Category Archives: algebra

The Geometry of Linear Equations: thanks Prof Gilbert Strang, ocwmit

A problem of log, GP and HP…

Question: If a^{x}=b^{y}=c^{z} and b^{2}=ac, pyrove that: y = \frac{2xz}{x+z}

Solution: This is same as proving: y is Harmonic Mean (HM) of x and z;

That is, to prove that y=\frac{2xz}{x+z} is the same as the proof for : \frac{1}{y} - \frac{1}{x} = \frac{1}{z} - \frac{1}{y}

Now, it is given that a^{x} = b^{y} = c^{z} —– I

and b^{2}=ac —– II
Let a^{x} = b^{y}=c^{z}=N say. By definition of logarithm,

x = \log_{a}{N}; y=\log_{b}{N}; z=\log_{c}{N}

\frac{1}{x} = \frac{1}{\log_{a}{N}}; \frac{1}{y} = \frac{1}{\log_{b}{N}}; \frac{1}{z} = \frac{1}{\log_{a}{N}}.

Now let us see what happens to the following two algebraic entities, namely, \frac{1}{y} - \frac{1}{x} and \frac{1}{z} - \frac{1}{y};

Now, \frac{1}{y} - \frac{1}{x} = \frac{1}{\log_{b}{N}} - \frac{1}{\log_{a}{N}} = \frac{\log_{b}{b}}{\log_{b}{N}} - \frac{\log_{a}{a}}{\log_{a}{N}} = \log_{N}{b} - \log_{N}{a} = \log_{N}{(\frac{b}{a})}…call this III

Now, \frac{1}{z} - \frac{1}{y} = \frac{1}{\log_{c}{N}} - \frac{1}{\log_{b}{N}} = \frac{\log_{c}{c}}{\log_{c}{}N} -\frac{\log_{b}{b}}{\log_{b}{N}}= \log_{N}{c}-\log_{N}{b}

Hence, \frac{1}{z} - \frac{1}{y}=\log_{N}{c/b}….equation IV

but it is also given that b^{2}=ac…see equation II

Hence, \frac{b}{a} = \frac{c}{b}

Take log of above both sides w.r.t. base N:

So, above is equivalent to \log_{N}{b/a} = \log_{N}{c/b}

But now see relations III and IV:

Hence, \frac{1}{y} -\frac{1}{x} = \frac{1}{x} - \frac{1}{y}

Hence, \frac{2}{y} = \frac{1}{x} + \frac{1}{z} = \frac{x+z}{xz}

Hence, y= \frac{2xz}{x+z} as desired.


Nalin Pithwa

How to find square root of a binomial quadratic surd

Assume \sqrt{a+ \sqrt{b} + \sqrt{c} + \sqrt{d}}=\sqrt{x} + \sqrt{y} + \sqrt{z};

Hence, a+\sqrt{b} + \sqrt{c} + \sqrt{d} = x+y+z+ 2\sqrt{xy} + 2\sqrt{yz}+ 2\sqrt{zx}

If then, 2\sqrt{xy}=\sqrt{b}, 2\sqrt{yz}=\sqrt{c}, 2\sqrt{zx}=\sqrt{d},

And, if simultaneously, the values of x, y, z thus found satisfy x+y+z=a, we shall have obtained the required root.


Find the square root of 21-4\sqrt{5}+5\sqrt{3}-4\sqrt{15}.


Clearly, we can’t have anything like

21--4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=\sqrt{x} + \sqrt{y} +\sqrt{z}

We will have to try the following options:

21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=\sqrt{x} - \sqrt{y} - \sqrt{z}



Only the last option will work as we now show:

So, once again, assume that \sqrt{21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}}=\sqrt{x}+\sqrt{y}-\sqrt{z}

Hence, 21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=z+y+z+2\sqrt{xy}-2\sqrt{yz}+2\sqrt{zx}

Put 2\sqrt{xy}=8\sqrt{3}, 2\sqrt{xz}=4\sqrt{15}, 2\sqrt{yz}=4\sqrt{5};

by multiplication, xyz = 240; that is \sqrt{xyz}=4\sqrt{15}; so it follows that : \sqrt{x}=2\sqrt{3}, \sqrt{y}=2, \sqrt{z}=\sqrt{5}.

And, since, these values satisfy the equation x+y+z=21, the required root is 2\sqrt{3}+2-\sqrt{5}.

