Category Archives: KVPY

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! ūüôā ūüôā ūüôā

Nalin Pithwa.

Can anyone have fun with infinite series?

Below is list of finitely many puzzles¬†on infinite series to keep you a bit busy¬†!! ūüôā Note that these puzzles do have an academic flavour, especially concepts of convergence and divergence of an infinite series.

Puzzle 1: A grandmother’s¬†vrat¬†(fast) requires her to keep odd number of lamps of finite capacity lit in a temple at any time during 6pm to 6am the next morning. Each oil-filled lamp lasts 1 hour and it burns oil at a constant rate. She is not allowed to light any lamp after 6pm but she can light any number of lamps before 6pm and transfer oil from some to the others throughout the night while keeping odd number of lamps lit all the time. How many fully-filled oil lamps does she need to complete her¬†vrat?

Puzzle 2: Two number theorists, bored in a chemistry lab, played a game with a large flask containing 2 liters of a colourful chemical solution and an ultra-accurate pipette. The game was that they would take turns to recall a prime number p such that p+2 is also a prime number. Then, the first number theorist would pipette out \frac{1}{p} litres of chemical and the second \frac{1}{(p+2)} litres. How many times do they have to play this game to empty the flask completely?

Puzzle 3: How farthest from the edge of a table can a deck of playing cards be stably overhung if the cards are stacked on top of one another? And, how many of them will be overhanging completely away from the edge of the table?

Puzzle 4: Imagine a tank that can be filled with infinite taps and can be emptied with infinite drains. The taps, turned on alone, can fill the empty tank to its full capacity in 1 hour, 3 hours, 5 hours, 7 hours and so on. Likewise, the drains opened alone, can drain a full tank in 2 hours, 4 hours, 6 hours, and so on. Assume that the taps and drains are sequentially arranged in the ascending order of their filling and emptying durations.

Now, starting with an empty tank, plumber A alternately turns on a tap for 1 hour and opens the drain for 1 hour, all operations done one at a time in a sequence. His sequence, by using t_{i} for i^{th} tap and d_{j} for j^{th} drain, can be written as follows: \{ t_{1}, d_{1}, t_{2}, d_{2}, \ldots\}_{A}.

When he finishes his operation, mathematically, after using all the infinite taps and drains, he notes that the tank is filled to a certain fraction, say, n_{A}<1.

Then, plumber B turns one tap on for 1 hour and then opens two drains for 1 hour each and repeats his sequence: \{ (t_{1},d_{1},d_{2}), (t_{2},d_{3},d_{4}), (t_{3},d_{4},d_{5}) \ldots \}_{B}.

At the end of his (B’s) operation, he finds that the tank is filled to a fraction that is exactly half of what plumber A had filled, that is, 0.5n_{A}.

How is this possible even though both have turned on all taps for 1 hour and opened all drains for 1 hour, although in different sequences?

I hope u do have fun!!

-Nalin Pithwa.

Pair of straight lines — quick review for IITJEE Math

Pair of Straight Lines:

I: Equation of Family of Lines:

A first degree equation ax+by+c=0 represents a straight line involving three constants a, b and c, which can be reduced to two by dividing both the sides of the equation by a non-zero constant. For instance, if a \neq 0 we can write the equation as

x+\frac{b}{a}y+\frac{c}{a}=0 or x+By+C=0

So, in order to determine an equation of a line, we need two conditions on the line to determine these constants. For instance, if we know two points on the line or a point on the line and its slope etc., we know the line. But, if we know just one condition, we have infinite number of lines satisfying the given condition. In this case, the equation of the line contains an arbitrary constant and for different values of the constant we have different lines satisfying the given condition and the constant is called a parameter.

