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首先貼下大陸高考past paper:

Given f(x) is a monotoinc increasing odd function. Consider f(k) > f(1/e^x - e^x - 2),

discuss the domain and range of k.

Hint: For odd function, f(-x) = -f(x)

Solution:

Denote t=e^x such that t>0

Now f(k) < f(t - 1/t + 2)

Since both t and -1/t are also monotonic increasing functions, we have:

1/t - t - 2 < 0 < k < t - 1/t + 2

Question:

Consider y=ln(|x|).

Explain why y' = 1/x for all real x =/= 0.

Solution:

First method: Use parity

ln(|-x|) = ln(x) = ln (|x|)

So ln (|x|) is an even function .

We can directly write ln'(|x|) = ln'(x) = 1/x

Second method: Use chain rule for ln(-x)

d(ln(-x))/dx

= [d(ln(-x))/d(-x)][d(-x)/dx]

= [1(-x)][-1] = 1/x

Third method: Use monotonicity

1/x is always monotonic decreasing for all x=/=0,

So ln'(-x) = ln'(x)

QED

Question:

Notice the following equations of straight line:

y = mx + c

y - y_0 = m (x - x_0)

Which one should be used when finding equations of a tangent and a normal?

Solution:

For y = mx +c, m=/=0 due to the definition of a polynomial that the leading coefficient cannot be zero. Therefore, this equation is not applicable for horizontal lines and vertical lines.

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