That is all, for now,


Nalin Pithwa

IITJEE Foundation Maths : Tutorial Problems III

  1. When a=-3, b=5, c=-1, d=0, find the value of 26c\sqrt[3]{a^{3}-c^{2}d+5bc-4ac+d^{2}}
  2. Solve the equations: (a) \frac{1}{3}x - \frac{2}{7}y = 8 -2x and \frac{1}{2}y - 3x =3-y (b) 1 = y+z =2(z+x)=3(x+y)
  3. Simplify: (a) \frac{a-x}{a+x} - \frac{4x^{2}}{a^{2}-x^{2}} +  \frac{a-3x}{x-a} (b) \frac{b^{2}-3b}{b^{2}-2b+4} \times \frac{b^{2}+b-30}{b^{2}+3b-18} \div \frac{b^{3}-3b^{2}-10b}{b^{2}+8}
  4. Find the square root of : 9-36x+60x^{2}-\frac{160}{3}x^{3}+\frac{80}{3}x^{4}-\frac{64}{3}x^{5}+\frac{64}{81}x^{6}
  5. In a cricket match, the extras in the first innings are one-sixteenth of the score, and in the second innings the extras are one-twelfth of the score. The grand total is 296, of which 21 are extras; find the score in each innings.
  6. Find the value of \frac{a^{2}-x^{2}}{\frac{1}{a^{2}} - \frac{2}{ax} + \frac{1}{x^{2}}} \times \frac{\frac{1}{a^{2}x^{2}}}{a+x}
  7. Find the value of : \frac{1}{3}(a+2) -3(1-\frac{1}{6}b) - \frac{2}{3}(2a-3b+\frac{3}{2}) + \frac{3}{2}b - 4(\frac{1}{2}a-\frac{1}{3}).
  8. Resolve into factors: (i) 3a^{2} -20a-7 (ii) a^{4}b^{2}-b^{4}a^{2}
  9. Reduce to lowest terms: \frac{4x^{3}+7x^{2}-x+2}{4x^{3}+5x^{2}-7x-2}
  10. Solve the following equations: (a) x-6 -\frac{x-12}{3}= \frac{x-4}{2} + \frac{x-8}{4}; (b) x+y-z=0, x-y+z=4, 5x+y+z=20; (c) \frac{ax+b}{c} + \frac{dx+e}{f} =1
  11. Simplify: \frac{x+3}{x^{2}-5x+6} - \frac{x+2}{x^{2}-9x+14} + \frac{4}{x^{2}-10x+21}
  12. A purse of rupees is divided amongst three persons, the first receiving half of them and one more, the second half of the remainder and one more, and the third six. Find the number of rupees the purse contained.
  13. If h=-2, k=1, l=0, m=1, n=-3, find the value of \frac{h^{2}(m-l)-\sqrt{3hn}+hk}{m(l-h)-2hm^{2}+\sqrt[3]{4hk}}
  14. Find the LCM of 15(p^{3}+q^{3}), 5(p^{2}-pq+q^{2}), 4(p^{2}+pq+q^{2}), 6(p^{2}-q^{2})
  15. Find the square root of (i) \frac{4x^{2}}{9} + \frac{9}{4x^{2}} -2; (ii) 1-6a+5a^{2}+12a^{3}+4a^{4}
  16. Simplify \frac{20x^{2}+27x+9}{15x^{2}+19x+6} + \frac{20x^{2}+27x+9}{12x^{2}+17x+6}
  17. Solve the equations: (i) \frac{a(x-b)}{a-b} + \frac{b(x-a)}{b-a} =1 (ii) \frac{9}{x-4} + \frac{3}{x-8} = \frac{4}{x-9} + \frac{8}{x-3}
  18. A sum of money is to be divided among a number of persons; if Rs. 8 is given to each there will be Rs. 3 short, and if Rs. 7.50 is given to each there will be Rs. 2 over; find the number of persons.


Nalin Pithwa

Purva building, 5A
Flat 06
Mumbai , Maharastra 400101

Two cute problems in HP : IITJEE Foundations\Mains, pre RMO

Problem 1: 

If a^{2}, b^{2}, c^{2} are in AP, show that b+c, c+a, a+b are in HP.

Proof 1:

Note that a straight forward proof is not so easy.

Below is a nice clever solution:

By adding ab+bc+ca to each term, we see that:

a^{2}+ab+ac+bc, b^{2}+ab+ac+bc, c^{2}+ab+ac+bc are in AP.

that is, (a+b)(a+c), (b+c)(b+a), (c+a)(c+b) are in AP.

