Question

This is not a question.

I wanted to request you something.

Can you send me some very tough iit questions from this and from quadratic equations.

Answer 1

1: The set of all real numbers x for which x² - |x + 2| + x > 0 is

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, ∞).

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

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 a² + 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.

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.

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 x² – 10ax -11b = 0

Hence, c² – 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

(c² - 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.