For a positive integer n if the quadratic equation

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Question 1:- For a positive integer n if the quadratic equation, x(x+1)+(x+1)(x+2) + . . . +[x+(n-1)][x+n] = 10n has two consecutive integral solution, then n is equal to

(a) 12           (b) 9

(c) 10            (d) 11

Solution:- Given quadratic equation is

x(x+1)+(x+1)(x+2) + . . . +[x+(n-1)][x+n] = 10n

⇒ (x² + x² + x² + . . . +x²) + [(1 + 3 + 5 +  . . . +(2n-1)]x + [(1.2 + 2.3 + . . . (n-1)n] = 10n

⇒ nx^2 + n^2x + \dfrac{n(n^2-1)}{3} 10n = 0

⇒ x^2 + nx + \dfrac{n^2-1}{3} - 10 = 0

⇒ 3x² + 3nx + n² -31 =0

Let α and β be the roots.

Since, α and β are consecutive.

|α – β| = 1

⇒ (α – β)² = 1

Again, (α – β)² = 1

⇒ (α + β)² – 4αβ = 1

⇒ (\dfrac{-3n}{3})^2 - 4(\dfrac{n^2 - 31}{3}) = 1

⇒ n^2 - \frac{4}{3}(n^2 - 31) = 1

⇒ 3n²- 4n² + 124 = 3

⇒ n² = 121

⇒ n = 11

Answer is (D) 11

Question 4:- If Re(z-1/2z+i) = 1, where z = x + iy, then the point (x, y) lies on a :

(A) Straight line whose slope is -2/3

(B) Straight line whose slope 3/2

(C) Circle whose diameter is √5/2

(D) Circle whose centre is at (-1/2, -3/2)

Solution:- See fulle solution

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