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Quadratic Equations - Solved Subjective Questions Part 1

Posted on - 05-05-2017

JEE Math QE

IIT JEE

Example 1

Find the least value of for real x.

Solution:

Let = y


&⇒
(6 – 5y)x2 – 2x(11 – 9y) + 21 – 17y = 0

Since x is real

4(11 – 9y)2 – 4(6 – 5y) (21 – 17y) ≥ 0


&⇒
-4y2 + 9y – 5 ≥ 0
&⇒
4y2 – 9y + 5 £ 0


&⇒
4(y – 1) (y – 5/4) £ 0
&⇒
1 £ y £ 5/4

Hence the least value of the given expression = 1.

Example 2

For what value of q, 1 lies between the roots of quadratic equation 3x2 - 3 sinq x - 2 cos2q = 0.

Solution:

Let f(x) = 3x2 – 3 sinq x – 2 cos2q

The coefficient of x2 > 0
&⇒
f(1) < 0


&⇒
3 - 3sinq - 2cos2q < 0
&⇒
2sin2q - 3sinq + 1 < 0


&⇒
(2sinq - 1) (sinq - 1) < 0
&⇒
< sinq < 1


&⇒
2np +< q < 2np +and q > (4n + 1)p/2, n ∈ I

Example 3

If a < b< c< d then prove that for any real l, the quadratic equation ( x - a) (x - c) +l( x-b ) (x-d) = 0 has real roots.

Solution:

Let f(x) = ( x-a) (x-c) +l( x-b ) (x-d)

Now f(a) =l (a-b) (a-d) and f(c) = l( c-b ) ( c-d)

So f(a).f(c) = l2 (a-b) (a-d)(c-b)(c-d) < 0 ( since a < b < c < d).

Hence f(x) = 0 has one real root between a and c . .

Since f(x) = 0 is a Quadratic equation with real coefficients,

therefore both the roots of f(x) = 0 will be real.

Example 4

Find the range of real values of x and y if 2x2 + 6xy + 5y2 = 1

Solution:

2x2 + 6xy + 5y2 = 1 …. (1).

Equation (1) can be rewritten as

2x2 + (6y) x + 5y2 – 1 = 0

Since x is real

36y2 – 8(5y2 –1) ≥ 0
&⇒
y2 £ 2
&⇒
£ y £

Equation (1) can also be rewritten as 5y2 +(6x)y +2x2-1 = 0

Since y is real, 36x2 –20 ( 2x2 –1) ≥ 0


&⇒
36x2 - 40x2 + 20 ≥ 0


&⇒
-4x2 ≥ -20
&⇒
x2 £ 5
&⇒
- £ x £ .

Note:

The given equation represents an ellipse, so xand y have limitations. .

Example 5

Find the value(s) of ‘a’ for which the inequality tan2 x + (a + 1) tan x – (a – 3) < 0, is true for at least one x ∈ (0, p/2).

Solution:

The required condition will be satisfied if

(i) The quadratic expression (quadratic in tanx)

f(x) = tan2x + (a+1) tanx – (a-3) has positive discrminant, and

(ii) At least one root of f(x) = 0, is positive, as tanx > 0, " x∈ (0, p/2)

For (i) Discriminant > 0
&⇒
(a + 1)2 + 4 (a-3) >0


&⇒
a > 2Ö5 - 3 or a < - (2Ö5 + 3) … (1)

For (ii), we first find the condition, that both the roots of t2 + (a + 1) t – (a-3) = 0

(t = tanx) are non-positive, for which

Sum of roots < 0 and product of roots ≥ 0


&⇒
– (a + 1) < 0 and – (a-3) ≥ 0
&⇒
– 1 < a £ 3

Condition (ii) will be fullfilled if a £ -1 or a > 3 …. (2).

Required values of a is given by intersection of (1) and (2). .

Hence a ∈ (– ¥, –(2Ö5 + 3)) È (3, ¥) .

 
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