Let f:R arrow R be a function defined by f(x)=-x3-3x2-6x+1. Number of integers in the solution set of x satisfying the inequality f(f (x3 + f (x)) ge f(f (- f (x) - x3))) is

Question

Let f:R arrow R be a function defined by f(x)=-x3-3x2-6x+1. Number of integers in the solution set of x satisfying the inequality f(f (x3 + f (x)) ge f(f (- f (x) - x3))) is (A) 3 (B) 4 (C) 5 (D) 6

Tardigrade Admin · Accepted Answer

f'(x) < 0 ∴ f(x3 + f (x) le f(- f (x) - x3)) ⇒ x3+f(x)≥ -f(x)-x3 ⇒ f(x)+x3≥ 0 ⇒ 3x2+6x-1≤ 0 As x∈ Z,x∈ .-2,-1,0.

Q. Let $f : R \to R$ be a function defined by $f (x) = - x^{3} - 3 x^{2} - 6 x + 1.$ Number of integers in the solution set of $x$ satisfying the inequality $f (f (x^{3} + f (x)) \geq f (f (- f (x) - x^{3})))$ is

$3$

$4$

$5$

$6$

$f^{^{'}} (x) < 0$
$∴ f (x^{3} + f (x) \leq f (- f (x) - x^{3}))$
$\Rightarrow x^{3} + f (x) \geq - f (x) - x^{3}$
$\Rightarrow f (x) + x^{3} \geq 0$
$\Rightarrow 3 x^{2} + 6 x - 1 \leq 0$
As $x \in Z, x \in {- 2, - 1, 0}$

Q. Let f:R→R be a function defined by f(x)=−x3−3x2−6x+1. Number of integers in the solution set of x satisfying the inequality f(f(x3+f(x))≥f(f(−f(x)−x3))) is

3

4

5

6

f′(x)<0 ∴f(x3+f(x)≤f(−f(x)−x3)) ⇒x3+f(x)≥−f(x)−x3 ⇒f(x)+x3≥0 ⇒3x2+6x−1≤0 As x∈Z,x∈{−2,−1,0}

Q. Let $f : R \to R$ be a function defined by $f (x) = - x^{3} - 3 x^{2} - 6 x + 1.$ Number of integers in the solution set of $x$ satisfying the inequality $f (f (x^{3} + f (x)) \geq f (f (- f (x) - x^{3})))$ is

$3$

$4$

$5$

$6$

$f^{^{'}} (x) < 0$
$∴ f (x^{3} + f (x) \leq f (- f (x) - x^{3}))$
$\Rightarrow x^{3} + f (x) \geq - f (x) - x^{3}$
$\Rightarrow f (x) + x^{3} \geq 0$
$\Rightarrow 3 x^{2} + 6 x - 1 \leq 0$
As $x \in Z, x \in {- 2, - 1, 0}$