Question

If \( y = m\log x + nx^2 + x \) has its extreme values at \( x = 2 \) and \( x = 1 \), then \( 2m + 10n = \)

• A. \( -1 \)
• B. \( -4 \)
• C. \( -2 \)
• D. \( 1 \)
• E. \( -3 \)

Hint:

For extrema problems, always use derivative = 0 and form equations using given points.

Answer 1

Answer: (E)

Concept:
• Extreme values (maxima or minima) occur where: \[ \frac{dy}{dx} = 0 \]
• These are called critical points.
• We use given points to form equations and solve for constants.

Step 1: Write the given function.
\[ y = m\log x + nx^2 + x \]

Step 2: Differentiate the function.
Differentiate term-by-term: \[ \frac{d}{dx}(m\log x) = \frac{m}{x} \] \[ \frac{d}{dx}(nx^2) = 2nx \] \[ \frac{d}{dx}(x) = 1 \] Thus: \[ \frac{dy}{dx} = \frac{m}{x} + 2nx + 1 \]

Step 3: Use condition for extreme points.
Given: \[ \text{Extrema at } x = 1 \text{ and } x = 2 \] So: \[ \frac{dy}{dx} = 0 \text{ at these points} \]

Step 4: Substitute \( x = 1 \).
\[ \frac{m}{1} + 2n(1) + 1 = 0 \] \[ m + 2n + 1 = 0 \] \[ m + 2n = -1 \quad \cdots (1) \]

Step 5: Substitute \( x = 2 \).
\[ \frac{m}{2} + 2n(2) + 1 = 0 \] \[ \frac{m}{2} + 4n + 1 = 0 \] Multiply entire equation by 2 to remove fraction: \[ m + 8n + 2 = 0 \] \[ m + 8n = -2 \quad \cdots (2) \]

Step 6: Solve simultaneous equations.
From (1): \[ m + 2n = -1 \] From (2): \[ m + 8n = -2 \] Subtract (1) from (2): \[ (m + 8n) - (m + 2n) = -2 - (-1) \] \[ 6n = -1 \] \[ n = -\frac{1}{6} \]

Step 7: Find \( m \).
Substitute into (1): \[ m + 2\left(-\frac{1}{6}\right) = -1 \] \[ m - \frac{1}{3} = -1 \] \[ m = -1 + \frac{1}{3} \] \[ m = -\frac{2}{3} \]

Step 8: Compute required value.
\[ 2m + 10n \] Substitute values: \[ 2\left(-\frac{2}{3}\right) + 10\left(-\frac{1}{6}\right) \] \[ = -\frac{4}{3} - \frac{10}{6} \] Convert to common denominator: \[ = -\frac{4}{3} - \frac{5}{3} \] \[ = -\frac{9}{3} = -3 \]

Step 9: Final Answer.
\[ \boxed{-3} \]