Rounding in Inequality

Example 1: 2015 P1 8ii

2015 P1 Q8

Solution

RI

Examiner’s report 

exam report

Example 2: 2015 P1 2ii

2015 P1 Q2

Examiner’s report

exact round

Comments

Graphical approach will yield  x> 1.732. Notice Cambridge answer  is not rounded to x>1.74 to satisfy the non-rounded inequality but rounded to 3 s.f.

Conclusion

Considering example 1 and example 2, we conclude that in real life applications questions, when rounding off your answer to 3 s.f, round off such that the rounded off answer still satisfies the non-rounded off inequality.

Inequalities involving logarithm

Students often ask why sometimes need to flip the inequality sign when dealing with logarithm.

This is a question asked by a student:

1

2

For 0<x<1, both ln x and lg x are negative. When we divide by negative number, we need to change the sign.

For x>1, both ln x and lg x are positive. There is no need to change the sign when divide by ln x or lg x.

Challenging Inequality problem 1

Solve the inequality

Tex2Img_1402449638

From G.C, 0.5<x< 2.81, x not equal to 2 (This is the easy part)

Hence solve

Tex2Img_1402449826

This is the HOT (Higher Order Thinking) part.

Method 1

Let f(x) =

Tex2Img_1402450007

Therefore f(x+2) =

Tex2Img_1402450329

f(x+2) is translation of f(x) by 2 units in the negative X direction.

Therefore, the solution is -1.5< x<0.81, x not equal to 0

This solution combines the concept of inequality and transformation of graph.

Method 2

Replace x by x+2.

Therefore 0.5<x+2< 2.81, x+2 not equal to 2

-1.5<x<0.81, x not equal to 0

Inequalities: Use And/OR correctly

Many students use the words “And/Or” indiscriminately in solving inequalities. However, there is a very important difference. ‘And‘ means you take the intersection of the solutions sets. ‘Or‘ means you take the union of the solutions sets.

Example

inequality

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