii)
2a
or a > 0, c > 0, b < 0 Since, Eqs. (i) and (ii) are indentical, therefore
∴ abc < 0 a 2 - 2b = a and b 2 - 2a = b
186 Textbook of Algebra

Eliminating b, we have ⇒ (7y - 1 ) (y - 7 ) ≤ 0

(a 2 - a ) 2 a2 - a ∴
1
≤y ≤7
- 2a =
4 2 7
1
⇒ a {a (a - 1 ) 2 - 8 - 2 (a - 1 )} = 0 ∴ G = 7 and L =
3 2
7
⇒ a (a - 2a - a - 6 ) = 0 ∴ GL = 1
⇒ a (a - 3 ) (a 2 + a + 2 ) = 0 G100 + L100 G100 + L100
Now, ≥ (GL )100 ⇒ ≥1
⇒ a = 0 or a = 3 or a 2 + a + 2 = 0 2 2
⇒ b = 0 or b = 3 ⇒ G100 + L100 ≥ 2
or b2 + b + 2 = 0 46. Least value of G100 + L100 is 2.
∴ a =b = 0 47. The quadratic equation having roots G and L, is
or a =b =3 x 2 - (G + L ) x + GL = 0
or a and b are roots of x 2 + x + 2 = 0. 50
⇒ x2 - x + 1 = 0
45. Here, D > 0 7
2
⇒ 7 x - 50 x + 7 = 0
a > 0
48. We have, L2 < λ < G 2
–2 2
X 2
α β  1 2
⇒   < λ <7
 7
1
⇒ < λ < 49
b 2 - 4ac > 0 49
or b 2 > 4ac …(i) ⇒ λ = 1, 2, 3,… , 48 as λ ∈ N
and f (0) < 0 48 × 49
∴ Sum of all values of λ = 1 + 2 + 3 +… + 48 = = 1176
⇒ c<0 …(ii) 2
f (1 ) < 0 Solutions (Q. Nos. 49 to 51)
⇒ a+b+c<0 …(iii) Let roots be α , β, γ, δ > 0.
f (- 1) < 0 ∴ α + β + γ + δ = 12
⇒ a -b + c < 0 …(iv) (α + β ) ( γ + δ ) + αβ + γδ = c
f (2 ) < 0 αβ ( γ + δ ) + γδ (α + β ) = - d
⇒ 4a + 2b + c < 0 …(v) α β γδ = 81
f (- 2) < 0 α+β+γ+δ
Q AM = =3
⇒ 4a - 2b + c < 0 …(vi) 4
From Eqs. (i) and (ii), we get and GM = (α β γ δ )1 /4 = (81 )1 /4 = 3
c < 0, b2 - 4ac > 0 ∴ AM = GM
From Eqs. (iii) and (iv), we get ⇒ α =β = γ =δ =3
a - |b | + c < 0 49. c = (α + β) ( γ + δ ) + αβ + γδ
and from Eqs. (v) and (vi), we get = (3 + 3 ) (3 + 3 ) + 3 ⋅ 3 + 3 ⋅ 3 = 36 + 18 = 54
4a - 2 | b | + c < 0 50. Qαβ ( γ + δ ) + γδ (α + β) = - d
Solutions (Q. Nos. 46 to 48) ∴ d = - {3 ⋅ 3 ⋅ (3 + 3 ) + 3 ⋅ 3 ⋅ (3 + 3)} = - 108
2x 2 - 3x + 2 d ( - 108 )
Let y = 2 51. Required root = - = - =1
2x + 3x + 2 2c 2 × 54
⇒ 2 x 2y + 3 xy + 2y = 2 x 2 - 3 x + 2 Solutions (Q. Nos. 52 to 54)
⇒ 2 (y - 1 ) x 2 + 3 (y + 1 ) x + 2 (y - 1 ) = 0 Given that, AC = 4 2 units
As x ∈ R AC
∴ AB = BC = = 4 units
∴ D≥0 2
⇒ 9 (y + 1 ) 2 - 4 ⋅ 2 (y - 1 ) ⋅ 2 (y - 1 ) ≥ 0 and OB = ( BC ) 2 - (OC ) 2
2 2
⇒ 9 (y + 1 ) - 16 (y - 1 ) ≥ 0  AC 
= ( 4 ) 2 - (2 2 ) 2 Q OC =
⇒ (3y + 3 ) 2 - ( 4y - 4 ) 2 ≥ 0  2 
⇒ (7y - 1 ) (7 - y ) ≥ 0 = 2 2 units
 Chap 02 Theory of Equations 187

∴ Vertices are A ≡ ( - 2 2, 0 ), Solutions (Q. Nos. 58 to 60)

B ≡ ( 0, -2 2 ) Let f ( x ) = ax 2 - bx + c has two distinct roots α and β. Then,
f ( x ) = a ( x - α ) ( x - β ). Since, f ( 0 ) and f (1 ) are of same sign.
and C ≡ (2 2, 0)
Therefore, c (a - b + c ) > 0
52. Since, y = f ( x ) = ax 2 + bx + c passes through A, B and C, then ⇒ c (a - b + c ) ≥ 1
0 = 8a - 2 2b + c - 2 2 = c ∴ a 2 αβ (1 - α ) (1 - β ) ≥ 1
and 0 = 8a + 2 2b + c 1 1 1
2

1 But α (1 - α ) = -  - α ≤
We get, b = 0, a = and c = - 2 2 4 2  4
2 2
a2
x2 ∴ a 2 αβ (1 - α ) (1 - β ) <
∴ y = f (x ) = -2 2 16
2 2 a2
2 ⇒ >1 ⇒ a > 4 [Q α ≠ β]
x 16
53. Minimum value of y = - 2 2 is at x = 0.
2 2 ⇒ a ≥ 5 as a ∈ I
∴ (y ) min = -2 2 Also, b 2 - 4ac ≥ 0
54. f ( x ) = 0 ⇒ b 2 ≥ 4ac ≥ 20
x2 ⇒ b ≥5
⇒ -2 2 = 0 ⇒ x = ± 2 2
2 2 Next, a ≥ 5, b ≥ 5, we get c ≥ 1
λ ∴ abc ≥ 25
Given, -2 2 < <2 2
2 ∴ log 5 abc ≥ log 5 25 = 2
or -4 2<λ<4 2 58. Least value of a is 5.
∴Initial values of λ are 59. Least value of b is 5.
-5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5. 60. Least value of logb abc is 2.
∴Number of integral values is 11.
Solutions. (Q. Nos. 61 to 63)
Solutions. (Q. Nos. 55 to 57) Let α , β and γ be the roots of 2 x 3 + ax 2 + bx + 4 = 0.
We have, (α - β ) = (α + k ) - ( β + k ) a
∴ α+β+γ=-
b 2 - 4c b 2 - 4c1 2
⇒ = 1
1 1 b
αβ + βγ + γα = and αβγ = - 2
⇒ b 2 - 4c = b12 - 4c1 …(i) 2
2
1 (b - 4c ) 1 61. Q AM ≥ GM
Given, least value of f ( x ) = - - =- (- α ) + (- β) + (- γ )
4 4 ×1 4 ∴ ≥ {( - α ) ( - β ) ( - γ )}1 /3
3
⇒ b 2 - 4c = 1
a
∴ b - 4c = 1 = b12 - 4c1
2
[from Eq. (i)] …(ii) 2 ≥ (2 )1/ 3
⇒
7 3
Also, given least value of g( x ) occurs at x = .
2 ∴ a ≥ 6 (2 )1/ 3 …(i)
b1 7 3
∴ - = or a ≥ 432
2 ×1 2
Hence, minimum value of a 3 is 432.
∴ b1 = - 7
62. Q AM ≥ GM
55. b 1 = - 7
(- α ) (- β) + (-β) (- γ ) + (- γ ) (- α )
b12 - 4c1 1 ∴
56. Least value of g ( x ) = - =- [from Eq. (ii)] 3
4 ×1 4
≥ {( - α ) ( - β ) ( - β ) ( - γ ) ( - γ ) ( - α )}1 / 3
57. Q g( x ) = 0 b/ 2
⇒ ≥ ( 4 )1/ 3
∴ x 2 + b1x + c1 = 0 3
- b1 ± b12 - 4c1 ⇒ b ≥ 6 ( 4 )1/ 3 …(ii)
⇒ x=
2 or b 3 ≥ 864
7±1 Hence, minimum value of b 3 is 864.
= = 3, 4
2 63. From Eqs. (i) and (ii), we get
∴ Roots of g( x ) = 0 are 3, 4. ab ≥ 6 (2 )1 / 3 ⋅ 6( 4 )1 / 3
188 Textbook of Algebra

⇒ ab ≥ 36 × 2 68. We have,
Q
a+b
≥ ab ≥ 6 2 ⇒
a+b
≥6 2 (5 + 2 ) x 2 - ( 4 + 5 ) x + 8 + 2 5 = 0
2 2
4+ 5
∴ a + b ≥ 12 2 ∴Sum of the roots =
5+ 2
or (a + b ) 3 ≥ 3456 2
8+2 5
Hence, minimum value of (a + b ) 3 is 3456 2. and product of the roots =
5+ 2
Solutions (Q. Nos. 64 to 66) ∴The harmonic mean of the roots
∴ α +β+γ+δ=-A …(i)
2 × Product of the roots 2 × (8 + 2 5 )
(α + β ) ( γ + δ ) + αβ + γδ = B …(ii) = = =4
Sum of the roots (4 + 5 )
αβ ( γ + δ ) + γδ (α + β ) = - C …(iii)
and αβγδ = D …(iv) 69. Let x 2 - ax + 30 = y
C αβ ( γ + δ ) + γδ (α + β ) ∴ y = 2 y + 15 …(i)
64. Q =
A α+β+γ+δ
k ( γ + δ ) + k (α + β ) ⇒ y 2 - 4y - 60 = 0
= [Qαβ = γδ =k]
α+β+γ+δ ⇒ (y - 10 ) (y + 6 ) = 0
=k …(v) ∴ y = 10, - 6
65. From Eq. (ii), we get ⇒ y = 10, y ≠ - 6 [Qy > 0]
(α + β ) ( γ + δ ) = B - (αβ + γδ ) = B - 2k [Qαβ = γδ = k] Now, x 2 - ax + 30 = 10
66. From Eq. (iv), we get
⇒ x 2 - ax + 20 = 0
αβγδ = D
⇒ k ⋅k = D [Qαβ = γδ = k] Given, αβ = λ = 20
2 α+β
C ∴ ≥ αβ = 20
⇒   =D [from Eq. (v)] 2
 A
⇒ α + β ≥ 2 20
∴ C 2 = A 2D
67. The given equation is | x - 2 | 2 + | x - 2 | - 2 = 0. or µ= 4 5

There are two cases: ∴ Minimum value of µ is 4 5.

