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For non-zero numbers, l, m, n and a, l...

For non-zero numbers, l, m, n and a, let f(x) = `lx^(3) + mx + n and f(a) = f(4a)` . Then the value
`x in [a, 4a] ` , at which the tangent to the curve y = f(x) is parallel to the x- axis is

A

`sqrt(5) a`

B

3a

C

2a

D

`sqrt(7)a`

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AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the value of \( x \) in the interval \([a, 4a]\) where the tangent to the curve \( y = f(x) \) is parallel to the x-axis. This occurs when the derivative \( f'(x) = 0 \). ### Step-by-Step Solution: 1. **Define the function**: \[ f(x) = lx^3 + mx + n \] 2. **Find the derivative**: \[ f'(x) = \frac{d}{dx}(lx^3 + mx + n) = 3lx^2 + m \] 3. **Set the derivative to zero for the tangent to be parallel to the x-axis**: \[ 3lx^2 + m = 0 \] 4. **Solve for \( x^2 \)**: \[ 3lx^2 = -m \implies x^2 = -\frac{m}{3l} \] 5. **Find \( x \)**: \[ x = \pm \sqrt{-\frac{m}{3l}} \] 6. **Check the conditions**: We need to ensure that \( x \) lies within the interval \([a, 4a]\). Therefore, we check both \( \sqrt{-\frac{m}{3l}} \) and \( -\sqrt{-\frac{m}{3l}} \). 7. **Determine the valid value of \( x \)**: Since \( l, m, n, a \) are non-zero, the sign of \( -\frac{m}{3l} \) will determine whether \( x \) is real and valid. We will consider the positive root \( x = \sqrt{-\frac{m}{3l}} \) since \( x \) must be non-negative. 8. **Check if \( x \) is within the interval**: We need to ensure: \[ a \leq \sqrt{-\frac{m}{3l}} \leq 4a \] This leads to the conditions: \[ a^2 \leq -\frac{m}{3l} \quad \text{and} \quad -\frac{m}{3l} \leq 16a^2 \] ### Conclusion: The value of \( x \) at which the tangent to the curve \( y = f(x) \) is parallel to the x-axis is: \[ x = \sqrt{-\frac{m}{3l}} \] provided that it lies within the interval \([a, 4a]\).
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