First problem:

Lets look at the first few terms of the Taylor series for f(x)=ln(1+x) centred

at 0.

f(0)=0

df/dx=1/(1+x),

so df/dx(0) = 1

d^2f/dx^2=-1/(1+x)^2

so d^2f/dx^2(0)=-1.

So the series starts:

f(x)=0+x-(1/2)x^2+...

Now look at your candidate answers. The first two can be ruled out as

the absolute value of the coefficient of x^2 is 2.

That leaves us with the second two. The fourth gives a -ve coefficient for

x, so it is ruled out, which leaves us with (c), which agrees with the terms

of the series that we know.

RonL