Category Archives: IITJEE Foundation mathematics

How to solve equations: Dr. Vicky Neale: useful for Pre-RMO or even RMO training

Dr. Neale simply beautifully nudges, gently encourages mathematics olympiad students to learn to think further on their own…

A nice dose of practice problems for IITJEE Foundation math and PreRMO

It is said that “practice makes man perfect”.

Problem 1:

Six boxes are numbered 1 through 6. How many ways are there to put 20 identical balls into  these boxes so that none of them is empty?

Problem 2:

How many ways are there to distribute n identical balls in m numbered boxes so that none of the boxes is empty?

Problem 3:

Six boxes are numbered 1 through 6. How many ways are there to distribute 20 identical balls between the boxes (this time some of the boxes can be empty)?

Finish this triad of problems now!

Nalin Pithwa.

IITJEE Foundation Math and PRMO (preRMO) practice: another random collection of questions

Problem 1: Find the value of \frac{x+2a}{2b--x} + \frac{x-2a}{2a+x} + \frac{4ab}{x^{2}-4b^{2}} when x=\frac{ab}{a+b}

Problem 2: Reduce the following fraction to its lowest terms:

(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \div (\frac{x+y+z}{x^{2}+y^{2}+z^{2}-xy-yz-zx} - \frac{1}{x+y+z})+1

Problem 3: Simplify: \sqrt[4]{97-56\sqrt{3}}

Problem 4: If a+b+c+d=2s, prove that 4(ab+cd)^{2}-(a^{2}+b^{2}-c^{2}-d^{2})^{2}=16(s-a)(s-b)(s-c)(s-d)

Problem 5: If a, b, c are in HP, show that (\frac{3}{a} + \frac{3}{b} - \frac{2}{c})(\frac{3}{c} + \frac{3}{b} - \frac{2}{a})+\frac{9}{b^{2}}=\frac{25}{ac}.

May u discover the joy of Math! 🙂 🙂 🙂

Nalin Pithwa.

A cute illustrative problem on basics of inequalities: IITJEE Foundation Maths


Graph the functions f(x)=\frac{3}{x-1} and \frac{2}{x+1} together to identify the values of x for which we can get \frac{3}{x-1} < \frac{2}{x+1}

Also, confirm your findings in the above graph algebraically/analytically.

Also, then find the points of intersection of these graphs using a TI graphing calculator and also algebraically/analytically.


Just a suggestion, it helps to have TI graphing calculator, a bit pricey, but …it might help a lot…If you wish, you can grab one from the following:

Of course, there might be excellent freeware software packages that plot graphs, but I am strongly biased: I like the TI graphing calculator toy very much!! I go to the extent of claiming that it is better for a kid to have a nice graphing calculator like a TI than a smartphone! 🙂 🙂 🙂

Later on, I hope to show how a detailed study of graphs helps right from high school level. In fact, an immortal Russian mathematician, I. M. Gel’fand had written a book titled “Functions and Graphs” for high-school children. You can check if it is available in Amazon India or Flipkart or Infibeam, etc.

Oh, one more thing: I love this topic of inequalities because of several reasons. Just now, one of my students, Darpan Gajra asked me certain questions about inequalities when he tried to solve the above problem —- I think his questions were good, and I hope this little explanation that I gave him helps many of other students also.


Consider the given inequality: \frac{3}{x-1} < \frac{2}{x+1}.

You might be tempted to solve this very fast in the following way:

Just cross-multiply: So, we get 3(x+1)<2(x-1), which in turn means 3x+3<2x-2, that is, x<-5. This is the answer by “fluke”!!!

In fact, the problem requires a detailed solution as follows:

Firstly, let us see what is wrong with the above fast solution:

Consider any inequality a<b. Now, we know that only if x>0, then ax<bx. But, if x<0, then ax>bx. Now, in the present question, if you simply cross-multiply, you are not clearing assuming whether (x-1)>0 or (x-1)<0; (x+1)>0 or (x+1)<0.

Also, one simple yet, I call it a golden rule is: even though it is not mentioned explicitly in the question, when you solve the question, immediately write the big restrictions: x \neq 1, x \neq -1 as these values make the denominators of the two fractions zero. Develop this good habit.

