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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. (-∞, -2) ∪ (2, ∞)
2. (-∞, -√2) ∪ (√2, ∞)
3. (-∞, -1) ∪ (1, ∞)
4. (√2, ∞)
Solution: The condition given in the question is x2 - |x + 2| + x > 0
Two cases are possible:
Case 1: When (x+2) ≥ 0.
Therefore, x2 - x - 2 + x > 0
Hence, x2 – 2 > 0
So, either x < - √2 or x > √2.
Hence, x ∈ [-2, -√2) ∪ (√2, ∞)     ……. (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 ≤ -2 or x ∈ (-∞, -2)       ...…… (2)
From equations (2) and (3) we get x ∈ (-∞, -√2) ∪ (√2, ∞).

Illustration 2: If a, b and c are the sides of a triangle ABC such that x2 – 2(a + b + c)x + 3μ (ab + bc + ca) = 0 has real roots, then (2006)
1. μ < 4/3
2. μ > 5/3
3. μ ∈ (4/3, 5/3)
4. μ ∈ (1/3, 5/3)
Solution: It is given in the question that the roots are real and hence, D ≥ 0
This gives, 4(a+b+c)2 - 12μ(ab+bc+ca) ≥ 0
Hence, (a+b+c)2 ≥ 3μ (ab+bc+ca)
So, a2+b2+c2 ≥ (ab+bc+ca)( 3μ-2)
Hence, 3μ-2 ≤ (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
3μ -2 < 2
So, this gives μ < 4/3.

Illustration 3: If x2 – 10ax -11b = 0 has ‘c’ and ‘d’ as its roots and the equation x2 – 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 – 10ax -11b = 0
And x2 – 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) ….. (1)
Since, ‘c’ is a root of x– 10ax -11b = 0
Hence, c– 10ac -11b = 0 …..(2)
Similarly, ‘a’ is a root of the equation x2 – 10cx -11d = 0
So, a2 – 10ca -11d = 0    …… (3)
Now, subtracting equation (3) from (2), we get
(c- a2) = 11(b-d)           …….. (4)
Therefore, (c+a) (c-a) = 11.11(c-a)   …. (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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