Number of solutions of the given equation

Here the equation to be solved is very simple |2x-[x]|=4. But unfortunately the traditional methods of solving will not apply! You have solved equations by methods like factorisation, vanishing method, hit and trial.... But in many equations these methods don't work or requires huge computation and guess work to solve it. Here is an equation which can be solved from the simple fact that if one side of an equation is an integer the we look to the other side only and see for what values its also an integer. This simple but powerful method can be used in many problems

A very hard integration problem! Can you solve it? ( Practice for IIT-JEE)

A very hard integration problem based on the method of substitution. It took almost an hour to figure this out! Can you solve it on your own?! This problem is well suited for students preparing for IIT-JEE.

Inequalities

Show that $|x|+|y| \leq |x+y|+|x-y| \quad and \quad \frac{|x+y|}{1+|x+y|} \leq \frac{|x|}{1+|x|} + \frac{|y|}{1+|y|} \quad for \quad x,y \in \mathbb{R} $ $$$$ We have $\quad|x+y+x-y| \leq |x+y|+|x-y| $ $$$$ $ \implies 2|x| \leq |x+y|+|x-y| \dots A$ $$$$ Now, $|x+y+y-x| \leq |x+y|+|y-x|$ $$$$ $ \implies 2|y| \leq |x+y|+|-1||x-y| $ $$$$ $\implies 2|y| \leq |x+y|+|x-y| \dots B$ $$$$ Adding $A$ and $B$ we get $ |x|+|y| \leq |x+y|+|x-y|$ The second inequality holds $$$$ $iff$ $\frac{(1+|x|)(1+|y|)|x+y|}{1+|x+y|} \leq |x|(1+|y|)+|y|(1+|x|)$ $$$$ $ iff \quad (1+|x|)(1+|y|)|x+y| \leq (|x|+|y|+2|xy|)(1+|x+y|)$ $$$$ (multiply and cancel out terms) $$$$ $iff \quad |x+y| \leq |x|+|y| +|xy|(2+|x+y|)$ $$$$ Now for any real $x$ and $y$, $|xy|(2+|x+y|) \geq 0$ and we know that $|x+y| \leq |x|+|y|$ $$$$ Thus the last inequality holds, which in turn proves the original ineqality.

Solved Trigonometry Problems: 10th Grade

Here is a list of few good problems at the 10th grade. Get the full solution below and if you want
more comment or mail us at maths.programming@gmail.com!


Download here

Circle: Geometry

Given below is a fine problem based on simple properties of circles ( tangent and chord ). Try yourself before looking at the solution!

Here is the solution!

Solved Problems : Logarithm

A collection of solved problems on logarithm meant for secondary students ( 9th and 10th grade). The collection covers almost all types of problems at the level mentioned.



Download the file here: Solved Logarithm Problems 

Solved Integration Problems for 10+2 Level - SN Dey

Hundreds of solved problems at 10+2 level or the final year at school from the topic of integration. The aim of the document to help students from different boards (ISC, CBSE and other State Boards) studying at Higher Secondary level to get access to wide variety of problems, that too solved! Most of the problems are from the books of SN Dey, which is the most sought after book in WBCHSE.

This should not be substituted for classroom teaching neither this documents aims to teach the theories of differentiation. Students are requested to go through the theory and the rules before looking at the problems. Any error in the document can be reported at: prime.maths@hotmail.com

Integration, Methods of Substitution, Integration by Parts, Partial Fractions, Special Integrations, Trigonometric Substitutions.

If you have any query, don't forget to comment below.

A  collection of completely solved special integrals for various entrance exams ( IIT and Indian Statistical Institute ) and boards exam (CBSE,ISC and other State Boards) at 10+2 level. 

Click below the links to download the files:

File 1 Introduction ( Basic Problems )

File 2 Special Integrals 

File 3 Method of Substitution- Set I

File 4 Method of Substitution- Set II

File 5 Method of Substitution- Set II

File 6 Problems based on Standard Integrals- Set I

File 7 Problems based on Standard Integrals- Set II 

File 8 Problems based on Standard Integrals- Set III

File 9 Problems based on Integration by Parts - Set I

File 10 Problems based on Integration by Parts - Set II

More solved problems to follow. Keep visiting this space.

Special Integration

Integration : A harder Problem

 Most of the studenst will fail to solve this particular integration problem. It is trickier but once you hit the right idea, you will be able to solve the integration problem easily.

A problem on inequality

Using simple formula to prove a strong inequality.