So, now, coming to the solution:

\frac{3}{x-1} < \frac{2}{x+1}, where x \neq \pm 1

\frac{3}{x-1} - \frac{2}{x+1}<0



Case I: (x^{2}-1)>0: Multiplying both sides of the above fraction inequality by (x^{2}-1) gives us:

x+5<0, that is, x<-5, but also (x^{2}-1)>0 is a restriction.  So, it means that x<-5 AND (x-1)(x+1)>0.

Subcase Ia: (x+1)>0 and (x-1)>0, which gives, x>-1 and x>1, which together imply that x>1, but we need x<-5 also as a restriction/assumption. So, in this subcase Ia, solution set is empty.

Subcase Ib: x+1<0 and x-1<0, which in turn imply, x<-1 and x<1, that is, x<-1, but we need x<-5 also. Hence, in subcase Ib: x<-5 and x<-1, we get x<-5 as the solution set.

Case II: (x^{2}-1)<0: Multiplying both sides of \frac{x+5}{x^{2}-1}<0 by (x^{2}-1), we get:

(x+5)>0, hence, x>-5. But, we also need the restriction/assumption: (x^{2}-1)<0, which implies that (x+1)(x-1)<0.

Subcase IIa: (x+1)>0 and (x-1)<0, that is, -1<x<1

So, in subcase IIa, we have (x>-5) \bigcap (-1<x<1), that is, (-1<x<1).

Subcase IIb: (x+1)<0, and (x-1)>0, that is, x<-1 and x>1. But, this itself is an empty set. Hence, in subcase IIb, the solution set is empty.

Now, the final solution is \{ case I\} OR \{ case II\}, that is, \{ case I\} \bigcup \{ case II\}, that is, \{ x<-5\}\bigcup {-1<x<1}


a) Graph the function f(x)=\frac{x}{2} and g(x)=1+\frac{4}{x} together to identify the values of x for which \frac{x}{2}>1+\frac{4}{x}

(b) Confirm your findings in (a) algebraically.

More later,

Nalin Pithwa.

Are kids no longer learning multiplication tables in school?

This topic is very close to my heart and head. I had written a remark about this in an earlier blog article. 

Today, I suddenly found some “echoes” or “resemblances” to my views. Kindly let me re-blog, or share the opinions about this from today’s EETimes, authored by Max Maxfield, Designline Editor. The URL is pasted below:;

Thanks Max and EETimes !

From Nalin Pithwa.

System of Equations — IITJEE Foundation Maths

Some problems are presented below as warm-up exercises:

Question 1:

Solve : x+y = 15 and xy=36

Question 2:

Solve: x-y=12 and xy=85

Question 3:

Solve: x^{2}+y^{2}=74 and xy=35

Question 4:

Solve : x^{2}+y^{2}=185 and x+y=17

Some other standard set of systems of equations and techniques for solving them are presented below:

Any pair of equations of the form

x^{2} \pm pxy + y^{2}=a^{2}….Equation I

x \pm y = b…Equation II

where p is any numerical quantity, can be reduced to one of the cases in the warm-up exercises; for, by squaring II and combining with I, an equation to find xy is obtained, the solution can then be completed by the aid of equation II.

Example 1:

Solve x^{3}-y^{3}=999…Equation 1

and x-y=3…Equation 2.

By division, x^{2}+xy+y^{2}=333…Equation 3.

From Equation 2, x^{2} - 2xy + y^{2} = 9;

by subtraction, 3xy=324 and xy=108…Equation 4.

From Equation (2) and (4), x=12, y=9 or x=9, y = -12

Example 2:

Solve: x^{4}+x^{2}y^{2}+y^{4}=2613…Equation I

and x^{2}+xy+y^{2}=67…Equation 2

Dividing (1) by (2), x^{2}-xy+y^{2}=39…Equation 3.

From Equation (2) and (3), by addition, x^{2}+y^{2}=33; by subtraction xy=14, hence, we get the following: x = \pm 7, x = \pm 2, or y = \pm 2, y = \pm 7

Example 3:

Solve: \frac{1}{x} - \frac{1}{y} = \frac{1}{3}…Equation (1)

and \frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{5}{9}…Equation (2)

From (1), by squaring, \frac{1}{x^{2}} - \frac{2}{xy} + \frac{1}{y^{2}} = \frac{1}{9},

and by subtraction, \frac{2}{xy} = \frac{4}{9}

adding to (2), \frac{1}{x^{2}} + \frac{2}{xy} + \frac{1}{y^{2}} = 1

and hence, \frac{1}{x} + \frac{1}{y} = \pm 1.