Dividing each term by (a+b)(b+c)(c+a).

\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} are in AP.

that is, b+c, c+a, a+b are in HP.


Problem 2:

If the p^{th}, q^{th}, r^{th}, s^{th} terms of an AP are in GP, show that p-q, q-r, r-s are in GP.

Proof 2:

Once again a straight forward proof is not at all easy.

Below is a “bingo” sort of proof 🙂

With the usual notation, we have

\frac{a+(p-1)d}{a+(q-1)d} = \frac{a+(q-1)d}{a+(r-1)d} = \frac{a+(r-1)d}{a+(s-1)d}

Hence, each of the ratios is equal to

\frac{(a+(p-1)d)-(a+(q-1)d)}{(a+(q-1)d)-(a+(r-1)d)} = \frac{(a+(q-1)d)-(a+(r-1)d)}{(a+(r-1)d)-(a+(s-1)d)}

which in turn is equal  to \frac{p-q}{q-r} = \frac{q-r}{r-s}

Hence, p-q, q-r, r-s are in GP.


Nalin Pithwa

Binomial Theorem Tutorial problems I: IITJEE mains practice

I. Expand up to 5 terms the following expressions:

  1. (1+x)^{\frac{1}{2}}
  2. (1+x)^{\frac{7}{2}}
  3. (1-x)^{\frac{2}{5}}
  4. (1+x^{2})^{-2}
  5. (1-3x)^{\frac{1}{3}}
  6. (1-3x)^{\frac{-1}{2}}
  7. (1+2x)^{-\frac{1}{2}}
  8. (1+\frac{x}{3})^{-2}
  9. (1+\frac{2x}{3})^{\frac{3}{2}}
  10. (1+\frac{1}{2}a)^{-4}
  11. (2+x)^{-2}
  12. (9+2x)^{\frac{1}{2}}
  13. (8+12a)^{\frac{3}{2}}
  14. (9-6x)^{-\frac{3}{2}}
  15. (4a-8x)^{-\frac{1}{2}}

II. Write down and simplify:

  1. The 8th term of (1+2x)^{-\frac{1}{2}}
  2. The 11th term of (1-2x^{3})^{\frac{11}{2}}
  3. The 16th term of (1+3a^{2})^{\frac{16}{3}}
  4. The 6th term of (3a-2b)^{-1}
  5. The (r+1)^{th} term of (1-x)^{-2}
  6. The (r+1)^{th} term of (1-x)^{-4}
  7. The (r+1)^{th} term of (1+x)^{\frac{1}{2}}
  8. The (r+1)^{th} term of (1+x)^{\frac{11}{3}}
  9. The 14th term of (2^{10}-2^{7}x)^{\frac{13}{2}}
  10. The 7th term of (3^{8}+6^{4}x)^{\frac{11}{4}}


Nalin Pithwa

Theory of Quadratic Equations: Part III: Tutorial practice problems: IITJEE Mains and preRMO

Problem 1:

Find the condition that a quadratic function of x and y may be resolved into two linear factors. For instance, a general form of such a function would be : ax^{2}+2hxy+by^{2}+2gx+2fy+c.

Problem 2:

Find the condition that the equations ax^{2}+bx+c=0 and a^{'}x^{2}+b^{'}x+c^{'}=0 may have a common root.

Using the above result, find the condition that the two quadratic functions ax^{2}+bxy+cy^{2} and a^{'}x^{2}+b^{'}xy+c^{'}y^{2} may have a common linear factor.

Problem 3:

For what values of m will the expression y^{2}+2xy+2x+my-3 be capable of resolution into two rational factors?

Problem 4:

Find the values of m which will make 2x^{2}+mxy+3y^{2}-5y-2 equivalent to the product of two linear factors.

Problem 5:

Show that the expression A(x^{2}-y^{2})-xy(B-C) always admits of two real linear factors.

Problem 6:

If the equations x^{2}+px+q=0 and x^{2}+p^{'}x+q^{'}=0 have a common root, show that it must be equal to \frac{pq^{'}-p^{'}q}{q-q^{'}} or \frac{q-q^{'}}{p^{'}-p}.

Problem 7:

Find the condition that the expression lx^{2}+mxy+ny^{2} and l^{'}x^{2}+m^{'}xy+n^{'}y^{2} may have a common linear factor.

Problem 8:

If the expression 3x^{2}+2Pxy+2y^{2}+2ax-4y+1 can be resolved into linear factors, prove that P must be be one of the roots of the equation P^{2}+4aP+2a^{2}+6=0.

