Example
1
Show that the value
of 
, whenever
defined, never lies between 1/3 and 3.
Solution:
Since tan2x
is positive, either y > 3 or y < 1/3. If y lies between 1/3 and 3 i.e.
if 1/3 < y < 3, tan2x is negative, which is not possible. .
Hence 
never lies
between 1/3 and 3.
Example
2
If x1, x2
are the roots of x2 - 3x + a = 0 and x3, x4
are those of the equation x2 - 12x + b = 0 and x1, x2,
x3, x4 are in increasing G.P., find a and b.
Solution:
Let x1 = A
and r be the common ratio. We have.
x1 = A, x2
= Ar, x3 = Ar2, x4 = Ar3
also, x1 +
x2 = 3 …(1)
x1 x2
= a …(2)
and x3 +
x4 = 12 …(3)
x3 x4
= b …(4)
From equation (1) and
(3), we have 
&⇒ r = ± 2
&⇒
r = 2 (since given sequence is an increasing G.P.).
Also A (1+ r) = 3
&⇒
A = 1
Therefore, x1=1,
x2 = 2, x3= 4, x4 = 8
Thus we have a = x1x2
= 2, b = x3x4= 32
Example
3
If x2 +
cx + ab = 0 and x2 + bx + ca = 0 have a common root, then show that
their other roots satisfy the equation x2 + ax + bc = 0.
Solution:
Let the equation x2
+ cx + ab = 0 …(1)
and x2
+ bx + ca = 0 …(2)
have the common root a
Therefore, a2 + ca + ab = 0
And a2 + ba + ca = 0
&⇒
&⇒
a
= - (b + c) and also a = a
&⇒ a + b + c = 0
Let the other root of
equations (1) and (2) be b and b1 respectively then a×b
= ab and ab1 = ca
But a
= a. Therefore b = b, b1 = c.
Now the equation whose
roots are b and b1 is
x2 – (b
+ b1)x
+ b×b1 = 0
&⇒ x2 – (b + c)x + bc = 0
&⇒
x2 + ax + bc = 0 (Since b + c = -a)
Hence other roots of
equation (1) and (2) are given by the equation
x2+ ax + bc
= 0
Example
4
If 2x3 +
ax2 +bx +4 = 0 ( a and b are positive real numbers ) has 3
real roots, then prove that a+ b ≥
6(21/3 + 41/3). .
Solution:
Let a, b,
g
be the roots of 2x3 + ax2 + bx + 4 = 0. Given that all
the coefficients are positive, so all the roots will be negative.
Let a1 = -a, b2 = -b, a3 = – r
&⇒
a1
+ a2
+ a3
= 
a1
a2
+ a2a3 + a3a1=
a1
a2
a3
= 2
Applying A.M. ≥
G.M. , we have.
≥ (a1 a2a3)1/3
&⇒
≥
2
&⇒ a ≥ 6 ´ 21/3
Also 
≥ (a1 a2 a3)2/.3
&⇒
≥
4
&⇒ b ≥ 6 ´41/3
Therefore a + b ≥
6 (21/3 + 41/3) = 6(21/3 + 41/3)
Example
5
Show that the
equation 
= k has no
imaginary root, where A, B, C ……,H and a, b, c ….., h and k ∈ R
Solution:
Suppose one root of the equation is
(u +iv) then other root would be u – iv
&⇒
….(1)
and 
….(2)
(1) –(2) , we
get
iv
This is possible only when v = 0 and
for this case there is no imaginary root.