Case I If x ≥ 2, then ( x - 2 ) 2 + x - 2 - 2 = 0 i.e., µ = 4 5 = 8.9 ⇒ (µ) = 9
6
⇒ x 2 - 3x = 0  1
70. Q N r =  x +  -  x 6 +
 1
 -2
⇒ x (x - 3) = 0
 x  x6
6 2 3
⇒ x = 0, 3  1  1   1  1 
=  x +  -  x 3 + 3  =   x +  +  x 3 + 3  
Here, 0 is not possible.  x  x   x  x 
∴ x =3 3
 1 1 
Case II If x < 2, then   x +  -  x 3 + 3 
 x   x 
(x - 2)2 - x + 2 - 2 = 0 
⇒ x 2 - 5x + 4 = 0   1
= Dr ⋅ 3  x +  
⇒ (x - 1) (x - 4) = 0   x
⇒ x = 1, 4 Nr  1
∴ = 3 x +  ≥ 6
Here, 4 is not possible. Dr  x
∴ x =1 Nr
Hence, minimum value of is 6.
∴The sum of roots = 1 + 3 = 4 Dr
71. a + b = 2c …(i)
Aliter ab = - 5d …(ii)
Let | x - 2 | = y . c + d = 2a …(iii)
Then, we get y2 + y -2 = 0 cd = - 5b …(iv)
⇒ (y - 1 ) (y + 2 ) = 0 ⇒ y = 1, - 2 From Eqs. (i) and (iii), we get
But - 2 is not possible. a + b + c + d = 2 (a + c )
Hence, | x - 2| = 1 ⇒ x = 1, 3 ∴ a + c =b +d …(v)
∴ Sum of the roots = 1 + 3 = 4
 Chap 02 Theory of Equations 189

From Eqs. (i) and (iii), we get 74. Q (1 + i ) x 2 + (1 - i ) x - 2i = 0

b - d = 3 (c - a ) …(vi) (1 - i ) 2i
2 ⇒ x2 + x- =0
Also, a is a root of x - 2cx - 5d = 0 (1 + i ) (1 + i )
∴ a 2 - 2ac - 5d = 0 …(vii) ⇒ x 2 - ix - (1 + i ) = 0
And c is a root of ∴ α + β = i , and αβ = - (1 + i )
c 2 - 2ac - 5b = 0 …(viii) ∴ α - β = (α + β ) 2 - 4 αβ = i 2 + 4 (1 + i ) = (3 + 4i )
From Eqs. (vii) and (viii), we get
| α - β| = 9 + 16 = 5
a 2 - c 2 - 5 (d - b ) = 0
∴ | α - β| 2 = 5
⇒ (a + c ) (a - c ) + 5 (b - d ) = 0
⇒ (a + c ) (a - c ) + 15 (c - a ) = 0 [from Eq. (vi)] 75. Q 4x 2 - 16 x + c = 0
⇒ (a - c ) (a + c - 15 ) = 0 c
⇒ x 2 - 4x + =0
∴ a + c = 15, a - c ≠ 0 4
c
From Eq. (v), we get b + d = 15 Let f (x ) = x 2 - 4x +
4
∴ a + b + c + d = a + c + b + d = 15 + 15 = 30
Then, the following cases arises:
⇒ Sum of digits of a + b + c + d = 3 + 0 = 3
x 2 - 3x + c
72. Q y =
x 2 + 3x + c
⇒ x 2(y - 1 ) + 3 x (y + 1 ) + c (y - 1 ) = 0 2
1 α β 3
Q x ∈R
∴ 9 (y + 1 ) 2 - 4c (y - 1 ) 2 ≥ 0 Case I D>0
⇒ 16 - c > 0
( 2 cy - 2 c ) 2 - (3y + 3 ) 2 ≤ 0
∴ c < 16
⇒ {( 2 c + 3 ) y - (2 c - 3 )} {(2 c - 3 )y - (2 c + 3 )} ≤ 0 Case II f (1 ) > 0
2 c -3 2 c +3 c
or ≤y ≤ ⇒ 1-4+ >0
2 c +3 2 c -3 4
c
2 c +3 ⇒ >3
But given, =7 4
2 c -3 ∴ c > 12
⇒ 2 c + 3 = 14 c - 21 Case III f (2 ) < 0
c
or 12 c = 24 or c =2 ⇒ 4 -8 + < 0
4
∴ c=4 c
⇒ <4
73. We have, x - (x - 1) + (x - 2)2 = 5
2 2 4
∴ c < 16
⇒ | x | - | x - 1| + | x - 2| = 5
Case IV f (3 ) > 0
Case I If x < 0, then c
⇒ 9 - 12 + > 0
- x + (x - 1) - (x - 2) = 5 4
x =1 - 5 c
⇒ >3
4
Case II If 0 ≤ x < 1, then
⇒ c > 12
x + (x - 1) - (x - 2) = 5
⇒ x = 5 - 1, which is not possible. Combining all cases, we get
Case III If 1 ≤ x < 2, then 12 < c < 16
x - (x - 1) - (x - 2) = 5 Thus, integral values of c are 13, 14 and 15.
⇒ x = 3 - 5, which is not possible. Hence, number of integral values of c is 3.
76. We have, r +s +t=0 …(i)
Case IV If x > 2, then
x - (x - 1) + (x - 2) = 5 1001
rs + st + tr = …(ii)
8
⇒ x =1 + 5 2008
and rst = - = - 251 …(iii)
Hence, number of solutions is 2. 8
190 Textbook of Algebra

Now, (r + s ) 3 + (s + t ) 3 + (t + r ) 3 = ( - t ) 3 + ( - r ) 3 + ( - s ) 3 Q a (b - c ) + b (c - a ) + c (a - b ) = 0
[Qr + s + t = 0] ∴ x = 1 is a root of
= - (t 3 + r 3 + s 3 ) = - 3 rst [Qr + s + t = 0] a (b - c ) x 2 + b (c - a ) x + c (a - b ) = 0 …(ii)
= - 3 ( - 251 ) = 753 Given, roots [Eq. (ii)] are equal.
c (a - b )
Now, 99 λ = (r + s ) 3 + (s + t ) 3 + (t + r ) 3 = 753 ∴ 1 ×1 =
a (b - c )
753
∴ λ= = 7.6 ⇒ a (b - c ) = c (a - b )
99
2ac
∴ [λ ] = 7 or b=
a+c
77. A Æ (r,s); B Æ (p, q, r,s, t); C Æ (p, q, t)
∴ a, b and c are in HP. …(iii)
x 2 - 2x + 4
(A) We have, y = From Eqs. (i) and (ii), we get
x 2 + 2x + 4
a =b =c
⇒ x 2 (y - 1 ) + 2 (y + 1 ) x + 4 (y - 1 ) = 0 ∴ a,b and c are in AP, GP and HP.
As x ∈ R , we get (B)Q x 3 - 3x 2 + 3x - 1 = 0
D≥0 ⇒ (x - 1)3 = 0
⇒ 4 (y + 1 ) - 16 (y - 1 ) 2 ≥ 0
2
∴ x = 1, 1, 1
⇒ 3y 2 - 10y + 3 ≤ 0 ⇒ Common root, x = 1
⇒ (y - 3 ) (3y - 1 ) ≤ 0 ∴ a (1 ) 2 + b (1 ) + c = 0
1 ⇒ a+b+c=0
⇒ ≤y ≤3
3
(C) Given, bx + ( (a + c ) 2 + 4b 2 ) x + (a + c ) ≥ 0
2
2x 2 + 4x + 1
(B) We have, y = 2 ∴ D≤0
x + 4x + 2
⇒ (a + c ) 2 + 4b 2 - 4b (a + c ) ≤ 0
⇒ x 2(y - 2 ) + 4(y - 1 ) x + 2y - 1 = 0
⇒ (a + c - 2b ) 2 ≤ 0
As x ∈ R , we get
D≥0 or (a + c - 2b ) 2 = 0
⇒ 16 (y - 1 ) 2 - 4 (y - 2 ) (2y - 1 ) ≥ 0 ∴ a + c = 2b
Hence a, b and c are in AP.
⇒ 4 (y - 1 ) 2 - (y - 2 ) (2y - 1 ) ≥ 0
79. A Æ (q,r,s,t); B Æ (q,r); C Æ (p,q)
⇒ 2y 2 - 3y + 2 ≥ 0
ax 2 + 3 x - 4
3 (A) We have, y =
⇒ y2 - y + 1 ≥ 0 3x - 4x 2 + a
2
3 7
2 ⇒ x 2 (a + 4y ) + 3 (1 - y ) x - (ay + 4 ) = 0

⇒ y -  + ≥0
 4 16 As x ∈R , we get
∴ y ∈R D≥0
2
2
x - 3x + 4 ⇒ 9 (1 - y ) + 4 (a + 4y ) (ay + 4 ) ≥ 0
(C) We have, y =
x -3 ⇒ ( 9 + 16a ) y 2 + ( 4a 2 + 46 )y + (9 + 16a ) ≥ 0, ∀ y ∈ R
⇒ x 2 - (3 + y ) x + 3y + 4 = 0 ⇒ If 9 + 16a > 0, then D ≤ 0
As x ∈ R , we get Now, D≤0
D ≥ 0 ⇒ (3 + y ) 2 - 4 (3y + 4 ) ≥ 0 ⇒ ( 4a + 46 ) - 4 ( 9 + 16a ) 2 ≤ 0
2 2

⇒ y 2 - 6y - 7 ≥ 0 ⇒ (y + 1 ) (y - 7 ) ≥ 0 ⇒ 4 [( 2a 2 + 23 ) 2 - ( 9 + 16a ) 2 ] ≤ 0

⇒ y ∈ ( - ∞, - 1 ] ∪ [7, ∞ ) ⇒ [(2 a + 23 ) + ( 9 + 16a )] [( 2a 2 + 23 ) - ( 9 + 16a )] ≤ 0