Pigeonhole Principle

The numbers 1 to 20 are placed in any order around a circle. Prove that the sum of some 3 consecutive numbers must be at least 32!

This problem uses the alternate form of pigeon hole principle which is as follows:
 If the average of n positive numbers is t, then at least one of the numbers is greater than or equal to t. Further, at least one of the numbers is less than or equal to t.

The proof is very simple, assume the contradiction and proceed!

#Solution https://youtu.be/GLQg6cSAbms

Sum of first 'n, natural numbers

A simple way to calculate the sum of first 'n' natural numbers witout the use of calculator. Infact the same procedure is used to calculate the sum of n terms of any A.P series. See it and try to obtain the formula yourself

Inequality

A challenging problem based on the inequality that square of a real number is always greater than equal to zero. Learn the trick and prepare yourself for more challenging problems based on the same ideology.

Divisibility of 2222^5555+5555^2222 by 7

For any prime p, square root of p is irrational.

Complex Numbers in Solving Trigonometrical problems

Show that $ \cos \frac{\pi}{11}+\cos \frac{3\pi}{11}+\cos \frac{5\pi}{11}+\cos 7\frac{\pi}{11}+\cos \frac{9\pi}{11}= \frac{1}{2}$ $$$$ I have used complex numbers to solve the problems. Standard trigonometric methods will be very much difficult to apply and will involve lots of calculation!!! Also it is important to link the different topics and see the beauty of mathematics! Important problem for students in 10+2 level $$$$ Also the idea used to solve the problem can be used to calculate the value of $ \sum_{i=1}^{\frac{n-1}{2}} \cos \frac{(2i-1) \pi}{n}$ where $n$ is an odd positive integer. $$$$ See the video below for the explanation. $$$$

Equation Solving

Find all real number(s) $x$ such that $2^x+3^x+6^x-4^x-9^x=1 $ $$$$ This problem can be solved by transforming the equation where it can be written as sum of squares of number equals zero. From there the result will follow. See the video below for the explanation. $$$$

Sophie Germain Identity

Here is a problem of primality testing of a number. As you can see a direct computation will not yield an effective result and will take a much longer time! So, the question arises how to solve the problem in the right way. Here comes the roll of simple and elegant identity known as Sophie Germain Identity. Watch the video below for the solution

For solution of the problem above watch the video below.

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Algebra

Let $a, b$ and $c$ be such that $a+b+c=0$ and \( P =\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\) is defined. What is the value of $P$? $$$$ $Solution: $ \( 2a^2+bc = a^2 + a^2 -(a+c)c\), since $b=-(a+c)$. $$$$ \( \implies 2a^2+bc = a^2 + a^2 -ac-c^2 = (a-c)(a+c)+a(a-c) = (a-c)(2a+c)=(a-c)(2a-a-b)=(a-c)(a-b)\) $$$$ Therefore \( \frac{a^2}{2a^2+bc} = \frac{a^2}{(a-c)(a-b)}\) $$$$ Similarly, \( \frac{b^2}{2b^2+ac} = \frac{b^2}{(b-a)(b-c)} \) and \( \frac{c^2}{2c^2+ab} = \frac{c^2}{(c-a)(c-b)} \) $$$$ Therefore \( P = \frac{a^2}{(a-c)(a-b)} +\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)} = \frac{1}{a-b} \big( \frac{a^2}{a-c} - \frac{b^2}{b-c} \big)+\frac{c^2}{(c-a)(c-b)}\) $$$$ \( \implies P = \frac{ab-ca-cb}{(c-a)(c-b)}+ \frac{c^2}{(c-a)(c-b)} = \frac{ab-ca-cb+c^2}{(c-a)(c-b)} = \frac{(c-a)(c-b)}{(c-a)(c-b)} = 1\)$$$$