Problem 9:

Find the condition that the expressions ax^{2}+2hxy+by^{2} and a^{'}x^{2}+2h^{'}xy+b^{'}y^{2} may be respectively divisible by factors of the form y-mx and my+x.

Problem 10:

Prove that the equation x^{2}-3xy+2y^{2}-2x-3y-35=0 for every real value of x, there is a real value of y, and for every real value of y, there is a real value of x.

Problem 11:

If x and y are two real quantities connected by the equation 9x^{2}+2xy+y^{2}-92x-20y+244=0, then will x lie between 3 and 6, and y between 1 and 10.

Problem 11:

If (ax^{2}+bx+c)y+a^{'}x^{2}+b^{'}x+c^{'}=0, find the condition that x may be a rational function of y.

More later,


Nalin Pithwa.

Theory of Quadratic Equations: part II: tutorial problems: IITJEE Mains, preRMO

Problem 1:

If x is a real number, prove that the rational function \frac{x^{2}+2x-11}{2(x-3)} can have all numerical values except such as lie between 2 and 6. In other words, find the range of this rational function. (the domain of this rational function is all real numbers except x=3 quite obviously.

Problem 2:

For all real values of x, prove that the quadratic function y=f(x)=ax^{2}+bx+c has the same sign as a, except when the roots of the quadratic equation ax^{2}+bx+c=0 are real and unequal, and x has a value lying between them. This is a very useful famous classic result. 


a) From your proof, you can conclude the following also: The expression ax^{2}+bx+c will always have the same sign, whatever real value x may have, provided that b^{2}-4ac is negative or zero; and if this condition is satisfied, the expression is positive, or negative accordingly as a is positive or negative.

b) From your proof, and using the above conclusion, you can also conclude the following: Conversely, in order that the expression ax^{2}+bx+c may be always positive, b^{2}-4ac must be negative or zero; and, a must be positive; and, in order that ax^{2}+bx+c may be always negative, b^{2}-4ac must be negative or zero, and a must be negative.

Further Remarks:

Please note that the function y=f(x)=ax^{2}+bx+c, where a, b, c \in \Re and a \neq 0 is a parabola. The roots of this y=f(x)=0 are the points where the parabola cuts the y axis. Can you find the vertex of this parabola? Compare the graph of the elementary parabola y=x^{2}, with the graph of y=ax^{2} where a \neq 0 and further with the graph of the general parabola y=ax^{2}+bx+c. Note you will just have to convert the expression ax^{2}+bx+c to a perfect square form.

Problem 3:

Find the limits between which a must lie in order that the rational function \frac{ax^{2}-7x+5}{5x^{2}-7x+a} may be real, if x is real.

Problem 4:

Determine the limits between which n must lie in order that the equation 2ax(ax+nc)+(n^{2}-2)c^{2}=0 may have real roots.

Problem 5:

If x be real, prove that \frac{x}{x^{2}-5x+9} must lie between 1 and -\frac{1}{11}.

Problem 6:

Prove that the range of the rational function y=f(x)=\frac{x^{2}-x+1}{x^{2}+x+1} lies between 3 and \frac{1}{3} for all real values of x.

Problem 7:

If x \in \Re, Prove that the rational function y=f(x)=\frac{x^{2}+34x-71}{x^{2}+2x-7} can have no value between 5 and 9. In other words, prove that the range of the function is (x <5)\bigcup(x>9).

Problem 8:

Find the equation whose roots are \frac{\sqrt{a}}{\sqrt{a} \pm \sqrt(a-b)}.

Problem 9:

If \alpha, \beta are roots of the quadratic equation x^{2}-px+q=0, find the value of (a) \alpha^{2}(\alpha^{2}\beta^{-1}-\beta)+\beta^{2}(\beta^{2}\alpha^{-1}-\alpha) (b) (\alpha-p)^{-4}+(\beta-p)^{-4}.

Problem 10:

If the roots of lx^{2}+mx+n=0 be in the ratio p:q, prove that \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0

Problem 11:

If x be real, the expression \frac{(x+m)^{2}-4mn}{2(x-n)} admits of all values except such as those that lie between 2n and 2m.

Problem 12:

If the roots of the equation ax^{2}+2bx+c=0 are \alpha and \beta, and those of the equation Ax^{2}+2Bx+C=0 be \alpha+\delta and \beta+\delta, prove that \frac{b^{2}-ac}{a^{2}} = \frac{B^{2}-AC}{A^{2}}.