2

78. A Æ (q,r,s); B Æ (p); C Æ (q) ⇒ ( 2a 2 + 16a + 32 ) ( 2a 2 - 16a + 14 ) ≤ 0

(A)Q(d + a - b ) 2 + (d + b - c ) 2 = 0 ⇒ 4 (a + 4 ) 2 (a 2 - 8a + 7 ) ≤ 0
which is possible only when ⇒ a 2 - 8a + 7 ≤ 0
d + a - b = 0, d + b - c = 0 ⇒ (a - 1 ) (a - 7 ) ≤ 0
⇒ b -a =c -b ⇒ 1 ≤a ≤7
⇒ 2b = a + c ∴ 9 + 16a > 0 and 1 ≤ a ≤ 7
∴ a, b and c are in AP. …(i) ⇒ 1 ≤a ≤7
 Chap 02 Theory of Equations 191

ax 2 + x - 2 f (1 ) > 0
(B) We have, y =
a + x - 2x 2 ⇒ 1 -6 + 9 + λ > 0
⇒ x 2 (a + 2y ) + x (1 - y ) - (2 + ay ) = 0 ⇒ λ>-4 …(ii)
and f (3 ) < 0
As x ∈R , we get
⇒ 27 - 54 + 27 + λ < 0
D≥0
2
⇒ λ<0 …(iii)
⇒ ( 1 - y ) + 4 ( 2 + ay ) (a + 2y ) ≥ 0
From Eqs. (i), (ii) and (iii), we get
⇒ ( 1 + 8a ) y 2 + ( 4a 2 + 14 ) y + (1 + 8a ) ≥ 0 -4<λ<0
⇒ If 1 + 8a > 0, then D ≤ 0 ⇒ -3 < λ + 1 < 1
⇒ ( 4a 2 + 14 ) 2 - 4 (1 + 8a ) 2 ≤ 0 ∴ [ λ + 1 ] = - 3, - 2, - 1, 0
⇒ 4 [(2a 2 + 7 ) 2 - (1 + 8a ) 2 ] ≤ 0 ∴ |[ λ + 1 ]| = 3, 2, 1, 0
(B)Q x 2 + x + 1 > 0, ∀ x ∈ R
⇒ [(2a 2 + 7 ) + (1 + 8a )] [(2a 2 + 7 ) - (1 + 8a )] ≤ 0
x 2 - λx - 2
⇒ (2a 2 + 8a + 8 ) (2a 2 - 8a + 6 ) ≤ 0 Given, -3 < <2
x2 + x + 1
⇒ 4 (a + 2 ) 2 (a 2 - 4a + 3 ) ≤ 0
⇒ - 3 x 2 - 3 x - 3 < x 2 - λx - 2 < 2 x 2 + 2 x + 2
⇒ a 2 - 4a + 3 ≤ 0
⇒ 4x 2 - (λ - 3) x + 1 > 0
⇒ (a - 1 ) (a - 3 ) ≤ 0
⇒ 1 ≤a ≤3 and x 2 + (λ + 2)x + 4 > 0
Thus, 1 + 8a > 0 and 1 ≤ a ≤ 3 ∴ (λ - 3)2 - 4 ⋅ 4 ⋅ 1 < 0
⇒ 1 ≤a ≤3 and (λ + 2)2 - 4 ⋅ 1 ⋅ 4 < 0
x 2 + 2x + a
(C) We have, y = 2 ⇒ (λ - 3)2 - 42 < 0
x + 4 x + 3a
and (λ + 2)2 - 42 < 0
⇒ x 2(y - 1 ) + 2 (2y - 1 ) x + a (3y - 1 ) = 0
⇒ - 4 < λ -3 < 4
As x ∈ R , we get and -4<λ+2<4
D≥0 or - 1 < λ <7
⇒ 4 (2y - 1 ) 2 - 4 (y - 1 ) a (3y - 1 ) ≥ 0 and -6 < λ <2
⇒ ( 4 - 3a ) y 2 - ( 4 - 4a )y + (1 - a ) ≥ 0 We get, -1 < λ <2
⇒ If 4 - 3a > 0, then D ≤ 0 ∴ [ λ ] = - 1, 0, 1
⇒ ( 4 - 4a ) 2 - 4 ( 4 - 3a ) (1 - a ) ≤ 0 ⇒ |[ λ ]| = 0, 1

⇒ 4 (2 - 2a ) 2 - 4 ( 4 - 3a ) (1 - a ) ≤ 0 (C)Q (b - c ) + (c - a ) + (a - b ) = 0
∴ x = 1 is a root of
⇒ 4 + 4a 2 - 8a - ( 4 - 7a + 3a 2 ) ≤ 0
(b - c ) x 2 + (c - a ) x + (a - b ) = 0
⇒ a2 - a ≤ 0
Also, x = 1 satisfies
⇒ a (a - 1 ) ≤ 0 x2 + λ x + 1 = 0
⇒ 0 ≤a ≤1
⇒ 1+λ+1=0
80. A Æ (p,q,r,s);B Æ (p,q); C Æ (s) ∴ λ = -2
(A) Let y = f ( x ) = x 3 - 6 x 2 + 9 x + λ Now, λ -1 = -3
[ λ - 1] = - 3
f ′ ( x ) = 3 x 2 - 12 x + 9 = 0
⇒ |[ λ - 1 ]| = 3
∴ x = 1, 3 81. If quadratic equation ax 2 + bx + c = 0 is satisfied by more than
f ′′ ( x ) = 6 x - 12 two values of x, then it must be an identity.
f ′′(1 ) < 0 and f ′′ (3 ) > 0 Therefore, a = b = c = 0
∴ Statement-2 is true.
But in Statement-1,
4 p - 3 = 4q - 3 = r = 0
3
0 3 Then, p =q = ,r = 0
1 4
λ
which is false.
Since, at one value of p or q or r, all coefficients at a time ≠ 0.
Also, f (0) < 0 ⇒ λ < 0 …(i) ∴ Statement-1 is false.
192 Textbook of Algebra

82. We have, x 2 + (2m + 1 ) x + (2n + 1 ) = 0 …(i)  4 4 

and z ∈ 2- ,2 +
 3 3 
m, n ∈ I
∴ D = b 2 - 4ac Since, Eqs. (i) and (ii) remains same, if x, y , z interchange their
positions.
= ( 2m + 1 ) 2 - 4 ( 2n + 1 )
Hence, both statements are true and Statement-2 is a correct
is never be a perfect square. explanation of Statement-1.
Therefore, the roots of Eq. (i) can never be integers. Hence, the 86. Let y = ax 3 + bx + c
roots of Eq. (i) cannot have any rational root as a = 1, b, c ∈ I .
dy
Hence, both statements are true and Statements –2 is a correct ∴ = 3ax 2 + b
explanation of Statement-1. dx
dy
83. Let α be one root of equation ax 2 + 3 x + 5 = 0. Therefore, For maximum or minimum = 0, we get
dx
1 5
α⋅ = b
α a x=± -
3a
5
⇒ 1= dy
a Case I If a > 0, b > 0, then >0
dx
⇒ a =5
In this case, function is increasing, so it has exactly one root
Hence, both the statements are true and Statement-2 is the
dy
correct explanation of Statement-1. Case II If a < 0, b < 0, then <0
dx
84. Let roots of Ax 3 + Bx 2 + Cx + D = 0 …(i)
In this case, function is decreasing, so it has exactly one root.
are α - β, α , α + β (in AP).
Case III a > 0, b < 0 or a < 0, b > 0, then y = ax 3 + bx + c is
B
Then, (α - β) + α + (α + β) = - maximum at one point and minimum at other point.
A
B Hence, all roots can never be non-negative.
⇒ , which is a root of Eq. (i).
α=- ∴Statement-1 is false. But
3A
Then, Aα 3 + Bα 2 + Cα + D = 0 Coefficient of x 2
Sum of roots = - =0
3 2 Coefficient of x 3
 B  B  B i.e., Statement-2 is true.
⇒ A -  + B -  + C -  +D=0
 3A   3A   3A 
87. Statement-2 is obviously true.
B3 B3 BC But y = ax 2 + bx + c
⇒ - 2
+ 2
- +D=0
27 A 9A 3 A
 b c
⇒ 2 B 3 - 9 ABC + 27 A 2D = 0 y = a x2 + x + 
 a a
Now, comparing with 2 B 3 + k1ABC + k2 A 2D = 0, we get  b
2
D 
= a  x +  - 2  [where, D = b 2 - 4ac]
k1 = - 9, k2 = 27  2a  4a 

∴ k2 - k1 = 27 - ( - 9 ) = 36 = 6 2 2
 b 1  D
Hence, both statements are true and Statement-2 is a correct ⇒  x +  = y + 
 2a  a  4a 
explanation of Statement-1.
b D
85. Q x, y , z ∈ R Let x+ = X and y + = Y.
2a 4a
x + y + z =6 …(i) 1
and xy + yz + zx = 8 …(ii) ∴ X2 = Y
a
⇒ xy + ( x + y ) {6 - ( x + y )} = 8 [from Eq. (i)] b
⇒ xy + 6 x + 6y - ( x 2 + 2 xy + y 2 ) = 8 Equation of axis, X = 0 i.e. x + =0
2a
or y 2 + (x - 6) y + x 2 - 6x + 8 = 0 or 2ax + b = 0
∴ ( x - 6 ) 2 - 4 ⋅ 1 ⋅ ( x 2 - 6 x + 8 ) ≥ 0, ∀ y ∈ R Hence, y = ax 2 + bx + c is symmetric about the line
2ax + b = 0.
⇒ - 3 x 2 + 12 x + 4 ≥ 0 or 3 x 2 - 12 x - 4 ≤ 0
∴ Both statements are true and Statement-2 is a correct
4 4
or 2- ≤ x ≤2 + explanation of Statement-1.
3 3
88. Q(1 + m ) x 2 - 2 (1 + 3m ) x + (1 + 8m ) = 0
 4 4 
or x∈ 2-

,2 + ∴ D = 4 (1 + 3m ) 2 - 4 (1 + m ) (1 + 8m ) = 4m (m - 3 )
3 3 
(i) Both roots are imaginary.
 4 4 
Similarly, y ∈ 2- ,2 + ∴ D<0
 3 3 
⇒ 4m (m - 3 ) < 0
 Chap 02 Theory of Equations 193

