Math Q & A

Q : If ax² + bx + 6 = 0 does not have two distinct real roots where a, b ∈ R, then the least value of 3a + b is

(A) 4

(B) -1

(C) 1

(D) -2

Ans: (D)

Sol. D ≤ 0 ⇒ f(x) ≥ 0 or, f(x) ≤ 0 for all real x, but f(0) = 6 > 0

⇒ f(x) ≥ 0 ∀ x ∈ R. In particular f(3) ≥ 0 ⇒ 9a + 3b + 6 ≥ 0

⇒ 3a + b ≥ – 2.

Q : The values of α and β such that equation x² + 2x + 2 + e^α – sinβ = 0 have a real solution is

(A) $\large \alpha , \beta \in R $

(B) $\large \alpha \in (0 , 1) \; , \beta \in (\pi/2 , 2 \pi) $

(C) $\large \alpha \in (0 , \infty) \; and \beta \in (\pi/2 , \pi) $

(D) None of these

Ans: (D)

Solution: x² + 2x + 2 + e^α – sinβ = 0 has real roots if D ≥ 0

⇒  1 – 2 – e^α + sinβ  ≥ 0

⇒ sinβ  ≥  1 + e^α

Hence no real values of α and β are possible.

Q : The quadratic equation whose roots are A.M. and H.M. between the roots of the equation ax² + bx + c = 0 is

(A) abx² + (b² + ac)x + bc = 0

(B) 2abx² + (b² + 4ac)x + 2bc = 0

(C) 2abx² + (b² + ac)x + bc = 0

(D) none of these

Ans: (B)

Solution: Let (α , β) be the roots of the given equation, then

$\large \alpha + \beta = -\frac{b}{a} \; , \alpha . \beta = \frac{c}{a} $

Required equation is

$\large x^2 – (\frac{\alpha + \beta}{2} + \frac{2 \alpha . \beta}{\alpha + \beta})x + (\frac{\alpha + \beta}{2}) (\frac{2 \alpha . \beta}{\alpha + \beta})) = 0 $

2abx² + (b² + 4ac)x + 2bc = 0