Mathematics Hothouse

On the value of time…essence of time management

November 27, 2018 – 10:40 pm

Time is Life.

Money lost can come back, but time lost can never come back. Time is more valuable than money. (Bertrand Russell).

By Nalin Pithwa | Posted in time management | Comments (0)

Mathematics versus Physics

November 27, 2018 – 10:36 pm

The object of pure Physics is the unfolding of the laws of the intelligible world; the object of pure Mathematic that of unfolding the laws of human intelligence. — J. J. Sylvester.

In my opinion, for example, Boole’s Laws of (Human) Thought.

By Nalin Pithwa | Posted in applications of maths, careers in mathematics, mathematicians, miscellaneous, motivational stuff | Comments (0)

An observation about Sir Isaac Newton

November 27, 2018 – 10:07 pm

Newton’s patience was limitless. Truth, he said much later, was the offspring of silence and meditation. And, he said, I keep the subject constantly before me and wait till the first dawnings open slowly, by little and little into a full and clear light.

By Nalin Pithwa | Posted in mathematicians, motivational stuff | Comments (0)

Applications of Derivatives IITJEE Maths tutorial: practice problems part IV

November 13, 2018 – 6:26 am

Question 1.

If the point on $y = x \tan {\alpha} - \frac{ax^{2}}{2u^{2}\cos^{2}{\alpha}}$ , where $\alpha>0$ , where the tangent is parallel to $y=x$ has an ordinate $\frac{u^{2}}{4a}$ , then what is the value of $\alpha$ ?

Question 2:

Prove that the segment of the tangent to the curve $y=c/x$ , which is contained between the coordinate axes is bisected at the point of tangency.

Question 3:

Find all the tangents to the curve $y = \cos{(x+y)}$ for $-\pi \leq x \leq \pi$ that are parallel to the line $x+2y=0$ .

Question 4:

Prove that the curves $y=f(x)$ , where $f(x)>0$ , and $y=f(x)\sin{x}$ , where $f(x)$ is a differentiable function have common tangents at common points.

Question 5:

Find the condition that the lines $x \cos{\alpha} + y \sin{\alpha} = p$ may touch the curve $(\frac{x}{a})^{m} + (\frac{y}{b})^{m}=1$ .

Question 6:

Find the equation of a straight line which is tangent to one point and normal to the point on the curve $y=8t^{3}-1$ , and $x=4t^{2}+3$ .

Question 7:

Three normals are drawn from the point $(c,0)$ to the curve $y^{2}=x$ . Show that c must be greater than 1/2. One normal is always the x-axis. Find c for which the two other normals are perpendicular to each other.

Question 8:

If $p_{1}$ and $p_{2}$ are lengths of the perpendiculars from origin on the tangent and normal to the curve $x^{2/3} + y^{2/3}=a^{2/3}$ respectively, prove that $4p_{1}^{2} + p_{2}^{2}=a^{2}$ .

Question 9:

Show that the curve $x=1-3t^{2}$ , and $y=t-3t^{3}$ is symmetrical about x-axis and has no real points for $x>1$ . If the tangent at the point t is inclined at an angle $\psi$ to OX, prove that $3t= \tan {\psi} +\sec {\psi}$ . If the tangent at $P(-2,2)$ meets the curve again at Q, prove that the tangents at P and Q are at right angles.

Question 10:

Find the condition that the curves $ax^{2}+by^{2}=1$ and $a^{'}x^{2} + b^{'}y^{2}=1$ intersect orthogonality and hence show that the curves $\frac{x^{2}}{(a^{2}+b_{1})} + \frac{y^{2}}{(b^{2}+b_{1})} = 1$ and $\frac{x^{2}}{a^{2}+b_{2}} + \frac{y^{2}}{(b^{2}+b_{2})} =1$ also intersect orthogonally.

More later,

Nalin Pithwa.

