**Problem**: Find the smallest value of k for which both the roots of the equation x

^{2 }- 8 k x+ 16 (k

^{2 }– k + 1) = 0 will be real roots ?

**Solution**: For both roots to be real first of all the Discriminant must be greater than zero

Hence, ∆ ≥ 0

64k

^{2}-4 (16(k^{2}-k+1) ) ≥064k

^{2 }-64k^{2 }+ 4k-4 ≥0k≥1

So the minimum value is k = 1

**Problem : Find the maximum and minimum value of x such that x**

^{2 }- 8 x + 6 < 0**Solution :**We can first find where the expression x

^{2 }- 8 x + 6 = 0 holds true, thereon we can better decide which direction to go.

x

^{2 }- 8 x + 6 = 0 gives us,x = -(-8) ± √64-4*6

x = 8 ± √64-16

x = 8 ± √48

x = 8 ± 4√3

So the minimum value of x is 8 – 4√3 and the maximum value is 8 + 4√3.

In between these two points the expression will be less than zero.

+ve

__‹‹‹‹ +ve ‹‹‹__(8- 4√3)__›››››› -ve ‹‹‹‹ (__8 + 4√3)__›››› +ve ››››__+ve**Problem:**Find the values of x satisfying the following,

x

^{2}-^{ }5x + 6 > 0, x^{2}-^{ }15x + 16 < 0 and x^{2}– 6x + 16 < 0**Solution:**We shall solve each of the inequations seperately and then take the common values in each.

First equation, x

^{2}-^{ }5x + 6 > 0On solving you will get, x >2 & x < 3

Second equation, x

^{2}-^{ }15x + 16 < 0x = [15 ± √225-64 ] /2

x = [15 ± √161] /2

Thus we get,

x > [15 - √161] /2 & x <[15 + √161] /2

Third equation, x

^{2}-6x - 16 < 0On solving we get, x > - 2 & x < 8

The confusing but not difficult part now, taking the common values. We will better do it on a number line to avoid any confusion.

Hence from above it is very clear that the common region is 2 < x < 3

IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010

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