Some Equations of Family of Lines:

  1. Family of lines with given slope. (Family of given lines) y = mx +k, where k is a parameter represents a family of lines in which each line has slope m.
  2. Family of lines through a point: y-y_{0}=k(x-x_{0}), where k is a parameter represents a family of lines in which each line passes through the point (x_{0},y_{0}).
  3. Family of lines parallel to a given line: ax + by + k=0, where k is a parameter represents a family of lines which are parallel to the line ax+by+c=0.
  4. Family of lines through intersection of two given lines:
    a_{1}x+b_{1}y+c_{1}+k(a_{2}x+b_{2}y+c_{2})=0, where k is a parameter represents a family of lines, in which each line passes through the point of intersection of two intersecting llines a_{1}x+b_{1}y+c_{1}=0 and a_{2}x+b_{2}y+c_{2}=0.
  5. Family of lines perpendicular to a given line:
    bx-ay+c=0 where k is a parameter represents a family of lines, in which each line is perpendicular to the line ax+by+c=0.
  6. Family of lines making a given intercept on axes:
    (a) \frac{x}{a}+\frac{y}{k}=1, where k is a parameter, represents a family of lines, in which each line makes an intercept a on the axis of x. and,

6(b) \frac{x}{k} + \frac{y}{b}=1, where k is a parameter, represents a family of lines, in which each line makes an intercept b on the axis of y.

7. Family of lines at a constant distance from the origin:

x\cos {\alpha}+y\sin{\alpha}=p, where \alpha is a parameter, represents a family of lines, in which each line is at a distance p from the origin.

II: Pair of Straight Lines:

A) Pair of straight lines through the origin: A homogeneous equation of the second degree ax^{2}+2hxy+by^{2}=0…call this equation 1; represents a pair of straight lines passing through the origin if and only if h^{2}-ab \geq 0. If b \neq 0, the given equation can be written as

y^{2}+\frac{2h}{b}xy+\frac{a}{b}x^{2}=0, or (\frac{y}{x})^{2}+\frac{2h}{b}(\frac{y}{x})+\frac{a}{b}=0, which, being quadratic in y/x, gives two values of y/x, say m_{1} and m_{2} and hence, the equations y=m_{1}x and y=m_{2}x of two straight lines passing through the origin. The slopes m_{1} and m_{2} of these straight lines are given by the relations

m_{1}+m_{2}=-\frac{2h}{b}…call this equation (2).

m_{1}m_{2}=\frac{a}{b}…call this equation (3).

B) Angle between the lines: represented by ax^{2}+2hxy+by^{2}=0. If \theta is an angle, between the lines represented by (1), then

\tan{\theta} = \frac{m_{1}-m_{2}}{1+m_{1}m_{2}}=\frac{\pm\sqrt{(m_{1}+m_{2})^{2}-4m_{1}m_{2}}}{1+m_{1}m_{2}} = \frac{\pm \sqrt{4h^{2}-4ab}}{a+b}

\Longrightarrow \theta = \arctan (\frac{\pm 2\sqrt{h^{2}-ab}}{a+b})

If (a+b)=0, the lines are perpendicular, and if h^{2}=ab, then the lines are coincident.

C) Equation of the bisectors of the angles between the lines ax^{2}+2hxy=by^{2}=0.

Let y=m_{1}x and y=m_{2}x be the equation of the straight lines represented by the given equation. Then, equations of the angle bisectors are:

\frac{y-m_{1}x}{\sqrt{1+m_{1}^{2}}} = \frac{y-m_{2}x}{\sqrt{1+m_{1}^{2}}}

The joint equation of  these bisectors can be written as

(\frac{y-m_{1}x}{\sqrt{1+m_{1}^{2}}} + \frac{y-m_{2}x}{1+m_{2}^{2}})(\frac{y-m_{1}x}{1+m_{1}^{2}} - \frac{y-m_{2}x}{1+m_{2}^{2}})=0

or, (1+m_{2}^{2})(y-m_{1}x)^{2} - (1+m_{1}^{2})(y-m_{2}x)^{2}=0

or, x^{2}(m_{1}^{2}-m_{2}^{2}) - 2xy(m_{1}-m_{2})(1-m_{1}m_{2}) + y^{2}(m_{2}^{2}-m_{1}^{2})=0

or, \frac{x^{2}-y^{2}}{2xy} = \frac{1-m_{1}m_{2}}{m_{1}+m_{2}} = \frac{a-b}{2h}….this comes from (2) and (3).

or, \frac{x^{2}-y^{2}}{a-b} = \frac{xy}{h}

D) The general equation of second degree:

ax^{2}+2hxy+by^{2}+2gx+2fy+c=0 represents a pair of straight lines if and only if

abc + 2fgh - af^{2} -bg^{2} - ch^{2} =0, that is, if

\left | \begin{array}{ccc}    a & h & g \\    h & b & f \\    g & f & c \end{array} \right |=0