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Calculus

If \( A = \int_{0}^{\pi} \frac{cos x}{(x+2)^2}dx\), then show that \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{(x+1)}dx = \frac{1}{2} \big( \frac{1}{2} +\frac{1}{\pi+2} - A\big)\). $$$$ Solution: \( A = \cos x \int_{0}^{\pi} \frac{dx}{(x+2)^2} - \int_{0}^{\pi} \bigg( \frac{d}{dx} \cos x \int \frac{dx}{(x+2)^2} \bigg) dx \) $$$$ \( \implies A = \frac{-\cos x}{x+2}\mid_{0}^{\pi} - \int_{0}^{\pi} \frac{\sin x}{(x+2)} dx \) $$$$ \( \implies A = \frac{1}{\pi +2} + \frac{1}{2} - \int_{0}^{\pi} \frac{2 \sin \frac{x}{2} \cos \frac{x}{2} }{2(1+\frac{x}{2})} dx \) $$$$ Putting $\frac{x}{2}=t$ in the last integral, we have $dx=2dt$ and $t$ varies from $0$ to $\frac{\pi}{2}$ $$$$ Therefore, \( A = \frac{1}{\pi +2} + \frac{1}{2} - 2 \int_{0}^{\frac{\pi}{2}} \frac{\sin t \cos t}{(1+t)}dt \)$$$$ \( \implies \int_{0}^{\frac{\pi}{2}} \frac{\sin t \cos t}{(1+t)}dt = \frac{1}{2} \bigg( \frac{1}{\pi +2} + \frac{1}{2} -A \bigg) \) $$$$ \( i.e., \) \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{(1+x)}dx = \frac{1}{2} \bigg( \frac{1}{\pi +2} + \frac{1}{2} -A \bigg) \) $$$$

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Calculus

Evaluate \( lim_{n \rightarrow \infty} \bigg(\prod_{r=1}^{r=n} (1+\frac{2r-1}{2n}) \bigg)^{\frac{1}{2n}} \) $$$$ Let \( A = \bigg(\prod_{r=1}^{r=n} (1+\frac{2r-1}{2n}) \bigg)^{\frac{1}{2n}} \) $$$$ \( \implies A = \frac{\bigg(\prod_{r=1}^{r=2n} (1+\frac{r}{2n}) \bigg)^{\frac{1}{2n}}}{\bigg(\prod_{r=1}^{r=n} (1+\frac{2r}{2n}) \bigg)^{\frac{1}{2n}}} \) $$$$ \( \implies ln A = \frac{1}{2n} \sum_{r=1}^{r=2n} ln ( 1+\frac{r}{2n})- \frac{1}{2n} \sum_{r=1}^{r=n} ln ( 1+\frac{r}{n})\) $$$$ \( \implies lim_{n \rightarrow \infty} ln A = \frac{1}{2} lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{r=2n} ln ( 1+\frac{r}{2n}) - \frac{1}{2} lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{r=n} ln ( 1+\frac{r}{n}) \) $$$$ \( \implies lim_{n \rightarrow \infty} ln A = \frac{1}{2} \int_{0}^{2} ln (1+ \frac{x}{2}) dx - \frac{1}{2} \int_{0}^{1} ln (1+ x ) dx \) $$$$ \( \implies lim_{n \rightarrow \infty} ln A = ln 4 - 1 - ( ln 2 - \frac{1}{2}) = ln 2 - \frac{1}{2}\)$$$$ \( \implies lim_{n \rightarrow \infty} A = e^{ln 2 - \frac{1}{2}} = \frac{2}{\sqrt{e}} \) $$$$ \( \implies lim_{n \rightarrow \infty} \bigg(\prod_{r=1}^{r=n} (1+\frac{2r-1}{2n}) \bigg)^{\frac{1}{2n}} = \frac{2}{\sqrt{e}} \) $$$$

ISC Mathematics Paper 2016 Solved

Get the solved ISC Mathematics Question paper for the year 2016. The solutions are short need not to be considered as the way of answering in board exams. Click Below 
ISC MATHEMATICS PAPER 2016


ISC MATHEMATICS PAPER 2015

Solved CBSE Mathematics Question paper 2015. 

JEE MAIN 2016 MATHEMATICS SOLVED PAPER

JEE MAIN 2015 MATHEMATICS SOLVED PAPER

JEE MAIN 2016 Mathematics

Solved Paper of JEE MAIN 2016 Mathematics.

