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# Notes | EduRev

## JEE Main Mock Test Series 2020 & Previous Year Papers

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## JEE : Notes | EduRev

The document Notes | EduRev is a part of the JEE Course JEE Main Mock Test Series 2020 & Previous Year Papers.
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Illustration 1: The set of all real numbers x for which x2 - |x + 2| + x > 0 is (2002)
1. (-âˆž, -2) âˆª (2, âˆž)
2. (-âˆž, -âˆš2) âˆª (âˆš2, âˆž)
3. (-âˆž, -1) âˆª (1, âˆž)
4. (âˆš2, âˆž)
Solution: The condition given in the question is x2 - |x + 2| + x > 0
Two cases are possible:
Case 1: When (x+2) â‰¥ 0.
Therefore, x2 - x - 2 + x > 0
Hence, x2 â€“ 2 > 0
So, either x < - âˆš2 or x > âˆš2.
Hence, x âˆˆ [-2, -âˆš2) âˆª (âˆš2, âˆž)     â€¦â€¦. (1)
Case 2: When (x+2) < 0
Then x2 + x + 2 + x > 0
So, x2 + 2x + 2 > 0
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 x2 â€“ 2(a + b + c)x + 3Î¼ (ab + bc + ca) = 0 has real roots, then (2006)
1. Î¼ < 4/3
2. Î¼ > 5/3
3. Î¼ âˆˆ (4/3, 5/3)
4. Î¼ âˆˆ (1/3, 5/3)
Solution: It is given in the question that the roots are real and hence, D â‰¥ 0
This gives, 4(a+b+c)2 - 12Î¼(ab+bc+ca) â‰¥ 0
Hence, (a+b+c)2 â‰¥ 3Î¼ (ab+bc+ca)
So, a2+b2+c2 â‰¥ (ab+bc+ca)( 3Î¼-2)
Hence, 3Î¼-2 â‰¤ (a2+b2+c2)/(ab+bc+ca)
Also, cos A = b2 + c2 - a2 /2bc
And so, cos A < 1
This means, b2 + c2 - a2 < 2bc
Similarly, c2 - a2 - b2 < 2ca
And a+ b2 - c2 < 2ab
So, a2 + b2 + c2 < 2(ab + bc + ca)
Therefore, (a2+b2+c2)/(ab+bc+ca) < 2
Hence, using the obtained equations, we get
3Î¼ -2 < 2
So, this gives Î¼ < 4/3.

Illustration 3: If x2 â€“ 10ax -11b = 0 has â€˜câ€™ and â€˜dâ€™ as its roots and the equation x2 â€“ 10cx -11d = 0 has its roots a and b , then find the value of a+b+c+d. (2006)
Solution: We have two equations x2 â€“ 10ax -11b = 0
And x2 â€“ 10ax -11b = 0
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 xâ€“ 10ax -11b = 0
Hence, câ€“ 10ac -11b = 0 â€¦..(2)
Similarly, â€˜aâ€™ is a root of the equation x2 â€“ 10cx -11d = 0
So, a2 â€“ 10ca -11d = 0    â€¦â€¦ (3)
Now, subtracting equation (3) from (2), we get
(c- a2) = 11(b-d)           â€¦â€¦.. (4)
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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