By Nalin Pithwa | Posted in applications of maths, calculus, IITJEE Mains, ISI Kolkatta Entrance Exam, KVPY | Comments (0)

Applications of Derivatives: IITJEE Maths tutorial problem set: III

October 23, 2018 – 9:43 pm

Slightly difficult questions, I hope, but will certainly re-inforce core concepts:

Prove that the segment of the tangent to the curve $y=c/x$ which is contained between the co-ordinate axes, is bisected at the point of tangency.
Find all tangents to the curve $y=\cos{(x+y)}$ for $-\pi \leq x \leq \pi$ that are parallel to the line $x+2y=0$ .
Prove that the curves $y=f(x)$ , where $f(x)>0$ and $y=f(x)\sin(x)$ , where $f(x)$ is a differentiable function, have common tangents at common points.
Find the condition that the lines $x\cos{\alpha} + y \sin{\alpha}=p$ may touch the curve $(\frac{x}{a})^{m} + (\frac{y}{b})^{m}=1$ .
If $p_{1}$ and $p_{2}$ are lengths of the perpendiculars from origin on the tangent and normal to the curve $x^{2/3} + y^{2/3}=a^{2/3}$ respectively, prove that $4p_{1}^{2}+p_{2}^{2}=a^{2}$ .
Show that the curve $x=1-3t^{2}$ , $y=t-3t^{3}$ is symmetrical about x-axis and has no real points for $x>1$ . If the tangent at the point t is inclined at an angle $\psi$ to OX, prove that $3t = \tan{\psi} + \sec{\psi}$ . If the tangent at $P(-2,2)$ meets the curve again at Q, prove that the tangents at P and Q are at right angles.
A tangent at a point $P_{1}$ other than $(0,0)$ on the curve $y=x^{3}$ meets the curve again at $P_{2}$ . The tangent at $P_{2}$ meets the curve at $P_{3}$ and so on. Show that the abscissae of $P_{1}, P_{2}, \ldots, P_{n}$ form a GP. Also, find the ratio of area $\frac{\Delta P_{1}P_{2}P_{3}}{area \hspace{0.1in} P_{2}P_{3}P_{4}}$ .
Show that the square roots of two successive natural numbers greater than $N^{2}$ differ by less than $\frac{1}{2N}$ .
Show that the derivative of the function $f(x) = x \sin {(\frac{\pi}{x})}$ , when $x>0$ , and $f(x)=0$ when $x=0$ vanishes on an infinite set of points of the interval $(0,1)$ .
Prove that $\frac{x}{(1+x)} < \log {(1+x)} < x$ for $x>0$ .

More later, cheers,

Nalin Pithwa.

By Nalin Pithwa | Posted in IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) | Comments (0)

Applications of Derivatives: Tutorial: IITJEE Maths: Part II

October 23, 2018 – 8:32 pm

Another set of “easy to moderately difficult” questions:

The function $y = \frac{}x{1+x^{2}}$ decreases in the interval (a) $(-1,1)$ (b) $[1, \infty)$ (c) $(-\infty, -1]$ (d) $(-\infty, \infty)$ . There are more than one correct choices. Which are those?
The function $f(x) = \arctan (x) - x$ decreases in the interval (a) $(1,\infty)$ (b) $(-1, \infty)$ (c) $(-\infty, -\infty)$ (d) $(0, \infty)$ . There is more than one correct choice. Which are those?
For $x>1$ , $y = \log(x)$ satisfies the inequality: (a) $x-1>y$ (b) $x^{2}-1>y$ (c) $y>x-1$ (d) $\frac{x-1}{x}<y$ . There is more than one correct choice. Which are those?
Suppose $f^{'}(x)$ exists for each x and $h(x) = f(x) - (f(x))^{2} + (f(x))^{3}$ for every real number x. Then, (a) h is increasing whenever f is increasing (b) h is increasing whenever f is decreasing (c) h is decreasing whenever f is decreasing (d) nothing can be said in general. Find the correct choice(s).
If $f(x)=3x^{2}+12x-1$ , when $-1 \leq x \leq 2$ , and $f(x)=37-x$ , when $2<x\leq 3$ . Then, (a) $f(x)$ is increasing on $[-1,2]$ (b) $f(x)$ is continuous on $[-1,3]$ (c) $f^{'}(2)$ doesn’t exist (d) $f(x)$ has the maximum value at $x=2$ . Find all the correct choice(s).
In which interval does the function $y=\frac{x}{\log(x)}$ increase?
Which is the larger of the functions $\sin(x) + \tan(x)$ and $f(x)=2x$ in the interval $(0<x<\pi/2)$ ?
Find the set of all x for which $\log {(1+x)} \leq x$ .
Let $f(x) = |x-1| + a$ , if $x \leq 1$ ; and, $f(x)=2x+3$ , if $x>1$ . If $f(x)$ has local minimum at $x=1$ , then $a \leq$ ?
There are exactly two distinct linear functions (find them), such that they map $[-1,1]$ and $[0,2]$ .

more later, cheers,

Nalin Pithwa.

