1. Problem -1: The set of values of p for which the roots of the equation 3x2
+2x +p(p-1) = 0 are of
opposite sign is
(A) (-, 0 ) (B) (0, 1)
(C) (1, ) (D) (0, )
Solution: Since the roots of the given equation are of opposite sign, product of the roots < 0

  0
3
1
p
p


 p(p-1) < 0  p (0, 1).
Hence (B) is the correct answer.
Problem -2: The number of real roots of (6 –x)4
+ (8 –x)4
= 16 is
(A) 0 (B) 2
(C) 4 (D) none of these
Solution: Let y = 7 – x. Then the given equation becomes
(y + 1)4
+ (y –1)4
= 16  y4
+ 6y2
–7 = 0
 (y2
–1) (y2
+ 7) = 0  y2
–1 = 0 [ y2
+ 7  0]
 y =  1  7 –x =  1  x = 6, 8
Hence (B) is the correct answer.
Problem -3: Given that tan A and tan B are the roots of x2
–px + q = 0, then the value of
sin2
(A + B) is
(A)
 2
2
2
q
1
p
p


(B) 2
2
2
q
p
p

(C)
 2
2
2
q
1
p
q


(D)
 2
2
q
p
p

Solution: We have: tan A + tan B = p and tan A tan B = q.
 tan (A + B) =
q
1
p
B
tan
A
tan
1
B
tan
A
tan




Now, sin2
(A + B)
=
2
1
[1 –cos 2(A + B)] =
 
 










B
A
tan
1
B
A
tan
1
1
2
1
2
2
=
 
 
 
 
 2
2
2
2
2
2
2
2
2
q
1
p
p
q
1
p
1
q
1
p
B
A
tan
1
B
A
tan










Hence (A) is the correct answer.
Problem -4: If p, q  {1, 2, 3, 4}, the number of equations of the form px2
+ qx + 1 = 0 having real
roots is
(A) 15 (B) 9
(C) 7 (D) 8
Solution: For real roots, we must have
Disc  0  q2
-4p  0  q2
 4p
If p = 1, then q2
 4p  q2
 4  q = 2, 3, 4
If p = 2, then q2
 4p  q2
 8  q = 3, 4
If p = 3, then q2
 4p  q2
 12  q = 4
If p = 4, then q2
 4p  q2
 16  q = 4
Thus, we see that there are 7 cases.
Hence (C) is the correct answer.

2. Problem -5: The harmonic mean of the roots of the equation (4 + 2 )x2
– (4 + 5 )x+ (8 +2 5 )=0 is
(A) 2 (B) 4
(C) 7 (D) 8
Solution: Let ,  be the roots of the given equation. Then,
 +  =
2
4
5
4


and  =
2
4
5
2
8


Let H be the H.M. of  and . Then,
H =
5
4
5
4
16
2







= 4.
Hence (B) is the correct answer.
Problem -6: In a triangle PQR, R =
2

. If tan 





2
P
and tan 





2
Q
are the roots of the equation
ax2
+ bx + c (a  0). Then
(A) a + b = c (B) b + c = 0
(C) a + c = b (D) b = c
Solution: R =
2

