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. $$$$ \( d_1+d_2+d_3+ \dots +d_k = 72 \implies \frac{1}{\lambda_1}+ \frac{1}{\lambda_2}+\dots +\frac{1}{\lambda_k} = \frac{72}{n} \implies \frac{1}{d_1}+ \frac{1}{d_2}+\dots +\frac{1}{d_k} = \frac{72}{n}\)

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

Among all the factors $4^6 6^7 21^8$ the number of factors which are perfect squares is? $$$$ \(4^6 6^7 21^8 = 2^{12} \times 2^7 \times 3^7 \times 3^8 \times 7^8 = 2^{19} \times 3^{15} \times 7^8\). Now a factor of $4^6 6^7 21^8$ must be of the form $2^i3^k7^k$ where $i,j,k$ are integers and \( 0 \leq i \leq 19, 0 \leq j \leq 15 \quad and \quad 0 \leq k \leq 8\), since the problem asks for divisors which are perfect squares $i,j,k$ must be even. Now conider the product \((2^0+ 2^2+ \dots + 2^{18})(3^0+ 3^2+ \dots + 3^{14})(7^0+ 7^2+ \dots + 7^8) \). Each term of the product satisfies the above two condtion, so the required number of factors which are perfect squares is equal to the number of terms of the above product, which in turn equals to \(10 \times 8 \times 5 = 400 \)