Find the least value of for real x.

Let = y

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

Since x is real

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

&⇒
-4y^{2} + 9y – 5 ≥ 0

&⇒ 4y^{2}
– 9y + 5 £ 0

&⇒
4(y – 1) (y – 5/4) £ 0

&⇒ 1 £
y £ 5/4

Hence the least value of the given expression = 1.

For
what value of q, 1 lies between the
roots of quadratic equation 3x^{2 }- 3 sinq x - 2 cos^{2}q = 0.

Let f(x) = 3x^{2} – 3 sinq
x – 2 cos^{2}q

The coefficient of x^{2} >
0

&⇒ f(1) < 0

&⇒
3 - 3sinq - 2cos^{2}q < 0

&⇒
2sin^{2}q - 3sinq + 1 < 0

&⇒
(2sinq - 1) (sinq - 1) < 0

&⇒
< sinq
< 1

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

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.

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) = l^{2} (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.

Find the range of real values of x and
y if 2x^{2} + 6xy + 5y^{2} = 1

2x^{2} + 6xy + 5y^{2}
= 1 …. (1).

Equation (1) can be rewritten as

2x^{2} + (6y) x + 5y^{2}
– 1 = 0

Since x is real

36y^{2} – 8(5y^{2}
–1) ≥ 0

&⇒ y^{2} £
2

&⇒ £
y £

Equation (1) can also be rewritten
as 5y^{2} +(6x)y +2x^{2}-1 = 0

Since y is real, 36x^{2}
–20 ( 2x^{2} –1) ≥ 0

&⇒
36x^{2} - 40x^{2} + 20 ≥ 0

&⇒
-4x^{2} ≥ -20

&⇒ x^{2}
£
5

&⇒ - £ x £
.

Note:

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

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

The required condition will be satisfied if

(i) The quadratic expression (quadratic in tanx)

f(x)
= tan^{2}x + (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 t^{2}
+ (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, ¥) .