⇒ 0 <m <3 Combining all cases, we get

or m ∈( 0, 3 )  1
m ∈  - 1, - 
(ii) Both roots are equal.  8
∴ D=0 (vii) Roots are equal in magnitude but opposite in sign, then
⇒ 4m (m - 3 ) = 0 Consider the following cases:
⇒ m = 0, 3 Case I Sum of the roots = 0
(iii) Both roots are real and distinct. 2 (1 + 3m )
⇒ =0
∴ D>0 (1 + m )
⇒ 4m (m - 3 ) > 0 1
⇒ m = - ,m ≠1
⇒ m < 0 or m > 3 3
∴ m ∈ ( - ∞, 0 ) ∪ (3, ∞ ) Case II D > 0 ⇒ 4m(m - 3 ) > 0
(iv) Both roots are positive. ⇒ m ∈ ( - ∞, 0 ) ∪ ( 3, ∞ )
Case I Sum of the roots > 0 Combining all cases, we get
2 (1 + 3m ) 1
⇒ >0 m=-
(1 + m ) 3
 1  (viii) Atleast one root is positive, then either one root is positive
⇒ m ∈ ( - ∞, - 1 ) ∪  - , ∞ or both roots are positive.
 3 
i.e. (d ) ∪ ( f )
Case II Product of the roots > 0
 1
(1 + 8m ) or m ∈ ( - ∞, - 1 ) ∪  - 1, -  ∪ [3, ∞ )
⇒ >0  8
(1 + m )
(ix) Atleast one root is negative, then either one root is
 1 
m ∈ ( - ∞, - 1 ) ∪  - , ∞ negative or both roots are negative.
 8 
 1
Case III D≥0 i.e. (e ) ∪ ( f ) or m ∈  - 1, - 
 8
⇒ 4m (m - 3 ) ≥ 0
(x) Let roots are 2α are 3α. Then,
m ∈ ( - ∞, 0 ] ∪ [ 3, ∞ )
Consider the following cases:
Combining all Cases, we get 2 (1 + 3m )
m ∈ ( - ∞, - 1 ) ∪ [ 3, ∞ ) Case I Sum of the roots = 2 α + 3 α =
(1 + m )
(v) Both roots are negative.
2 (1 + 3m )
Consider the following cases: ⇒ α=
5 (1 + m )
2 (1 + 3m )
Case I Sum of the roots < 0 ⇒ <0 (1 + 8m )
(1 + m ) Case II Product of the roots = 2 α ⋅ 3 α =
(1 + m )
 1
⇒ m ∈  - 1, -  (1 + 8m )
 3 ⇒ 6α 2 =
(1 + m )
(1 + 8m )
Case II Product of the roots > 0 ⇒ >0 From Eqs. (i) and (ii), we get
(1 + m ) 2
 1  2 (1 + 3m )  (1 + 8m )
⇒ m ∈ ( - ∞, 1 ) ∪  - , ∞ 6  =
 8   5 (1 + m )  (1 + m )
Case III D ≥ 0 ⇒ 24 (1 + 3m ) 2 = 25 (1 + 8m ) (1 + m )
4m (m - 3 ) ≥ 0 ⇒ m ∈ ( - ∞, 0 ] ∪ [3, ∞ ) ⇒ 24 ( 9m 2 + 6m + 1 ) = 25 ( 8m 2 + 9m + 1 )
Combining all cases, we get
16m 2 - 81m - 1 = 0
m ∈φ
(vi) Roots are opposite in sign, then 81 ± ( - 81 ) 2 + 64
or m=
Case I Consider the following cases: 32
Product of the roots < 0 81 ± 6625
⇒ m=
(1 + 8m ) 32
⇒ <0
(1 + m ) 89. Q2x 2 - 2 (2m + 1) x + m (m + 1 ) = 0 [Q m ∈ R ]
 1 ∴ 2
D = [ - 2 (2m + 1 )] - 8m (m + 1 ) [ D = b 2 - 4ac ]
m ∈  - 1, - 
 8
= 4 {(2m + 1 ) 2 - 2m (m + 1 )}
Case II D > 0 ⇒ 4m (m - 3 ) > 0
⇒ m ∈ ( - ∞, 0 ) ∪ (3, ∞ ) = 4 (2m 2 + 2m + 1 )
194 Textbook of Algebra

 2 Combining all cases, we get

 1 1 1
= 8  m 2 + m +  = 8  m +  +  > 0
 2  2 4  7 + 33 
 m ∈ , ∞
 2 
or D > 0, ∀ m ∈ R …(i)
(iii) Both roots lie in the interval (2, 3).
b 2 (2m + 1 )  1
x -coordinate of vertex = - = = m +  …(ii) Consider the following cases:
2a 4  2
and let
1
f ( x ) = x 2 - (2m + 1 ) x + m (m + 1 ) …(iii)
2 f(2) f(3)
(i) Both roots are smaller than 2.
X
2 α β 3

Case I D ≥ 0
∴ m ∈R [from Eq. (i)]
X Case II f (2 ) > 0
α β 2  7 - 33   7 + 33 
∴ m ∈  - ∞,  ∪ , ∞ [from part (a)]
Consider the following cases:  2   2 
Case I D ≥ 0 Case III f (3 ) > 0
∴ m ∈R [from Eq. (i)] 1
Case II x -coordinate of vertex < 2. ⇒ 9 - 3 (2m + 1 ) + m (m + 1 ) > 0
2
1
⇒ m + <2 [from Eq. (ii)] or m 2 - 11m + 12 > 0
2
3  11 - 73   11 + 73 
or m< ∴ m ∈  - ∞,  ∪ , ∞
2  2   2 
Case III f (2 ) > 0 Case IV 2 < x -coordinate of vertex < 3
1 1
⇒ 4 - (2m + 1 ) 2 + m (m + 1 ) > 0 ⇒ 2 <m + <3
2 2
⇒ m 2 - 7m + 4 > 0 or
3 5  3 5
< m < or m ∈  , 
2 2  2 2
 7 - 33   7 + 33 
∴ m ∈  - ∞,  ∪ , ∞ Combining all cases, we get
 2   2 
m ∈φ
Combining all cases, we get (iv) Exactly one root lie in the interval (2,3) .
 7 - 33  Consider the following cases:
m ∈  - ∞, 
 2  Case I D > 0
∴ m ∈R [from Eq. (i)]
(ii) Both roots are greater than 2.
Consider the following cases:

3
f(2) X
2 α β
X
2 α β
Case II f (2 ) f (3 ) < 0
Case I D ≥ 0  1 
 4 - 2 ( 2m + 1 ) + m ( m + 1 )
∴ m ∈R [from Eq. (i)]  2 
Case II x -coordinate of vertex > 2  1 
1 9 - 3 ( 2m + 1 ) + m ( m + 1 ) < 0
⇒ m + >2 [from Eq. (ii)]  2 
2
⇒ ( m 2 - 7m + 4 ) ( m 2 - 11m + 12 ) < 0
3
∴ m>  7 - 33   7 + 33 
2 ⇒ m -  m - 
Case III f (2 ) > 0  2   2 
 7 - 33   33   11 - 73   11 + 73 
m ∈  - ∞,  ∪ 7 + , ∞ [from part (a)] m -  m -  <0
 2   2   2   2 
 Chap 02 Theory of Equations 195

7 + √33  11 - 73 11 + 73 
∴ m ∈ , 
+ + 2 +  2 2 
7 – √33 – 7 – √73 – 11 + √73
Combining all cases, we get
2 2 2
 7 - 33 7 + 33 
m ∈ , 
 7 - 33 11 - 73   7 + 33 11 + 73   2 2 
∴ m ∈ ,  ∪ , 
 2 2   2 2  (vii) Atleast one root lies in the interval (2, 3).
Combining all cases, we get i.e. (d ) ∪ (c )
 7 - 33 11 - 73   7 + 33 11 + 73   7 - 33 11 - 73   7 + 33 11 + 73 
m ∈ ,  ∪ ,  ∴ m ∈ ,  ∪ , 
 2 2   2 2   2 2   2 2 
(v) One root is smaller than 1 and the other root is greater (viii) Atleast one root is greater than 2.
than 1.
i.e. (Exactly one root is greater than 2) ∪ (Both roots are
Consider the following cases: greater than 2)

1 X 2
α β X
α β
Case I D > 0
∴ m ∈R [from Eq. (i)] or(Exactly one root is greater than 2) ∪ (b ) …(I)
Case II f (1 ) < 0 Consider the following cases:
1 Case I D > 0
⇒ 1 - (2m + 1 ) + m (m + 1 ) < 0 [from Eq. (iii)]
2 ∴ m ∈R [from Eq. (i)]
⇒ m 2 - 3m < 0 Case II f (2 ) < 0
⇒ m (m - 3 ) < 0 ⇒ m 2 - 7m + 4 < 0
∴ m ∈( 0, 3 )
 7 - 33 7 + 33 
Combining both cases, we get ∴ m ∈ , 
m ∈( 0, 3 )  2 2 
(vi) One root is greater than 3 and the other root is smaller Combining both cases, we get
than 2.  7 - 33 7 + 33 
Consider the following cases: m ∈ ,  …(II)
 2 2 
Finally from Eqs. (I) and (II), we get
 7 - 33 7 + 33   7 + 33 
m ∈ ,  ∪ , ∞
 2 2   2 
2 3 X
α β (ix) Atleast one root is smaller than 2.
i.e. (Exactly one root is smaller than 2) ∪(Both roots are
smaller than 2)
Case I D > 0
or (h) (II) ∪ (a)
∴ m ∈R [from Eq. (i)]
 7 - 33   7 - 33 7 + 33 
Case II f (2 ) < 0 We get, m ∈  - ∞,  ∪ , 
⇒ m 2 - 7m + 4 < 0  2   2 2 

7 - 33 7 + 33 (x) Both 2 and 3 lie between α and β.

∴ <m < Consider the following cases:
2 2
 7 - 33 7 + 33  Case I D > 0
∴ m ∈ ,  ∴ m ∈R [from Eq. (i)]
 2 2 
Case III f (3 ) < 0
⇒ m 2 - 11m + 12 < 0
11 - 73 11 + 73 2 3 X
∴ <m < α β
2 2
196 Textbook of Algebra

