Question

If function f(x) = 2x² + 3x - m log x is monotonic decreasing in the interval (0, 1), then the least value of the parameter mis-

• A. 7
• B. 15 2
• C. 31 4
• D. 8

Answer 1

Answer: (A)

7

Answer 2

To determine the least value of the parameter $m$ for which the function $f(x) = 2x^2 + 3x - m \log x$ is monotonically decreasing in the interval $(0, 1)$, we analyze its first derivative.

1. Find the first derivative of $f(x)$: Given $f(x) = 2x^2 + 3x - m \log x$.
   Differentiating with respect to $x$:
   $f'(x) = \frac{d}{dx}(2x^2) + \frac{d}{dx}(3x) - \frac{d}{dx}(m \log x)$
   $f'(x) = 4x + 3 - m \cdot \frac{1}{x}$
   $f'(x) = 4x + 3 - \frac{m}{x}$

2. Apply the condition for monotonic decreasing function:
   A function $f(x)$ is monotonically decreasing in an interval if $f'(x) \le 0$ for all $x$ in that interval.
   So, for $f(x)$ to be monotonically decreasing in $(0, 1)$, we must have:
   $4x + 3 - \frac{m}{x} \le 0$ for all $x \in (0, 1)$.

3. Rearrange the inequality to isolate $m$:
   $4x + 3 \le \frac{m}{x}$
   Since $x \in (0, 1)$, $x$ is positive ($x > 0$). Therefore, we can multiply both sides of the inequality by $x$ without changing the direction of the inequality sign:
   $x(4x + 3) \le m$
   $4x^2 + 3x \le m$

   This inequality means that $m$ must be greater than or equal to the expression $4x^2 + 3x$ for all $x$ in the interval $(0, 1)$. To satisfy this condition, $m$ must be greater than or equal to the maximum value of the expression $4x^2 + 3x$ in the interval $(0, 1)$.

4. Find the maximum value of $g(x) = 4x^2 + 3x$ in the interval $(0, 1)$:
   Let $g(x) = 4x^2 + 3x$.
   To find its maximum value, we can find its derivative:
   $g'(x) = \frac{d}{dx}(4x^2 + 3x) = 8x + 3$.

   Now, we analyze the sign of $g'(x)$ in the interval $(0, 1)$:
   For $x \in (0, 1)$, $x > 0$.
   So, $8x > 0$.
   Therefore, $8x + 3 > 3$.
   This means $g'(x) > 0$ for all $x \in (0, 1)$.
   Since $g'(x)$ is positive throughout the interval $(0, 1)$, the function $g(x)$ is strictly increasing in this interval.

   For an increasing function, the maximum value in a given interval occurs at the right endpoint of the interval. In this case, the maximum value of $g(x)$ in $(0, 1)$ occurs at $x = 1$.
   $g_{max} = g(1) = 4(1)^2 + 3(1) = 4 + 3 = 7$.

5. Determine the least value of $m$:
   From step 3, we have $m \ge 4x^2 + 3x$.
   Since the maximum value of $4x^2 + 3x$ in $(0, 1)$ is 7, we must have:
   $m \ge 7$.

   The least value of the parameter $m$ that satisfies this condition is 7.