Combining with (1), \frac{1}{x} = \frac{2}{3}, or -\frac{1}{3}; \frac{1}{y} = - \frac{1}{3}, or -\frac{2}{3}; so, the solutions are: x = \frac{3}{2}, -3 and y =3, -\frac{3}{2}.

The following method of solution may always be used when the equations are of the same degree and homogeneous.


Solve x^{2}+ xy + 2y^{2} = 74 …Equation 1

and 2x^{2} + 2xy + y^{2} = 73…Equation 2.

Put y = mx, and substitute in both equations. Thus,

x^{2}(1+m+2m^{2}) =74…Equation 3.

x^{2}(2 + 2m + m^{2}) =73…Equation 4.

By division, \frac{1+m+2m^{2}}{2+2m+m^{2}} = \frac{74}{73}

73 + 73m + 146m^{2} = 148 + 148m + 74m^{2}

Hence, 72m^{2} -75m  - 75=0; or 24m^{2} - 25m - 25 = 0; hence, (8m + 5)(3m - 5) = 0; so m = -\frac{5}{8} or \frac{5}{3}.

Case (i):

Take m= -\frac{5}{8} and substitute in either (3) or (4).

From (3): x^{2}(1- \frac{5}{8} + \frac{50}{64}) = 74

x^{2} = 64, so x = \pm 8; so y = mx = - \frac{5}{8}x = \mp 5.

Case (ii):

Take m = \frac{5}{3}, then from (3) x^{2}(1 + \frac{5}{3} + \frac{50}{9}) = 74, so we get x^{2} = 9, so x = \pm 3; and hence, y = mx = \frac{5}{3}x = \pm 5.

When one of the equations is of first degree and the other of a higher degree, we may from the simple equation, find the value of one of the unknowns in terms of the other, and substitute in the second equation.


Solve 3x - 4y = 5…Equation (1)

and 3x^{2} - xy - 3y^{2} = 21…Equation (2)

From (1), we have x = \frac{5+4y}{3}

and substituting in (2), 3\frac{(5+4y)^{2}}{9} - \frac{y(5+4y)}{3} - 3y^{2}= 21

75 + 120y +48y^{2} - 15y - 12y^{2} - 27y^{2} = 189

9y^{2} + 105y -114 = 0

3y^{2} + 35y - 38 = 0

(y-1)(3y+31) = 0

Hence, y = 1, -\frac{38}{3}, and x = 3, -\frac{137}{9}.

The examples we have presented above will be sufficient as a general explanation of the methods to be employed; but, in some cases, special artifices are necessary.


Solve: x^{2} + 4xy + 3x = 40 - 6y -4y^{2}…Equation (1)

and 2xy - x^{2} = 3…Equation (2)

From (1) and (2), we have x^{2} + 4xy + 4y^{2} + 3x + 6y = 40;

that is, (x+2y)^{2} + 3(x+2y) - 40 = 0;

or, ((x+2y)+8)((x+2y)-5) = 0, hence, x+2y = -8, 5


Combining x+2y =5 with (2), we obtain 2x^{2} - 5x + 3=0, so x = 1, \frac{3}{2}; and by substituting in x + 2y = 5, we get y = 2, -\frac{7}{4}.


Combining  x+2y = -8 with (2), we obtain 2x^{2} + 8x + 3 = 0; hence, x = \frac{-4 \pm \sqrt{10}}{2}, and y = \frac{-12 \mp \sqrt{10}}{4}


Solve: x^{2}y^{2} - 6x = 34 - 3y…Equation (1)

and 3xy + y = 2((9+x)…Equation (2)

From (1), we have x^{2}y^{2} -6x + 3y = 34;

From (2), we have 9xy - 6x + 3y = 54;

By subtraction, x^{2}y^{2} - 9xy + 20 = 0; that is, (xy - 5)(xy -4) = 0; so xy = 5, 4.


