Match the statements of Column I with values of Column II. Column I Column II A) The least positive integral value of λ for which P) 3 ( λ − 2 ) x 2 + 8 x + ( λ + 4 ) 0 , for all real x is B) The eq

Question

Match the statements of Column I with values of Column II.

Column I Column II

A) The least positive integral value of $λ$ for which P) 3

$(λ - 2) x^{2} + 8 x + (λ + 4) > 0$ , for all real $x$ is

B) The equation $x^{2} + 2 (a^{2} + 1) x + (a^{2} - 14 a + 48) = 0$ Q) 5

possesses roots of opposite signs, then the value of 'a' can be

C) If the equation $a x^{2} + 2 b x + 4 c = 16$ has no real roots andR) 7

$a + c > b + 4$ , then the integral value of $c$ can be equal to

D) If $N$ be the number of solutions of the equation S) 12

$x^{2} - x - 6 ∣ = x + 2$ , then the value of $N$ is

T) 20

TestHub · Accepted Answer

A) If λ > 2 , then 64 − 4 ( λ − 2 ) ( λ + 4 ) < 0 . This implies ( λ + 6 ) ( λ − 4 ) > 0 , so λ < − 6 or λ > 4 . Therefore, the least positive integral value of λ is 5. B) If roots are of opposite signs, then a 2 − 14 a + 48 < 0 . This means ( a − 6 ) ( a − 8 ) < 0 , so a can be 7. The equation is x 2 + 100 x − 1 = 0 . The discriminant E = 10 0 2 + 4 > 0 . Therefore, the roots are real. C) Let f ( x ) = a x 2 + 2 b x + 4 c − 16 . Clearly, f ( − 2 ) = 4 a − 4 b + 4 c − 16 = 4 ( a − b + c − 4 ) > 0 . Since f ( x ) > 0 for all x ∈ R , it implies f ( 0 ) > 0 , so 4 c − 16 > 0 . Thus, c > 4 . D) Given ∣ x 2 − x − 6∣ = x + 2 . This can be written as ∣ ( x − 3 ) ( x + 2 ) ∣ = x + 2 . ∣ ( x − 3 ) ( x + 2 ) ∣ = x + 2 This leads to: ⇒ ∣ x − 3∣∣ x + 2∣ = x + 2 ⇒ ⎩ ⎨ ⎧ ( x − 3 ) ( x + 2 ) = x + 2 , − ( x − 3 ) ( x + 2 ) = x + 2 , ( x − 3 ) ( x + 2 ) = x + 2 , x < − 2 − 2 ≤ x < 3 x > 3 ⇒ ⎩ ⎨ ⎧ x = 4 , x = − 2 , 2 x = 4 , x < − 2 − 2 ≤ x < 3 x < 3 Hence, x = − 2 , 2 , 4 N = 3

Mathematics - Quadratic Equation Question with Solution | TestHub

Options:

Doubts & Discussion