Download Link : Click Here

Binomial Series

Find the sum of the series \( \sum_{r=0}^{n} (-1)^r \binom{n}{r} \big( \frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+ \dots \) to m terms \( \big) \) $$$$ Given series is equal to \( \sum_{r=0}^{n} (-1)^r \binom{n}{r} \sum_{k=1}^{m} \frac{(2^k -1)^r}{2^{kr}} \) $$$$ \( = \sum_{r=0}^{n} \sum_{k=1}^{m} (-1)^r \binom{n}{r} \frac{(2^k -1)^r}{2^{kr}} \) $$$$ \( = \sum_{r=0}^{n} \sum_{k=1}^{m} (-1)^r \binom{n}{r} \lambda^r \) where $\lambda = \frac{(2^k -1)}{2^{k}}$ $$$$ \( = \sum_{k=1}^{m} \sum_{r=0}^{n} (-1)^r \binom{n}{r} \lambda^r \) Interchanging the summation $$$$ \( = \sum_{k=1}^{m} (1- \lambda)^n \), Substituting the value of $\lambda$ we have, $$$$ \( = \sum_{k=1}^{m} \frac{1}{2^{nk}} = \frac{1}{2^n} \frac{\bigg(1- \big(\frac{1}{2^n}\big)^m \bigg)}{1 - \frac{1}{2^n} } \) $$$$ \( = \frac{1}{2^n} \frac{2^n(2^{nm}-1)}{2^{nm}(2^n-1)} = \frac{2^{nm}-1}{2^{nm}(2^n-1)} \)

Integration

Evaluate \(\int \frac{dx}{e^x \big(1+e^{2008x}\big)^{\frac{2007}{2008}}} \) $$$$ \(\int \frac{dx}{e^x \big(1+e^{2008x}\big)^{\frac{2007}{2008}}} = \int \frac{dx}{e^x \big(e^{2008x}(1+e^{-2008x})\big)^{\frac{2007}{2008}}}\) $$$$ \(=\int \frac{dx}{e^x e^{2007x}\big(1+e^{-2008x}\big)^{\frac{2007}{2008}}} = \int \frac{e^{-2008x}dx}{\big(1+e^{-2008x}\big)^{\frac{2007}{2008}}} \) \(= \frac{-1}{2008}\int \frac{d(1+e^{-2008x})}{\big(1+e^{-2008x}\big)^{\frac{2007}{2008}}} = \frac{\frac{-1}{2008}\big(1+e^{-2008x}\big)^{-\frac{2007}{2008}+1}}{-\frac{2007}{2008}+1} + c \)

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Algebra

Let $a, b, c$ be real numbers greater than $1.$ Let $S$ denote the sum $S = log_{a}{bc} + log_{b}{ca} + log_{c}{ab}.$ $$$$ Find the smallest possible value of $S.$ $$$$ \( S =log_{a}{\frac{abc}{a}} + log_{b}{\frac{bca}{b}} + log_{c}{\frac{abc}{c}} = log_{a}{abc} + log_{b}{abc} + log_{c}{abc}-3 \) $$$$ \( = log_{a}{e} \times log_{e}{abc} + log_{b}{e} \times log_{e}{abc} + log_{c}{e} \times log_{e}{abc}-3 \) $$$$ \( = log_{e}{abc} \bigg( \frac{1}{log_{e}{a}} +\frac{1}{log_{e}{b}}+\frac{1}{log_{e}{c}} \bigg) -3 \) $$$$ \( = (log_{e}{a}+log_{e}{b}+log_{e}{c}) \bigg( \frac{1}{log_{e}{a}} +\frac{1}{log_{e}{b}}+\frac{1}{log_{e}{c}} \bigg) -3 \) $$$$ Now Using the inequality $A.M \times H.M \geq n^2 $ for $n$ positive real numbers, we see that $$$$ \( (log_{e}{a}+log_{e}{b}+log_{e}{c}) \bigg( \frac{1}{log_{e}{a}} +\frac{1}{log_{e}{b}}+\frac{1}{log_{e}{c}} \bigg) \geq 3^2 = 9 \) $$$$ Thus \( S \geq 9-3 = 6 \). Note \( log_{e}{a},log_{e}{b},log_{e}{c} \) are all positive since $a,b,c > 1$.

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Algebra

Show that the polynomial $x^8 − x^7 + x^2 − x + 15$ has no real root. $$$$ Let $f(x)=x^8 − x^7 + x^2 − x + 15$, we will show that $f(x) >0$ for all $ x \in \mathbb{R}$. $$$$ \( f(x) = x^7(x-1)+x(x-1)+15 = (x-1)x(x^6+1)+15 \) $$$$ Now, \(\mathbb{R} = (- \infty , 0] \cup (0,1] \cup (1, \infty) \). Note that \( f(0)=f(1)=15\) $$$$ When $ x \in (1, \infty), $ \( x,x-1 \quad and \quad x^6+1 > 0 \implies f(x) > 15 \quad \forall \quad x \in (1, \infty)\) $$$$ When $ x \in (- \infty , 0), $ \( x \quad and \quad x-1 < 0 \quad thus \quad x(x-1) > 0\) since $x^6+1 > 0$ for any $x$, $f(x) > 15$ in this case too. $$$$ When $ x \in (0,1) $, $1 < x^6+1 < 2$ and $ 0 < x < 1 $. Since both of them are positive $ 0 < x(x^6+1) < 2$. Further $ -1 < x-1 < 0 $, thus $x(x-1)(x^6+1) < 0$. Again $|x-1| < 1$ this implies $ -2 < x(x-1)(x^6+1) < 0 $. Thus \(f(x) > 13 > 0\). $$$$ Combining all the cases we see that \( f(x) > 0 \quad \forall \quad x \in \mathbb{R} \) which shows $f(x)$ has no real root.