By Nalin Pithwa | Posted in applications of maths, calculus, IITJEE Mains | 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 | Posted in applications of maths, calculus, Inequalities, ISI Kolkatta Entrance Exam, KVPY | Comments (0)

Primality testing: recap

September 11, 2018 – 4:13 am

via Primality testing: recap

By Nalin Pithwa | Posted in IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) | Comments (0)

Dear IIT Students: Address to IIT Hyderabad by present President of India, Shri Ram Nath Kovind

August 6, 2018 – 6:55 am

Dear IIT students

By Nalin Pithwa | Posted in miscellaneous, motivational stuff | Comments (0)

Applications of Derivatives: A Quick Review

July 17, 2018 – 12:19 am

Section I:

The Derivative as a Rate of Change

In case of a linear function $y=mx+c$ , the graph is a straight line and the slope m measures the steepness of the line by giving the rate of climb of the line, the rate of change of y with respect to x.

As x changes from $x_{0}$ to $x_{1}$ , y changes m times as much:

$y_{1}-y_{0}=m(x_{1}-x_{0})$

Thus, the slope $m=(y_{1}-y_{0})(x_{1}-x_{0})$ gives the change in y per unit change in x.

In more general case of differentiable function $y=f(x)$ , the difference quotient

$\frac{f(x+h)-f(x)}{x+h-x} = \frac{f(x+h)-f(x)}{h}$ , where $h \neq 0$

give the average rate of change of y (or f) with respect to x. The limit as h approaches zero is the derivative $dy/dx = f^{'}(x)$ , which can be interpreted as the instantaneous rate of change of f with respect to x. Since, the graph is a curve, the rate of change of y can vary from point to point.

Velocity and Acceleration:

Suppose that an object is moving along a straight line and that, for each time t during a certain time interval, the object has location/position $x(t)$ . Then, at time $t+h$ the position of the object is $x(t+h)$ and $x(t+h)-x(t)$ is the change in position that the object experienced during the time period t to $t+h$ . The ratio

$\frac{x(t+h)-x(t)}{t+h-t} = \frac{x(t+h)-x(t)}{h}$

gives the average velocity of the object during this time period. If

$\lim_{h \rightarrow 0} \frac{x(t+h)-x(t)}{h}=x^{'}(t)$

exists, then $x^{'}(t)$ gives the instantaneous rate of change of position with respect to time. This rate of change of position is called the velocity of the object. If the velocity function is itself differentiable, then its rate of change with respect to time is called the acceleration; in symbols,

$a(t) = v^{'}(t) = x^{''}(t)$

The speed is by definition the absolute value of the velocity: speed at time t is $|v(t)|$

If the velocity and acceleration have the same sign, then the object is speeding up, but if the velocity and acceleration have opposite signs, then the object is slowing down.

A sudden change in acceleration is called a jerk. Jerk is the derivative of acceleration. If a body’s position at the time t is $x(t)$ , the body’s jerk at time t is

$j = \frac{da}{dt} = \frac{d^{3}x}{dt^{3}}$

Differentials

Let $y = f(x)$ be a differentiable function. Let $h \neq 0$ . The difference $f(x+h) - f(x)$ is called the increment of f from x to $x+h$ , and is denoted by $\Delta f$ .

$\Delta f = f(x+h) - f(x)$

The product $f^{'}(x)h$ is called the differential of f at x with increment h, and is denoted by $df$

$df = f^{'}(x)h$

The change in f from x to $x+h$ can be approximated by $f^{'}(x)h$ :

$f(x+h) - f(x) = f^{'}(x)h$

Tangent and Normal

Let $y = f(x)$ be the equation of a curve, and let $P(x_{0}, y_{0})$ be a point on it. Let PT be the tangent, PN the normal and PM the perpendicular to the x-axis.

