Question:
If the equation \(ax^2+4xy-2y^2+4x+8y+1 = 0\) represents a pair of straight lines, then the coordinates of their point of intersection are.........
If the equation \(ax^2+4xy-2y^2+4x+8y+1 = 0\) represents a pair of straight lines, then the coordinates of their point of intersection are.........
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Use the differential about the nearby perfect cube 0.008.
Updated On: Oct 1, 2026
- \((\frac{1}{2},-\frac{3}{2})\)
- \((-\frac{3}{2},\frac{1}{2})\)
- \((\frac{1}{2},\frac{3}{2})\)
- \((-\frac{1}{2},\frac{3}{2})\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Use \(f(x + \Delta x) \approx f(x) + f'(x)\,\Delta x\) with \(f(x) = x^{1/3}\), taking \(x = 0.008\) and \(\Delta x = -0.001\).
Step 2: Compute:
\(f(0.008) = 0.2\) and \(f'(x) = \frac{1}{3}x^{-2/3}\), so \(f'(0.008) = \frac{1}{3}\cdot\frac{1}{(0.2)^2} = \frac{1}{3\times 0.04} = \frac{25}{3}\).
\[ (0.007)^{1/3} \approx 0.2 + \frac{25}{3}(-0.001) = 0.2 - \frac{1}{120} = \frac{24 - 1}{120} = \frac{23}{120} \]
This is about 0.1917.
Final Answer:
The approximate value is \(\frac{23}{120}\), option (B). \[ \boxed{\frac{23}{120}} \]
Use \(f(x + \Delta x) \approx f(x) + f'(x)\,\Delta x\) with \(f(x) = x^{1/3}\), taking \(x = 0.008\) and \(\Delta x = -0.001\).
Step 2: Compute:
\(f(0.008) = 0.2\) and \(f'(x) = \frac{1}{3}x^{-2/3}\), so \(f'(0.008) = \frac{1}{3}\cdot\frac{1}{(0.2)^2} = \frac{1}{3\times 0.04} = \frac{25}{3}\).
\[ (0.007)^{1/3} \approx 0.2 + \frac{25}{3}(-0.001) = 0.2 - \frac{1}{120} = \frac{24 - 1}{120} = \frac{23}{120} \]
This is about 0.1917.
Final Answer:
The approximate value is \(\frac{23}{120}\), option (B). \[ \boxed{\frac{23}{120}} \]
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