 P + Q =
2


4
2
Q
2
P 


 tan 






2
Q
2
P
= tan
4


2
Q
tan
2
P
tan
1
2
Q
tan
2
P
tan


= 1
 1
a
c
1
a
b













a
c
2
Q
tan
2
P
tan
and
a
b
2
Q
tan
2
P
tan


a
b
a
c
1 

  a –c = -b  a + b = c.
Hence (A) is the correct answer.
Problem -7: If the roots of the equation x2
–2ax + a2
+ a –3 = 0 are real and less than 3, then
(A) a < 2 (B) 2  a  3
(C) 3 < a  4 (D) a > 4
Solution: Let f (x) = x2
–2ax + a2
+ a –3 = 0. Since f (x) has real roots both less than 3.
Therefore, Disc > 0 and f (3) > 0
 a2
–(a2
+ a –3) > 0 and a2
–5a + 6 > 0  a < 3 and (a –2) (a –3) > 0
 a < 3 and a < 2 or a > 3  a < 2.
Hence (A) is the correct answer.
Problem -8: If  and  are the roots of the equation, 2x2
–3x –6 =0, then equation whose roots are
2
+2, 2
+2 is
(A) 4x2
+ 49x +118 = 0 (B) 4x2
- 49x +118 = 0
(C) 4x2
- 49x –118 = 0 (D) x2
- 49x +118 = 0
Solution:  +  = 3/2 ,  = -6/2 = -3
S = 2
+2
+4 = (+)2
-2 +4 =
4
49

3. P = 2
2
+2 (2
+2
) +4 = 2
2
+4 +2 [( +)2
-2 ] =
4
118
.
Therefore, the equation is x2
- 0
4
118
x
4
49














 4x2
– 49x +118 = 0
Hence (B) is the correct answer.
Problem -9: If the roots of the equation x2
–px + q = 0 differ by unity then
(A) p2
= 1- 4q (B) p2
= 1+ 4q
(C) q2
= 1- 4p (D) q2
= 1+ 4p
Solution: Suppose the equation x2
– px + q = 0 has the roots  + 1
and  then  +1+ = p  2 = p –1 . . . . (1)
and (+1)  = q  2
+  = q . . . .. (2)
Putting the value of  from (1) in (2) , we get
  q
2
1
p
4
1
p
2




 (p-1)2
+ 2(p-1) = 4q  p2
–1 =4q  p2
= 4q +1 .
Hence (B) is the correct answer.
Problem -10: If p and q are the roots of the equation x2
+px +q = 0 , then
(A) p =1, q = -2 (B) p =0 , q = 1
(C) p = –2, q = 0 (D) p = –2, q = 1
Solution: Since p and q are roots of the equation x2
+px + q = 0 ,
p + q = - p and pq = q
pq = q  q = 0 or p = 1
if q = 0, then p = 0 and if p =1, then q = –2
Hence (A) is the correct answer.
Problem -11: Let ,  be the roots of the equation (x - a) (x - b) = c , c  0. Then the roots of the
equation (x - )(x - ) + c = 0 are
(A) a, c (B) b, c
(C) a, b (D) a+c , b+c
Solution: By given condition
(x-a) (x-b) – c  (x-)(x-) or (x-)(x-) +c  (x-a) (x-b).
This shows that the roots of ( x- ) ( x-) + c = 0 are a and b.
Hence (C) is the correct answer.
Problem -12 x4
–4x –1 = 0 has
(A) atmost one positive real root (B) atmost one negative real root
(C) atmost two real roots (D) none of these .
Solution: Let f (x) = x4
–4x –1
+ – –
 atmost one positive real root.
f (–x) = x4
+ 4x –1
+ + –
 atmost one negative real root.
 atmost two real root.
Hence (C) is the correct answer.
Problem -13 If x2
+ax +b is an integer for every integer x then
(A) ‘a’ is always an integer but ‘b’ need not be an integer
(B) ‘b’ is always an integer but ‘a’ need not be an integer
(C) a+b is always an integer
(D) a and b are always integers.

4. Solution: Let f(x) = x2
+ax +b
Clearly, f(0) = b  b is an integer .
Now f(1) = 1+ a+ b  a is an integer.
Hence (C) and (D) are the correct answers.
Problem -14 Sum of the real roots of the equation x2
+5|x| +6 = 0
(A) equals to 5 (B) equals to 10
(C) equals to –5 (D) does not exit.
Solution: Since x2
, 5|x| and 6 are positive so x2
+5|x| +6 = 0
does not have any real root . Therefore sum does not exist.
Hence (D) is the correct answer.
Problem -15 If c > 0 and 4a +c < 2b then ax2
–bx +c = 0 has a root in the interval
(A) (0, 2) (B) (2, 4)
(C) (0, 1) (D) (-2, 0)
Solution: Let f(x) = ax2
–bx +c
f(0) = c > 0 and f(2) = 4a –2b +c < 0 so that f(x) =0 has a root in the interval (0, 2).
Hence (A) is the correct answer.
Problem -16 The largest negative integer which satisfies
  
0
3
x
2
x
1
x 2




is
(A) - 4 (B) –3
(C) –1 (D) –2
Solution: By wavy curve method .
  
