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The number of non-zero integral solution of K for which the equation `(x^(3))/(3) - 4x= K` has three distinct solution is

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To solve the problem, we need to find the number of non-zero integral solutions for \( K \) such that the equation \[ \frac{x^3}{3} - 4x = K \] has three distinct solutions. ### Step 1: Define the function Let \[ f(x) = \frac{x^3}{3} - 4x \] ### Step 2: Differentiate the function We differentiate \( f(x) \) to find its critical points: \[ f'(x) = \frac{d}{dx}\left(\frac{x^3}{3}\right) - \frac{d}{dx}(4x) = x^2 - 4 \] ### Step 3: Find critical points Set the derivative equal to zero to find the critical points: \[ x^2 - 4 = 0 \] This gives us: \[ x^2 = 4 \implies x = -2 \text{ and } x = 2 \] ### Step 4: Analyze the sign of the derivative To determine the nature of these critical points, we analyze the sign of \( f'(x) \): - For \( x < -2 \): \( f'(x) > 0 \) (increasing) - For \( -2 < x < 2 \): \( f'(x) < 0 \) (decreasing) - For \( x > 2 \): \( f'(x) > 0 \) (increasing) This indicates that \( x = -2 \) is a local maximum and \( x = 2 \) is a local minimum. ### Step 5: Evaluate the function at critical points Next, we evaluate \( f(x) \) at the critical points: \[ f(-2) = \frac{(-2)^3}{3} - 4(-2) = \frac{-8}{3} + 8 = \frac{-8 + 24}{3} = \frac{16}{3} \] \[ f(2) = \frac{(2)^3}{3} - 4(2) = \frac{8}{3} - 8 = \frac{8 - 24}{3} = \frac{-16}{3} \] ### Step 6: Determine the range for \( K \) For the function \( f(x) \) to have three distinct solutions, \( K \) must lie between the maximum and minimum values: \[ f(2) < K < f(-2) \implies \frac{-16}{3} < K < \frac{16}{3} \] ### Step 7: Find the integer solutions Now we need to find the integer values of \( K \) that satisfy: \[ -5.33 < K < 5.33 \] The integers in this range are: \[ -5, -4, -3, -2, -1, 1, 2, 3, 4, 5 \] ### Step 8: Count the non-zero integral solutions The non-zero integers are: \[ -5, -4, -3, -2, -1, 1, 2, 3, 4, 5 \] This gives us a total of 10 non-zero integral solutions for \( K \). ### Final Answer Thus, the number of non-zero integral solutions for \( K \) is \[ \boxed{10} \]
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