The slope of the tangent to the curve $y = f(x)$ at P is given by $(\frac{dy}{dx})_{(x_{0}, y_{0})}$

Thus, the equation of the tangent to the curve $y = f(x)$ at $(x_{0}, y_{0})$ is $y - y_{0} = (\frac{dy}{dx})_{(x_{0}, y_{0})}(x-x_{0})$

Since PM is perpendicular to PT, it follows that if $(\frac{dy}{dx})_{(x_{0}, y_{0})} \neq 0$ , the slope of PN is

$- \frac{1}{(\frac{dy}{dx})_{(x_{0}, y_{0})}} = - (\frac{dx}{dy})_{(x_{0}, y_{0})}$

Hence, the equation of the normal to the curve $y = f(x)$ at $(x_{0}, y_{0})$ is

$y - y_{0} = - (\frac{dx}{dy})_{(x_{0}, y_{0})}(x-x_{0})$

The equation of the normal parallel to the x-axis is $y = y_{0}$ , that is, when $(\frac{dy}{dx})_{(x_{0}, y_{0})} = 0$ . The length of the tangent at $(x_{0}, y_{0})$ is PT, and it is equal to

$y_{0}\csc{\theta} = y_{0}\sqrt{1+\cot^{2}{\theta}} = y_{0}\sqrt{1+[(\frac{dx}{dy})_{(x_{0}, y_{0})}]^{2}}$

The length of the normal is PN and it is equal to $y_{0}\sec {\theta} = y_{0}\sqrt{1 + [(\frac{dy}{dx})_{(x_{0}, y_{0})}]^{2}}$

If the curve is represented by $x = f(t)$ and $y = g(t)$ , that is, parametric equations in t, then

$\frac{dy}{dx} = \frac{g^{'}(t)}{f^{'}(t)}$ where $g^{'}(t)= \frac{dy}{dt}$ and $f^{'}(t) = \frac{dx}{dt}$ . In this case, the equations of the tangent and the normal are given by

$y - g(t) = \frac{g^{'}(t)}{f^{'}(t)}[x - f(t)]$ and $[y-g(t)] g^{'}(t) + [x-f(t)]f^{'}(t) = 0$ respectively.

The Angle between Two Curves

The angle of intersection of two curves is defined as the angle between the two tangents at the point of intersection. Let $y = f(x)$ and $y=g(x)$ be two curves, and let $P(x_{0}, y_{0})$ be their point of intersection. Also, let $\psi$ and $\phi$ be the angles of inclination of the two tangents with the x-axis, and let $\theta$ be the angle between the two tangents. Then,

$\tan {\theta} = \frac{\tan{\phi}-\tan{\psi}}{1+\tan{\phi}\tan{\psi}} = \frac{g^{'}(x) - f^{'}(x)}{1+f^{'}(x)g^{'}(x)}$

Example 1:

Write down the equations of the tangent and the normal to the curve $y = x^{3} - 3x + 2$ at the point $(2,4)$ .

Solution 1:

$\frac{dy}{dx} = 3x^{2}-3 \Longrightarrow \frac{dy}{dx}_{(2,4)} = 3.4 - 3 = 9$ .

Hence, the equation of the tangent at $(2,4)$ is given by $y-4 = 9(x-2) \Longrightarrow 9x-y-14=0$ and the equation of the normal is $y - 4 = (-1/9)(x-2) \Longrightarrow x+9y -38=0$ .

Rolle’s Theorem and Lagrange’s Theorem:

Rolle’s Theorem:

Let $f(x)$ be a function defined on a closed interval $[a,b]$ such that (i) f(x) is continuous on $[a,b]$ , (ii) f(x) is derivable on $(a,b)$ , and (iii) f(a) = f(b). Then, there exists a $c \in (a,b)$ such that $f^{'}(x)=0$ .