If h^{2}=ab, the co-ordinates (x_{1},y_{1}) of the point of intersection of these lines are obtained by solving the equations

ax_{1}+hy_{1}+x=0 and hx_{1}+by_{1}+f=0

\Longrightarrow x_{1}=\frac{hf-bg}{ab-h^{2}} and y_{1}=\frac{gh-af}{ab-h^{2}}

E) Equation ax^{2}+2hxy +by^{2}=0 represents a pair of straight lines through the origin parallel to the lines given by the general equation of second degree given in (4) and hence, the angle between the lines given in (4) is same as in (2), that is,

\theta = \arctan {\frac{\pm {h^{2}-ab}}{a+b}}

so that if a+b=0, the lines are perpendicular and if h^{2}=ab, the lines are parallel.

F) Equation of the lines joining the origin to the points of intersection of a line and a conic:

Let L \equiv lx + my + n=0

and S \equiv ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c=0

be the equations of a line and a conic, respectively. Writing the equation of the line as

\frac{lx + my}{-n} = 1 and making S=0 homogeneous with its help, we get S=ax^{2}+2hxy+by^{2}+2(gx +fy)

(\frac{lx+my}{-n}) + c(\frac{lx+my}{-n})^{2}=0, which being a homogeneous equation of second degree, represents a pair of straight lines through the origin and passing through the points common to S=0 and L=0.

G) Some important results regarding pair of lines:

(i) Equation of the pair of lines through the origin perpendicular to the pair of lines ax^{2}+2hxy+by^{2}=0 is bx^{2}+2hxy +ay^{2}=0.

(ii) The product of the perpendicular lines drawn from the point (x_{1}, y_{1}) on the lines ax^{2}+2hxy+by^{2}=0 is |\frac{ax_{1}^{2}+2hx_{1}y_{1}+by_{1}^{2}}{\sqrt{(a-b)^{2}+4h^{2}}}|

(iii) The product of the perpendiculars drawn from the origin to the lines ax^{2}+2hxy + by^{2}+2gx+2fy+c=0 is \frac{c}{\sqrt{(a-b)^{2}+4h^{2}}}

(iv) Equations of the bisectors of the angles between the lines represented by ax^{2}+2hxy + by^{2} +2gx + 2fy + c=0 are given by

\frac{(x-x_{1})^{2}-(y-y_{1})^{2}}{a-b} = \frac{(x-x_{1})(y-y_{1})}{h} where (x_{1}, y_{1}) is the point of intersection of the lines represented by the given equation.

(v) If ax^{2}+2hxy + by^{2}+2gx + 2fy + c=0 represents two parallel straight lines, then the distance between them is 2\sqrt{\frac{g^{2}-ac}{a(a+b)}}

Every nth degree homogeneous equation a_{0}x^{n}+a_{1}x^{n-1}y + a_{2}x^{n-2}y + \ldots + a_{n-1}xy^{n-1}+a_{n}y^{n}=0, then

m_{1}+m_{2}+ \ldots + m_{n} = - \frac{a_{n-1}}{a_{n}}

and m_{1}m_{2}\ldots m_{n}=(-1)^{n}\frac{a_{0}}{a_{n}}

To be continued later,

-Nalin Pithwa.

Lagrange’s Mean Value Theorem and Cauchy’s Generalized Mean Value Theorem

Lagrange’s Mean Value Theorem:

If a function f(x) is continuous on the interval [a,b] and differentiable at all interior points of the interval, there will be, within [a,b], at least one point c, a<c<b, such that f(b)-f(a)=f^{'}(c)(b-a).

Cauchy’s Generalized Mean Value Theorem:

If f(x) and phi(x) are two functions continuous on an interval [a,b] and differentiable within it, and phi(x) does not vanish anywhere inside the interval, there will be, in [a,b], a point x=c, a<c<b, such that \frac{f(b)-f(a)}{phi(b)-phi(a)} = \frac{f^{'}(c)}{phi^{'}(c)}.

Some questions based on the above:

Problem 1:

Form Lagrange’s formula for the function y=\sin(x) on the interval [x_{1},x_{2}].

Problem 2:

Verify the truth of Lagrange’s formula for the function y=2x-x^{2} on the interval [0,1].

