Posted on - 22-05-2017

IIT JEE

The set of
values of p for which the roots of the equation

3x^{2} +2x +p(p-1) = 0 are of opposite sign is

(A) (-¥, 0 )

(B) (0, 1)

(C) (1, ¥)

(D) (0, ¥)

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.

If a_{1},
a_{2}, a_{3} (a_{1} > 0) are in G. P. with common
ratio r, then the value of r, for which the inequality 9a_{1} + 5 a_{3}
> 14 a_{2} holds, can not lie in the interval.

(A) [1, ¥)

(B) [1, 9/5]

(C) [4/5, 1]

(D) [5/ 9, 1]

Since a_{1}, a_{2}, a_{3}
(a_{1} > 0) are in G.P. .

So,
a_{2} = a_{1} r ; a_{3} = a_{1} r^{2}

Given inequality

9a_{1}
+ 5 a_{3} > 14 a_{2}

9
a_{1}+ 5 a_{1} r^{2} > 14 a_{1}r

5r^{2}
– 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.

Consider
the equation x^{2} +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

x^{2} +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 +l^{2} = l( l+1)
= even number

&⇒ n can be 2, 6, 12, 20, 30, 42, 56,
72, 90. .

Hence (C) is the correct answer.

The least value of
the expression x^{2} + 4y^{2} + 3z^{2} – 2x – 12y – 6z
+14 is

(A) 0

(B) 1

(C) no least value

(D) none of these.

Let f(x, y, z) = x^{2} + 4y^{2}+
3z^{2} – 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.

If the roots of the equation x^{2}
–px + q = 0 differ by unity then

(A) p^{2 }= 1- 4q

(B) p^{2 }= 1+ 4q

(C) q^{2 }= 1- 4p

(D) q^{2 }= 1+ 4p

Suppose the equation x^{2} –
px + q = 0 has the roots a + 1

and a then a
+1+a = p

&⇒ 2a = p –1 .
. . . (1) .

and (a+1) a
= q

&⇒ a^{2 }+ a = q .
. . .. (2).

Putiting the value of a from (1) in (2) , we get

_{}

&⇒
(p-1)^{2} + 2(p-1) = 4q

&⇒
p^{2} –1 =4q

&⇒ p^{2}= 4q +1 .

Alternative Solution:

Let a and b
be the roots. |a –b| =1

&⇒
(a
+ b)^{2} - 4ab
=1 .

&⇒
p^{2} – 4q = 1 , or p^{2} = 1+ 4q

Hence (B) is the correct answer.

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

We know that ax^{2} +bx
+c ≥ 0 if a > 0 and b^{2} –4ac £
0. .

So, mx - 1 +_{}
≥
0

&⇒ _{}

&⇒
mx^{2} -x +1 ≥ 0 as x > 0

Now, mx^{2} –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.

If both the roots of the equation x^{2}
– (p – 4) x + 2 e^{2lnp} – 4 =0 are negative then p belongs to

(A) _{}

(B) _{}

(C) _{}

(D) _{}

Both roots negative

&⇒
sum of roots < 0 and product of roots > 0

&⇒ p – 4 < 0 and _{}- 4
> 0

&⇒ p < 4 and p^{2} > 2

&⇒ p ∈ (_{}, 4) .

Hence (B) is the correct answer.

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.

Let f(x) = ax^{2} + bx + c be
a quadratic expression such that f(x) > 0 for all x∈R.
Then a > 0 and b^{2} – 4ac < 0.

Now, g(x) = f(x) + f¢(x) + f¢¢(x)

&⇒ g(x) = ax^{2} + x (b + 2a) +
(b + 2a + c)

Discriminant of g(x) is

D
= (b + 2a)^{2} – 4a (b + 2a + c)

=
b^{2} – 4a^{2} – 4ac

=
(b^{2} – 4ac) – 4a^{2} < 0 (as b^{2} – 4ac < 0)

Thus D < 0 and a > 0

Therefore, g(x) > 0 for all x∈R .

Hence (B) is the correct answer.

The value of ‘p’ for
which the sum of the square of the roots of

2x^{2} - 2(p -2)x - p -1= 0 is least, is

(A) 1

(B) 3/2

(C) 2

(D) –1

Let a and b be two roots of the equation

2x^{2}
– 2(p – 2) x – p – 1 = 0. Then.

a + b = p –2 and ab = _{}

Let
S = a^{2} + b^{2} = (a + b)^{2} –
2 a b

=
(p – 2)^{2} + p + 1 = p^{2} – 3p + 5 = ( p – 3/2) ^{2}
+ _{}

which is least when p = 3/2 . .

Hence (B) is the correct answer.

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

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. .