Posted on - 24-05-2017

IIT JEE

If x^{2} - 4x +log_{1/2}a
= 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

Since
x^{2} - 4x +log_{1/2}a = 0 does not have two distinct
real roots,

discriminant £ 0

&⇒ 16 - 4 log_{1/2} a £
0

&⇒ log_{1/2} a ≥ 4

&⇒
a £ 1/16

Hence (B) is the correct answer.

The set of values of ‘a’ for which the
equation x^{3} –3x +a has three distinct real roots , is

(A) ( -¥, ¥ )

(B) (-2, 2)

(C) ( -1, 1)

(D) none of these

Let f(x) = x^{3} – 3x +a

f¢(x)
= 3x^{2} –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.

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

x
= _{}

&⇒ x^{3} = 7 + 49 + 3_{}

&⇒ x^{3} – 21 x – 56 = 0

&⇒
Product of root = 56 .

Hence (C) is the correct answer.

If a
and b are the roots of
the equation, 2x^{2} –3x –6 =0, then equation whose roots are a^{2}+2, b^{2} +2 is

(A) 4x^{2}+ 49x +118 = 0

(B) 4x^{2}- 49x +118 = 0

(C) 4x^{2}- 49x –118 = 0

(D) x^{2}- 49x +118 = 0

a + b = 3/2 , ab = -6/2 = -3

S = a^{2} +b^{2} +4 = (a+b)^{2}
-2ab +4 = _{}

P = a^{2} b^{2} +2 (a^{2} +b^{2}) +4

= a^{2} b^{2} +4 +2 [(a
+b)^{2}
-2ab ] = _{}.

Therefore, the equation is x^{2}
- _{}

&⇒
4x^{2} – 49x +118 = 0

Let y = x^{2} +2 , then 2x^{2}
– 3x – 6 = 0

&⇒
(3x)^{2} = (2x^{2}– 6)^{2}

&⇒
[2 ( y – 2) – 6]^{2} = 9 ( y-2)

&⇒ 4 y^{2} – 49y +118 =0 . .

Hence (B) is the correct answer.

If a, b, c are in G. P , then the
equations ax^{2} +2bx +c = 0 and

dx^{2} +2ex +f = 0 have a common root if _{}are
in

(A) A.P. .

(B) G.P.

(C) H.P. .

(D) none of these.

a,b, c are in G. P

&⇒
b^{2} = ac .

Now the equation ax^{2}
+2bx +c = 0 can be rewritten as

ax^{2} +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.

Number of positive integers n for which
n^{2} + 96 is a perfect square is

(A) 4

(B) 8

(C) 12

(D) infinite

Let m be a positive integer for
which n^{2} + 96 = m^{2}

&⇒ m^{2} –n^{2} = 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.

If x^{2} +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.

Let
f(x) = x^{2} +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.

The equations ax^{2}
+ bx + a = 0, x^{3} – 2x^{2} + 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

We have x^{3} – 2x^{2}
+ 2x – 1 = 0

&⇒ (x – 1) (x^{2} – x + 1) = 0

&⇒ x = 1 or x = -w,
-w^{2}

since
ax^{2} + bx + a = 0 and x^{3} – 2x^{2} + 2x –1 = 0 have
two roots in common, therefore –w and –w^{2} are common roots. (as either
both the roots of a quadratic equation are real or non-real). .

Now
- w is a root of ax^{2} + bx + a = 0

&⇒
a ( 1+ w^{2}) - bw = 0

&⇒ a(-w) - bw
= 0 ( as 1 + w + w^{2} = 0)

&⇒ a + b = 0

Hence (C) is the correct answer.

If p and q are the roots of the
equation x^{2} +px +q = 0 , then

(A) p =1, q = -2

(B) p =0 , q = 1

(C) p = –2, q = 0

(D) p = –2, q = 1

Since p and q are roots of
the equation x^{2} +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.

If the roots of the equation x^{2}
– 2ax + a^{2} +a -3 = 0 are less than 3 then

(A) a < 2

(B) 2 £ a £ 3

(C) 3 < a £ 4

(D) a> 4 .

Since roots are less than a real number, roots must be real

&⇒ 4a^{2} - 4 (a^{2}
+ a –3 ) ≥ 0

&⇒ a £ 3 , …..
(1).

Let
f(x) = x^{2} – 2ax + a^{2} +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.