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

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 $n$.