Date | May Example questions | Marks available | 6 | Reference code | EXM.1.AHL.TZ0.2 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Express | Question number | 2 | Adapted from | N/A |
Question
Consider the integral t∫1−1x+x2 dx for t>1.
Very briefly, explain why the value of this integral must be negative.
Express the function f(x)=−1x+x2 in partial fractions.
Use parts (a) and (b) to show that ln(1+t)−lnt<ln2.
Markscheme
The numerator is negative but the denominator is positive. Thus the integrand is negative and so the value of the integral will be negative. R1AG
[1 mark]
−1x+x2=−1(1+x)x≡A1+x+Bx M1M1A1
⇒−1≡Ax+B(1+x)⇒A=1,B=−1 M1A1
−1x+x2≡11+x+−1x A1
[6 marks]
t∫111+x+−1xdx=[ln(1+x)−lnx]t1=ln(1+t)−lnt−ln2 M1A1A1
Hence ln(1+t)−lnt−ln2<0⇒ln(1+t)−lnt<ln2 R1AG
[4 marks]