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Quadratic Equations - Solved Objective Questions Part 2 for Conceptual Clarity

Posted on - 24-05-2017

JEE Math QE

IIT JEE

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.

 
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