If kx 2+2 kx +(1/2)0, ∀ x ∈ R, then complete set of values of k is

Question

If kx 2+2 kx +(1/2)>0, ∀ x ∈ R, then complete set of values of k is (A) (0, (1/2)) (B) (-∞, 0) ∪((1/2), ∞) (C) ((1/2), ∞) (D) [0, (1/2))

Tardigrade Admin · Accepted Answer

text Case-I: k ≠ 0 k >0 D <0 4 k 2-4( k ) ⋅ (1/2)<0 ⇒ 4 k 2-2 k <0 ⇒ 2 k (2 k -1)<0 ∴ k ∈(0, (1/2)) text Case-II: k =0 (1/2)>0 text , which is true ∴ text solution set for k

Q. If $k x^{2} + 2 k x + \frac{1}{2} > 0, \forall x \in R$ , then complete set of values of $k$ is

$(0, \frac{1}{2})$

$(- \infty, 0) \cup (\frac{1}{2}, \infty)$

$(\frac{1}{2}, \infty)$

$[0, \frac{1}{2})$

$Case-I: k \neq = 0$
$k > 0 D < 0$
$4 k^{2} - 4 (k) \cdot \frac{1}{2} < 0$
$\Rightarrow 4 k^{2} - 2 k < 0$
$\Rightarrow 2 k (2 k - 1) < 0$
$∴ k \in (0, \frac{1}{2})$
$Case-II: k = 0$
$\frac{1}{2} > 0, which is true$
$∴ solution set for k$

Q. If kx2+2kx+21​>0,∀x∈R, then complete set of values of k is

(0,21​)

(−∞,0)∪(21​,∞)

(21​,∞)

[0,21​)