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 \( \bigg( \frac{5 \times \frac {b^2}{4} \cot \frac{\pi}{5}}{5 \times \frac {a^2}{4} \cot \frac{\pi}{5}} \bigg) = \frac{b^2}{a^2} = \frac{(2 R \sin \frac{\pi}{5})^2} {(2 r \tan \frac{\pi}{5})^2} = \cos^2 \frac{\pi}{5}\)

Indian Statistical Institute B.Math & B.Stat : Trigonometry

Let \( \theta_1 = \frac{2 \pi}{3}, \theta_2 = \frac{4 \pi}{7}, \theta_3 = \frac{7 \pi}{3} \). Then show that \( ( \sin \theta_1)^{ \sin \theta_1} < ( \sin \theta_3)^{ \sin \theta_3} < ( \sin \theta_2)^{ \sin \theta_2} \). $$$$ First note that \( \pi > \theta_1 > \theta_3 > \theta_2 > 0\) and all of them belong to the $second$ quadrant. $Sine$ function strictly decreases from $1$ to $0$ in the $second$ quadrant. Also \( \sin \theta_1 \neq \sin \theta_2 \neq \sin \theta_3 \neq 0 \) and each of them are posititve. $$$$ Using the strictly decreasing property of $Sine$ in the second quadrant we have \( \sin \theta_1 < \sin \theta_3 < \sin \theta_2 \). Now the result follows the standard inequality \( x^c < y^d \) for \( x,y,c,d > 0 \quad where \quad x < y, \quad c < d \).

Indian Statistical Institute B.Math & B.Stat : Trigonometry

Find the value of the sum \( \cos \frac{2 \pi}{1000} + \cos \frac{4 \pi}{1000} + \dots + \cos \frac{1998 \pi}{1000} \). $$$$ Let \( z = \cos \frac{\pi}{1000} + i \sin \frac{\pi}{1000} = \cos \theta + i \sin \theta \) where $ \theta = \frac{\pi}{1000}$ . It is easy to see that $ z \neq 1,-1$. $$$$ Consider the sum \( 1 +z^2+z^4+ \dots + z^{1998} \), $ z \neq 1,-1$. Putting $w = z^2$ the sum reduces to \( 1 +w+w^2+ \dots + w^{999} \), $ w \neq 1 $. $$$$ Now, \( 1 +w+w^2+ \dots + w^{999} = \frac{w^{1000}-1}{w-1}\) $$$$ Substituting back $w$ we have the following identity \( 1 +z^2+z^4+ \dots + z^{1998} = \frac{z^{2000}-1}{z^2-1}\),$ z \neq 1,-1$. $$$$ Using $De-Moivre's$ theorem we have \( z^n = \cos n \theta + i \sin n \theta \) for \( n \in \mathbb{N} \). $$$$ Substituting back in the above identity we have, \( \big(1+ \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta \big) + i \big(1+ \sin 2 \theta + \sin 4 \theta + \dots + \sin 1998 \theta \big) = \frac{\cos 2000 \theta + i \sin 2000 \theta -1}{\cos 2 \theta + i \sin 2 \theta -1} \) $$$$ Equating the real part from both side we have. \( 1+ \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta = Re \bigg( \frac{\cos 2000 \theta + i \sin 2000 \theta -1}{\cos 2 \theta + i \sin 2 \theta -1} \bigg) = Re \bigg( \frac{\cos 2 \pi + i \sin 2 \pi -1 }{\cos 2 \theta + i \sin 2 \theta -1}\bigg) \), since $ \theta = \frac{\pi}{1000}$. $$$$ Therefore \( 1+ \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta = Re (0) = 0 \implies \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta = -1\). $$$$ \( \implies \cos \frac{2 \pi}{1000} + \cos \frac{4 \pi}{1000} + \dots + \cos \frac{1998 \pi}{1000} = -1 \)