Substituting xy = 5 in (2), gives y -2x = 3. From these equations, we obtain x =1 or -\frac{5}{2}; y = 5, or -2.


Substituting xy =4 in (2), gives y - 2x = 6. From these equations, we obtain x = \frac{-3 \pm \sqrt{17}}{2}, and y = 3 \pm \sqrt{17}.

Hope you enjoyed it,

Nalin Pithwa.










Some problems for IITJEE Foundation mathematics

Problem 1.

A cistern can be filled by two pipes in 33\frac{1}{3}; if the larger pipe takes 15 minutes less than the smaller to fill the cistern, find in what time it will be filled by each pipe singly.

Problem 2.

By rowing half the distance and walking the other half, a man can travel 24 km. on a river in 5 hours with the stream, and in 7 hours against the stream. If there were no current, the journey would take 5\frac{2}{3} hours; find the rate of his walking, and rowing and the rate of the stream.

Problem 3.

Factorize: 2a^{2}x^{2}-2(3b-4c)(b-c)y^{2}+abxy

Problem 4:

Find the fourth root of 81x^{4}-216x^{3}y+216x^{2}y^{2}-96x^{3}y+16y^{4}

Problem 5:

Find the sixth root of


More interesting stuff on IITJEE foundation maths later,

Nalin Pithwa




A little note : IITJEE foundation maths


Solve 2(x^{2}-6)=3(x-4)


We have 2x^{2}-12=3x-12

that is, 2x^{2}=3x call this as Equation (1)

Transposing, 2x^{2}-3x=0


Hence, x=0, or 2x-3=0

Thus, the roots are 0 and \frac{3}{2}


In equation (1), above, we might have divided both sides by x and obtained the simple equation 2x=3, hence, x=\frac{3}{2}, which is one of the solutions of the given equation. But, the student must be particularly careful to notice that whenever an x, or a factor containing x, is removed by division from every term of an equation, it must not be neglected, since the equation is satisfied by x=0, which is therefore, one of the roots.

More on IITJEE foundation maths later,

Nalin Pithwa

Cyclic Fractions for IITJEE foundation maths

Consider the expression


Here, in finding the LCM of the denominators, it must be observed that there are not six different compound factors to be considered; for, three of them differ from the other three only in sign.


(a-c)  =  -(c-a)

(b-a) = -(a-b)

(c-b) = -(b-c)

Hence, replacing the second factor in each denominator by its equivalent, we may write the expression in the form

-\frac{1}{(a-b)(c-b)}-\frac{1}{(b-c)(a-b)}-\frac{1}{(c-a)(b-c)} call this expression 1

Now, the LCM is (b-c)(c-a)(a-b)

and the expression is \frac{-(b-c)-(c-a)-(a-b)}{(b-c)(c-a)(a-b)}=0.,

Some Remarks:

There is a peculiarity in the arrangement of this example, which is desirable to notice. In the expression 1, the letters occur in what is known as cyclic order; that is, b follows a, a follows c, c follows b. Thus, if a, b, c are arranged round the circumference of a circle, if we may start from any letter and move round in the direction of  the arrows, the other letters follow in cyclic  order; namely, abc, bca, cab.

The observance of this principle is especially important in a large class of examples in which the differences of three letters are involved. Thus, we are observing cyclic order when we write b-c, c-a, a-b, whereas we are violating order by the use of arrangements such as b-c, a-c, a-b, etc. It will always be found that the work is rendered shorter and easier by following cyclic order from the beginning, and adhering to it throughout the question.


(1) Find the value of \frac{a}{(a-b)(a-c)} + \frac{b}{(b-c)(b-a)} + \frac{c}{(c-a)(c-b)}

2) Find the value of \frac{b}{(a-b)(a-c)} + \frac{c}{(b-c)(b-a)} + \frac{a}{(c-a)(c-b)}

3) Find the value of \frac{z}{(x-y)(x-z)} + \frac{x}{(y-z)(y-x)} + \frac{y}{(z-x)(z-y)}

4) Find the value of \frac{y+z}{(x-y)(x-z)} + \frac{z+x}{(y-z)(y-x)} + \frac{x+y}{(z-x)(z-y)}

5) Find the value of \frac{b-c}{(a-b)(a-c)} + \frac{c-a}{(b-c)(b-a)} + \frac{a-b}{(c-a)(c-b)}

More later,

Nalin Pithwa

Some properties of continuous functions


If g,f : [a,b] \rightarrow \Re are continuous functions and c is a constant, then

a) f+g is a continuous functions.

b) f-g is a continuous functions.

c) cf is a continuous function.

d) fg is a continuous function.


