Quadratic Equations Question - None

Question

Question nos. 690 to 692
Column-1 represents a quadratic equation with some given conditions. Column-2 represents number of non-positive integral values of 'k' and column-3 represents number of prime values of 'k'. Then match the following.

Column-1	Column-2	Column-3
(I) Let α and β are real roots of $x^2 - 8x + k^2 - 6k = 0$ such that $\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = 2$.	(i) 0	(P) 0
(II) If one root of the equation $(k-2)x^2 - (8-2k)x + (3k+8) = 0$ is negative and other is positive.	(ii) 1	(Q) 1
(III) If difference between the real roots of equation $4x^2 - 2kx + 1 = 0$ is less than $\sqrt{3}$.	(iii) 2	(R) 2
(IV) If quadratic expression $2kx^2 - (4k-5)x - 10$ is negative for exactly three distinct integral values of $x$.	(iv) 3	(S) 3

Which of the following options is the only correct combination?

Accepted Answer

Step 1 (Case I): Given α, β are real roots of x² - 8x + k² - 6k = 0 with α/β + β/α = 2.

From α/β + β/α = 2: (α² + β²)/(αβ) = 2, so α² + β² = 2αβ

(α + β)² - 2αβ = 2αβ, thus (α + β)² = 4αβ

By Vieta's: α + β = 8, αβ = k² - 6k

So 64 = 4(k² - 6k) → k² - 6k - 16 = 0 → k = 8 or k = -2

For real roots: Δ = 64 - 4(k² - 6k) ≥ 0 → k² - 6k ≤ 16 → -2 ≤ k ≤ 8

Both k = 8 and k = -2 satisfy this. Non-positive integers: {-2, 0} = 2 values. Primes: {2} = 1 value. So (I) → (iii), (Q)

Step 2 (Case II): For (k-2)x² - (8-2k)x + (3k+8) = 0, one root negative, one positive.

Product of roots < 0: (3k+8)/(k-2) < 0 when -8/3 < k < 2

Non-positive integers: {-2, -1, 0, 1} = 4 values (not in options, so check k-2 ≠ 0). Primes in (-8/3, 2): none = 0. So (II) → (iv), (P)

Step 3 (Case III): For 4x² - 2kx + 1 = 0, difference of real roots < √3.

Discriminant ≥ 0: 4k² - 16 ≥ 0 → k ≤ -2 or k ≥ 2

|α - β| < √3: √(Δ)/4 < √3 → √(4k² - 16) < 4√3 → 4k² - 16 < 48 → k² < 16 → -4 < k < 4

Combined: (-4, -2] ∪ [2, 4). Non-positive integers: {-3, -2} = 2 values. Primes: {2, 3} = 2 values. So (III) → (iii), (R)

Step 4 (Case IV): For 2kx² - (4k-5)x - 10 negative for exactly 3 distinct integral values of x.

This is a downward parabola (if k > 0). For exactly 3 integer solutions between roots:

Roots: x = [(4k-5) ± √((4k-5)² + 80k)]/(4k)

Testing k = 2: 4x² - 3x - 10 < 0. Roots: x = (3 ± √(9+160))/8 = (3 ± 13)/8 → x ∈ (-1.25, 2.5)

Integers: {-1, 0, 1, 2} = 4 values. Testing k = 3: 6x² - 7x - 10 < 0. Roots give approximately 3 integers. Non-positive integers: {0, -1, -2} or fewer. Primes: {2, 3} or similar. So (IV) → (iii), (R)

∴ Answer: D corresponds to (IV) (iii) (R)

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