Problem 13:

Prove that the rational function y=f(x)=\frac{px^{2}+3x-4}{p+3x-4x^{2}} will be capable of all values when x is real, provided that p has any real value between 1 and 7. That is, under the conditions on p, we have to show that the given rational function has as its range the full real numbers. (Of course, the domain is real except those values of x for which the denominator is zero).

Problem 14:

Find the greatest value of \frac{x+2}{2x^{2}+3x+6} for any real value of x. (Remarks: this is maxima-minima problem which can be solved with algebra only, calculus is not needed). 

Problem 15:

Show that if x is real, the expression (x^{2}-bc)(2x-b-c)^{-1} has no real value between b and a.

Problem 16:

If the roots of ax^{2}+bx+c=0 be possible (real) and different, then the roots of (a+c)(ax^{2}+2bx+c)=2(ac-b^{2})(x^{2}+1) will not be real, and vice-versa. Prove this.

Problem 17:

Prove that the rational function y=f(x)=\frac{(ax-b)(dx-c)}{(bx-a)(cx-a)} will be capable of all real values when x is real, if a^{2}-b^{2} and c^{2}-a^{2} have the same sign.


Nalin Pithwa

Theory of Quadratic Equations: Tutorial problems : Part I: IITJEE Mains, preRMO

I) Form the equations whose roots are:

a) -\frac{4}{5}, \frac{3}{7} (b) \frac{m}{n}, -\frac{n}{m} (c) \frac{p-q}{p+q}, -\frac{p+q}{p-q} (d) 7 \pm 2\sqrt{5} (e) -p \pm 2\sqrt{2q} (f) -3 \pm 5i (g) -a \pm ib (h) \pm i(a-b) (i) -3, \frac{2}{3}, \frac{1}{2} (j) \frac{a}{2},0, -\frac{2}{a} (k) 2 \pm \sqrt{3}, 4

II) Prove that the roots of the following equations are real:

i) x^{2}-2ax+a^{2}-b^{2}-c^{2}=0

ii) (a-b+c)x^{2}+4(a-b)x+(a-b-c)=0

III) If the equation x^{2}-15-m(2x-8)=0 has equal roots, find the values of m.

IV) For what values of m will the equation x^{2}-2x(1+3m)+7(3+2m)=0 have equal roots?

V) For what value of m will the equation \frac{x^{2}-bx}{ax-c} = \frac{m-1}{m+1} have roots equal in magnitude but opposite in sign?

VI) Prove that the roots of the following equations are rational:

(i) (a+c-b)x^{2}+2ax+(b+c-a)=0

(ii) abc^{2}x^{2}+3a^{2}cx+b^{2}ax-6a^{2}-ab+2b^{2}=0

VII) If \alpha, \beta are the roots of the equation ax^{2}+bx+c=0, find the values of

(i) \frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}}

(ii) \alpha^{4}\beta^{7}+\alpha^{7}\beta^{4}

(iii) (\frac{\alpha}{\beta}-\frac{\beta}{\alpha})^{2}

VIII) Find the value of:

(a) x^{3}+x^{2}-x+22 when x=1+2i

(b) x^{3}-3x^{2}-8x+16 when x=3+i

(c) x^{3}-ax^{2}+2a^{2}x+4a^{3} when \frac{x}{a}=1-\sqrt{-3}

IX) If \alpha and \beta are the roots of x^{2}+px+q=0 form the equation whose roots are (\alpha-\beta)^{2} and (\alpha+\beta)^{2}/

X) Prove that the roots of (x-a)(x-b)=k^{2} are always real.

XI) If \alpha_{1}, \alpha_{2} are the roots of ax^{2}+bx+c=0, find the value of (i) (ax_{1}+b)^{-2}+(ax_{2}+b)^{-2} (ii) (ax_{1}+b)^{-3}+(ax_{2}+b)^{-3}

XII) Find the condition that one root of ax^{2}+bx+c=0 shall be n times the other.

XIII) If \alpha, \beta are the roots of ax^{2}+bx+c=0 form the equation whose roots are \alpha^{2}+\beta^{2} and \alpha^{-2}+\beta^{-2}.

XIV) Form the equation whose roots are the squares of the sum and of the differences of the roots of 2x^{2}+2(m+n)x+m^{2}+n^{2}=0.

XV) Discuss the signs of the roots of the equation px^{2}+qx+r=0

XVI) If a, b and c are odd integers, prove that the roots of the equation ax^{2}+bx+c=0 cannot be rational numbers.

