Let $ f $ be a differentiable function on $ \mathbb{R} $ such that $ f(2) = 4 $. Let $ \lim_{x \to 0} \left( f(2+x) \right)^{3/x} = e^\alpha $. Then the number of times the curve $ y = 4x^3 - 4x^2 - 4(\alpha - 7)x - \alpha $ meets the x-axis is:
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- 2
- 1
- 0
- 3
The Correct Option is A
Approach Solution - 1
Thus, \( \alpha = 1 + 4x \). Substituting this into the equation of the curve: \[ y = 4x^3 - 4x^2 - 4(1 + 4x - 7)x - (1 + 4x) \] Simplifying: \[ y = 4x^3 - 4x^2 - 4(-6x) - 1 - 4x = 4x^3 - 4x^2 + 24x - 1 - 4x \] \[ y = 4x^3 - 4x^2 + 20x - 1 \] Now, to find the number of times the curve meets the x-axis, we solve for \( y = 0 \): \[ 4x^3 - 4x^2 + 20x - 1 = 0 \] Using a numerical method or approximation, we find that the cubic equation has two real roots.
Therefore, the curve meets the x-axis twice.
Thus, the correct answer is \( 2 \).
Approach Solution -2
Step 1: Analyzing the given information. The given limit is: \[ \lim_{x \to 0} \frac{(f(2 + x))^3}{x} = e^{\alpha}. \] Taking the cube root of both sides, we get: \[ \lim_{x \to 0} \frac{f(2 + x)}{x^{1/3}} = e^{\alpha/3}. \] Step 2: Investigating the equation of the curve. The curve equation is given by: \[ y = 4x^3 - 4x^2 - 4(\alpha - 7)x - \alpha. \] To find the points where the curve meets the x-axis, we set \( y = 0 \): \[ 4x^3 - 4x^2 - 4(\alpha - 7)x - \alpha = 0. \] Step 3: Solving the cubic equation. The cubic equation will give the number of times the curve intersects the x-axis. The number of real roots of the cubic equation determines the answer.
Step 4: Conclusion. Given that the function is cubic, it will have 2 real roots. Therefore, the curve meets the x-axis 2 times. Final Answer: \[ \boxed{2}. \]
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