Indian Statistical Institute B.Math & B.Stat : Quadratic Equations

Consider the function \(f(x) = ax^3 + bx^2 + cx + d\), where $a, b, c$ and $d$ are real numbers with $a > 0$. If $f$ is strictly increasing, then show that the function \( g(x) =f′(x) − f′′(x) + f′′′(x) \) is positive for all \( x \in \mathbb{R} \). $$$$ First we calculate the derivatives up to the third order. \(f'(x) = 3ax^2+2bx+c, \quad f''(x) = 6ax+2b \quad and \quad f'''(x) = 6a \). $$$$ It is given that $f$ is strictly increasing which implies \( f' > 0 \) which in turn implies \( 3ax^2+2bx+c > 0\). $$$$ Let \(y = 3ax^2+2bx+c \) It is easy to see that \( y = 3a \big( x + \frac{b}{3a} \big)^2 + \frac{3ac-b^2}{3a} \). Since $y > 0$ and $a$ is given to be positive $3ac$ must be strictly greater than $b^2$. Note $\big( x + \frac{b}{3a} \big)^2$ is always non-negative. $$$$ Now \( g(x) = f′(x) − f′′(x) + f′′′(x) = 3ax^2+2bx+c -(6ax+2b) + 6a = 3ax^2 + 2x(b-3a)+(c-2b+6a) \) \( = 3a\big( x^2 + 2 x \frac{b-3a}{3a} + \frac{(b-3a)^2}{9a^2}+ \frac{(c-2b+6a)}{3a} - \frac{(b-3a)^2}{9a^2} \big) \) = \( 3a\bigg( x^2 + 2 x \frac{b-3a}{3a} + \frac{(b-3a)^2}{9a^2} \bigg) + 3a \bigg( \frac{(c-2b+6a)}{3a} - \frac{(b-3a)^2}{9a^2}\bigg) \) \( = 3a \big( x + \frac{b-3a}{3a} \big)^2 + \frac{9a^2+3ac-b^2}{3a}\) $$$$ \( 3a \big( x + \frac{b-3a}{3a} \big)^2 \geq 0 \) for all \( x \in \mathbb{R} \). (since $a$ is given to be positive) We have already shown that $3ac > b^2$ therefore \(\frac{9a^2+3ac-b^2}{3a} > 0 \). Thus \( g(x) > 0 \) for all \( x \in \mathbb{R} \).

Indian Statistical Institute B.Math & B.Stat : Real Analysis

Suppose $f$ is a differentiable and increasing function on $[0, 1]$ such that \(f(0) < 0 < f(1)\). Let \(F(t) = \int_{0}^{t} f(x) dx \). Then show that $F$ has a unique minimum in the open interval $(0, 1)$. $$$$ Since $f$ is differentiable, $f$ is continuous. Therefore $F(t)$ is differentiable and \( F'(t) = f(t) , t \in [0,1] \) ( students interested in the proof are advised to read INTRODUCTION TO REAL ANALYSIS, Author : ROBERT G. BARTLE, DONALD R. SHERBERT ). $$$$ Given that $f$ is a differentiable and increasing function on $[0, 1]$ such that \(f(0) < 0 < f(1)\) it follows that $F'(t)$ satisfies the same conditions of $f$ in $[0,1]$ since $F'(t) = f(t)$. Now see that \( F'(0)F'(1) = f(0)f(1) < 0 \), continuity of $F'$ implies $\exists$ \( c \quad \in (0,1) \) such that $F'(c)=0$. Again since $F'$ is increasing, the points at which $F'$ vanishes forms a sub-interval $(c,d)$ of $[0,1]$ (i.e, the points at which the function $F'$ vanishes forms a connected set !) where \( 0 < c < d < 1 \). $$$$ Thus $F'(t) < 0$ for \( t \in [0,c) \) and $F'(t) > 0$ for \( t \in (d,1] \) which implies $F$ is strictly decreasing in $[0,c)$, strictly increasing in $(d,1]$ and $F$ is constant on $[c,d]$ which in turn implies $F$ has a unique minimum on $[c,d]$. $$$$ Note that if it was given $f$ to be strictly increasing then $c=d$ and the unique minimum will be attained at a unique point.

Indian Statistical Institute B.Math & B.Stat : Complex Numbers

Show that the set of complex numbers $z$ satisfying the equation \( (3+7i)z+(10-2i)\overline{z}+100 = 0 \) represents, in the Argand plane, a point. $$$$ Let $z=x+iy$, taking the conjugate of the given equation we have \( (3-7i)\overline{z}+(10+2i)z+100 = 0 \) $$$$ Adding the two equations we get, \( 26x-18y+200 = 0\) (do the calculations yourself!), this shows that $z$ lies on the line $26x-18y+200 = 0$ $$$$ Subtracting the two equations we get, \( 10x-4y = 0 \), this again shows that that $z$ lies on the line $10x-4y = 0$ $$$$ Thus $z$ satisfies both the equations $26x-18y+200 = 0$ and $10x-4y = 0$, thus $z$ represents a point in the Argand Plane.