Indian Statistical Institute (ISI) B.Math & B.Stat : Combinatorics

For $ k \geq 1$, find the value of \[ \binom{n}{0}+ \binom{n+1}{1}+ \binom{n+2}{2}+ \dots + \binom{n+k}{k}\] Using the identity \( \binom{n}{r} = \binom{n}{n-r} \), \( \binom{n}{0}+ \binom{n+1}{1}+ \binom{n+2}{2}+ \dots + \binom{n+k}{k}\) reduces to \[ \binom{n}{n}+ \binom{n+1}{n}+ \binom{n+2}{n}+ \dots + \binom{n+k}{n}\] = Coefficient of $x^n$ in $(1+x)^n$ + Coefficient of $x^n$ in $(1+x)^{n+1}$ + $\dots$ + Coefficient of $x^n$ in $(1+x)^{n+k}$ $$$$ = Coefficient of $x^n$ in \( (1+x)^n + (1+x)^{n+1} + \dots + (1+x)^{n+k} \) $$$$ = Coefficient of $x^n$ in \( (1+x)^n \frac{(1+x)^{k+1}-1}{1+x-1} = \frac{(1+x)^{n+k+1}-(1+x)^n}{x}\) $$$$ = Coefficient of $x^{n+1}$ in $(1+x)^{n+k+1}$ $= \binom{n+k+1}{n+1}$ $$$$

Indian Statistical Institute (ISI) B.Math & B.Stat : Algebra

If \(p,q,r\) are positive real numbers such that $pqr=1$, then find the value of \( \frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}\). $$$$ Throught the simplification we will use \( 1=pqr,q^{-1}=pr,r^{-1}=pq \quad and \quad p^{-1} = qr \) $$$$ Given expression is \[ \frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}\] \[= \frac{pqr}{pqr+p+pr}+\frac{pqr}{pqr+q+pq}+\frac{1}{1+r+p^{-1}} \] \[= \frac{qr}{qr+1+r}+\frac{pr}{pr+1+p}+\frac{1}{1+r+p^{-1}} \] \[= \frac{qr}{p^{-1}+1+r}+\frac{pr}{pr+pqr+p}+\frac{1}{1+r+p^{-1}} \] \[= \frac{qr}{p^{-1}+1+r}+\frac{r}{r+qr+1}+\frac{1}{1+r+p^{-1}} \] \[= \frac{qr}{p^{-1}+1+r}+\frac{r}{r+p^{-1}+1}+\frac{1}{1+r+p^{-1}} \] \[= \frac{qr+r+1}{p^{-1}+1+r} \] \[= \frac{p^{-1}+r+1}{p^{-1}+1+r} \] \[= 1\]

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Number Theory

Consider the equation $x^2 + y^2 = 2007$. How many solutions $(x, y)$ exist such that $x$ and $y$ are positive integers? $$$$ \( 2007 = 2000 + 7 \equiv 0 + 3 \equiv 3 (mod \quad 4) \). Now we know that square on an integer is either divisible by $4$ or leaves a remainder $1$ when divided by $4$, said otherwise \( x \in \mathbb{Z} \implies x^2 \equiv 0 \quad or \quad 1 (mod \quad 4) \). Thus for integers $x$ and $y$, \( x^2+y^2 \equiv 0 \quad or \quad 1 \quad or \quad 2 (mod \quad 4) \). Since we have different remainders $mod \quad 4$ on the two sides, it follows there cannot be any solution in $\mathbb{Z}$ hence no solution in $\mathbb{Z^+}$

