Monthly Archives: October 2016

Teaching methods fail kids on Mathematics and Science

{Ref: DNA newspaper, print edition, dated Oct 31 2016, Mumbai; section on Education; author: Kritika Sharma; E-mail: kritika.sharma@dnaindia.net)

A lack of stamina in Indian students to learn mathematics and the applied sciences is preventing government initiatives from India into a pivot of innovation and global manufacturing. A study conducted by NCERT has found that 53 percent of students in Class 10 levels are unable to pass simple competitive exams in these subjects, both of which play a significant role in Prime Minister Modi’s Make in India initiative.

Countries like South Korea, Singapore, Taiwan, Hong Kong, Japan, Russia, Israel, Finland, United States, and England top the list in mathematics. In Applied Science, the top spot is occupied by Singapore, followed by Taiwan, South Korea, Japan, Finland, Slovenia, Russia, Hong Kong, England, United States, Hungary and Australia.

A senior official at the NCERT told DNA that the lack of interest in these subjects is a worrisome situation. “It shows that students are learning everything just by rote and reproducing them in examination papers. They are not applying logic while solving problems,” he said. A primary reason for this non-performance was that students lacked basic conceptual knowledge of the subject.

“Students could not even attempt basic formula-based questions in Science and Mathematics that we had prepared to test their competence. Our survey found that only 45 per cent of students could answer a maximum of questions correctly in Science when basic questions related to acceleration and velocity were asked,” said NCERT official Y Sreekanth, who headed the survey. “Similarly, for Mathematics, we picked up questions from Geometry and Mensuration, and the performance was equally bad,” he added. According to him, the best performance was in languages like Hindi, and other vernaculars, where no logic was required.

The survey was conducted on a sample comprising 277,416 students in 2216 schools across 33 states/Union Territories and Boards. It is a part of the National Assessment Survey, which was done for Class 10 for the first time, and a detailed report recently released thereafter.

Dr. B. S. Rawat, a mathematician, who has been part of these NCERT surveys said, “The problem with Mathematics starts at the basic level. Most of the teachers do not explain the basics of the subject to students in a proper way and once the foundation is spoiled, it is difficult to do well in the subject for the rest of the classes.”

“Also students are busy memorizing things instead of understanding their concepts. The fault is not completely theirs. Teaching methods also need to be changed. Students should get more practical exposure so that they can understand concepts. Science and Mathematics cannot  be taught without practical exposure”.

NCERT in its report has emphasized that there is a need to change the way children at class 10 level are being taught as it is important to build a foundation that may in the future help them build a career.

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(my idea was to share this news/report with my students/their parents, and other readers of my blog(s))

(Thanks Ms. Kritika Sharma and DNA!)

Nalin Pithwa

Miscellaneous examples of Algebra: part 2 for IITJEE Mains

Problem 1:

Prove that $(a+b)^{5} - a^{5} - b^{5} = 5ab(a+b)(a^{2} + ab+b^{2})$.

Proof I:

First carry out the following steps:

(a) Is the given expression symmetric?

(b) Is the given expression alternating?

(c) Is the given expression cyclic?

(d) Is the given expression homogeneous?

So here goes the proof:

Denote the expression by E. Clearly, the substitutions $a=0$, $b=0$ give us $E=0$. Hence, two of the factors are $a, b$. Similarly, the substitution $a=-b$ give us $E=0$. So the other factor is $(a+b)$.

The degree of  the given expression E is 5. E is symmetric w.r.t. a and b; it is homogeneous. The factors we found above are $a, b, (a+b)$. So $E=kab(a+b)F(x,y)$ where F(x,y) ought to be of degree 2, homogeneous and symmetric. So, let $F(x, y) = Aa^{2}+Bab+Bb^{2}$; thus, we have

$(a+b)^{5}-a^{5}-b^{5}=kab(a+b)(Aa^{2}+Bab+Ab^{2})$. Now, find k, A, and B by the undetermined coefficients.

Hence, you will get $(a+b)^{5}-a^{5}-b^{5}=5ab(a+b)(a^{2}+ab+b^{2})$.

Problem 2:

Find the factors of $(b^{3}+c^{3})(b-c)+(c^{3}+a^{3})(c-a)+(a^{3}+b^{3})(a-b)$.

Solution 2:

First carry out the following steps:

(a) Is the given expression symmetric w.r.t. a and b; b and c; c and a?

(b) Is the given expression alternating w.r.t. a and b; b and c; c and a?

(c) Is the given expression cyclic w.r.t. a, b and c?

(d) Is the given expression homogeneous and if so, what is the degree of each expression?

