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 \(10 \times 8 \times 5 = 400 \)

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

What is the number of ordered triplets $(a,b,c)$ where $a,b,c$ are positive integers ( not necessarily distinct ) such that $abc=1000.$ $$$$ Since $1000=2^3 5^3$ any ordered triplet $(a,b,c)$ must be of the form \((2^i5^p,2^j5^q,2^k5^r)\) where $i+j+k=3$, $p+q+r=3$ and $i,j,k,p,q,r$ are non-negative integers. Number of solutions to the equation $i+j+k=3$ and $p+q+r=3$ is given by \( \binom{3+3-1}{3-1} =10 \). Now for each set of values of $i,j,k$ there are $10$ possible combinations of $p,q,r$. Thus giving a total of \(10 \times 10 =100\) ordered triplets. $$$$ Note the number of solutions to the equation \(x_1+x_2+\dots+x_r=n, n \in \mathbb{N}\) in positive integers is \( \binom {n-1}{r-1} \) and in non-negative integers is \( \binom {n+r-1}{r-1} \).

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

Let \( S = \{(a_1,a_2,a_3)\} | \quad 0 \leq a_i \leq 9 \quad and \quad a_1+a_2+a_3 \quad is \quad divisible \quad by \quad 3 \} \). Find the number of elements in $S$. $$$$ We divide the integers \( 0 \leq a_i \leq 9 \) into three groups having the property that each element of the same group leaves the same remainder on being divided by $3$. The groups are \( \{0,3,6,9\}, \{1,4,7\} \quad and \quad \{2,5,8\} \). Now a $three-tuple$ \((a_1,a_2,a_3)\) will be divisible by $3$ iff and only if each of the co-ordinate belongs to the same group or each of them belongs to the different groups, giving a total of \( 4^3+3^3+3^3+ 4 \times 3 \times 3 \times 3! = 334\) possible $three-tuples$ which are divisible by $3$.

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

Evaluate \[ \int_{0}^{2 \pi} |1+2 \sin x| dx\] when \( x \in [0,\pi] \quad \sin x \geq 0 \quad \implies |1+2 \sin x| = 1+2 \sin x \) $$$$ Now note that when \( -1 \leq \sin x < - \frac{1}{2} \quad \implies -2 \leq 2 \sin x < -1 \quad \implies -1 \leq 1+ 2 \sin x < 0 \quad \implies |1+2 \sin x| = -(1+2 \sin x) \) $$$$ In the $third-quadrant$ sine function continuously decreases (strictly) from $0$ to $-1$ and \( \sin \frac{7 \pi}{6} = - \frac{1}{2} \) and at $x=\frac{3 \pi}{2}$ it takes the value $-1$. After that int the $fourth-quadrant$ sine function continuously increases (strictly) from $-1$ to $0$ and at $x=\frac{11 \pi}{6}$ it takes the value $- \frac{1}{2}$. So we conclude that in the $third$ and the $fourth-quadrant,$ \( -1 \leq \sin x < - \frac{1}{2} \quad when \quad x \in [\frac{7 \pi}{6},\frac{11 \pi}{6}] \). $$$$ So,\[ \int_{0}^{2 \pi} |1+2 \sin x| dx\ = \int_{0}^{\frac{7 \pi}{6}} (1+2 \sin x) dx+ \int_{\frac{11 \pi}{6}}^{2 \pi} (1+2 \sin x) dx- \int_{\frac{7 \pi}{6}}^{\frac{11 \pi}{6}} (1+2 \sin x) dx\] \[= [x-2 \cos x]_{0}^{\frac{7 \pi}{6}}+ [x-2 \cos x]_{\frac{11 \pi}{6}}^{2 \pi}- [x-2 \cos x]_{\frac{7 \pi}{6}}^{\frac{11 \pi}{6}} \] \[=\frac{7 \pi}{6}+ \sqrt{3}+2 + 2\pi -2 - \frac{11 \pi}{6} + \sqrt{3}- \frac{11 \pi}{6} + \sqrt{3} + \frac{7 \pi}{6} + \sqrt{3} \] \[= \frac{2\pi}{3} + 4 \sqrt{3}\]

