If the roots α, β of the quadratic equation ( sin 6+ sin 10+ cos 10) x2+x+k2-3 k+2=0, is such that α

Question

If the roots α, β of the quadratic equation ( sin 6+ sin 10+ cos 10) x2+x+k2-3 k+2=0, is such that α<0<β, then the true set of values of k is, (A) (-∞, 1) ∪(2, ∞) (B) (1,2) (C) (-∞,-2) ∪(-1, ∞) (D) (-2,-1)

Tardigrade Admin · Accepted Answer

text Note: sin 6+ sin 10+ cos 10<0 text So, f(0)>0 ⇒ (k-1)(k-2)>0 ⇒ k ∈(-∞, 1) ∪(2, ∞)

Q. If the roots $α, β$ of the quadratic equation $(sin 6 + sin 10 + cos 10) x^{2} + x + k^{2} - 3 k + 2 = 0$ , is such that $α < 0 < β$ , then the true set of values of $k$ is,

$(- \infty, 1) \cup (2, \infty)$

$(1, 2)$

$(- \infty, - 2) \cup (- 1, \infty)$

$(- 2, - 1)$

$Note: sin 6 + sin 10 + cos 10 < 0$
$So, f (0) > 0 \Rightarrow (k - 1) (k - 2) > 0$
$\Rightarrow k \in (- \infty, 1) \cup (2, \infty)$

Q. If the roots α,β of the quadratic equation (sin6+sin10+cos10)x2+x+k2−3k+2=0, is such that α<0<β, then the true set of values of k is,

(−∞,1)∪(2,∞)

(1,2)

(−∞,−2)∪(−1,∞)

(−2,−1)