Denote the expression by E; then, E is a function of a which vanishes when $a=b$, $b=c$, $c=a$, so the following are certainly factors of E: $(a-b), (b-c), (c-a)$, and hence, E contains $(a-b)(b-c)(c-a)$ as a factor.

Now note that E is of fourth degree and symmetric, so the remaining factor is of the form : $M(a+b+c)$, where M is a coefficient to be determined.

So, we have now: $E=M(a-b)(b-c)(c-a)(a+b+c)$, and you can easily find that $M=1$, and hence,

$(b^{3}+c^{3})(b-c)+(c^{3}+a^{3})(c-a)+(a^{3}+b^{3})(a-b) = (a-b)(b-c)(c-a)(a+b+c)$

More mathematical miscellany for IITJEE mathematics to follow!

Nalin Pithwa

Miscellaneous examples of Algebra Part I: IIT JEE Mains

Example 1:

Find the roots of $4x^{3}-16x^{2}-9x+36=0$, given that one root is the negative of the another.

Solution 1:

If the roots are a, b, and c, we have $b=-a$, say. So, using Viete’s relations,

$a-a+c=4$

$-a^{2}+ac-ac=-9/4$

$-a^{2}c = -9$.

Hence, $c = 4$ and $a= 3/2 = -b$. Hence, the roots are $\pm \frac{3}{2}, 4$.

Example 2:

Let $a, b, c \in \Re$, with $a \neq 0$ such that a and $4a+3b+2c$ have the same sign. Show that the equation $ax^{2}+bx+c=0$ cannot have both roots in the interval $(1,2)$.

Solution 2:

Let $\alpha, \beta$ be roots of the given quadratic equation. We have $0 \leq \frac{4a+3b+2c}{a}$

which in turn equals

$= 4 + 3 \frac{b}{a} + 2\frac{c}{a} = 4 - 3(\alpha + \beta) + 2\alpha. \beta = (\alpha-1)(\beta-2) + (\alpha -2)(\beta -1)$.

If $\alpha, \beta$ both belong to $(1,2)$ then each term of  the sum will be negative, which is a contraction. Hence, the proof.

Example 3:

Consider all lines which meet the graph of $y = 2x^{4}+7x^{3}+3x-5$ in four distinct points, say $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}), (x_{4}, y_{4})$. Then, show that $\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}$ is independent of the line and find its value.

Solution 3:

Let $y=mx+c$ be any line which intersects the graph $y = 2x^{4}+7x^{3}+3x-5$ at $(x_{i}, y_{i})$, where $1 \leq i \leq 4$. Then, $x_{i}$ are the roots of

$mx + c = 2x^{4} + 7x^{3} + 3x -5$.

(Note that $x_{i}'s$ are distinct as $x_{i} = x_{j}$ would imply $y_{i}=y_{j}$).

The above equation reduces to an equation of degree 4, namely,

$2x^{4}+7x^{3}+(3-m)x - 5-c=0$. Hence, by Viete’s relations,

$\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4} = - \frac{7}{8}$.

Example 4:

The product of two of the four roots of $x^{4}-20x^{3}+kx^{2}+590x-1992=0$ is 24. Find k.

Solution 4:

Let the given equation be written as $f(x)=0$, and let the roots of  the equation be $r_{1}, r_{2}, r_{3}, r_{4}$ with $r_{1}r_{2}=24$. Now,

$r_{1}r_{2}r_{3}r_{4}= -1992$, so $r_{3}r_{4} = -1992/24 = -83$. Also,

$f(x) = (x-r_{1})(x-r_{2})(x-r_{3})(x-r_{4}) = (x^{2}-cx+r_{1}r_{2})(x^{2}-dx+r_{3}r_{4}) = (x^{2}-cx+24)(x^{2}-dx-83)$

with $c = r_{1} + r_{2}$, $d = r_{3} + r_{4}$. Comparing coefficients of $x^{3}$ and x we get $c+d=20$ and $83c - 24d = 590$. This gives $c = 10, d = 10$. Comparing coefficients of $x^{2}$, $k = cd - 83 + 24 = 100 -83 +24 =41$.

Example 5:

If $\alpha$ and $\beta$ are roots of $x^{2}+px+q=0$, where p and q are integers with $q|p^{2}$, then show that

(i) $\alpha^{n} + \beta^{n}$ is an integer. ($n \geq 1$)

(ii) $\alpha^{n} + \beta^{n}$ is an integer divisible by q. ($n \geq 2$).

