Now , so k is on left hand side of the point
Also f (k)>0, so it is possible only when k<a.

So,a, b>k.
1)      If D³0, f (k)>0 and  then both roots of f (x) =0 are less than k. Why?
As D³0 and a>0 so shape of parabola is like this

Fig (9)
Now f (k)>0 so k>b or k<a
And also

So, k<a is ruled out

Therefore k>b

Hence, k>a,b.
2)      If D>0 and f (k) <0 then k lies between root of f (x) = 0 why?
D>0 and a>0

So the shape of parabola is

Fig (10)
Now f (k) <0, so clearly a<k<b.
4) If D>0 and f (k1)´f (k2) <0, then exactly one root of equation f (x) = 0 lies in                interval (k1, k2). When (k2>k1) Why?
Since f (k1)´f (k2) <0

So, f (k1) and f (k2) should be of opposite sign.

Fig (11)
1)      Now let f(k1)<0 and f(k2)>0

So, a<k1<b and k2>b (because k2>k1)

And hence b lies in interval (k1, k2).  
2)      Second case is f (k1) >0 and f (k2) <0

So, a< k2<b and k1<a (because k2>k1)

Hence a lies in interval (k1, k2).
So exactly one root lies in interval (k1, k2).
5) If D³0, f (p)>0 and f (q)>0, q>p then both the roots of the equation f(x) = 0 will                  
lie between p and q, if  why?
Suppose both roots lie in interval (p, q)
Now D³0, a>0 therefore shape of parabola is

Fig (12)
a, b lies in interval (p, q)
So, b<q and p<a
And hence p<a<b<q


6) If D³0,  and  then both roots of the equation f(x) = 0 are positive. Why?

Also  so, product of root is positive, and hence clearly both roots are positive.

Illustration 8:

For what values of ‘a’ exactly one root of the equation  lies between 1 and 2.
Solution: Since exactly one root of given equation lie between 1 and 2,

So f (1)´f (2) <0

Here f(x) =

So, f (1)´f (2) <0




Some important results using differentiability:

1)      If f(x) = 0 has a real root a of multiplicity g (g>1) then f(x) = (x-a)gg (a) where g(a)¹0. Also f(x) = 0 has a as a real root with multiplicity (g-1)
2)      If f(x) = 0 has n real roots then f (x) = 0 has (n-1) real roots.


Illustration 9:

A polynomial p(x) has  as one of the factors, other roots of this polynomial lie in the range  Prove that g(x) where g(x) = p1(x) has at least one positive root.
Solution:



Now p(x) has a root 2 with multiplicity 2.

So the derivative of p(x), p1(x) or g(x) must has 2 as root with multiplicity 2-1=1.

So g(x) has at least one positive root which is 2.
Easy (Quadratic)

Q-1: The roots of the quadratic equation  are a and b2 Where  then Show that  has roots (a+ib)100 and (a-ib)100
Solution:



Product = 1.

Required equation is
Q-2: If a, b are the roots of the equation  then prove that


Q-3: If   then find value of expression .
Solution:

Q-4: Solve the equation

Solution:

Since x=0 is not a solution of given equation. Dividing by x2 in both sides of (1) we get,

Putting  in equation (2) then (2) reduces in the form,