Example 1: 2015 P1 8ii
Example 2: 2015 P1 2ii
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.
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.
Students often ask why sometimes need to flip the inequality sign when dealing with logarithm.
This is a question asked by a student:
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.
RI H2 Math Inequalities Tutorial Section C (Challenging Questions)
Consider the following 3 cases:
Given that k>0, solve the inequality
Express your answer in terms of k.
First, we determine the roots of
Then we need to determine whether the roots are bigger or smaller than k, so we can sketch the number line.
Since k> 0
Solve the inequality
From G.C, 0.5<x< 2.81, x not equal to 2 (This is the easy part)
This is the HOT (Higher Order Thinking) part.
Let f(x) =
Therefore f(x+2) =
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.
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
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.