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**Illustration 1: The set of all real numbers x for which x ^{2} - |x + 2| + x > 0 is (2002)**

Two cases are possible:

Therefore, x

Hence, x

So, either x < - âˆš2 or x > âˆš2.

Hence, x âˆˆ [-2, -âˆš2) âˆª (âˆš2, âˆž) â€¦â€¦. (1)

Then x

So, x

This gives (x+1)2 + 1 > 0 and this is true for every x

Hence, x â‰¤ -2 or x âˆˆ (-âˆž, -2) ...â€¦â€¦ (2)

From equations (2) and (3) we get x âˆˆ (-âˆž, -âˆš2) âˆª (âˆš2, âˆž).

**Illustration 2: If a, b and c are the sides of a triangle ABC such that ****x ^{2} â€“ 2(a + b + c)x + 3Î¼ (ab + bc + ca) = 0 has real roots, then (2006)**

This gives, 4(a+b+c)

Hence, (a+b+c)

So, a

Hence, 3Î¼-2 â‰¤ (a

Also, cos A = b

And so, cos A < 1

This means, b

Similarly, c

And a

So, a

Therefore, (a

Hence, using the obtained equations, we get

3Î¼ -2 < 2

So, this gives Î¼ < 4/3.

**Illustration 3: If x ^{2} â€“ 10ax -11b = 0 has â€˜câ€™ and â€˜dâ€™ as its roots and the equation x^{2} â€“ 10cx -11d = 0 has its roots a and b , then find the value of a+b+c+d. (2006)**

And x

Now using the concepts of sum of roots we obtain the following relations,

a+b = 10c and c+d = 10a

So, we get (a-c) + (b-d) = 10(c-a)

This gives, (b-d) = 11(c-a) â€¦.. (1)

Since, â€˜câ€™ is a root of x

Hence, c

Similarly, â€˜aâ€™ is a root of the equation x

So, a

Now, subtracting equation (3) from (2), we get

(c

Therefore, (c+a) (c-a) = 11.11(c-a) â€¦. (From eq(1))

Hence, this implies c+a = 121

Therefore, a+b+c+d = 10c + 10a

= 10(c+a)

= 1210.

Hence, the required value of (a+b+c+d) = 1210.

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