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, ¥) .
