Show that the value of , whenever defined, never lies between 1/3 and 3.

Since tan^{2}x
is positive, either y > 3 or y < 1/3. If y lies between 1/3 and 3 i.e.
if 1/3 < y < 3, tan^{2}x is negative, which is not possible. .

Hence never lies between 1/3 and 3.

If x_{1}, x_{2}
are the roots of x^{2 }- 3x + a = 0 and x_{3}, x_{4}
are those of the equation x^{2 }- 12x + b = 0 and x_{1}, x_{2},
x_{3}, x_{4} are in increasing G.P., find a and b.

Let x_{1} = A
and r be the common ratio. We have.

x_{1} = A, x_{2}
= Ar, x_{3} = Ar^{2}, x_{4} = Ar^{3}

also, x_{1} +
x_{2} = 3 …(1)

x_{1} x_{2}
= a …(2)

and x_{3} +
x_{4} = 12 …(3)

x_{3} x_{4}
= 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, x_{1}=1,
x_{2} = 2, x_{3}= 4, x_{4} = 8

Thus we have a = x_{1}x_{2}
= 2, b = x_{3}x_{4}= 32

If x^{2 }+
cx + ab = 0 and x^{2 }+ bx + ca = 0 have a common root, then show that
their other roots satisfy the equation x^{2 }+ ax + bc = 0.

Let the equation x^{2}
+ cx + ab = 0 …(1)

and x^{2}
+ bx + ca = 0 …(2)

have the common root a

Therefore, a^{2} + ca + ab = 0

And a^{2} + 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 b_{1} respectively then a×b
= ab and ab_{1} = ca

But a
= a. Therefore b = b, b_{1} = c.

Now the equation whose
roots are b and b_{1} is

x^{2} – (b
+ b_{1})x
+ b×b_{1} = 0

&⇒ x^{2} – (b + c)x + bc = 0

&⇒
x^{2} + ax + bc = 0 (Since b + c = -a)

Hence other roots of equation (1) and (2) are given by the equation

x^{2}+ ax + bc
= 0

If 2x^{3} +
ax^{2} +bx +4 = 0 ( a and b are positive real numbers ) has 3
real roots, then prove that a+ b ≥
6(2^{1/3} + 4^{1/3}). .

Let a, b,
g
be the roots of 2x^{3} + ax^{2} + bx + 4 = 0. Given that all
the coefficients are positive, so all the roots will be negative.

Let a_{1} = -a, b_{2} = -b, a_{3} = – r

&⇒
a_{1}
+ a_{2}
+ a_{3}
=

a_{1}
a_{2}
+ a_{2}a_{3} + a_{3}a_{1}=

a_{1}
a_{2}
a_{3}
= 2

Applying A.M. ≥ G.M. , we have.

≥ (a_{1} a_{2}a_{3})^{1/3 }

&⇒
≥
2

&⇒ a ≥ 6 ´ 2^{1/3}

Also ≥ (a_{1} a_{2} a_{3})^{2/.3}

&⇒
≥
4

&⇒ b ≥ 6 ´4^{1/3}

Therefore a + b ≥
6 (2^{1/3} + 4^{1/3}) = 6(2^{1/3} + 4^{1/3})

Show that the equation = k has no imaginary root, where A, B, C ……,H and a, b, c ….., h and k ∈ R

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.