Solution 5:

Since $\alpha, \beta$ are the roots of $x^{2}+px+q=0$, we get

$\alpha + \beta = -p$ —– Equation A

$\alpha \beta = q$ —– Equation B

Note that $\alpha^{2} = -p \alpha - q$. For $n \geq 2$, multiplying this equation by $\alpha^{n-2}$, we get $\alpha^{n} = -p \alpha^{n-1} - q\alpha^{n-2}$. Similarly, $\beta^{n} = -p\beta^{n-1} -q\beta^{n-2}$. Hence,

$\alpha^{n} + \beta^{n} = -p(\alpha^{n-1} + \beta^{n-1}) -q(\alpha^{n-2} +\beta^{n-2})$ Eqn C

Also, for $n=2$, $\alpha^{2} + \beta^{2} = (-p)^{2} -2q = p^{2} -2q$ Equation D

1.By  equation A and D, $\alpha + \beta$ and $\alpha^{2} + \beta^{2}$ are both integers. Hence, by Eqn C, it follows by induction on n that $\alpha^{n} + \beta^{n}$ is an integer for $n \geq 1$.

2. Since $q|p^{2}$, equation D shows that $q|\alpha^{2} + \beta^{2}$. Further, $\alpha^{3} + \beta^{3} = -p(\alpha^{2} + \beta^{2}) - q(\alpha + \beta)$ and so $q|(\alpha^{3} + \beta^{3})$. Hence, by C, it follows by induction on n that $q|(\alpha^{n}+\beta^{n})$ for $n \geq 2$.

Example 6:

Find all integers a such that the equation $x^{3} - 3x +a=0$ has three integer roots.

Solution 6:

Let the integer roots of the given equation be $\alpha, \beta, \gamma$. Then,

$\alpha + \beta + \gamma = 0, \alpha \beta + \beta \gamma + \gamma \alpha = -3$, $\alpha \beta \gamma = -a$Let this be Equation I.

Hence, $\alpha^{2} + \beta^{2} + \gamma^{2} = (\alpha + \beta + \gamma)^{2} - 2(\alpha \beta + \beta \gamma + \gamma \alpha) = 6$Let this be Equation II.

So, $0 \leq \alpha^{2}, \beta^{2}, \gamma^{2} \leq 6$ and so the solution of Equation II are essentially the following:

$\alpha = 2, \beta = -1, \gamma = -1$ —- Call this ***

$\alpha = -2, \beta =1, \gamma = 1$ —- Call this *****

Both these sets satisfy Equation I. Hence, the required values of a are $a = -2, 2$ corresponding to the roots in (***) and (*****) respectively.

Example 7:

Find the remainder when $(x+1)^{n}$ is divided by $(x-1)^{3}$.

Solution 7:

Dividing $(x+1)^{n}$ by $(x-1)^{3}$, we get $(x+1)^{n} = f(x)(x-1)^{3}$, we get

$(x+1)^{n} = f(x) (x-1)^{3} + Ax^{2} + Bx + C$.

Put $x-1=y$ or $x=y+1$. Hence,

$(y+2)^{n} = f(y+1)y^{3} + A(y+1)^{2} + B(y+1) + C$

Using Binomial theorem, we get

$y^{n} + \ldots + y^{2}(\frac{n(n-1)}{2}2^{n-2}) + y(n2^{n-1}) + 2^{n}$, which in turn equals

$= f(y+1)y^{3}+ Ay^{2} + (2A + B)y + A + B + C$ Call this Equation @@@

Now, equating coefficients of $y^{2}, y^{1}, y^{0}$ we get

$A = n(n-1)2^{n-3}$, $2A + B = n.2^{n-1}$, $A+B+C=2^{n}$.

Solving these equations, we get

$A = n(n-1)2^{n-3}$, $B = n(3-n)2^{n-2}$, $C = (n^{2}-5n+8)2^{n-3}$

Hence, the remainder is

$n(n-1)2^{n-3}x^{2} + n(3-n)2^{n-2}x + (n^{2}-5n+8)2^{n-3}$

More algebraic stuff in the pipeline!!

Nalin Pithwa

Beating all odds, Midnapur lad makes it to IIT Bombay

(From a recent print edition of the DNA newspaper, Mumbai).

Siliguri (West Bengal):

Chandan Roy, belonging to a poor household in West Bengal’s Midnapur district, has made his family proud after earning a seat in the coveted IIT Bombay.

His father is a priest who struggles to make ends meet. But, this did not stop Chandan from achieving his dream. “I did not take any coaching. The only help  I got was from my school and tuition teacher. I have a dream to become a scientist. My father is a Brahmin priest and owns a small grocery shop, where I help my father,” Chandan said.

