Let f(x)=(2 x3/3)-(x ln |x|-x)-k x+10. If f(x) is strictly increasing for every x ∈ R- 0 then the maximum integral value of k is

Question

Let f(x)=(2 x3/3)-(x ln |x|-x)-k x+10. If f(x) is strictly increasing for every x ∈ R- 0 then the maximum integral value of k is (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

f prime( x )=2 x 2- ln | x |- k ≥ 0 ∀ x ∈ R - 0 k ≤ 2 x 2- ln | x | ∀ x ∈ R - 0 Hence, k ≤ minimum value of 2 x 2- ln | x | ∴ k ≤ (1/2)+ ln 2 Hence, maximum integral value of k is 1 .

Q. Let $f (x) = \frac{2 x ^{3}}{3} - (x ln ∣ x ∣ - x) - k x + 10$ . If $f (x)$ is strictly increasing for every $x \in R - {0}$ , then the maximum integral value of $k$ is

0

1

2

3

$f^{'} (x) = 2 x^{2} - ln ∣ x ∣ - k \geq 0\forall x \in R - {0}$
$k \leq 2 x^{2} - ln ∣ x ∣\forall x \in R - {0}$
Hence, $k \leq$ minimum value of $2 x^{2} - ln ∣ x ∣$
$∴ k \leq \frac{1}{2} + ln 2$
Hence, maximum integral value of $k$ is 1 .

Q. Let f(x)=32x3​−(xln∣x∣−x)−kx+10. If f(x) is strictly increasing for every x∈R−{0}, then the maximum integral value of k is

0

1

2

3

f′(x)=2x2−ln∣x∣−k≥0∀x∈R−{0} k≤2x2−ln∣x∣∀x∈R−{0} Hence, k≤ minimum value of 2x2−ln∣x∣ ∴k≤21​+ln2 Hence, maximum integral value of k is 1 .

Q. Let $f (x) = \frac{2 x ^{3}}{3} - (x ln ∣ x ∣ - x) - k x + 10$ . If $f (x)$ is strictly increasing for every $x \in R - {0}$ , then the maximum integral value of $k$ is

$f^{'} (x) = 2 x^{2} - ln ∣ x ∣ - k \geq 0\forall x \in R - {0}$
$k \leq 2 x^{2} - ln ∣ x ∣\forall x \in R - {0}$
Hence, $k \leq$ minimum value of $2 x^{2} - ln ∣ x ∣$
$∴ k \leq \frac{1}{2} + ln 2$
Hence, maximum integral value of $k$ is 1 .