Indian Statistical Institute ( ISI ) B.Math & B.Stat :Complex Numbers

Let $z$ be a non-zero complex number such that \( |z −\frac{1}{z}| = 2.\) What is the maximum value of $|z|$? $$$$ Given \( 2 = |z −\frac{1}{z}| \geq \big||z|- |\frac{1}{z}|\big| \) $$$$ Let $t = |z|$ \( \implies \big|t- \frac{1}{t}\big| \leq 2 \) $$$$ \( \implies -2 \leq t- \frac{1}{t} \leq 2 \) $$$$ \( \implies -2t \leq t^2- 1 \leq 2t \) $$$$ \( \implies t^2 +2t- 1 \geq 0 \quad and \quad t^2 -2t- 1 \leq 0 \) $$$$ The first inequality gives \( t \in ( - \infty, -1-\sqrt{2}] \cup [\sqrt{2}-1, \infty)\). Since $t \geq 0$ \(\implies t \in [\sqrt{2}-1, \infty) \). $$$$ The second inequality gives \( t \in [1-\sqrt{2} , 1+\sqrt{2}] \). Again since $t \geq 0$ \(\implies t \in [0,1+\sqrt{2}] \) $$$$ Combining the two case we see \( t \in [\sqrt{2}-1,\sqrt{2}+1] \implies |z| \in [\sqrt{2}-1,\sqrt{2}+1] \). Thus the maximum value of $|z|$ is $\sqrt{2}+1$. $$$$ $Practice-Problem$ Let $z$ be a non-zero complex number such that \( |z +\frac{1}{z}| = a, a \in \mathbb{R^+}.\) What is the maximum and minimum value of $|z|$? $$$$

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Co-ordinate Geometry

Let $A$ be the set of all points $(h, k)$ such that the area of the triangle formed by \((h, k), (5, 6)\) and \((3, 2)\) is $12$ square units. What is the least possible length of a line segment joining $(0, 0)$ to a point in $A$? $$$$ Take the base of the triangle to be the line segment obtained by joining the points $(5,6)$ and $(3,2)$. Equation of the base is \( 2x-y-4 =0 \). Length of the base is \( \sqrt{(5-3)^2+(6-2)^2} = 2 \sqrt{5}. \) Let $p$ be the length of the perpendicular from the point $(h,k)$ onto the base. ( Note that the point $(h,k)$ cannot lie on the base. Why?) Since the area is given to be $12$, \( 12 = \frac{1}{2} \times p \times 2 \sqrt{5} \implies p = \frac{12}{\sqrt{5}}\). Therefore the point $(h,k)$ lies at a distance of $\frac{12}{\sqrt{5}}$ units from the base on both sides. Thus $A$ is the set of all points on the line $parallel$ to the base and at a distance $\frac{12}{\sqrt{5}}$ units away from the base. In the diagram, the lines colored green represents the set $A$. Clearly the least possible length of a line segment joining $(0, 0)$ to a point in $A$? is the distance between the point $(0,0)$ and the line drawn parallel to the base and to the left side of the base. Let $XY$ be the line segment perpendicular to the base and the line and passing through the orgin as shown in the diagram. Required distance is $OX$ and \( OX = XY - OY = \frac{12}{\sqrt{5}} - \big|\frac{-4}{\sqrt{2^2+(-1)^2}}\big| = \frac{12}{\sqrt{5}} - \frac{4}{\sqrt{5}} = \frac{8}{\sqrt{5}} \)

Indian Statistical Institute (ISI) B.Math & B.Stat : Trigonometry

Find the ratio of the areas of the regular pentagons inscribed and circumscribed around a given circle. $$$$ Let $a$ be the side of the circumscribed pentagon and $b$ be that of the inscribed pentagon. $$$$ First note that for the circle is inscribed for the exterior pentagon and circumscribed for the interior pentagon. Therefore the $in-radius$ of the exterior polygon, say $r$ is equal to the $circum-radius$, say $R$ of the interior pentagon, i.e., $R=r$. See the figure below. Using standard formula, \[ a = 2 r \tan \frac{\pi}{5}, \quad b = 2 R \sin \frac{\pi}{5} \]. Area of a regular polygon having $n$ sides is \( n \times \frac {(side)^2}{4} \cot \frac{\pi}{n} \). $$$$ Therefore the required ratio is \( \bigg( \frac{5 \times \frac {b^2}{4} \cot \frac{\pi}{5}}{5 \times \frac {a^2}{4} \cot \frac{\pi}{5}} \bigg) = \frac{b^2}{a^2} = \frac{(2 R \sin \frac{\pi}{5})^2} {(2 r \tan \frac{\pi}{5})^2} = \cos^2 \frac{\pi}{5}\)