For details, the very beautiful, lucid, accessible explanation in Wikipedia:

https://en.wikipedia.org/wiki/Rolle%27s_theorem

Lagrange’s theorem:

Let $f(x)$ be a function defined on a closed interval $[a,b]$ such that (i) $f(x)$ is continuous on $[a,b]$ , and (ii) $f(x)$ is derivable on $(a,b)$ . Then, there exists a $c \in [a,b]$ such that

$f^{'}(c) = \frac{f(b)-f(a)}{b-a}$

Example 2:

The function $f(x) = \log {\sin(x)}$ satisfies the conditions of Rolle’s theorem on the interval $[\frac{\pi}{6}, \frac{5\pi}{6}]$ , as the logarithmic function and $\sin (x)$ are continuous and differentiable functions and $\log {\sin (\frac{5\pi}{6})} = \log {\sin (\pi - \frac{\pi}{6})} = \log{\sin{(\frac{\pi}{6})}}$ .

The conclusion of Rolle’s theorem is given at $c=\frac{\pi}{2}$ , for which $f^{'}(c) = \cot (c) = \cot (\pi/2) =0$ .

Rolle’s theorem for polynomials:

If $\phi(x)$ is any polynomial, then between any pair of roots of $\phi(x)=0$ lies a root of $\phi^{'}(x)=0$ .

Monotonicity:

A function $f(x)$ defined on a set D is said to be non-decreasing, increasing, non-increasing and decreasing respectively, if for any $x_{1}, x_{2} \in D$ and $x_{1} < x_{2}$ , we have $f(x_{1}) \leq f(x_{2})$ , $f(x_{1}) < f(x_{2})$ , $f(x_{1}) \geq f(x_{2})$ and $f(x_{1}) > f(x_{2})$ respectively. The function $f(x)$ is said to be monotonic if it possesses any of these properties.

For example, $f(x) = e^{x}$ is an increasing function, and $f(x)=\frac{1}{x}$ is a decreasing function.

Testing monotonicity:

Let $f(x)$ be continuous on $[a,b]$ and differentiable on $(a,b)$ . Then,

(i) for $f(x)$ to be non-decreasing (non-increasing) on $[a,b]$ it is necessary and sufficient that $f^{'}(x) \geq 0$ ( $f^{'}(x) \leq 0$ ) for all $x \in (a,b)$ .

(ii) for $f(x)$ to be increasing (decreasing) on $[a,b]$ it is sufficient that $f^{'}(x)>0$ ( $f^{'}(x)<0$ ) for all $x \in (a,b)$ .

(iii) If $f^{'}(x)=0$ for all x in $(a,b)$ , then f is constant on $[a,b]$ .

Example 3:

Determine the intervals of increase and decrease for the function $f(x)=x^{3}+2x-5$ .

Solution 3:

We have $f^{'}(x) = 3x^{2}+2$ , and for any value of x, $3x^{2}+2>0$ . Hence, f is increasing on $(-\infty, -\infty)$ . QED.

The following is a simple criterion for determining the sign of $f^{'}(x)$ :

If $a,b \geq 0$ , then $(x-a)(x-b)>0$ iff $x > \max (a,b)$ or $x < \min(a,b)$ ;

$(x-a)(x-b)<0$ if and only if $\min(a,b) < x < \max(a,b)$

Maxima and Minima:

A function has a local maximum at the point $x_{0}$ if the value of the function $f(x)$ at that point is greater than its values at all points other than $x_{0}$ of a certain interval containing the point $x_{0}$ . In other words, a function $f(x)$ has a maximum at $x_{0}$ if it is possible to find an interval $(\alpha, \beta)$ containing $x_{0}$ , that is, with $\alpha < x_{0} < \beta$ , such that for all points different from $x_{0}$ in $(\alpha, \beta)$ , we have $f(x) < f(x_{0})$ .

A function $f(x)$ has a local minimum at $x_{0}$ if there exists an interval $(\alpha, \beta)$ containing $x_{0}$ such that $f(x) > f(x_{0})$ for $x \in (\alpha, \beta)$ and $x \neq x_{0}$ .

One should not confuse the local maximum and local minimum of a function with its largest and smallest values over a given interval. The local maximum of a function is the largest value only in comparison to the values it has at all points sufficiently close to the point of local maximum. Similarly, the local minimum is the smallest value only in comparison to the values of the function at all points sufficiently close to the local minimum point.

