Test: Quadratic Equations: Nature of Roots (June 8) - JEE MCQ

Concept Used:

If two numbers are opposite in sign then their product is always less than zero.

Calculations:

Given function is f(x) = 3x² - 5x + p

Put x = 0 and x = 1 in f(x) to find f(0) and f(1).

Put x = 0

⇒ f(0) = p

Put x = 1

⇒ f(1) = 3(1)² - 5(1) + p

⇒ f(1) = - 2 + p

Here, f(0) and f(1) are opposite in sign.

So, f(0) × f(1) < 0

⇒ p × (-2 + p) < 0

⇒ p (p - 2) < 0

∴ p ∈ (0, 2) [p ∈ (0, 1) lies inside the (0, 2)]

Hence, f(x) = 3x² - 5x + p and f(0) and f(1) are opposite in sign then 0 < p < 1 is correct statement.

Concept:

If ax² + bx + c = 0 is a quadratic equation with D = b² - 4ac as its discriminant then,

• D > 0 then the quadratic equation has real and distinct roots
• D = 0 then the quadratic equation has real and repeated roots
• D < 0 then the quadratic equation has complex and conjugate roots

Calculation:

Given: The quadratic equation (k - 2)x² + 8x + k + 4 = 0 has both real, distinct and negative roots

As we know that, for a quadratic equation ax² + bx + c = 0 if the discriminant D > 0 then roots are real and distinct.

Here, a = k - 2, b = 8 and c = k + 4

⇒ 64 - 4 × (k - 2) × (k + 4) > 0

⇒ 16 - k² - 2k + 8 > 0

⇒ k² + 2k - 24 < 0

⇒ (k + 6) (k - 4) < 0

⇒ k ∈ (- 6, 4)----------(1)

For both the roots to be negative - b/2a < 0

⇒ k ∈ (2 , ∞) --------(2)

Now from (1) and (2), we get k ∈ (2, 4)

Hence, option C is the correct answer.

Concept:

For a quadratic equation ax² + bx + c = 0,

Discriminant (D) = b² - 4ac

For equal roots, D = 0

Formula used:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Calculation:

Given that,

x² - ax - bx - cx + bc + ca = 0

⇒ x² - (a + b + c)x + bc + ca = 0

For equal roots, D = 0

⇒ [-(a + b +  c)]²  - 4(bc + ca) = 0

⇒ a² + b² + c² + 2ab + 2bc + 2ca - 4bc - 4ca = 0

⇒ a² + b² + c² + 2ab - 2bc - 2ca = 0

Using the above formula, we can write

(a + b - c)² = 0

∴  a + b - c = 0

Concept:

Non-Collinear Points: The set of points that do not lie on the same line are called non-collinear points. We cannot draw a single straight line through these points.

Formula Used:

a³ - b³ = (a - b) (a² + ab + b²)

Calculation:

Given equation x³ - 8 = 0

⇒ x³ - 2³ = 0

⇒ (x - 2) (x² + 2x + 4) = 0

⇒ x - 2 = 0 or x² + 2x + 4 = 0 

⇒ x = 2 or x = -1 + i√3 or x = -1 - i√3

We can take these three points on complex plane 

So, we can write 1st point as 2 + i.0, -1 + i√3, -1 - i√3

The three points on complex plain is (2, 0), (-1,√3), (-1, -√3)

Case I. 

Equation of line passes through (-1,√3), (-1, -√3) is

(y - √3) = (2√3/0) (x + 1)

⇒ x + 1 = 0 

And this line does not passes through (2, 0).

∴ These points (roots) are not collinear.

Case 2. 

We know that these points are non-collinear it means they will definitely pass through a circle 

Let the equation of circle be (x - a)² + (y - b)² = r²

Now, According to our assumption 

(2 - a)² + (0 - b)² = r²     ----(i)

(-1 - a)² + (√3 - b)² = r²     ----(ii)

(-1 - a)² + (-√3 - b)² = r²     ----(iii)

On solving (i), (ii) and (iii), we get 

a = 0, b = 0 and r = 2

Hence the equation of circle is (x - 0)² + (y - 0)² = 22

∴ The roots represents a circle but radius of circle is 2.