1
Case II f (2 ) < 0
cn + 1
⇒ m 2 - 7m + 4 < 0 ⇒ α= 
a
 7 - 33 7 + 33  ∴ From Eq. (i), we get
∴ m ∈ , 
 2 2  1 n
cn + 1 cn + 1 b
Case III f (3 ) < 0   +  =-
a a a
⇒ m 2 - 11m + 12 < 0
1 1 1 n
- +1 - +1
 11 - 73 11 + 73  ⇒ (c )n + 1 ⋅ a n+1
+ (c n )n + 1 ⋅ a n+1
+b=0
∴ m ∈ , 
 2 2  1 n 1 1
n+1 n+1 n n+1 n +1
Combining all cases, we get ⇒ c ⋅a + (c ) ⋅a +b=0
 11 - 73 7 + 33  1 1
m ∈ , 
 2 2  ⇒ (anc )n + 1 + (c na )n + 1 + b = 0
α b
90. Q =r 93. We have, α+β=-
β a
c m n
α+β r +1 αβ = ⇒ γ+δ=- and γδ =
⇒ = a l l
α -β r -1
Now, sum of the roots
[using componendo and dividendo method]
= (αγ + βδ ) + (αδ + βγ ) = (α + β ) γ + (α + β ) δ
- b /a r + 1
⇒ = ⇒ b (1 - r ) = (1 + r ) D = (α + β ) ( γ + δ )
D r -1
 b   m  mb
a = -  -  =
 a  l  al
On squaring both sides, we get
and product of the roots
⇒ b 2(1 - r ) 2 = (1 + r ) 2 (b 2 - 4ac ) = (αγ + βδ ) (αδ + βγ )
(1 + r ) 2 b 2 = (α 2 + β 2 ) γδ + αβ ( γ 2 + δ 2 )
or (1 + r ) 2 ⋅ 4ac = b 2( 4r ) or =
r ac = {(α + β ) 2 - 2αβ} γδ + αβ {( γ + δ ) 2 - 2 γδ }
1 1 1
91. We have, + =  b  2 2c  n c  m  2 2n 
x+p x+q r =  -  -  +  -  - 
  a  l a  l 
(x + q ) + (x + p ) 1  a l 
⇒ =
x 2 + ( p + q ) x + pq r b 2 - 2ac  n c m 2 - 2nl  (b 2 - 2ac ) ln + ( m 2 - 2nl ) ac
= 2  +  2 =
⇒ x 2 + ( p + q - 2r ) x + pq - ( p + q ) r = 0  a l a  l  a 2l 2
Now, since the roots are equal in magnitudes, but opposite in ∴ Required equation is
sign. Therefore,  mb  (b 2 - 2ac ) ln + (m 2 - 2nl ) ac
Sum of the roots = 0 x2 -   x + =0
 al  a 2l 2
⇒ p + q - 2r = 0
⇒ a 2l 2x 2 - mbalx + (b 2 - 2ac ) ln + (m 2 - 2nl ) ac = 0
⇒ p + q = 2r …(i)
and product of the roots = pq - ( p + q ) r 94. Since, the roots are equal.
∴ D=0
 p + q
= pq - ( p + q )   [from Eq. (i)] ⇒ 4 (b 2 - ac ) 2 - 4 (a 2 - bc ) (c 2 - ab ) = 0
 2 
⇒ (b 2 - ac ) 2 - (a 2 - bc ) (c 2 - ab ) = 0
2 pq - p 2 - q 2 - 2 pq
= ⇒ b (a 3 + b 3 + c 3 - 3abc ) = 0
2
p2 + q2 ⇒ b = 0 or a 3 + b 3 + c 3 - 3abc = 0
=-
2 95. Let α and β be the roots of x 2 - px + q = 0. Then,
92. Let α be one root of the equation ax 2 + bx + c = 0. α+β=p …(i)
Then, other root be α n . αβ = q …(ii)
b 1
∴ α + αn = - …(i) And α and be the roots of x 2 - ax + b = 0. Then,
a β
c 1
and α ⋅ αn = α + =a …(iii)
a β
c α
⇒ αn + 1 = =b ... (iv)
a β
 Chap 02 Theory of Equations 197

Now, LHS = (q - b ) 2 ∴ λ = 10,

1
2 10
 α 2
= αβ -  [from Eqs. (ii) and (iv)] ⇒ (2 + 3 ) x - 2x
= 10, 10 - 1
 β
2 2 ⇒ x 2 - 2 x = log 2 + 3
10, - log 2 + 3
10
21   1 
= α β -  = α 2 (α + β ) - α +   2
 β   β  ⇒ ( x - 1 ) = 1 + log 2 + 3
10, 1 - log 2 + 3
10
2 2
= α ( p - a ) [from Eqs. (i) and (iii)] ∴ ( x - 1 ) 2 = 1 + log 2 + 3
10
α
= αβ ⋅ ( p - a ) 2 2
[Q ( x - 1 ) ≠ 1 - log 2 + 3
10 ]
β
= bq ( p - a ) 2 [from Eqs. (ii) and (iv)] ⇒ x = 1 ± (1 + log 2 + 3
10 )
= RHS ⇒ x 1 = 1 + (1 + log 2 + 3
10 )
2
96. Since, roots of x - 2px + q = 0 are equal.
x 2 = 1 - (1 + log 2 + 3
10 )
∴ D=0
2
i.e., ( -2 p ) - 4q = 0 or p 2 = q
2
…(i)  x 
99. We have, x2+   =8
Now, (1 + y ) x 2 - 2 ( p + y ) x + (q + y ) = 0  x - 1
2
∴ Discriminant = 4 ( p + y ) 2 - 4 (1 + y ) (q + y )  x  x
⇒ x +  - 2⋅ x ⋅ =8
= 4 ( p 2 + 2 py + y 2 - q - y - qy - y 2 )  x - 1 (x - 1)
2
= 4 [(2 p - q - 1 ) y + p 2 - q ]  x2   x2 
⇒   -2   -8 = 0 …(i)
= 4 [(2 p - p 2 - 1 ) y + 0 ] [from Eq. (i)]  x - 1  x - 1
= - 4 ( p - 1 ) 2y x2
Let y = . Then, Eq. (i) reduces to
>0 [Qy < 0 and p ≠ 1] x -1
Hence, roots of (1 + y ) x 2 - 2 ( p + y ) x + (q + y ) = 0 are real y 2 - 2y - 8 = 0
and distinct. ⇒ (y - 4 ) (y + 2 ) = 0
2
97. x log x ( x + 3) = 16 …(i) ∴ y = 4, - 2
Equation is defined, when x2
If y = 4, then 4=
x > 0, x ≠ 1, x ≠ - 3, x -1
Then, (x + 3)2 = 42 [by property] or x 2 - 4x + 4 = 0
⇒ x+3=±4
or (x - 2)2 = 0
∴ x = 1 and x = - 7
But x ≠ 1, x ≠ - 7 or x =2
i.e. no solution. ∴ x1 = 2
∴ x ∈φ x2
and if y = - 2, then -2 =
2
- 2x + 1 2
- 2x - 1 101 x -1
98. Q(2 + 3 ) x + (2 - 3 ) x =
10 (2 - 3 ) or x 2 + 2x - 2 = 0
2
- 2x
⇒ (2 + 3 )x ⋅ (2 + 3 ) (2 - 3 ) - 2 ± (4 + 8)
101 ∴ x=
x2 - 2x - 1 2
+ (2 - 3 ) ⋅ (2 - 3 ) =
10 ⇒ x = -1 ± 3
x2 - 2x x2 - 2x 101
⇒ (2 + 3 ) + (2 - 3 ) = ∴ x 2 = - 1 + 3, x 3 = - 1 - 3
10
x2 - 2x 1 101 100. We have, x + 8 + 2 ( x + 7 ) + ( x + 1 ) - ( x + 7 ) = 4 …(i)
or (2 + 3 ) + 2
= …(i)
(2 + 3 ) x - 2 x 10
Let (x + 7) = λ …(ii)
 1  2
Q 2 - 3 = 2 + 3  or x = λ -7
  Then, Eq. (i) reduces to
2
- 2x
Let (2 + 3 ) x = λ, then Eq. (i) reduces to
( λ2 - 7 + 8 + 2 λ ) + ( λ2 - 7 + 1 - λ ) = 4
1 101
λ+ = ⇒ ( λ + 1 ) + ( λ2 - λ - 6 ) = 4
λ 10
⇒ 10 λ2 - 101 λ + 10 = 0 or ( λ2 - λ - 6 ) = 3 - λ
or ( λ - 10 ) (10 λ - 1 ) = 0
198 Textbook of Algebra

On squaring both sides, we get Case I If 0 < x 2 + 2 x - 3 < 1

λ2 - λ - 6 = 9 + λ2 - 6 λ ⇒ 4 < x 2 + 2x + 1 < 5
⇒ 5 λ = 15 ⇒ 4 < (x + 1)2 < 5
∴ λ =3
⇒ - 5 < ( x + 1 ) < - 2 or 2 < x + 1 < 5
⇒ (x + 7) = 3 [from Eq. (ii)]
⇒ - 5 - 1 < x < - 3 or 1 < x < 5 - 1
or x + 7 =9
∴ x =2 ∴ x ∈ ( - 5 - 1, - 3 ) ∪ (1, 5 - 1 ) …(ii)
and x = 2 satisfies Eq. (i). | x + 4| - | x |
Then, <1
Hence, x 1 = 2 (x - 1)
2 2
101. We have, 4 x + 2 (2a + 1 ) 2 x + 4a 2 - 3 > 0 …(i) - (x + 4) + x
Now, x < - 4, then <1
(x - 1)
x2
Putting t = 2 in the Eq. (i), we get
4
⇒ 1+ >0
t 2 + 2 (2a + 1 ) t + 4a 2 - 3 > 0 x -1
2
Let f (t ) = t 2 + 2 (2a + 1 ) t + 4a 2 - 3 [Q t > 0,∴2 x > 0] (x + 3)
⇒ >0
Q f (t ) > 0 (x - 1)
∴ x ∈ ( - ∞, - 3 ) ∪ (1, ∞ )
⇒ x ∈ ( - ∞, - 4 ) [Q x < - 4] …(iii)
x+4+x
- 4 ≤ x < 0, then -1 < 0
(x - 1)
(x + 5)
⇒ <0
T-axis (x - 1)
Consider the following cases: ∴ x ∈ ( - 5, 1 )
Case I Sum of the roots > 0 ⇒ x ∈ [ - 4, 0 ) [Q - 4 ≤ x < 0] …(iv)
( 2a + 1 ) (x + 4) - x
-2 >0 and x ≥ 0, then <1
1 (x - 1)
 1 4
∴ a ∈  - ∞, -  ⇒ 1- >0
 2 x -1
Case II Product of the roots > 0 (x - 5)
⇒ >0
4a 2 - 3 (x - 1)
⇒ >0
1 ∴ x ∈ ( - ∞, 1 ) ∪ ( 5, ∞ )
3 ⇒ x ∈ [ 0, 1 ) ∪ ( 5, ∞ ) [Q x ≥ 0] …(v)
or a2 >
4 From Eqs. (iii), (iv) and (v), we get
 3  3  x ∈ ( - ∞, 1 ) ∪ ( 5, ∞ ) …(vi)
or a ∈  - ∞, -  ∪  , ∞
 2   2  Now, common values in Eqs. (ii) and (iv) is
x ∈ ( - 5 - 1, - 3 ) …(vii)
Case III D<0
2
⇒ 4 ( 2a + 1 ) - 4 ⋅ 1 ⋅ ( 4a 2 - 3 ) < 0
2 Case II If x + 2x - 3 > 1
⇒ 4a + 4 < 0 ⇒ x 2 + 2x + 1 > 5 ⇒ (x + 1)2 > 5
∴ a < -1 ⇒ x+1<- 5
or a ∈ ( - ∞, - 1 ) or x+1> 5
Combining all cases, we get
∴ x ∈ ( - ∞, - 1 - 5 ) ∪ ( 5 - 1, ∞ ) …(viii)
 3 
a ∈ ( - ∞, - 1 ) ∪  , ∞ | x + 4| - | x |
 2  Then, >1
(x - 1)
 | x + 4| - | x | -4
102. We have, log x 2 + 2 x - 3   >0 Now, x < - 4, then >1
 x -1  x -1
The given inequation is valid for 4
| x + 4| - | x | ⇒ 1+ <0
>0 x -1
(x - 1) x+3
⇒ <0
and x 2 + 2 x - 3 > 0, ≠ 1 …(i) x -1
Now, consider the following cases: ∴ x ∈ ( - 3, 1 )
 Chap 02 Theory of Equations 199