The general term for the maximum and minimum of a function is extremum, or the extreme values of the function. A necessary condition for the existence of an extremum at the point $x_{0}$ of the function $f(x)$ is that $f^{'}(x_{0})=0$ , or $f^{'}(x_{0})$ does not exist. The points at which $f^{'}(x)=0$ or $f^{'}(x)$ does not exist, are called critical points.

First Derivative Test:

(i) If $f^{'}(x)$ changes sign from positive to negative at $x_{0}$ , that is, $f^{'}(x)>0$ for $x < x_{0}$ and $f^{'}(x)<0$ for $x > x_{0}$ , then the function attains a local maximum at $x_{0}$ .

(ii) If $f^{'}(x)$ changes sign from negative to positive at $x_{0}$ , that is, $f^{'}(x)<0$ for $x<x_{0}$ , and $f^{'}(x)>0$ for $x > x_{0}$ , then the function attains a local minimum at $x_{0}$ .

(iii) If the derivative does not change sign in moving through the point $x_{0}$ , there is no extremum at that point.

Second Derivative Test:

Let f be twice differentiable, and let c be a root of the equation $f^{'}(x)=0$ . Then,

(i) c is a local maximum point if $f^{''}(c)<0$ .

(ii) c is a local minimum point if $f^{''}(c)>0$ .

However, if $f^{''}(c)=0$ , then the following result is applicable. Let $f^{'}(c) = f^{''}(c) = \ldots = f^{n-1}(c)=0$ (where f^{r} denotes the rth derivative), but $f^{(n)}(c) \neq 0$ .

(i) If n is even and $f^{(n)}(c)<0$ , there is a local maximum at c, while if $f^{(n)}(c)>0$ , there is a local minimum at c.

(ii) If n is odd, there is no extremum at the point c.

Greatest/Least Value (Absolute Maximum/Absolute Minimum):

Let f be a function with domain D. Then, f has a greatest value (or absolute maximum) at a point $c \in D$ if $f^(x) \leq f(c)$ for all x in D and a least value (or absolute minimum) at c, if $f(x) \geq f(c)$ for all x in D.

If f is continuous at every point of D, and $D=[a,b]$ , a closed interval, the f assumes both a greatest value M and a least value m, that is, there are $x_{1}, x_{2} \in [a,b]$ such that $f(x_{1})=M$ and $f(x_{2})=m$ , and $m \leq f(x) \leq M$ for every $x \in [a,b]$ .

Example 4:

a) $y=x^{2}$ , with domain $(-\infty, \infty)$ . This has no greatest value; least value at $x=0$

b) $y=x^{2}$ with domain $[0,2]$ . This has greatest value at $x=2$ and least value at $x=0$ .

c) $y=x^{2}$ with domain $(0,2]$ . This has greatest value at $x=2$ and no least value.

d) $y=x^{2}$ with domain $(0,2)$ . This has no greatest value and no least value.

Some other remarks:

The greatest (least) value of continuous function $f(x)$ on the interval $[a,b]$ is attained either at the critical points or at the end points of the interval. To find the greatest (least) value of the function, we have to compute its values at all the critical points on the interval $(a,b)$ , and the values $f(a), f(b)$ of the function at the end-points of the interval, and choose the greatest (least) out of the values so obtained.

We will continue with problems on applications of derivatives later,

Nalin Pithwa.

By Nalin Pithwa | Posted in applications of maths, calculus, IITJEE Mains | Comments (0)

Mathematics Hothouse

On the value of time…essence of time management

Mathematics versus Physics

An observation about Sir Isaac Newton

Applications of Derivatives IITJEE Maths tutorial: practice problems part IV

Applications of Derivatives: IITJEE Maths tutorial problem set: III

Applications of Derivatives: Tutorial: IITJEE Maths: Part II

Applications of Derivatives: Tutorial Set 1: IITJEE Mains Maths

Primality testing: recap

Dear IIT Students: Address to IIT Hyderabad by present President of India, Shri Ram Nath Kovind

Applications of Derivatives: A Quick Review

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