Inequalities « Mathematics Hothouse

Category Archives: Inequalities

best explanation of epsilon delta definition

February 10, 2020 – 8:35 am

Refer any edition of (i) Calculus and Analytic Geometry by Thomas and Finney (ii) recent editions which go by the title “Thomas’ Calculus”. If you need, you will have to go through the previous stuff (given in the text) on “preliminaries” and/or functions also. For Sets, Functions and Relations, I have also presented a long series of articles on this blog.

Ref:

https://www.amazon.in/Thomas-Calculus-George-B/dp/9353060419/ref=sr_1_1?crid=3F1XO0L9KBT1F&keywords=thomas+calculus&qid=1581323971&s=books&sprefix=Thomas+%2Caps%2C265&sr=1-1

By Nalin Pithwa | Also posted in calculus, Cnennai Math Institute Entrance Exam, IITJEE Advanced Mathematics, IITJEE Mains, INMO, ISI Kolkatta Entrance Exam, KVPY, pure mathematics | Comments (0)

Rules for Inequalities

September 12, 2019 – 1:25 pm

If a, b and c are real numbers, then

$a < b \Longrightarrow a + c< b + c$
$a < b \Longrightarrow a - c < b - c$
$a < b \hspace{0.1in} and \hspace{0.1in}c > 0 \Longrightarrow ac < bc$
$a < b \hspace{0.1in} and \hspace{0.1in}c < 0 \Longrightarrow bc < ac$ special case: $a < b \Longrightarrow -b < -a$
$a > 0 \Longrightarrow \frac{1}{a} > 0$
If a and b are both positive or both negative, then $a < b \Longrightarrow \frac{1}{b} < \frac{1}{a}$ .

Remarks:

Notice the rules for multiplying an inequality by a number: Multiplying by a positive number preserves the inequality; multiplying by a negative number reverses the inequality. Also, reciprocation reverses the inequality for numbers of the same sign.

Regards,

Nalin Pithwa.

By Nalin Pithwa | Also posted in calculus, IITJEE Foundation mathematics, ISI Kolkatta Entrance Exam, KVPY, Pre-RMO, RMO | Comments (0)

Applications of Derivatives: Tutorial Set 1: IITJEE Mains Maths

October 19, 2018 – 6:33 pm

“Easy” questions:

Question 1:

Find the slope of the tangent to the curve represented by the curve $x=t^{2}+3t-8$ and $y=2t^{2}-2t-5$ at the point $(2,-1)$ .

Question 2:

Find the co-ordinates of the point P on the curve $y^{2}=2x^{3}$ , the tangent at which is perpendicular to the line $4x-3y+2=0$ .

Question 3:

Find the co-ordinates of the point $P(x,y)$ lying in the first quadrant on the ellipse $x^{2}/8 + y^{2}/18=1$ so that the area of the triangle formed by the tangent at P and the co-ordinate axes is the smallest.

Question 4:

The function $f(x) = \frac{\log (\pi+x)}{\log (e+x)}$ , where $x \geq 0$ is

(a) increasing on $(-\infty, \infty)$

(b) decreasing on $[0, \infty)$

(d) decreasing on $[0, \pi/e)$ and increasing on $[\pi/e, \infty)$ .

Fill in the correct multiple choice. Only one of the choices is correct.

Question 5:

Find the length of a longest interval in which the function $3\sin(x) -4\sin^{3}(x)$ is increasing.

Question 6:

Let $f(x)=x e^{x(1-x)}$ , then $f(x)$ is

(a) increasing on $[-1/2, 1]$

(b) decreasing on $\Re$

(d) decreasing on $[-1/2, 1]$ .

Fill in the correct choice above. Only one choice holds true.

Question 7:

Consider the following statements S and R:

S: Both $\sin(x)$ and $\cos (x)$ are decreasing functions in the interval $(\pi/2, \pi)$ .

R: If a differentiable function decreases in the interval $(a,b)$ , then its derivative also decreases in $(a,b)$ .

Which of the following is true?

(i) Both S and R are wrong.

(ii) Both S and R are correct, but R is not the correct explanation for S.

(iii) S is correct and R is the correct explanation for S.

(iv) S is correct and R is wrong.

Indicate the correct choice. Only one choice is correct.

Question 8:

For which of the following functions on $[0,1]$ , the Lagrange’s Mean Value theorem is not applicable:

(i) $f(x) = 1/2 -x$ , when $x<1/2$ ; and $f(x) = (1/2-x)^{2}$ , when $x \geq 1/2$ .

(ii) $f(x) = \frac{\sin(x)}{x}$ , when $x \neq 0$ ; and $f(x)=1$ , when $x=0$ .

(iii) $f(x)=x |x|$

(iv) $f(x)=|x|$ .

Only one choice is correct. Which one?

Question 9:

How many real roots does the equation $e^{x-1}+x-2=0$ have?

Question 10:

What is the difference between the greatest and least values of the function $f(x) = \cos(x) + \frac{1}{2}\cos(2x) -\frac{1}{3}\cos(3x)$ ?

More later,

Nalin Pithwa.

By Nalin Pithwa | Also posted in applications of maths, calculus, ISI Kolkatta Entrance Exam, KVPY | Comments (0)

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

April 25, 2017 – 6:32 am

Problem:

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.

Solution:

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:

http://www.amazon.in/TEXAS-GRAPHIC-CALCULATOR-NSPIRE-CX/dp/B00A49F98U/ref=sr_1_fkmr0_1?s=electronics&ie=UTF8&qid=1493101690&sr=1-1-fkmr0&keywords=TI+graphing+calculator+Non+CAS

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.

Explanation:

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 $x \neq \pm 1$

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

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

$\frac{x+5}{x^{2}-1}<0$

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, $\{ x<-5\}\bigcup {-1<x<1}$