which is false. [Qx < - 4] ⇒ x2 - x - 1 = 0

2x + 4 ∴ x =1
- 4 ≤ x < 0, then -1 > 0
(x - 1) 1± 5
∴ x=
(x + 5) 2
⇒ >0
(x - 1) 1± 5
∴ x= fail
∴ x ∈ ( - ∞, - 5 ) ∪ (1, ∞ ) 2
which is false. [Q- 4 ≤ x < 0 ] 1- 5
⇒ x= [Q x < 0]
4 2
and x ≥ 0, then >1
x -1 1- 5 1- 5
⇒ x= , 1, then y = ,0
4 2 2
⇒ 1- <0 1 - 5 1 - 5
x -1 ∴ Solutions are  ,  and (1, 0).
x -5  2 2 
⇒ <0
x -1 1 - 5 1 - 5
Hence, all pairs ( 0, 1 ), (1, 0 ) and  ,  are solutions
∴ x ∈(1, 5 ) …(ix)  2 2 
which is false. [Qx ≥ 0] of the original system of equations.
Now, common values in Eq. (viii) and (ix) is 104. Given, α , β and γ are the roots of the cubic equation
∴ x ∈ ( 5 - 1, 5 ) …(x) x 3 - px 2 + qx - r = 0 …(i)
Combining Eqs. (viii) and (x), we get ∴ α + β + γ = p, αβ + βγ + γα = q , αβγ = r
x ∈ ( - 5 - 1, - 3 ) ∪ ( 5 - 1, 5 ) 1
(i) Let y = βγ +
103. Let y ≥ 0, then | y | = y α
αβγ + 1 r + 1
and then given system reduces to ⇒ y = =
α α
| x 2 - 2x | + y = 1 …(i) r +1
2 ∴ α=
and x + y =1 …(ii) y
From Eqs. (i) and (ii), we get From Eq. (i), we get
x 2 = | x 2 - 2x | α 3 - pα 2 + qα - r = 0
⇒ x 2 = | x | | x - 2| (r + 1 ) 3 p (r + 1 ) 2 q (r + 1 )
⇒ - + -r = 0
y3 y2 y
Now, x < 0, 0 ≤ x < 2, x ≥ 2
x 2 = x ( x - 2 ), x 2 = - x ( x - 2 ) or ry 3 - q(r + 1 )y 2 + p (r + 1 ) 2y - (r + 1 ) 3 = 0

x 2 = x (x - 2) (ii) Let y = β + γ - α = (α + β + γ ) - 2α = p - 2 α
p -y
∴ x=0 ∴ α=
2
⇒ x (x + x - 2) = 0
From Eq. (i), we get
∴ x=0 α 3 - pα 2 + qα - r = 0
fail ∴ x = 0, 1 fail
(p - y ) 3 p (p - y )2 q (p - y )
⇒ x = 0, 1, then y = 1, 0 ⇒ - + -r = 0
∴Solutions are (0, 1) and (1, 0). 8 4 2
If y < 0 then | y | = - y and then given system reduces to or y 3 - py 2 + ( 4q - p 2 )y + ( 8r - 4 pq + p 3 ) = 0
| x 2 - 2x | + y = 1 …(iii) Also product of roots = - (8r - 4 pq + p 3 )
and x2 - y = 1 …(iv) 105. Assume α + iβ is a complex root of the given equation, then
From Eqs. (iii) and (iv), we get conjugate of this root, i.e. α - iβ is also root of this equation.
| x 2 - 2x | + x 2 = 2 On putting x = α + iβ and x = α - iβ in the given equation, we
get
⇒ | x | | x - 2| + x 2 = 2
2
A12 A22 A3 An2
Now, x < 0, 0 ≤ x < 2, x ≥ 2 + + +… +
α + iβ - a1 α + iβ - a 2 α + iβ - a 3 α + iβ - an
x (x - 2) + x 2 = 2
- x (x - 2) + x 2 = 2 = ab 2 + c 2(α + iβ ) + ac …(i)
2
x (x -2 ) + x = 2 A12 A22 A32 An2
and + + +… +
2 α - iβ - a1 α - iβ - a 2 α - iβ - a 3 α - iβ - an
⇒ 2x - 2x - 2 = 0 ⇒ 2x = 2
⇒ x2 - x - 1 = 0 = ab 2 + c 2(α - iβ ) + ac …(ii)
200 Textbook of Algebra

On subtracting Eq. (i) from Eq. (ii), we get ∴ (a + (a 2 + 1 ) ) 2 - 4 > 0

 A12 A22 A32
2iβ  2 2
+ 2 2
+ ⇒ (a + (a 2 + 1 ) + 2 ) (a + (a 2 + 1 ) - 2 ) > 0
( α - a1 ) + β (α - a2 ) + β (α - a 3 )2 + β2
Q a + (a 2 + 1 ) + 2 > 0
An2 
+… + + c2 = 0
2
(α - an ) + β 2
 ∴ a + (a 2 + 1 ) - 2 > 0
The expression in bracket ≠ 0 ⇒ (a 2 + 1 ) > 2 - a
∴ 2iβ = 0 ⇒ β = 0
a ≥ 2
Hence, all roots of the given equation are real.  2 2
106. Given equation is or a + 1 > (2 - a ) , if a < 2
x 4 + 2ax 3 + x 2 + 2ax + 1 = 0 …(i) a ≥ 2

2
⇒ or a > 3 , if a < 2
On dividing by x , we get 
2a 1 4
x 2 + 2ax + 1 + + 2 =0
x x a ≥ 2

 2 1  1 ⇒ or 3 < a < 2
⇒  x + 2  + 2a  x +  + 1 = 0  4
 x   x
 1
2
 1 3 3 
⇒  x +  - 2 + 2a  x +  + 1 = 0 Hence, < a < ∞ or a ∈  , ∞
   4 4 
x x
1
2
1 107. We have, [2 x ] - [ x + 1 ] = 2 x
 
or  x +  + 2a  x +  - 1 = 0
 x   x Since, LHS = Integer
1 ∴ RHS = 2x = Integer
or y 2 + 2ay - 1 = 0, where y = x + ⇒ [2 x ] = 2 x
x
Now, - [ x + 1] = 0
- 2a ± ( 4a 2 + 4 )
∴ y = = - a ± (a 2 + 1 ) ⇒ [ x + 1] = 0
2
or 0 ≤ x + 1 <1
Taking ‘+’ sign, we get
or -1 ≤ x < 0
y = - a + (a 2 + 1 ) or - 2 ≤ 2x < 0
1 ∴ 2 x = - 2, - 1
⇒ x+ = - a + (a 2 + 1 )
x 1
or x = - 1, -
or x 2 + (a - (a 2 + 1 ) ) x + 1 = 0 …(ii) 2
1
Taking ‘-’ sign, we get y = - a - (a 2 + 1 ) or x 1 = - 1, x 2 = -
2
1 108. We have, (a 2 + 3 ) x 2 + (a + 2 ) x - 6 < 0
⇒ x+ = - a - (a 2 + 1 )
x
or x 2 + (a + (a 2 + 1 ) ) x + 1 = 0 …(iii)
Let α , β be the roots of Eq. (ii) and γ, δ be the roots of Eq. (iii).
Then, α + β = (a 2 + 1 ) - a
X
and αβ = 1
and γ + δ = - (a 2 + 1 ) - a Let f ( x ) = (a 2 + 3 ) x 2 + (a + 2 ) x - 6
and γδ =1 Q (a 2 + 3 ) > 0 and f ( x ) < 0
Clearly, α + β > 0 and αβ > 0 ∴ D>0
∴Either α , β will be imaginary or both real and positive ⇒(a + 2 ) 2 + 24 (a 2 + 3 ) > 0 is true for all a ∈ R .
according to the Eq. (i) has atleast two distinct negative roots.
Therefore, both γ and δ must be negative. Therefore, 109. We have, 6 x 2 - 77[ x ] + 147 = 0
(i) γδ > 0, which is true as γ δ = 1. 6 x 2 + 147
⇒ = [x ]
(ii) γ+δ<0 77
⇒ - (a + (a 2 + 1 ) ) < 0 ⇒ (0.078) x 2 = [x ] - 1.9
⇒ a + (a 2 + 1 ) > 0, which is true for all a. Q (0.078 ) x 2 > 0 ⇒ x 2 > 0

∴ a ∈R ∴ [ x ] - 1.9 > 0
(iii) D>0 or [ x ] > 1.9
 Chap 02 Theory of Equations 201

∴ [ x ] = 2, 3, 4, 5,…
If [ x ] = 2, i. e. 2 ≤ x < 3
2 - 1 .9
Then, x2 = = 1. 28
0.078 α β
X
γ δ
∴ x = 1.13 [fail]
If [ x ] = 3, i. e. 3 ≤ x < 4
Let f ( x ) = x 2 - 2(a + 1 ) x + a (a - 1 ), thus the following
3 - 1.9
Then, x2 = = 14.1 conditions hold good:
0.078
Consider the following cases:
∴ x = 3.75 [true]
Case I D>0
If [ x ] = 4, i.e. 4 ≤ x < 5 2
4 - 1.9 ⇒ 4 (a + 1 ) - 4a (a - 1 ) > 0
Then, x2 = = 26.9
0.078 ⇒ 3a + 1 > 0
∴ x = 5.18 [fail] 1
∴ a>-
If [ x ] = 5, i.e. 5 ≤ x < 6 3
5 - 1.9 Case II f (α ) < 0
Then, x2 = = 39.7
0.078 ⇒ f (1 + a ) < 0
∴ x = 6.3 [fail] ⇒ (1 + a ) 2 - 2 (1 + a ) (1 + a ) + a (a - 1 ) < 0
If [ x ] = 6, i. e. 6 ≤ x < 7 ⇒ - (1 + a ) 2 + a (a - 1 ) < 0
6 - 1.9 4.1
Then, x2 = = = 52.56 ⇒ - 3a - 1 < 0
0.078 0.078 1
∴ x = 7.25 [fail] ⇒ a>-
3
If [ x ] = 7, i. e. 7 ≤ x < 8 Case III f (s ) = 0
7 - 1.9 5.1
Then, x2 = = = 65.38 ⇒ f (1 - a ) < 0
0.078 0.078
⇒ (1 - a ) 2 - 2 (a + 1 ) (1 - a ) + a (a - 1 ) < 0
∴ x = 8.08 [fail]
⇒ ( 4a + 1 ) (a - 1 ) < 0
If [ x ] = 8, i. e. 8 ≤ x < 9
1
8 - 1.9 6.1 ∴ - <a <1
Then, x2 = = = 78.2 4
0.078 0.078
Combining all cases we get
 1 
∴ x = 8.8 [true] a ∈  - , 1
 4 
If [ x ] = 9, i. e. 9 ≤ x < 10
9 - 1.9 7.1 111. pr = ( - p ) (- r )
Then, x2 = = = 91.03
0.078 0.078 = ( α + β + γ + δ ) ( αβγ + αβδ + γδα + γδβ )
∴ x = 9.5 [true] = α 2 βγ + α 2 βδ + α 2 γδ + αβγδ + β 2 γα
If [ x ] = 10, i. e.10 ≤ x < 11
+ β 2 αδ + αβγδ + β 2 γδ + γ 2 αβ + αβγδ
10 - 1.9 8.1
Then, x2 = = = 103.8
0.078 0.078 + γ 2 δα + γ 2δβ + αβγδ + αβδ 2 + γαδ 2 + γβδ 2
∴ x = 10.2 [true] Q AM ≥ GM
If [ x ] = 11, i.e. 11 ≤ x < 12 pr
⇒ ≥ (α 16 β16 γ16 δ16 )1/6 = α β γδ = 5
11 - 1.9 16
Then, x2 =
0.078 pr
⇒ ≥5
9.1 16
= = 116.7
0.078 or pr ≥ 80
∴ x = 10.8 [fail] ∴ Minimum value of pr is 80.
Other values are fail. 112. (α 2 + β 2 ) 2 = (α + β ) (α 3 + β 3 )
Hence, number of solutions is four.
⇒ {(α + β ) 2 - 2 αβ } 2 = (α + β ) {(α + β) 3 - 3 αβ (α + β)}
110. Since, the given equation is 2
 b 2 2c   b  - b
3
3bc 
x 2 - 2x - a 2 + 1 = 0 ⇒  2 -  = -   3 + 2 
a a  a  a a 
⇒ (x - 1)2 = a 2
2
∴ x - 1 ≠ a or x = 1 ± a  b 2 - 2ac  3
 - b   - b + 3abc 
⇒  2
 =   
⇒ α = 1 + a and β = 1 - a  a   a   a3 
202 Textbook of Algebra