Indian Statistical Institute B.Math & B.Stat : Polynomials

If the roots of the equation \(x^4 + ax^3 + bx^2 + cx + d = 0\) are in geometric progression then show that $c^2 = a^2d$.$$$$ Let the roots of the equation be \(x_1,x_2,x_3,x_4\). Since the roots are in geometric progression we have \(x_1x_4 = x_2x_3 \). Also using Vieta's Formulas ( relation between roots and coefficients ) we have $$$$ \[ x_1+x_2+x_3+x_4 = - a \] \[ (x_1+x_4)(x_2+x_3)+x_1x_4+x_2x_3 = b \] \[ x_1x_4(x_2+x_3)+x_2x_3(x_1+x_4) = -c\] \[ x_1x_2x_3x_4 = d\] Since \(x_1x_4 = x_2x_3 \) and \( x_1x_4(x_2+x_3)+x_2x_3(x_1+x_4) = -c\) we have \( x_1x_4(x_2+x_3+x_1+x_4) = -c\). Now using \( x_1+x_2+x_3+x_4 = - a \) we have \( x_1x_4 \times -a = -c \implies x_1x_4 = \frac{c}{a} \). Thus \( x_1x_4 = x_2x_3 = \frac{c}{a} \). Again since \( x_1x_2x_3x_4 = d \implies \frac{c}{a} \times \frac{c}{a} = d \implies c^2 = da^2 \)

Indian Statistical Institute B.Math & B.Stat : Combinatorics

Let \( A = \{1,2,3,4,5,6\} \). Find the number of functions $f$ from $A$ to $A$ such that range of $f$ contains exactly $5$ elements. $$$$ $5$ elements of the range can be selected in $\binom {6}{5} = 6$ ways. Now we will find the number of onto functions (since the range of $f$ contains exactly $5$ elements) for each such case.$$$$ Thus the problem reduces to finding the number of onto functions from a set containing $6$ elements to a set containing $5$ (\( \{ 2,3,4,5,6\} \quad say\)) elements. $$$$ Let $T_1$ be the set of all functions with the property that the element $2$ is not in the range of the function.$$$$ Let $T_2$ be the set of all functions with the property that the element $3$ is not in the range of the function. $$$$ \[ \dots\] Let $T_5$ be the set of all functions with the property that the element $6$ is not in the range of the function. $$$$ Now the function will be onto ( i.e., the range of f will contain exactly $5$ elements ) $iff$ none of the above properties hold. Number of such functions is \( |T_1 \cup T_2 \cup \dots \cup T_5| \). Using the principle of $Exclusion-Inclusion$ we have $$$$ \( |T_1 \cup T_2 \cup \dots \cup T_5| = \sum_{i=1}^{5} |T_i|- \sum_{1 \leq i < j < \leq 5}^{} |T_i \cap T_j| + \dots + |T_1 \cap T_2 \cap \dots \cap T_5| \) $$$$ \( = \binom{5}{1} 4^6 - \binom{5}{2} 3^6 + \binom{5}{3} 2^6 - \binom{5}{4} 1^6 + \binom{5}{5} 0^6 = 13825\) $$$$ Total number of functions = \(5^6 = 15625 \) $$$$ Therefore number of functions with required condition \( = (15625 - 13825) \times \binom {6}{5} = 10800 \)

Indian Statistical Institute B.Math & B.Stat : Polynomials

Let \( 0 < a_0 < a_1 < a_2 < \dots < a_n \) be real numbers. Suppose $p(t)$ is a real valued polynomial of degree n such that \[ \int_{a_j}^{a_{j+1}} p(t) dt =0 \quad \forall \quad 0 \leq j \leq n-1. \] Show that, for \(0 \leq j \leq n-1\), the polynomial $p(t)$ has exactly one root in the interval \((a_j ,a_{j+1}) \). $$$$ Let \( p(t) = b_0 t^n+b_1t^{n-1}+\dots+b_{n-1}t+b_n \) and let \( g(t) = \frac{b_0}{n+1}t^{n+1}+\frac{b_1}{n}t^{n}+\dots+\frac{b_{n-1}}{2}t^{2}+\frac{b_n}{1}t^{1}\) $$$$ Note that \(g'(t) = p(t) \dots (A)\). Now consider the interval \( [a_j,a_{j+1}] \quad where \quad 0 \leq j \leq n-1. \) $g(t)$ is continuous and differentiable in $[a_j,a_{j+1}]$ and $(a_j,a_{j+1})$ respectively ( $g(t)$ being a polynomial.) $$$$ It is given that \( \int_{a_j}^{a_{j+1}} p(t) dt =0 \implies g(a_{j+1})-g(a_j) = 0 \) ( using $(A)$ ) $$$$ \( \implies g(a_{j+1}) =g(a_j) \) which in turn shows that $g(t)$ satisfies all the conditions of $Rolle's-Theorem$ on \( [a_j,a_{j+1}] \). $$$$ \( \implies g'(t) = 0 \) for at least one $t$ in \( (a_j,a_{j+1}) \) for all $j \in \{0,1,2,\dots,n-1\}$. $$$$ \( \implies p(t) = 0 \) ( using $(A)$ )for at least one $t$ in \( (a_j,a_{j+1}) \) for all $j \in \{0,1,2,\dots,n-1\}$. $$$$ \( \implies p(t) \) has at least one real root in \( (a_j,a_{j+1}) \) for all $j \in \{0,1,2,\dots,n-1\}$. $$$$ Since $degree(p(t))=n$ it has $n$ number of roots ( counting multiplicity ). Since there are $n$ interval of the form $(a_j,a_{j+1})$ and each of them contains at lest one root of $p(t)$, each of them must contain exactly one root of $p(t)$ otherwise number of roots will exceed $n$.

