Q1.
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
&⇒ 
&⇒ p(p-1) <
0
&⇒ p∈ (0, 1).
Hence
(B) is the correct answer.
Q2.
If a1,
a2, a3 (a1 > 0) are in G. P. with common
ratio r, then the value of r, for which the inequality 9a1 + 5 a3
> 14 a2 holds, can not lie in the interval.
(A) [1, ¥)
(B) [1, 9/5]
(C) [4/5, 1]
(D) [5/ 9, 1]
Solution:
Since a1, a2, a3
(a1 > 0) are in G.P. .
So,
a2 = a1 r ; a3 = a1 r2
Given
inequality
9a1
+ 5 a3 > 14 a2
9
a1+ 5 a1 r2 > 14 a1r
5r2
– 14 r +9 > 0
&⇒ (r – 1) ( r – 9/5) > 0
r
> 9/ 5 and r < 1
r
∈
[1, 9/ 5]. .
Hence
(B) is the correct answer.
Q3.
Consider
the equation x2 +x – n = 0, where n is an integer lying between
1 to 100. Total number of different values of ‘n’ so that the equation
has integral roots is .
(A) 6
(B) 4
(C) 9
(D)
None of these
Solution:
x2 +x –n = 0,
discriminent = 1+ 4n = odd number = D(say)
Now
given equation would have a integral soluton if D is a perfect square
.
Let
D = ( 2l +1)2
&⇒ n = l +l2 = l( l+1)
= even number
&⇒ n can be 2, 6, 12, 20, 30, 42, 56,
72, 90. .
Hence
(C) is the correct answer.
Q4.
The least value of
the expression x2 + 4y2 + 3z2 – 2x – 12y – 6z
+14 is
(A) 0
(B) 1
(C) no least
value
(D) none of these.
Solution:
Let f(x, y, z) = x2 + 4y2+
3z2 – 2x –12y – 6z + 14
=
( x- 1)2 + (2y – 3)2 + 3( z- 1)2 + 1
For
least value of f ( x, y, z)
x-1
= 0 ; 2y – 3 =0 and z – 1 =0
\ x = 1 ; y = 3/2 ; z = 1
Hence
least value of f(x, y, z) is f( 1, 3/2, 1) = 1 .
Hence
(B) is the correct answer.
Q5.
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 a + 1
and a then a
+1+a = p
&⇒ 2a = p –1 .
. . . (1) .
and (a+1) a
= q
&⇒ a2 + a = q .
. . .. (2).
Putiting the value of a
from (1) in (2) , we get
&⇒
(p-1)2 + 2(p-1) = 4q
&⇒
p2 –1 =4q
&⇒ p2= 4q +1 .
Alternative Solution:
Let a and b
be the roots. |a –b| =1
&⇒
(a
+ b)2 - 4ab
=1 .
&⇒
p2 – 4q = 1 , or p2 = 1+ 4q
Hence
(B) is the correct answer.
Q6.
If the expression 
is
non-negative for all positive real x , then the minimum value of m must
be
(A) –1/2
(B) 0
(C) 1/4
(D) 1/2
Solution:
We know that ax2 +bx
+c ≥ 0 if a > 0 and b2 –4ac £
0. .
So, mx - 1 +
≥
0
&⇒ 
&⇒
mx2 -x +1 ≥ 0 as x > 0
Now, mx2 –x +1 ≥
0 if m > 0 and 1-4m £ 0
Or if m > 0 and m ≥
1/4.
Thus , the minimum value of m
is 1/4.
Hence
(C) is the correct answer.
Q7.
If both the roots of the equation x2
– (p – 4) x + 2 e2lnp – 4 =0 are negative then p belongs to
(A) 
(B) 
(C) 
(D) 
Solution:
Both roots negative
&⇒
sum of roots < 0 and product of roots > 0
&⇒ p – 4 < 0 and 
- 4
> 0
&⇒ p < 4 and p2 > 2
&⇒ p ∈ (
, 4) .
Hence
(B) is the correct answer.
Q8.
Suppose that f(x) is a quadratic expression
positive for all real x.
If g(x) = f(x) + f '(x)+f"(x), then for any real x.
(A) g(x)<0,
(B) g(x)>0,
(C) g(x)=0,
(D) g(x)≥0.
Solution:
Let f(x) = ax2 + bx + c be
a quadratic expression such that f(x) > 0 for all x∈R.
Then a > 0 and b2 – 4ac < 0.
Now,
g(x) = f(x) + f¢(x) + f¢¢(x)
&⇒ g(x) = ax2 + x (b + 2a) +
(b + 2a + c)
Discriminant
of g(x) is
D
= (b + 2a)2 – 4a (b + 2a + c)
=
b2 – 4a2 – 4ac
=
(b2 – 4ac) – 4a2 < 0 (as b2 – 4ac < 0)
Thus
D < 0 and a > 0
Therefore,
g(x) > 0 for all x∈R .
Hence
(B) is the correct answer.
Q9.
The value of ‘p’ for
which the sum of the square of the roots of
2x2 - 2(p -2)x - p -1= 0 is least, is
(A) 1
(B) 3/2
(C) 2
(D) –1
Solution:
Let a and b
be two roots of the equation
2x2
– 2(p – 2) x – p – 1 = 0. Then.
a + b = p –2 and ab = 
Let
S = a2 + b2 = (a + b)2 –
2 a b
=
(p – 2)2 + p + 1 = p2 – 3p + 5 = ( p – 3/2) 2
+ 
which
is least when p = 3/2 . .
Hence
(B) is the correct answer.
Q10.
Let a,
b be the roots of
the equation ( x - a) (x-b) =c , c >
0. Then the roots of the equation (x-a)
(x-b) + 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 º (x-a)(x-b)
or (x-a)(x-b) +c º (x-a) (x-b).
This
shows that the roots of ( x-a ) ( x-b) + c = 0 are
a and b.
Hence
(C) is the correct answer. .