⇒ 4a 2c 2 = acb 2 Combining all cases, we get

⇒ 2
ac (b - 4ac ) = 0 k ∈ ( -∞,4 )
117. We have, a + b = 10c, ab = - 11d
As a≠0
⇒ c∆ = 0 and c + d = 10a, cd = - 11b
113. Let P ( x ) = bx 2 + ax + c ∴ a + b + c + d = 10 (a + c )
and abcd = 121 bd
As P (0) = 0
⇒ b + d = 9 (a + c )
⇒ c=0
and ac = 121
As P (1 ) = 1
Next, a 2 - 10 ac - 11d = 0
⇒ a + b =1
P ( x ) = ax + (1 - a ) x 2 and c 2 - 10ac - 11b = 0
Now, P ′ ( x ) = a + 2 (1 - a ) x ⇒ a 2 + c 2 - 20 ac - 11 (b + d ) = 0
As P ′ ( x ) > 0 for x ∈( 0, 1 ) ⇒ (a + c ) 2 - 22 × 121 - 99 (a + c ) = 0
Only option (d) satisfies above condition. ⇒ a + c = 121 or - 22
114. Let the roots are α and α + 1, where α ∈ I . If a + c = - 22 ⇒a = c , rejecting these values, we have
Then, sum of the roots = 2 α + 1 = b a + c = 121
Product of the roots = α (α + 1 ) = c ∴ a + b + c + d = 10 (a + c ) = 1210
118. D≥0
Now, b 2 - 4c = ( 2α + 1 ) 2 - 4α (α + 1 ) 4 (a + b + c ) 2 - 12 λ (ab + bc + ca ) ≥ 0
2 2
= 4α + 1 + 4α - 4α - 4α = 1 (a 2 + b 2 + c 2 ) - (3 λ - 2 ) (ab + bc + ca ) ≥ 0
2
∴ b - 4c = 1 (a 2 + b 2 + c 2 )
n n- 1 ∴ (3 λ - 2 ) ≤
115. Let f ( x ) = an x + an - 1 x +… + a1 x, (ab + bc + ca )
f ( 0 ) = 0;f (α ) = 0 Since, |a - b | < c
⇒ f ′( x ) = 0 has atleast one root between ( 0, α ). ⇒ a + b - 2ab < c 2
2 2
(i)
i.e. Equation |b - c | < a
nan xn - 1 + (n - 1 ) an - 1xn - 2 +… + a1 = 0 ⇒ b + c 2 - 2bc < a 2
2
…(ii)
has a positive root smaller than α. | c - a| < b
116. Let f ( x ) = x 2 - 2kx + k 2 + k - 5 ⇒ c + a - 2ca < b 2
2 2
…(iii)
Consider the following cases: From Eqs. (i), (ii) and (iii), we get
a2 + b2 + c2
<2 …(iv)
ab + bc + ca
f(5) From Eqs. (i) and (iv), we get
4
α X 3λ - 2 < 2 ⇒ λ <
P S 3
119. Q x 2 - 2mx + m 2 - 1 = 0
Case I D≥0 ⇒ (x - m )2 = 1
2 2
⇒ 4k - 4 .1(k + k - 5 ) ≥ 0 ∴ x - m = ± 1 or x = m - 1, m + 1
⇒ -4(k - 5 ) ≥ 0 According to the question,
⇒ k -5 ≤ 0 m - 1 > - 2, m + 1 > - 2
⇒ k ≤ 5 or k ∈ ( -∞, 5 ] ⇒ m > - 1, m > - 3
Case II x-Coordinate of vertex x < 5 Then, m > -1 …(i)
2k and m - 1 < 4, m + 1 < 4
⇒ <5
2 ⇒ m < 5, m < 3 and m < 3 …(ii)
⇒ k < 5 or k ∈ ( -∞, 5 ) From Eqs. (i) and (ii), we get - 1 < m < 3
Case III f (5 ) > 0 120. x 2 + px + q = 0
⇒ 25 - 10k + k 2 + k - 5 > 0 Sum of the roots = tan 30 °+ tan 15 °= - p
2 Product of the roots = tan 30 °⋅ tan 15 °= q
⇒ k - 9k + 20 > 0
tan 30 °+ tan 15 °
⇒ (k - 4 )(k - 5 ) > 0 or k ∈ ( -∞,4 ) ∪ (5, ∞ ) tan 45 °= tan (30 °+ 15 °) =
1 - tan 30 °⋅ tan 15 °
 Chap 02 Theory of Equations 203

-p 125. Let f ( x ) = x 7 + 14x 5 + 16x 3 + 30x - 560

⇒ 1= ⇒ - p =1 -q
1 -q
∴ f ′( x ) = 7 x 6 + 70 x 4 + 48 x 2 + 30 > 0, ∀ x ∈ R
⇒ q- p =1
∴ 2 + q - p =3 ⇒ f ( x ) is an increasing function, for all x ∈ R
121. The equation x 2 - px + r = 0 has roots (α , β) and the equation Hence, number of real solutions is 1.
α  126. Let f ( x ) = x 3 - px + q
x 2 - qx + r has roots  , 2β .
2 
∴ f ′( x ) = 3 x 2 - p
α
⇒ r = αβ and α + β = p and + 2β = q ⇒ f ′′( x ) = 6 x
2
2q - p 2 (2 p - q )
⇒ β= and α = p
3 3 √3
2
⇒ αβ = r = (2q - p ) (2 p - q ) –
p
9 √3
122. α + β = -a
| α - β | < 5 ⇒ (α - β ) 2 < 5 For maxima or minima, f ′( x ) = 0
⇒ a 2 - 4 < 5 ⇒ a ∈ ( - 3, 3 ) ∴ x=±
p
3
123. Suppose roots are imaginary, then β = α
 p  p
1 ⇒ f ′′   =6   > 0
and =α  3   3
β
1  p p
⇒ β=- [not possible] and f ′′  -  = -6 <0
β  3 3
⇒ Roots are real ⇒( p 2 - q ) (b 2 - ac ) ≥ 0 p
Hence, given cubic minima at x = and maxima at
⇒ Statement -1 is true. 3
2b 1 p
- =α + x=- .
a β 3
α c 127. Let f ( x ) = x 2 - 8kx + 16 (k 2 - k + 1)
and = , α + β = - 2 p, αβ = q
β a
If β = 1, then α = q
⇒ c = qa [not possible]
- 2b
Also, α+1=
a
X
- 2b 4
⇒ - 2p =
a ∴ D>0
⇒ b = ap [not possible] 2 2
⇒ 64k - 4 ⋅ 16 (k - k + 1 ) > 0
⇒ Statement -2 is true but it is not the correct explanation of ⇒ k >1 …(i)
Statement-1. -b 8k
124. Let α,4β be roots of x 2 - 6x + a = 0 and α , 3 β be the roots of ⇒ >4 ⇒ >4
2a 2
x 2 - cx + 6 = 0. ⇒ k >1 …(ii)
Then, α + 4 β = 6 and 4αβ = a …(i) and f (4) ≥ 0
α + 3 β = c and 3α β = 6 …(ii) ⇒ 16 - 32k + 16 (k 2 - k + 1 ) ≥ 0
From Eqs. (i) and (ii), we get ⇒ k 2 - 3k + 2 ≥ 0
a = 8, αβ = 2
⇒ (k - 1 ) (k - 2 ) ≥ 0
Now, first equation becomes ⇒ k ≤ 1 or k ≥ 2 …(iii)
x 2 - 6x + 8 = 0 From Eqs. (i), (ii) and (iii), we get
⇒ x = 2, 4 k ≥2
If α = 2, 4 β = 4, then 3 β = 3 kmin = 2
3 128. Since, roots of bx 2 + cx + a = 0 are imaginary.
If α = 4, 4 β = 2, then 3 β = [non-integer]
2 ∴ c 2- 4ab < 0
∴Common root is x = 2. ⇒ - c 2 > - 4ab …(i)
204 Textbook of Algebra

Let f ( x ) = 3b 2x 2 + 6bcx + 2c 2 132. Let α be the common root.