Indian Statistical Institute B.Math & B.Stat : Polynomials

If a polynomial $P$ with integer coefficients has three distinct integer zeroes, then show that $P(n) \neq 1$ for any integer. $$$$ Let \( \alpha, \beta\) and $\gamma$ be the distinct integer zeroes of the polynomial $P$. If possible let $P(m)=1$ where $ m \in \mathbb{Z}$. Since $P \in \mathbb{Z}(x)$ we have \( \alpha - m | P(\alpha)-P(m) \implies \alpha - m | (-1) \). Similarly $\beta - m | (-1)$ and $\gamma - m | (-1)$. Since \( \alpha, \beta\) and $\gamma$ are distinct \( \alpha-m, \beta-m\) and $\gamma-m$ are distinct. This shows that \( \alpha-m, \beta-m\) and $\gamma-m$ are distinct factors of $1$, which is impossible! So the assumption that $P(m)=1$ is not tenable for any integer $m$.

Indian Statistical Institute B.Math & B.Stat : Number Theory

Let \( a_1,a_2,_3, \dots ,a_n\) be integers. Show that there exists integers $k$ and $r$ such that the sum \[ a_k+a_{k+1}+\dots +a_{k+r} \] is divisible by $n.$ $$$$ We construct a finite sequence of partial sum of the given finite sequence as follows, \[s_1 = a_1\] \[s_2 = a_1+a_2\] \[s_3 = a_1+a_2+a_3\] \[ \dots \] \[s_n = a_1+a_2+ \dots +a_n \] If \( s_i \equiv 0(mod \quad n) \) for any admissible value of $i$ then we are done with \( k= 1 \quad and \quad r= i-1\). Therefore assume \( s_i \not\equiv 0(mod \quad n) \quad \forall \quad i \in \{1,2,\dots,n\} \implies s_i \equiv k(mod \quad n)\) where $1 \leq k \leq n-1, k \in \mathbb{N} .$ Since there are $n$ such congruences and $n-1$ possible values of $k$, $Pigeon-Hole$ principle asserts that at least two different partial sums have the same remainder, i.e, \( s_i \equiv k(mod \quad n) \quad s_j \equiv k(mod \quad n)\) for $i \neq j$. $$$$ Therefore \( s_i \equiv s_j(mod \quad n)\) (Without loss of generality assume that $ i > j $ ) $$$$ \( \implies s_i-s_j \equiv 0(mod \quad n)\) $$$$ \( \implies s_{j+1}+s_{j+2}+\dots +s_{j+i-j} \equiv 0(mod \quad n)\) This is what was asked!

Indian Statistical Institute B.Math & B.Stat : Number Theory

Let \( d_1,d_2,d_3, \dots ,d_k\) be all the factors of a positive integer $n$ including $1$ and $n$. Suppose \( d_1+d_2+d_3+ \dots +d_k = 72\). Then find the value of \[ \frac{1}{d_1}+ \frac{1}{d_2}+\dots +\frac{1}{d_k}\] Since $n$ is positive and $d_i$ is a factor of $n$ for each \( i \in \{1,2,3,\dots,k\} \quad \exists \quad \lambda_i > 0 \quad \lambda_i \in \mathbb{N}\) such that $n=d_i \lambda_i$. This also shows that $\lambda_i$ is a factor of $n$. We now show that $\lambda_i \neq \lambda_j$ for $i \neq j$. If possible let $\lambda_i = \lambda_j$ for some $i,j$ where $i \neq j$. \( \implies d_i \lambda_i = d_j \lambda_i \implies d_i = d_j \), a contradiction since $d_i$ and $d_j$ are distinct factors of $n$.Thus \( \lambda_1,\lambda_2,\lambda_3, \dots ,\lambda_k\) are also the possible factors of $n$ \( \implies \{ d_1,d_2,d_3, \dots ,d_k\}=\{\lambda_1,\lambda_2,\lambda_3, \dots ,\lambda_k \} \) in some order. $$$$ \( d_1+d_2+d_3+ \dots +d_k = 72 \implies \frac{1}{\lambda_1}+ \frac{1}{\lambda_2}+\dots +\frac{1}{\lambda_k} = \frac{72}{n} \implies \frac{1}{d_1}+ \frac{1}{d_2}+\dots +\frac{1}{d_k} = \frac{72}{n}\)