Problem 3:

Applying Lagrange’s theorem, prove the inequalities: (i) e^{x} \geq 1+x (ii) \ln (1+x) <x, for x>0. (iii) b^{n}-a^{n}<ab^{n-1}(b-a) for b>a. (iv) \arctan(x) <x.

Problem 4:

Write the Cauchy formula for the functions f(x)=x^{2}, phi(x)=x^{3} on the interval [1,2] and find c.

More churnings with calculus later!

Nalin Pithwa.



Some questions based on Rolle’s theorem

Problem 1:

Verify the truth of Rolle’s theorm for the following functions:

(a) y=x^{2}-3x+2 on the interval [1,2].

(b) y=x^{3}+5x^{2}-6x on the interval [0,1].

(c) y=(x-1)(x-2)(x-3) on the interval [1,3].

(d) y=\sin^{2}(x) on the interval [0,\pi].

Problem 2:

The function f(x)=4x^{3}+x^{2}-4x-1 has roots 1 and -1. Find the root of the derivative f^{'}(x) mentioned in Rolle’s theorem.

Problem 3:

Verify that between the roots of the function y=\sqrt[3]{x^{2}-5x+6} lies the root of its derivative.

Problem 4:

Verify the truth of Rolle’s theorem for the function y=\cos^{2}(x) on the interval [-\frac{\pi}{4},+\frac{\pi}{4}].

Problem 5:

The function y=1-\sqrt[5]{x^{4}} becomes zero at the end points of the interval [-1,1]. Make it clear that the derivative of the function does not vanish anywhere in the interval (-1,1). Explain why Rolle’s theorem is NOT applicable here.

Calculus is the fountainhead of many many ideas in mathematics and hence, technology. Expect more beautiful questions on Calculus !

-Nalin Pithwa

Some Applications of Derivatives — Part II

Derivatives in Economics.

Engineers use the terms velocity and acceleration to refer to the derivatives of functions describing motion. Economists, too, have a specialized vocabulary for rates of change and derivatives. They call them marginals.

In a manufacturing operation, the cost of production c(x) is a function of x, the number of units produced. The marginal cost of production is the rate of change of cost (c) with respect to a level of production (x), so it is dc/dx.

For example, let c(x) represent the dollars needed needed to produce x tons of steel in one week. It costs more to produce x+h units, and the cost difference, divided by h, is the average increase in cost per ton per week:

\frac{c(x+h)-c(x)}{h}= average increase in cost/ton/wk to produce the next h tons of steel

The limit of this ratio as h \rightarrow 0 is the marginal cost of producing more steel when the current production level is x tons.

\frac{dc}{dx}=\lim_{h \rightarrow 0} \frac{c(x+h)-c(x)}{h}= marginal cost of production

Sometimes, the marginal cost of production is loosely defined to be the extra cost of producing one unit:

\frac{\triangle {c}}{\triangle {x}}=\frac{c(x+1)-c(x)}{1}

which is approximately the value of dc/dx at x. To see why this is an acceptable approximation, observe that if the slope  of c does not change quickly near x, then the difference quotient will be close to its limit, the derivative dc/dx, even if \triangle {x}=1. In practice, the approximation works best for large values of x.

Example: Marginal Cost

Suppose it costs c(x)=x^{3}-6x^{2}+15x  dollars to produce x radiators when 8 to 30 radiators are produced. Your shop currently produces 10 radiators a day. About how much extra cost will it cost to produce one more radiator a day?

Example : Marginal tax rate

To get some feel for the language of marginal rates, consider marginal tax rates. If your marginal income tax rate is 28% and your income increases by USD 1000, you can expect to have to pay an extra USD 280 in income taxes. This does not mean that you pay 28 percent of your entire income in taxes. It just means that at your current income level I, the rate of increase of taxes I with respect to income is dT/dI = 0.28. You will pay USD 0.28 out of every extra dollar you earn in taxes. Of course, if you earn a lot more, you may land in a higher tax bracket and your marginal rate will increase.

Example: Marginal revenue:

If r(x) = x^{3}-3x^{2}+12x gives the dollar revenue from selling x thousand candy bars, 5<= x<=20, the marginal revenue when x thousand are sold is

r^{'}(x) = \frac{d}{dx}(x^{3}-3x^{2}+12x)=3x^{2}-6x+12.

