Let f:R→R be a differentiable function such that f(2)=1, f′(2)=4. If x→0lim(f(2+x))3/x=eα, and y=4x3−4x2−4(α−7)x−α, then the number of times the curve y meets the x-axis is:
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Official solution
Since f(2)=1, limx→0(f(2+x))3/x is a 1∞ form:
\[e^\alpha=e^{\lim_{x\to0}\frac3x(f(2+x)-1)}=e^{3f'(2)}=e^{12}\implies\alpha=12\]
Then y=4x3−4x2−4(5)x−12=4x3−4x2−20x−12=4(x3−x2−5x−3)=4(x+1)2(x−3).
The roots are x=−1 (a double, or tangential, root) and x=3 (a simple root). So the curve meets the x-axis at 2 distinct points: x=−1 and x=3.