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Quadratic Equations - Solved Objective Questions Part 1 for Conceptual Clarity

Posted on - 22-05-2017

JEE Math QE

IIT JEE

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. .

 
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