We shall only prove statement d. Choose and fix any \varepsilon >0. Since is continuous at x_{0}, we have that for the positive number \frac{\varepsilon}{(2|g(x_{0})|+1} there exists a \delta_{1}>0 such that

|f(x)-f(x_{0}|< \frac{\varepsilon}{2(|g(x_{0}|+1)} whenever |x-x_{0}|<\delta_{1}

Since ||f(x)|-|f(x_{0})|| \leq |f(x)-f(x_{0})|, we conclude that

|f(x)|<|f(x_{0})|+\frac{\varepsilon}{2(|g(x_{0})|+1)}, whenever |x-s_{0}|<\delta_{1}. Let |f(x_{0})|+\varepsilon 2(|g(x_{0})|+1)=M. Also, since g is continuous at x_{0}, for the positive number \frac{\varepsilon}{2M}, there is a \delta_{2}>0 such that |g(x)-g(x_{0})|< \frac{\varepsilon}{2M} whenever |x-x_{0}|<\delta_{2}. Put \delta=min(\delta_{1},\delta_{2}). Then, whenever |x-x_{0}|<\delta, we have

|f(x)g(x)-f(x_{0})g(x_{0})| equals


\leq |f(x)||g(x)-g(x_{0})|+|g(x_{0})(f(x)-f(x_{0}))|

which equals


< M.\frac{\varepsilon}{2M}+|g(x_{0})|.\frac{\varepsilon}{2(|g(x_{0}|+1)}

which equals \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon

Observe that we have not claimed that the quotient of two continuous functions is continuous. The problem is obvious: \frac{f(x)}{g(x)} cannot have any meaning at x for which g(x)=0. So, the question would be, if g(x) \neq 0 for every x \in [a,b], is the function h:[a,b] \rightarrow \Re, defined by h(x)=\frac{f(x)}{g(x)}, continuous? The answer is yes. For a proof, we need a preliminary result.


if g:[a,b] \rightarrow \Re is continuous and g(x_{0}) \neq 0, then there is an m > 0 and \delta >0 such that if x_{0}-\delta<x<x_{0}+\delta, then |g(x)|>m.


Let |g(x_{0})|=2m. Now, m>0. By continuity of g, there is a \delta>0 such that

|g(x)-g(x_{0})|<m for x_{0}-\delta<x<x_{0}+\delta

But, |g(x)-g(x_{0})| \geq ||g(x)|-|g(x_{0})|| and hence, $-m<|g(x)|-|g(x_{0})|<m$, giving us

m=|g(x_{0})|-m<|g(x)| for x_{0}-\delta<x<x_{0}+\delta. Hence, the proof.

The lemma says that if a continuous function does not vanish at a point, then there is an interval containing it in which it does not vanish at any point.


If f,g :[a,b] \rightarrow \Re are continuous and g(x) \neq 0 for all x, then h:[a,b] \rightarrow \Re defined by h(x)=\frac{f(x)}{g(x)} is continuous.

The proof of the above theorem using the lemma above is left as an exercise.


a) f:\Re \rightarrow \Re defined by f(x)=a_{0} for all x \in \Re, where a_{0} is continuous.

b) f:\Re \rightarrow \Re defined by f(x)=x is continuous.

c) g:\Re \rightarrow \Re defined by g(x)=x^{2} is a continuous function because g(x)=f(x)f(x), where f(x)=x. Since f is continuous by (b), g must be continuous.

d) h:\Re \rightarrow \Re by h(x)=x^{n}, n being a positive integer, is continuous by repeated application of the above reasoning.