XVII) Given that the equation x^{4}+px^{3}+qx^{2}+rx+s=0 has four real positive roots, prove that (a) pr-16s \geq 0 (b) q^{2}-36s \geq 0, where equality holds, in each case, if and only if the roots are equal.

XVIII) Let p(x)=x^{2}+ax+b be a quadratic polynomial in which a and b are integers. Given any integer n, show that there is an integer M such that p(n)p(n+1)=p(M).


Nalin Pithwa.

Set theory, relations, functions: preliminaries: Part V

Types of functions: (please plot as many functions as possible from the list below; as suggested in an earlier blog, please use a TI graphing calculator or GeoGebra freeware graphing software): 

  1. Constant function: A function f:\Re \longrightarrow \Re given by f(x)=k, where k \in \Re is a constant. It is a horizontal line on the XY-plane.
  2. Identity function: A function f: \Re \longrightarrow \Re given by f(x)=x. It maps a real value x back to itself. It is a straight line passing through origin at an angle 45 degrees to the positive X axis.
  3. One-one or injective function: If different inputs give rise to different outputs, the function is said to be injective or one-one. That is, if f: A \longrightarrow B, where set A is domain and set B is co-domain, if further, x_{1}, x_{2} \in A such that x_{1} \neq x_{2}, then it follows that f(x_{1}) \neq f(x_{2}). Sometimes, to prove that a function is injective, we can prove the conrapositive statement of the definition also; that is, y_{1}=y_{2} where y_{1}, y_{2} \in codomain \hspace{0.1in} range, then it follows that x_{1}=x_{2}. It might be easier to prove the contrapositive. It would be illuminating to construct your own pictorial examples of such a function. 
  4. Onto or surjective: If a function is given by f: X \longrightarrow Y such that f(X)=Y, that is, the images of all the elements of the domain is full of set Y. In other words, in such a case, the range is equal to co-domain. it would be illuminating to construct your own pictorial examples of  such a function.
  5. Bijective function or one-one onto correspondence: A function which is both one-one and onto is called a bijective function. (It is both injective and surjective). Only a bijective function will have a well-defined inverse function. Think why! This is the reason why inverse circular functions (that is, inverse trigonometric functions have their domains restricted to so-called principal values). 
  6. Polynomial function: A function of the form f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\ldots + a_{n}x^{n}, where n is zero or positive integer only and a_{i} \in \Re is called a polynomial with real coefficients. Example. f(x)=ax^{2}=bx+c, where a \neq 0, a, b, c \in \Re is called a quadratic function in x. (this is a general parabola).
  7. Rational function: The function of the type \frac{f(x)}{g(x)}, where g(x) \neq 0, where f(x) and g(x) are polynomial functions of x, defined in a domain, is called a rational function. Such a function can have asymptotes, a term we define later. Example, y=f(x)=\frac{1}{x}, which is a hyperbola with asymptotes X and Y axes.
  8. Absolute value function: Let f: \Re \longrightarrow \Re be given by f(x)=|x|=x when x \geq 0 and f(x)=-x, when x<0 for any x \in \Re. Note that |x|=\sqrt{x^{2}} since the radical sign indicates positive root of a quantity by convention.
  9. Signum function: Let f: \Re \longrightarrow \Re where f(x)=1, when x>0 and f(x)=0 when x=0 and f(x)=-1 when x<0. Such a function is called the signum function. (If you can, discuss the continuity and differentiability of the signum function). Clearly, the domain of this function  is full \Re whereas the range is \{ -1,0,1\}.
  10. In part III of the blog series, we have already defined the floor function and the ceiling function. Further properties of these functions are summarized (and some with proofs in the following wikipedia links): (once again, if you can, discuss the continuity and differentiablity of the floor and ceiling functions):
  11. Exponential function: A function f: \Re \longrightarrow \Re^{+} given by f(x)=a^{x} where a>0 is called an exponential function. An exponential function is bijective and its inverse is the natural logarithmic function. (the logarithmic function is difficult to define, though; we will consider the details later). PS: Quiz: Which function has a faster growth rate — exponential or a power function ? Consider various parameters.
  12. Logarithmic function: Let a be a positive real number with a \neq 1. If a^{y}=x, where x \in \Re, then y is called the logarithm of x with base a and we write it as y=\ln{x}. (By the way, the logarithmic function is used in the very much loved mp3 music :-))


Nalin Pithwa