QuizSolver
  • Bank PO
  • CBSE
  • IIT JEE
 
 

Quadratic Equations - Solved Subjective Questions Part 4

Posted on - 14-05-2017

JEE Math QE

IIT JEE

Example 1

Consider the inequation x2 + |x + a| –9 < 0, find the values of the real parameter 'a' so that the given inequation has atleast one negative solution.

Solution:

9 – x2 > |x + a|

As per the question, we have to make sure that for atleast one negative x, graph of y = 9 – x2 must lie above the graph of y = |x + a|

Case-I: a ≥ 0, Let us take up the limiting case when y= |x + a| touches the parabola

y = 9 – x2
&⇒
9 – x2 = x + a should have equal roots


&⇒
x2 + x + (a – 9) = 0 has equal roots


&⇒
1 – 4(a – 9) = 0
&⇒


&⇒
0 £ a <

Case-II: a < 0

In the limiting case the left branch of y = | x + a |

i.e. y = – x – a will pass through the vertex of the parabola.


&⇒
9 = 0 – a
&⇒
a = –9


&⇒
a > – 9

That means required set of values of a is

Alternative Solution:

x2 + |x +a| - 9 < 0

Case –I Let x +a ≥ 0
&⇒
For all x < 0, a > 0

Also x2 +x +a – 9 < 0


&⇒
a < 9 – x2 –x


&⇒
a <
&⇒
a <

i.e. 0 < a < . . . (1)

Case –II Let x +a < 0
&⇒
a£ 0

Also x2 –x –a –9 < 0
&⇒
a > x2 –x – 9


&⇒
a > - 9 ,

so, -9 < a £ 0 . . . . (2).

From (1) and (2), -9 < a < .

Example 2

Show that if x is real, the expression has no value lying between b and c.

Solution:

Let
&⇒
x2 – 2xy + (b + c)y – bc = 0

Now x is real
&⇒
Discriminant ≥ 0


&⇒
4y2 – 4[(b + c)y – bc] ≥ 0
&⇒
(y – b) (y – c) ≥ 0


&⇒
y has no value between b and c

Example 3

Solve the equation log­4 (2x2 + x + 1) – log2 (2x –1) = 1.

Solution:

log4(2x2 + x + 1) – log2(2x –1) = 1


&⇒


&⇒
loge


&⇒
(2x2 + x + 1) = 4(4x2 – 4x + 1)


&⇒
14 x2 - 17x + 3 = 0
&⇒
(14x - 3) (x - 1) = 0


&⇒
x = , x = 1

But x = does not lies in the domain of function.

Hence x = 1 is only solution. .

Example 4

Let a,b,c be real. If ax2+bx+c=0 has two real roots a and b where a < - 1 and b > 1, then show that

Solution:

a < -1


&⇒
a + E­1 = -1, where E1 > 0

Also b > 1


&⇒
b - E2 = 1, where E2 > 0.

= 1-1-E1-E2-E1E2+½E2-E1½ = -E1-E2-E1E2+E2-E1 , if E2 > E1

= -E1-E2-E1E2+E1-E2 if E1 > E2. .

Hence L.H.S. = -2E1-E1E2 or -2E2-E1E2.

In both the cases . (E1, E2 > 0)

Alternative Method

Let f(x) = x2 + x +

from graph f(-1) < 0 and f(1) < 0


&⇒
1+< 0 and 1 + + < 0
&⇒
1 + + < 0.

Example 5

Find all the real values of the parameter a for which the equation x(x+1)(x+a)(x+a+1)=a2 has four real roots.

Solution:

Rewritting the given equation as (x2 +ax + x )(x2+ax + x+a) = a2 and put x2 + ax + x= y, so that y(y +a) = a2

or y2 + ay - a2 = 0 whose solution is

y =

There will then be four real roots if each of the following two equations x2 + x(a + 1) = a has two real roots which implies (a +1)2 + 4
&⇒
a2 +2a +1 –2a ± 2a ≥ 0


&⇒
a2 ± 2a +1 ≥ 0
&⇒
(a + )2 ≥ 4 and (a -)2 ≥ 4

This leads to the four possibilities

a +£ -2, a +≥ 2, a - £ -2 or a - ≥ 2

In other words, a £ -2 - , a ≥ 2 - , a £ -2 or a ≥ +2 .

It is clear that 2 - £ a £ -2 , a £ -2 -or a ≥ 2 +

Whence |a| £ -2 or |a| ≥ 2 + .

 
Quadratic Equations - Solved Objective Questions Part 2 for Conceptual Clarity
Quadratic Equations - Solved Objective Questions Part 1 for Conceptual Clarity
Solved Objective Question on Probability Set 2
Solved Objective Question on Probability Set 1
Solved Objective Question on Progression and Series Set 2
Solved Objective Question on Permutations and Combinations Set 3
Solved Objective Question on Permutations and Combinations Set 2
Solved Objective Question on Progression and Series Set 1
Solved Objective Question on Permutations and Combinations Set 1
Quadratic Equations - Solved Subjective Questions Part 2
Quadratic Equations - Solved Subjective Questions Part 3
Quadratic Equations - Solved Subjective Questions Part 1
Solved Subjective Questions on Circle Set 9
Solving Equations Reducible to Quadratic Equations
Theory of Polynomial Equations and Remainder Theorem
Solved Subjective Questions on Circle Set 8
Solving Quadratic Inequalities Using Wavy Curve Methods
Division and Distribution of Objects - Permutation and Combination
Basics of Quadratic Inequality or Inequations
Basic Concepts Of Combinations for IIT and Other Engineering Exams

Comments