e) p: \Re \rightarrow \Re defined by p(x)=a_{0}+a_{1}x+\ldots +a_{n}x^{n}, where a_{0}, a_{1}, \ldots , a_{n} is also continuous. This is because of the fact that if

f_{1}, f_{2}, f_{3} \ldots, f_{n}:\Re \rightarrow \Re are defined by f_{1}(x)=x, f_{2}=x^{2}, …, f_{n}=x^{n}, then a_{1}f_{1}, a_{2}f_{2}, \ldots a_{n}f_{n} are also continuous functions. Hence,

a_{0}+a_{1}f_{1}+ \ldots +a_{n}f_{n}=p is also a continuous function as the sum of continuous functions is a continuous function. Thus, we have shown that a polynomial is a continuous function.

f) Let p and q be polynomials. Let \alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} \in \Re be such that q(\alpha_{1})=q(\alpha_{2})=\ldots=q(\alpha_{n})  and q(\alpha) \neq 0 for \alpha \neq \alpha_{1}, \alpha \neq \alpha_{2}, \ldots, \alpha neq \alpha_{n}.

Now, let D = \Re - \{ \alpha_{1}, \alpha_{2}, \ldots , \alpha_{n}\}.

Then, h:D \rightarrow \Re defined by h(x)=\frac{p(x)}{q(x)} is a continuous function. What we have said is that a rational function which is defined everywhere except on the finite set of zeroes of the denominator is continuous.

g) f:\Re \rightarrow \Re defined by f(x)=\sin{x} is continuous everywhere. Indeed, f(x)-f(x_{0})=\sin{x}-\sin{x_{0}}=2\sin{\frac{x-x_{0}}{2}}\cos{\frac{x+x_{0}}{2}}. Therefore,

|f(x)-f(x_{0})|=2|\sin{\frac{(x-x_{0})}{2}}| |\cos{\frac{(x+x_{0})}{2}}|\leq |x-x_{0}| (because |\sin{x}| \leq |x|, where x is measured in radians)

h) f:\Re \rightarrow \Re defined by f(x)=\cos{x} is continuous since

|f(x)-f(x_{0})|=|\cos{x}-\cos{x_{0}}|=2|\sin{\frac{(x_{0}-x)}{2}}\sin{\frac{x+x_{0}}{2}}| \leq |x-x_{0}|

i) f:\Re - \{ (2n+1)\frac{\pi}{2}: n \in \mathbf{Z}\} \rightarrow \Re defined by f(x)=\tan{x} is continuous. We had to omit numbers like \ldots, \frac{-3\pi}{2}, \frac{-\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \ldots from the domain of f as \tan{x} cannot be defined for these values of x.

j) f:\Re_{+} \rightarrow \Re defined by f(x)=x^{1/n} is a continuous function. Indeed,

f(x)-f(a)=x^{1/n}-a^{1/m} which equals

\frac{(x-a)}{x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}}\frac{1}{a^{n}}+\ldots +a^{\frac{n-1}{n}} }

Choose |x-a|<|a/2| to start with, so that |a/2|<|x|<(3/2)|a|. Thus,

|x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}}a^{1/n}+\ldots+a^{\frac{n-1}{n}}|>|a|^{\frac{n-1}{n}} \times ((1/2)^{\frac{n-1}{n}}+(1/2)^{\frac{n-2}{n}}+\ldots+1)

Given an \varepsilon >0, let

\delta=min\{\frac{|a|}{2}, \varepsilon \times |a|^{\frac{n-1}{n}} \times \left( (1/2)^{\frac{n-1}{n}}+\ldots+1 \right)\}.

Then, for |x-a|<\delta, we have

|f(x)-f(a)|=\frac{|x-a|}{|x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}} \times a^{1/n}+\ldots+a^{\frac{n-1}{n}}|}< \varepsilon.

It can be shown that f defined by f(x)=x^{r} is also a continuous function for every real r \in \Re.

k) Consider the function f:\Re \rightarrow \Re defined by f(x)=a^{x}. Is f a continuous function? This is left as an exercise. (Hint: It will suffice to prove continuity at x=0. This would follow from \lim_{m \rightarrow \infty}a^{1/m}).

k) Suppose f:\Re \rightarrow \Re is defined by f(x)=1/x, if x \neq 0 and f(0)=0. We can see that f is not continuous at 0 as f(x) changes abruptly when x goes over from negative to positive values.

More later,

Nalin Pithwa