 Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        can anyone give me the rules for SIGN SCHEME FOR RATIONAL EXPRESSION`
7 years ago

```							Dear Anindita,
Rational inequality is an inequality involving rational expression. There are four forms of inequality. Corresponding to each of these forms, there are four rational inequality forms. These inequality forms essentially compare a rational expression, f(x), with zero. The four inequalities are :
f x < 0 f x ≤ 0 f x > 0 f x ≥ 0
We need to note two important aspects of these inequalities. Solution of inequalities, in general, are not discrete values but set of “x” values in the form of interval or union of intervals. Generally, the inequality holds for a continuum of values. Second aspect is about the basic nature of inequality. We know that zero has special significance in real number system. It divides real number system in positive and negative segments. Therefore, solution of these inequalities is about knowing the sign of function values for different intervals in the domain of the function. Corresponding to four inequalities, we need to know intervals in which rational function is (i) negative (ii) non-positive (iii) positive and (iv) non-negative. In the following section, we shall devise a technique to determine sign of rational expression in different internals.

Sign scheme or diagram for rational function

Sign scheme or diagram is representation of sign in different intervals along real number line. This gives a visual idea about the sign of function. Graphically, sign of function changes when graph crosses x-axis. This means that sign of function changes about the zeroes of function i.e. about real roots of a function. However, rational function is ratio of two functions. A change of sign of either numerator or denominator affects sign of rational function.
We consider here only integral rational functions such that expressions in numerator and denominator can be decomposed into linear factors. Equating each of the linear factors, we determine points about which either or both of numerator and denominator functions change sign. We should understand that each of the linear factors is a potential source of sign change as the value of x changes in the domain. This means that each of the points so determined plays a critical role in deciding the sign of function. For this reason, we call these points as “critical points”.
Let us consider an example here :
f x = x 2 − x − 2 x 2 − 3 x − 8 = x + 1 x − 2 x + 1 x − 4
Critical points are -1, 2, -1 and 4. There are two important things to realize here. First, we can not cancel common linear factors as this will result in loosing undefined points and will loose information on sign change. The marking on real number line is as shown here :

Figure 1: Sign scheme/ diagram

Sign scheme/ diagram Second, the fact that function may change its sign in the domain has an interesting consequence. It can be better understood in terms of function graph, which is essentially a curve. The event of crossing of x-axis by the graph records the event of change of sign. Another change in the sign of graph warrants that curve should cross x-axis again. This corresponds to reversal of sign. It is not possible to change sign of function without crossing x-axis. This means that function will change sign at critical points. Equivalently, we say that sign of function alternates in consecutive sub-intervals. Now, these considerations set up the first two steps of sign diagram :
1: Decompose both numerator and denominator into linear factors. Find critical points by equating linear factors individually to zero.
2: Mark critical points on a real number line. If n be the numbers of critical points, then real number line is divided into (n+1) sub-intervals.
The question however remains that we should know sign of function in at least one interval. We determine the same by testing function value for an intermediate x-value in any of the sub-intervals. Though it is not a rule, we consider a test point in the right most interval, which extends to positive infinity. This helps us to assign signs in the intervals left to it by alternating signs. Sometimes, it may, however, be easier to evaluate function value at x= 0, 1 or -1, provided they are not the critical points. This has the advantage that calculation of function value is easier. Now, these consideration set up the next step of sign diagram :
3: Test sign of function in a particular interval. Assign alternate signs in adjacent sub-intervals.
For the example case, let us put x=0,
⇒ f 0 = 0 − 0 − 2 0 − 0 − 8 = 1 4 > 0
Thus, sign of function in the interval between -1 and 2 is positive. The signs of function alternate in adjacent sub-intervals.

Figure 2: Sign scheme/ diagram

Sign scheme/ diagram We have noted that sign of each linear factor combines to determine the sign of rational function. This fact is reflected as sign alternates in adjacent sub-intervals. However, we need to consider the effect of case in which a linear factor is repeated. If a linear factor evaluates to a positive number in an interval and is repeated, then there is no effect on the sign of function. If a linear factor evaluates to a negative number in an interval and is repeated even times, then there is no effect on the sign of function. The product of negative sign repeated even times yield a positive sign and as such does not affect the sign of function. However, if a linear factor evaluates to a negative number in an interval and is repeated odd times, then sign of function changes. Product of negative sign repeated odd times yield a negative sign and as such sign of function changes.
We conclude that if a linear factor is repeated even times, then sign of function will not alternate about the critical point corresponding to linear factor in question. On the other hand, if a linear factor is repeated odd times, then sign of function will alternate as before. Now, these consideration set up the next step of sign diagram :
4: If a linear factor is repeated even times, then sign of function will not alternate about the critical point corresponding to linear factor in question.
In the example case, the linear factor (x+1) is repeated even times (count both in numerator and denominator). As such, sign of function will not change about critical point “-1”. Thus, sign diagram drawn as above need to be modified as :

Figure 3: Sign scheme/ diagram

Sign scheme/ diagram We can verify modification due to repeated linear factors by putting x = -2 in the function :
⇒ f − 2 = − 2 2 − − 2 − 2 − 2 2 − 3 − 2 − 8 = 4 + 2 − 1 4 + 6 − 8 = 5 2 > 0
We summarize steps for drawing sign scheme/ diagram as :
1: Decompose both numerator and denominator into linear factors. Do not cancel common linear factors. Find critical points by equating linear factors individually to zero.
2: Mark distinct critical points on a real number line. If n be the numbers of distinct critical points, then real number line is divided into (n+1) sub-intervals.
3: Test sign of function in a particular interval. Assign alternate signs in adjacent sub-intervals.
4: If a linear factor is repeated even times, then sign of function will not alternate about the critical point corresponding to linear factor in question.

Solution of rational inequalities using sign scheme or diagram

An important point about interpreting sign diagram is that sign of function relates to non-zero values of function. Note that zero does not have sign. The critical points corresponding to numerator function are zeroes of rational function. As such, the graph of function is continuous at these critical points and these critical points can be included in the sub-interval. On the other hand, the rational function is not defined for critical points corresponding to denominator function (as denominator turns zero). We, therefore, conclude that an interval can include critical points corresponding to numerator function, but not the critical points corresponding to denominator function. In case, there are common critical points between numerator and denominator, then those critical points can not be included in the sub-interval.
We can interpret sign diagram in two ways. Either we determine the solution of a given quadratic inequality or we determine intervals of all four types of inequalities for a given quadratic expression. We shall illustrate these two approaches by working with the example case.

Determining solution of a given quadratic inequality

Let us consider that we are required to solve rational inequality
f x = x 2 − x − 2 x 2 − 3 x − 8 ≥ 0
The sign diagram as drawn earlier for the given rational function is shown here :

Figure 4: Sign scheme/ diagram

Sign scheme/ diagram We need to interpret signs of different intervals to find the solution of a given rational inequality.
Clearly, solution of given inequality is :
x ∈ ( - ∞ , 2 ] U 4, ∞ − { − 1 , 4 }
Note that we need to remove -1 and 4 from the solution set as function is not defined for this x – value. However, inequality involved “greater than or equal to” is not strict inequality. It allows equality to zero. As such, we include critical point “2” belonging to numerator function. Further, we can also write the solution set in alternate form as :
x ∈ - ∞ , - 1 U ( - 1,2 ] U 4, ∞

Determining interval of four quadratic inequalities

Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple  to download the toolbar….
So start the brain storming…. become a leader with Elite Expert League ASKIITIANS
Thanks
Aman Bansal

```
7 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Discuss with colleagues and IITians

View all Questions »  ### Course Features

• 728 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions