The smallest natural number 'n' so that (2-n) x2-8 x-4-n

Question

The smallest natural number 'n' so that (2-n) x2-8 x-4-n<0, ∀ x ∈ R is equal to (A) 2 (B) 3 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

text We have ( n -2) x 2+8 x +( n +4)>0 . ∀ x ∈ R ⇒ n -2>0 text and D <0 ⇒ 64-4( n -2)( n +4)<0 ⇒ n 2+2 n -24>0 ⇒( n +6)( n -4)>0 ⇒ n >4 text as n ∈ N ∴ n text smallest =5

Q. The smallest natural number 'n' so that $(2 - n) x^{2} - 8 x - 4 - n < 0, \forall x \in R$ is equal to

2

3

4

5

$We have (n - 2) x^{2} + 8 x + (n + 4) > 0.\forall x \in R$
$\Rightarrow n - 2 > 0 and D < 0$
$\Rightarrow 64 - 4 (n - 2) (n + 4) < 0$
$\Rightarrow n^{2} + 2 n - 24 > 0$
$\Rightarrow (n + 6) (n - 4) > 0$
$\Rightarrow n > 4 as n \in N$
$∴ n_{smallest} = 5$

Q. The smallest natural number 'n' so that (2−n)x2−8x−4−n<0,∀x∈R is equal to

2

3

4

5

We have (n−2)x2+8x+(n+4)>0.∀x∈R ⇒n−2>0 and D<0 ⇒64−4(n−2)(n+4)<0 ⇒n2+2n−24>0 ⇒(n+6)(n−4)>0 ⇒n>4 as n∈N ∴nsmallest ​=5

Q. The smallest natural number 'n' so that $(2 - n) x^{2} - 8 x - 4 - n < 0, \forall x \in R$ is equal to

$We have (n - 2) x^{2} + 8 x + (n + 4) > 0.\forall x \in R$
$\Rightarrow n - 2 > 0 and D < 0$
$\Rightarrow 64 - 4 (n - 2) (n + 4) < 0$
$\Rightarrow n^{2} + 2 n - 24 > 0$
$\Rightarrow (n + 6) (n - 4) > 0$
$\Rightarrow n > 4 as n \in N$
$∴ n_{smallest} = 5$