Let f(.x.)=[(1/c o s . x . )] and g(.x.)=2x2-3x(.k+1.)+k(.3k+1.) (where [.x]. .x. denote the greatest integer and fractional part function respectively) are two real valued function. If g(.f(.x.).)

Question

Let f(.x.)=[(1/c o s . x . )] and g(.x.)=2x2-3x(.k+1.)+k(.3k+1.) (where [.x]. .x. denote the greatest integer and fractional part function respectively) are two real valued function. If g(.f(.x.).) < 0, ∀ x∈ R , then the number of integral value of k is : (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

∵ f(x)=[(1/ cos x )]=1 ∴ g(f(x))=g(1) < 0 ⇒ 2-3(k+1)+k(3 k+1)<0 3 k2-2 k-1<0 ⇒ k ∈(-(1/3), 1) ∴ The integral value of k is 0

Q. Let $f (x) = [\frac{1}{cos { x }}]$ and $g (x) = 2 x^{2} - 3 x (k + 1) + k (3 k + 1)$ (where $[x] & {x}$ denote the greatest integer and fractional part function respectively) are two real valued function. If $g (f (x)) < 0,$ $\forall x \in R$ , then the number of integral value of $k$ is :

$∵ f (x) = [\frac{1}{c o s { x }}] = 1$
$∴ g (f (x)) = g (1) < 0$
$\Rightarrow 2 - 3 (k + 1) + k (3 k + 1) < 0$
$3 k^{2} - 2 k - 1 < 0$
$\Rightarrow k \in (- \frac{1}{3}, 1)$
$∴$ The integral value of $k$ is $0$

Q. Let f(x)=[cos{x}1​] and g(x)=2x2−3x(k+1)+k(3k+1) (where [x]&{x} denote the greatest integer and fractional part function respectively) are two real valued function. If g(f(x))<0, ∀x∈R , then the number of integral value of k is :

∵f(x)=[cos{x}1​]=1 ∴g(f(x))=g(1)<0 ⇒2−3(k+1)+k(3k+1)<0 3k2−2k−1<0 ⇒k∈(−31​,1) ∴ The integral value of k is 0

Q. Let $f (x) = [\frac{1}{cos { x }}]$ and $g (x) = 2 x^{2} - 3 x (k + 1) + k (3 k + 1)$ (where $[x] & {x}$ denote the greatest integer and fractional part function respectively) are two real valued function. If $g (f (x)) < 0,$ $\forall x \in R$ , then the number of integral value of $k$ is :

$∵ f (x) = [\frac{1}{c o s { x }}] = 1$
$∴ g (f (x)) = g (1) < 0$
$\Rightarrow 2 - 3 (k + 1) + k (3 k + 1) < 0$
$3 k^{2} - 2 k - 1 < 0$
$\Rightarrow k \in (- \frac{1}{3}, 1)$
$∴$ The integral value of $k$ is $0$