0
3
x
2
x
1
x2




 x  (-, -1)  (1, 2)  ( 3,  )
Therefore largest negative integer is -2 .
Hence (D) is the correct answer.
Problem -17 If x2
–4x +log1/2a = 0 does not have two distinct real roots, then maximum value of a
is
(A) 1/4 (B) 1/ 16
(C) –1/4 (D) none of these
Solution: Since x2
–4x +log1/2a = 0 does not have two distinct real roots,
discriminant  0
 16 – 4 log1/2 a  0  log1/2 a  4  a  1/16
Hence (B) is the correct answer.
1. If |x-2|+|x-9|=7, then the set values of x is
(A) {2, 9} (B) (2, 7)
(C) {2} (D) [2, 9]
1. D
2. If (m2
-3) x2
+ 3mx + 3m + 1= 0 has roots which are reciprocals of each other, then the value
of m equals to
(A) 4 (B) –3
(C) 2 (D) None of these
2. A
3. If ax2
+bx + 6 =0 does not have two distinct real roots, then the least value of 3a+b is
(A) 2 (B) –2
(C) 1 (D) -1
3. B

5. 4. If the quadratic equation x2
+x+a2
+b2
+c2
– ab – bc – ca = 0 has imaginary roots, then
(A) 2 ( - ) +(a - b)2
+(b - c)2
+ (c - a)2
> 0 (B) 2 ( - ) +(a - b)2
+(b - c)2
+ (c - a)2
< 0
(C) 2 ( - ) +(a - b)2
+(b - c)2
+ (c - a)2
= 0 (D) none of these .
4. A
5. The roots  and  of the quadratic equation ax2
+bx +c = 0 are real and of opposite
sign. Then the roots of the equation (x-)2
+ (x-)2
=0 are
(A) positive (B) negative
(C) Real and of opposite sign (D) imaginary
5. C
6. If p, q, r are real and p q, then the roots of the equation (p -q)x2
+5(p +q)x -2(p -q)=0 are
(A) real and equal (B) complex
(C) real and unequal (D) none of these.
6. C
8. The real roots of the equation |x|3
– 3x2
+3|x| - 2 =0 are
(A) 0, 2 (B)  1
(C)  2 (D) 1,2
8. C
9. If the equations x2
+ax+b=0 and x2
+bx+a=0 have exactly one common root, then the
numerical value of a + b is
(A) 1 (B) –1
(C) 0 (D) none of these
9. B
10. If r be the ratio of the roots of the equation ax2
+ bx + c = 0, then
r
)
1
r
( 2

is equal to
(A)
bc
a2
(B)
ac
b2
(C)
ab
c2
(D) none of these
10. B
11. The set of values of a for which the inequality x2
+ax+a2
+6a<0 is satisfied for all
x  (1, 2) lies in the interval
(A) (1, 2) (B) [1, 2]
(C) [-7, 4] (D) None of these
11. D
12. If the product of the roots of the equation x2
-3kx+2e2lnk
-1=0 is 7, then for real roots the value
of k is equal to
(A) 1 (B) 2
(C) 3 (D) 4
12. B
13. If 2 + i3 is a root of x2
+ px + q = 0, where p, q are real, then (p, q) is equal to
(A) (-4, 7) (B) (4, - 7)
(C) (-7 4) (D) (4, 7)
13. A
14. The equations ax2
+ bx + a = 0, x3
– 2x2
+ 2x – 1 = 0 have two roots in common. Then a +
b must be equal to
(A) 1 (B) –1
(C) 0 (D) none of these
15. C