Concept:

For any quadratic equation, ax² + bx + c = 0. We have discriminant, D = b² – 4ac then the given quadratic equation has:

I. Distinct and real roots if D > 0.

II. Real and repeated roots, if D = 0.

III. Complex roots and conjugate of each other, if D < 0.

Calculation:

In the given quadratic equation px² + qx + r = 0, coefficients p, q, r are positive with roots a and b.

⇒ a + b = - q/p and a × b = r/p.

From the condition: a × b = r/p, we can say that there are two possibilities which are:

Case 1: Both roots are negative so that the product is positive i.e a < 0 and b < 0.

Case 2: Both roots are positive so that the product is positive i.e a > 0 and b > 0.

From the relation a + b = - q/p, we can say that the sum of roots is negative.

Out of the two cases, Case 1 is a valid case satisfying both relations a + b = - q/p and a × b = r/p.

Given:

The equation 2x² + 5x + 5 = 0

Concept Used:

The quadratic equation Ax² + Bx + C = 0 

If roots are imaginary, B² - 4ac < 0 

Calculation:

Comparing of given equation

Now, A = 2, B = 5 & C = 5 

⇒ (5)² - 4 × 2 × 5 < 0

⇒ 25 - 40 < 0 

⇒ -15 < 0 

∴ The roots are imaginary and distinct. 

Concept:

For standard quadratic equation:

ax² + bx + c = 0

Sum of root =−b/a

Product of roots = c/a

Analysis:

equation x² – px + q = 0

Since roots are α and β

α + β = −(p)/1 = p

α β = q

α² + β² = (α + β)² – 2 α β

α² + β² = p² – 2q

Let the two roots be a and b.

Given: a < 5 < b

So, from the above diagram, we can conclude that D > 0 and F(5) < 0.

F(5) < 0

5² + 2(k - 3)5 + 9 < 0

25 + 10k – 30 + 9 < 0

10k + 4 < 0

k < (- 4)/10

So,

k ε (- ∞, (- 4)/10)

Concept:

Let us consider the standard form of a quadratic equation, ax² + bx + c =0

Discriminant = D = b² – 4ac

• If the Discriminant > 0 then the roots are real and distinct.
• If the Discriminant = 0 then the roots are real and equal.
• If the Discriminant < 0 then the roots are Imaginary.

Logarithmic identities:

log_b(x) = c ⇔ b^c = x

Calculation:

Given: x² – 2x – log₂ K = 0

General form of equation: ax² + bx + c = 0

On comparing a = 1, b = - 2 c = -  log₂ K 

Given that the roots are real

So, Determinant ≥ 0

⇒ b² – 4ac ≥ 0

⇒ (-2)² - 4 × 1 × (- log₂ K) ≥  0

⇒ 4 + 4 log₂ K ≥  0

⇒ 4 log₂ K ≥  -4

⇒ log₂ K ≥  -1

⇒ K  ≥  (2)^-1

∴ K ≥  1/2

The minimum value of K is 1/2.

Concept:

If two roots of the equation ax² + bx + c = 0 are equal, then 

b² - 4ac = 0

Calculation:

If the equation x² - (2 + m)x + (m² - 4m + 4) = 0 in x has equal roots

⇒ (2 + m)² - 4(1)(m² - 4m + 4) = 0

⇒ 4 + 4m + m² - 4m² + 16m - 16 = 0

⇒ 3m² - 20m + 12 = 0

⇒ 3m2 - 18m - 2m + 12 = 0

⇒ 3m(m - 6) - 2(m - 6) = 0

⇒ (m - 6)(3m - 2) = 0

⇒ m = 6, 2/3

∴ The correct answer is option (2).