Q1.
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.
Q2.
The set of values of ‘a’ for which the
equation x3 –3x +a has three distinct real roots , is
(A) ( -¥,
¥ )
(B) (-2, 2)
(C) ( -1, 1)
(D) none of these
Solution:
Let f(x) = x3 – 3x +a
f¢(x)
= 3x2 –3.
For
three distinct real roots (i) f¢(x) = 0 should have two
distinct real roots a and b
and
(ii) f(a) f(b) < 0
Here
a
= 1, b = -1 . Now f(a) f(b) < 0 .
&⇒ ( 1- 3+a) (-1+ 3+a) < 0
&⇒
( a – 2) ( a + 2) < 0
&⇒ -2 < a < 2.
Hence
(B) is the correct answer.
Q3.
Let
p(x) = 0 be a polynomial equation of least possible degree, with rational
coefficients, having 
as
one of its roots. Then the product of all the roots of p(x) = 0 is
(A) 7
(B) 49
(C) 56
(D) 63
Solution:
x
=
&⇒ x3 = 7 + 49 + 3
&⇒ x3 – 21 x – 56 = 0
&⇒
Product of root = 56 .
Hence
(C) is the correct answer.
Q4.
If a
and b are the roots of
the equation, 2x2 –3x –6 =0, then equation whose roots are a2+2, b2 +2 is
(A) 4x2+ 49x +118 = 0
(B) 4x2- 49x +118 = 0
(C) 4x2- 49x –118 = 0
(D) x2- 49x +118 = 0
Solution:
a
+ b = 3/2 , ab
= -6/2 = -3
S = a2 +b2 +4 = (a+b)2
-2ab +4 = 
P = a2 b2 +2 (a2 +b2) +4
= a2 b2 +4 +2 [(a
+b)2
-2ab ] = 
.
Therefore, the equation is x2
- 
&⇒
4x2 – 49x +118 = 0
Alternate:
Let y = x2 +2 , then 2x2
– 3x – 6 = 0
&⇒
(3x)2 = (2x2– 6)2
&⇒
[2 ( y – 2) – 6]2 = 9 ( y-2)
&⇒ 4 y2 – 49y +118 =0 . .
Hence
(B) is the correct answer.
Q5.
If a, b, c are in G. P , then the
equations ax2 +2bx +c = 0 and
dx2 +2ex +f = 0 have a common root if 
are
in
(A) A.P. .
(B) G.P.
(C) H.P. .
(D) none of these.
Solution:
a,b, c are in G. P
&⇒
b2 = ac .
Now the equation ax2
+2bx +c = 0 can be rewritten as
ax2 +2
x +c = 0
&⇒
(
x +
)2
= 0
&⇒ x = –
If the two given equations have a
common root, then this root must be - 
.
Thus d
&⇒
&⇒
are
in A. P.
Hence
(A) is the correct answer.
Q6.
Number of positive integers n for which
n2 + 96 is a perfect square is
(A) 4
(B) 8
(C) 12
(D) infinite
Solution:
Let m be a positive integer for
which n2 + 96 = m2
&⇒ m2 –n2 = 96
&⇒
(m+n)(m –n) = 96
&⇒ (m+n) {(m+ n) – 2n } = 96
&⇒ m + n and m –n must be both
even
96
= 2 ´ 48 or 4 ´ 24 or 6 ´
16 or 8 ´ 12
Number
of solution = 4 .
Hence
(A) is the correct answer.
Q7.
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 need not be an integer
(C) a+b is always an integer
(D) a and b are always integers.
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.
Q8.
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
Solution:
We have x3 – 2x2
+ 2x – 1 = 0
&⇒ (x – 1) (x2 – x + 1) = 0
&⇒ x = 1 or x = -w,
-w2
since
ax2 + bx + a = 0 and x3 – 2x2 + 2x –1 = 0 have
two roots in common, therefore –w and –w2 are common roots. (as either
both the roots of a quadratic equation are real or non-real). .
Now
- w is a root of ax2 + bx + a = 0
&⇒
a ( 1+ w2) - bw = 0
&⇒ a(-w) - bw
= 0 ( as 1 + w + w2 = 0)
&⇒ a + b = 0
Hence
(C) is the correct answer.
Q9.
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.
Q10.
If the roots of the equation x2
– 2ax + a2 +a -3 = 0 are less than 3 then
(A) a < 2
(B) 2 £
a £ 3
(C) 3 < a £ 4
(D) a> 4 .
Solution:
Since
roots are less than a real number, roots must be real
&⇒ 4a2 - 4 (a2
+ a –3 ) ≥ 0
&⇒ a £ 3 , …..
(1).
Let
f(x) = x2 – 2ax + a2 +a - 3. Since 3 lie outside the
roots, .
f(3)
> 0
&⇒ a< 2 or a> 3 . . . .(2).
Sum
of the roots must be less than 6
2a
< 6
&⇒ a < 3 . . . . (3).
From
(1), (2) and (3) we have a< 2.
Hence (A) is
the correct answer.