As with marginal cost, the marginal revenue function estimates the increase in revenue that will result from selling one additional unit. If you currently sell 10 thousand candy bars a week, you can expect your revenue to increase by about r^{'}(10) = 3(100) -6(10) +12=252 USD, if you increase sales to 11 thousand bars a week.

Choosing functions to illustrate economics.

In case, you are wondering why economists use polynomials of low degree to illustrate complicated phenomena like cost and revenue, here is the rationale: while formulae for real phenomena are rarely available in any given instance, the theory of  economics can still provide valuable guidance. the functions about which theory speaks can often be illustrated with low degree polynomials on relevant intervals. Cubic polynomials provide a good balance between being easy to work with and being complicated enough to illustrate important points.

Ref: Calculus and Analytic Geometry by G B Thomas.

More later,

Nalin Pithwa


Could a one-sided limit not exist ?

Here is basic concept of limit :

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

A problem on solutions of triangles (ambiguous case)

Problem 1:

Given b=16, c=25, B= 33 \deg 13^{'}, prove that the triangle is ambiguous and find the other angles, using five figure tables (log/trig).

Kindly send your answers, comments, etc.

Nalin Pithwa

PS: Can you check/prove/verify the ambiguous case of solutions of triangles using plane geometry of your high school days?

Solution of Triangles (Ambiguous Cases) : IIT JEE Maths

The three sides a, b, c and the three angles A, B, C are called the elements of the triangle ABC. When any three of these six elements (except all the three angles) of a triangle are given, the triangle is known completely; that is, the other three elements can be expressed in terms of the given elements and can be evaluated. This process is called the solution of triangles.

  • If the three sides a, b, ¬†and c are given, angle A is obtained from \tan{(A/2)}= \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} or \cos{A}=\frac{b^{2}+c^{2}-a^{2}}{2bc}. B and C can be obtained in a similar way.
  • If two sides b and c and the included angle A are given, then \tan{\frac{B-C}{2}}=\frac{b-c}{b+c}\cot{(A/2)} gives \frac{B-C}{2}. Also, \frac{B-C}{2}=90 \deg - \frac{A}{2}, so that B and C can be evaluated. The third side is given by a=b \frac{\sin{A}}{\sin{B}}, or, a^{2}=b^{2}+c^{2}-2bc \cos{A}.
  • If two sides b and c and the angle B (opposite to side b) are given, then \sin{C}=\frac{c}{b}\sin{B}. And, A=180\deg -(B+C), and a=\frac{b \sin{A}}{\sin{B}} give the remaining elements.

By applying the cosine rule, we have:

\cos{B}=\frac{a^{2}+c^{2} - b^{2}}{2ac}, or if we manipulate this, we get


or, a=c \cos{B} \pm \sqrt{b^{2}-(c\sin{B})^{2}}

This equation leads to the following cases:

Case I:

If b<c\sin{B}, no such triangle is possible.

Case II:

Let b=c\sin{B}. There are further following two cases:

Sub-case II a:

B is an obtuse angle, that is, \cos{B} is negative. There exists no such triangle.

Sub-case II b:

B is an acute angle, that is, \cos {B} is positive. There exists only one such triangle.

Case III:

Let b >c \sin{B}. There are following two cases further here also:

Sub-case IIIa:

B is an acute angle, that is, \cos {B} is positive. In this case, two values of a will exist if and only if c\cos{B} > \sqrt{b^{2}-(c \sin{B})^{2}} or, c>b, which means two such triangles are possible. If c<b, only one such triangle is possible.

Sub-case IIIb:

B is an obtuse angle, that is, \cos{B} is negative. In this case, triangle will exist if and only if \sqrt{b^{2}-(c \sin{B})^{2}} > c |\cos{B}| \Longrightarrow b > c. So, in this case, only one such triangle is possible. If b <c, there exists no such triangle.


If one side a and angles B and C are given, then A=180 \deg -(B+C), and b=a \frac{\sin{B}}{\sin{A}} and c=a\frac{\sin{C}}{\sin{A}}.

If the three angles A, B and C are given, we can only find the ratios of the three sides a, b, and c by using the sine rule(since there are infinite number of similar triangles possible).

More theory later,

Nalin Pithwa