Through his sheer hard work and dedication, Chandan secured 314 in JEE Advanced (All India Rank) and 78 in JEE Main examination. Chandan said he received Rs. 50000/- as scholarship and the rest of the amount was paid with the help of his teachers. His father, Suchinta Roy, is worried as he is cash strapped and wonders how he would pay for his education.

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Mathematics that we can taste, touch, see, smell, and hear !!

In some blog article earlier, I had claimed that we can see, taste, touch, smell, and hear mathematics every where around us! Well, to be precise, I had explained that ‘sweetness’ or ‘sourness’ of what we eat or drink can be measured with the pH value, which uses the tool of common logarithms !

Now, we all love mp3 music. It uses a special algorithm to measure the sensitivity of the human ear to various frequencies and intensities. It masks out certain pieces of the sound or music signal (called information signal, in the language of engineering) and thus achieves “compression” of the music signal. So, the creators of the mp3 algorithm actually created a mathematical model of the human ear.

And, of course, the loudness of sound is measured in terms of decibels, which is again a logarithmic quantity.

Architects use “golden ratio” to create a majestic structure that we can see/marvel at. This golden ratio is a special number.

The pain that human skin/body feels is measured in terms of “bols”. ( Go ahead and google this term).

And, of course, smell, obnoxious or aromatic, is measured in terms of the concentration of special molecules in the immediate vicinity.

Nature also loves math! Do you know that the spirals of a daisy flower are arranged like Fibonacci numbers/sequences?

More on each of these topics later,

Nalin Pithwa

Some good trig practice problems for IITJEE Mains Mathematics

Problem:

Find the value of $\cos^{4}\frac{\pi}{8} + \cos^{4}\frac{3\pi}{8}+\cos^{4}\frac{5\pi}{8}+\cos^{4}\frac{7\pi}{8}$.

Problem:

Find the value of $\cos{\theta}\cos{2\theta}\cos{4\theta}\ldots\cos{2^{n-1}}{\theta}$

Problem:

Find the value of $\cos{(2\pi/15)}\cos{(4\pi/15)}\cos{(8\pi/15)}\cos{(14\pi/15)}$

Problem:

Prove that $(\frac{\cos{A}+\cos{B}}{\sin{A}-\sin{B}})^{n}+(\frac{\sin{A}+\sin{B}}{\cos{A}-\cos{B}})^{n} = 2\cot^{n}(\frac{A-B}{2})$, if n is even, and is zero, if n is odd.

More later,

Nalin Pithwa

Matches

The man emptied a box of matches on the table and divided them into three heaps.

“You aren’t going to start a bonfire, are you?” someone quipped.

“No, they are for a brain teaser. Here it is —- three uneven heaps. There are altogether 48 matches. I won’t tell how many there are in each heap. Look well. If I take as many matches from the first heap as there are in the second and add them to the second, and then take as many from the second as there are in the third, and add them to the third, and finally, take as many from the third as there are in the first, and add them to the first — well, if I do all this, the heaps will all have the same number of matches. How many were there originally in each heap?”

More later,

Nalin Pithwa

“My puzzle,” asked the next man, “is also about a dirigible. What’s longer, the dirigible or its perfect shadow?”

“Is that all?”

“It is.”

“Well, then. The shadow is naturally longer than the dirigible: sunrays spread fan-like, don’t they?”

“I wouldn’t say so,” another interjected. “Sun rays are parallel to each other and that being so, the dirigible and its shadow are of the same size.”

“No, they aren’t. Have a  you ever seen rays spreading from behind a cloud? If you have, you have probably noticed how much they spread. The shadow of the dirigible must be considerably bigger than the dirigible itself, just as the shadow of a cloud is bigger than the cloud itself.”

“Then, why is it that people say that sun rays are parallel to each other? Sea-men, astronomers for instance.”

The professor put a stop to the argument by asking the next person to go ahead with his conundrum.

Cheers,

Nalin Pithwa

Dirigible’s flight

“A dirigible took off from Leningrad in a northerly direction. Five hundred km away it turned and flew 500 km eastward. After that it turned south and covered another 500 km. Then, it flew 500 km in a westerly direction, and landed. The question is: where did it land: west, east, north, or south of Leningrad?”

“That’s an easy one,” someone said. “Five hundred steps forward, 500 to the right, 500 back and 500 to the left, and you are naturally back, where you  had started from!”

“Easy! Well then, where did the dirigible land?”

“In Leningrad, of course. Where else?”

“Wrong!”

“Then, I don’t understand.”

“Yes, there is some catch to  this puzzle,” another joined in. “Didn’t the dirigible land in Leningrad?”