Then, α 2 + bα - 1 = 0 and α 2 + α + b = 0
Since, 3b 2 > 0
2
1 b b -1 -1 1
and D = (6bc ) 2 - 4 (3b 2 ) (2c 2 ) = 12b 2c 2 ⇒ × =
1 1 1 b b 1
D 12b 2c 2
∴ Minimum value of f (x ) = - =- = -c 2 > -4ab ⇒ (1 - b ) (b + 1 ) = ( - 1 - b ) 2
2
4a 4 (3b 2 )
⇒ b 3 + 3b = 0
α β α 2 + β 2 (α + β ) 2 - 2 αβ
129. + = = …(i) ∴ b = 0, i 3, - i 3 , where i = - 1.
β α αβ αβ
and given, 3 3
α + β = q, α + β = - p 133. Let f ( x ) = x 4 - 4 x 3 + 12 x 2 + x - 1
⇒ (α + β ) 3 - 3 αβ (α + β ) = q ∴ f ′( x ) = 4 x 3 - 12 x 2 + 24 x + 1
⇒ - p 3 + 3 pαβ = q ⇒ f ′′( x ) = 12 x 2 - 24 x + 24
q + p3 = 12 ( x 2 - 2 x + 2 )
or αβ =
3p = 12 [( x - 1 ) 2 + 1 ] > 0
∴ From Eq. (i), we get
2 (q + p 3 ) i.e. f ′′( x ) has no real roots.
p2 - Hence, f ( x ) has maximum two distinct real roots, where
α β 3p p 3 - 2q
+ = 3
= f ( 0 ) = - 1.
β α (q + p ) (q + p 3 )
134. Given, p ( x ) = f ( x ) - g ( x )
3p
⇒ p( x ) = (a - a1 ) x 2 + (b - b1 ) x + (c - c1 )
α β
and product of the roots = ⋅ = 1 It is clear that p( x ) = 0 has both equal roots - 1, then
β α
(b - b1 )
 p 3 - 2q  -1 -1 = -
∴ Required equation is x 2 -   x+1=0 (a - a1 )
 q + p3 
c - c1
and - 1 × -1 =
or (q + p 3 ) x 2 - ( p 3 - 2q ) x + (q + p 3 ) = 0 a - a1
130. Since, f ′ ( x ) = 12 x 2 + 6 x + 2 ⇒ b - b1 = 2 (a - a1 ) and c - c1 = (a - a1 ) …(i)
Here, D = 6 2 - 4 ⋅ 12 ⋅ 2 = 36 - 96 = - 60 < 0 Also given, p( -2 ) = 2
∴ f ′( x ) > 0, ∀ x ∈ R ⇒ 4 (a - a1 ) - 2 (b - b1 ) + (c - c1 ) = 2 …(ii)
⇒ Only one real root for f ( x ) = 0 From Eqs. (i) and (ii), we get
Also, f ( 0 ) = 1, f ( - 1 ) = - 2 4 (a - a1 ) - 4 (a - a1 ) + (a - a1 ) = 2
⇒ Root must lie in ( - 1, 0 ). ∴ (a - a1 ) = 2 …(iii)
 1 1 ⇒ b - b1 = 4 and c - c1 = 2 [from Eq. (i)] …(iv)
Taking average of 0 and ( - 1 ), f  -  = Now, p(2 ) = 4 (a - a1 ) + 2 (b - b1 ) + (c - c1 )
 2 4
= 8 + 8 + 2 = 18 [from Eqs. (iii) and (iv)]
 1
⇒ Root must lie in  - 1, -  . 135. Let the quadratic equation be
 2
ax 2 + bx + c = 0
 3 1
Similarly, f  -  =- Sachin made a mistake in writing down constant term.
 4  2
∴ Sum of the roots is correct.
 3 1
⇒ Root must lie in  - , -  . i.e. α + β =7
 4 2
Rahul made a mistake in writing down coefficient of x .
131. Qα 2 - 6 α - 2 = 0 ⇒ α 2 - 2 = 6 α …(i)
∴ Product of the roots is correct.
2 2
and β - 6β - 2 = 0 ⇒ β - 2 = 6β …(ii) i.e. αβ = 6
10 10 8 8
a10 - 2a 8 ( α - β ) - 2 (α - β ) ⇒ Correct quadratic equation is
∴ =
2a 9 2 (α 9 - β9 ) x 2 - (α + β ) x + αβ = 0
α 8( α 2 - 2 ) - β 8 ( β 2 - 2 ) ⇒ x 2 - 7x + 6 = 0
=
2 (α 9 - β9 ) ⇒ ( x - 6 )( x - 1 ) = 0 ⇒ x = 6,1
α 8 ⋅ 6α - β8 ⋅ 6β Hence, correct roots are 1 and 6.
= [from Eqs. (i) and (ii)]
2 (α 9 - β9 )
6 (α 9 - β9 )
136. Let a + 1 = h 6
= =3
2 (α 9 - β9 ) ∴ (h 2 - 1 ) x 2 + (h 3 - 1 ) x + (h - 1 ) = 0
 Chap 02 Theory of Equations 205

 h 2 - 1 2  h 3 - 1 q
⇒  x +   x+1=0 -
 h -1  h -1 p
⇒ =4
r
As a Æ 0, then h Æ 1
p
 h 2 - 1 2  h 3 - 1
lim   x + lim   x+1=0 ⇒ q = - 4r ... (i)
h Æ 1 h - 1  h Æ 1 h - 1 
Also, given p, q, r are in AP.
⇒ 2x 2 + 3x + 1 = 0 ∴ 2q = p + r
2
⇒ 2x + 2x + x + 1 = 0 ⇒ p = - 9r [from Eq. (i)] …(ii)
⇒ (2 x + 1 ) ( x + 1 ) = 0 D Q for ax 2 + bx + c = 0, α - β = D 
1 Now, | α - β | =
∴ x = - 1 and x = - | a|  a 
2
137. Let e sin x = t …(i) (q 2 - 4 pr )
=
Then, the given equation can be written as | p|
1 (16r 2 + 36r 2
t - - 4 = 0 ⇒ t 2 - 4t - 1 = 0 = =
52 | r |
[from Eqs. (i) and (ii)]
t 9 |r | 9 |r |
4 ± (16 + 4 )
∴ t= 2 13
2 =
9
⇒ e sin x = (2 + 5 ) [Qe sin x > 0,∴taking + ve sign]
141. f ( x ) = x 5 - 5x and g( x ) = - a
⇒ sin x = loge ( 2 + 5 ) …(ii)
∴ f ′( x ) = 5 x 4 - 5
Q (2 + 5 ) > e [Qe = 2.71828… ]
⇒ loge ( 2 + 5 ) > 1 …(iii) 4
From Eqs. (ii) and (iii), we get
sin x > 1 [which is impossible]
Hence, no real root exists. –1 1
138. Given equations are
–4
ax 2 + bx + c = 0 …(i)
and x 2 + 2x + 3 = 0 …(ii) = 5 (x 2 + 1) (x - 1) (x + 1)
Clearly, roots of Eq. (ii) are imaginary, since Eqs. (i) and (ii) Clearly, f ( x ) = g( x ) has one real root, if a > 4 and three real
have a common root, therefore common root must be roots, if | a| < 4.
imaginary and hence both roots will be common. Therefore,
Eqs. (i) and (ii) are identical. 142. Since, b = 0 for p ( x ) = ax 2 + bx + c, as roots are pure
a b c imaginary.
∴ = = or a : b : c = 1 : 2 : 3
1 2 3 (- c ± i c )
⇒x = ± , which are clearly neither pure real nor
139. Q x - [ x ] = { x } [fractional part of x ] a
pure imaginary, as c ≠ 0.
For no integral solution, { x } ≠ 0
∴ a≠ 0 …(i) 143. Qαx 2 - x + α = 0 has distinct real roots.
The given equation can be written as ∴ D>0
3 { x } 2 - 2{ x } - a 2 = 0 2  1 1
⇒ 1 - 4α > 0 ⇒ α ∈  - ,  ...(i)
 2 2
2 2
2 ± ( 4 + 12a ) 1 + (1 + 3a )
⇒ {x } = = [Q 0 < { x } < 1 ] Also, | x1 - x 2 | < 1 ⇒ | x1 - x 2 | 2 < 1
6 3
D 1 - 4α 2 1
1 + (1 + 3a 2 ) ⇒ <1 ⇒ < 1 ⇒α 2 >
⇒ 0< < 1 ⇒ (1 + 3a 2 ) < 2 a 2
α2 5
3
⇒ a2 < 1 ⇒ - 1 < a < 1 …(ii)  1   1 
⇒ α ∈  -∞,-  ∪  , ∞ ...(ii)
 5  5 
From Eqs. (i) and (ii), we get
a ∈ ( - 1, 0 ) ∪ ( 0, 1 ) From Eqs. (i) and (ii), we get
1 1 α+β  1 1   1 1
140. Q + = 4 ⇒ =4 S =  - ,-  ∪  , 
α β αβ  2 5   5 2
206 Textbook of Algebra

2
144. ( x 2 - 5x + 5) x + 4 x - 60
=1 Q α 1, β1 are roots of x 2 - 2 x sec θ + 1 = 0 and α 1 > β1
Case I ∴ α 1 = sec θ - tan θ and β1 = sec θ + tan θ
x 2 - 5 x + 5 = 1 and x 2 + 4 x - 60 can be any real number ⇒ α 2, β 2 are roots of x 2 + 2 x tan θ - 1 = 0
⇒ x = 1, 4 and α 2 > β2
Case II
∴ α 2 = - tan θ + sec θ
x 2 - 5 x + 5 = - 1 and x 2 + 4 x - 60 has to be an even number
and β 2 = - tan θ - sec θ
⇒ x = 2, 3
For x = 3, x 2 + 4 x - 60 is odd, ∴ x ≠ 3 Hence, α 1 + β 2 = - 2 tan θ

Hence, x =2 146. Q x ( x + 1) + ( x + 1) ( x + 2) + .... + ( x + n - 1) ( x + n ) = 10n

Case III x 2 - 5 x + 5 can be any real number and
⇒ nx 2 + x (1 + 3 + 5 + K + (2n - 1 )) + (1 . 2 + 2 . 3
x 2 + 4 x - 60 = 0
+ ...+ (n - 1 ) . n ) = 10n
⇒ x = - 10, 6 1
⇒ Sum of all values of x = 1 + 4 + 2 - 10 + 6 = 3 or nx 2 + n 2x + (n - 1 ) n (n + 1 ) = 10n
3
145. Q x 2 - 2x sec θ a + 1 = 0 ⇒ x = sec θ ± tan θ
or 3 x 2 + 3nx + (n 2 - 1 ) = 30 (Q n ≠ 0)
π π
and - <θ < - or 3 x 2 + 3nx + (n 2 - 31 ) = 0
6 12
 π  π Q |α -β| =1
⇒ sec  -  > sec θ > sec  - 
 6  12
or (α - β ) 2 = 1
 π  π
or sec   > sec θ > sec   D
 6  12 or =1
a2
 π  π
and tan  -  < tan θ < tan  -  or D = a2
 6  12
 π  π or 9n 2 - 12 . (n 2 - 31 ) = 9
⇒ - tan   < tan θ < - tan  
 6  12
or n 2 = 121
 π  π
or tan   > - tan θ > tan   ∴ n = 11
 6  12