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

Let $n$ be positive integers and $[.]$ be the floor function. Evaluate the integral \[ \int_{0}^{n^{\frac{1}{k}}} [ x^k + n] dx\] Since \( 0<1<2<3< \dots < n-1 < n \) we have \( 0^{\frac{1}{k}}<1^{\frac{1}{k}}<2^{\frac{1}{k}}<3^{\frac{1}{k}}< \dots < (n-1)^{\frac{1}{k}} < n^{\frac{1}{k}} \) $$$$ Therefore \[ \int_{0}^{n^{\frac{1}{k}}} [ x^k + n] dx = \sum_{i=0}^{n-1} \int_{i^{\frac{1}{k}}}^{(i+1){\frac{1}{k}}} [x^k + n] dx \] Now for a fixed $i$ in the range, consider the integral \[\int_{i^{\frac{1}{k}}}^{(i+1)^{\frac{1}{k}}} [x^k + n] dx \] \[ i^{\frac{1}{k}} < x < (i+1)^{\frac{1}{k}} \implies i < x^k < i+1 \implies x^k = i + \epsilon \quad where \quad 0 < \epsilon < 1\] \[ therefore \quad [x^k +n] = [i + \epsilon+n]= [i+n+ \epsilon] = i+n \] \[ therefore \quad \int_{i^{\frac{1}{k}}}^{(i+1)^{\frac{1}{k}}} [x^k + n] dx = \int_{i^{\frac{1}{k}}}^{(i+1)^{\frac{1}{k}}} (i+n) dx = (i+n)\bigg((i+1)^{\frac{1}{k}}-i^{\frac{1}{k}}\bigg) \] \[ Thus \quad \int_{0}^{n^{\frac{1}{k}}} [ x^k + n] dx = \sum_{i=0}^{n-1} \int_{i^{\frac{1}{k}}}^{(i+1){\frac{1}{k}}} [x^k + n] dx = \sum_{i=0}^{n-1} (i+n)\bigg((i+1)^{\frac{1}{k}}-i^{\frac{1}{k}}\bigg) \] \[ = (n+0)\bigg(1^{\frac{1}{k}}-0^{\frac{1}{k}}\bigg) + (n+1)\bigg(2^{\frac{1}{k}}-1^{\frac{1}{k}}\bigg)+ \dots +(n+n-2)\bigg((n-1)^{\frac{1}{k}}-(n-2)^{\frac{1}{k}}\bigg) + (n+n-1)\bigg(n^{\frac{1}{k}}-(n-1)^{\frac{1}{k}}\bigg) \] \( = n \bigg( 1^{\frac{1}{k}}-0^{\frac{1}{k}} +2^{\frac{1}{k}}-1^{\frac{1}{k}}+\dots+(n-1)^{\frac{1}{k}}-(n-2)^{\frac{1}{k}}+n^{\frac{1}{k}}-(n-1)^{\frac{1}{k}} \bigg) \) \(+ 1\bigg(2^{\frac{1}{k}}-1^{\frac{1}{k}}\bigg)+2\bigg(3^{\frac{1}{k}}-2^{\frac{1}{k}}\bigg)\dots+(n-2)\bigg((n-1)^{\frac{1}{k}}-(n-2)^{\frac{1}{k}}\bigg)+(n-1)\bigg(n^{\frac{1}{k}}-(n-1)^{\frac{1}{k}}\bigg) \) $$$$ \( = n \times n^{\frac{1}{k}}+2^{\frac{1}{k}}-1^{\frac{1}{k}}+2 \times 3^{\frac{1}{k}}-2 \times 2^{\frac{1}{k}}+ \dots +(n-2)(n-1)^{\frac{1}{k}}-(n-2)(n-2)^{\frac{1}{k}}+ (n-1)n^{\frac{1}{k}}-(n-1)(n-1)^{\frac{1}{k}} \) $$$$ \( =n^{\frac{1+k}{k}}-1^{\frac{1}{k}}-2^{\frac{1}{k}}-\dots-(n-1)^{\frac{1}{k}}+(n-1)n^{\frac{1}{k}}\) $$$$ \( =n^{\frac{1+k}{k}}-1^{\frac{1}{k}}-2^{\frac{1}{k}}-\dots-(n-1)^{\frac{1}{k}}-n^{\frac{1}{k}}+n \times n^{\frac{1}{k}}\) $$$$ \( =n^{\frac{1+k}{k}}-1^{\frac{1}{k}}-2^{\frac{1}{k}}-\dots-(n-1)^{\frac{1}{k}}-n^{\frac{1}{k}}+n^{\frac{1+k}{k}}\) $$$$ \( =2n^{\frac{1+k}{k}}-1^{\frac{1}{k}}-2^{\frac{1}{k}}-\dots-(n-1)^{\frac{1}{k}}-n^{\frac{1}{k}}\) $$$$ \( = 2n^{\frac{1+k}{k}} - \sum_{i=1}^{n} i^\frac{1}{k} \)

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

Prove that \[ \frac{1}{2}. \frac{3}{4}.\frac{5}{6}. \frac{7}{8} \dots \frac{2n-1}{2n} < \frac{1}{\sqrt{2n+1}} \] Let \(a,b > 0 \quad and \quad a < b \), we first show that \( \frac{a}{b} < \frac{a+1}{b+1}\). Now \( \frac{a+1}{b+1} - \frac{a}{b} = \frac{b-a}{b(b+1)} > 0 \quad since \quad b-a,a,b >0\). $$$$ Using the above result we see that \[\frac{1}{2} < \frac{2}{3} \] \[\frac{3}{4} < \frac{4}{5} \] \[\dots\] \[ \frac{2n-1}{2n} < \frac{2n}{2n+1} \] Let \( x_n = \frac{1}{2}. \frac{3}{4}.\frac{5}{6}. \frac{7}{8} \dots \frac{2n-1}{2n} \quad and \quad y_n = \frac{2}{3}. \frac{4}{5}.\frac{6}{7}. \frac{8}{9} \dots \frac{2n}{2n+1} \). Using the above result it is easy to see that \( x_n < y_n \). $$$$ Now \[ x_n^2 = x_n . x_n < x_n .y_n = \frac{1}{2}. \frac{3}{4}.\frac{5}{6}. \frac{7}{8} \dots \frac{2n-1}{2n} . \frac{2}{3}. \frac{4}{5}.\frac{6}{7}. \frac{8}{9} \dots \frac{2n}{2n+1} = \frac{1}{2n+1} \] \[ \implies x_n < \frac{1}{\sqrt{2n+1}} \quad i.e., \quad \frac{1}{2}. \frac{3}{4}.\frac{5}{6}. \frac{7}{8} \dots \frac{2n-1}{2n} < \frac{1}{\sqrt{2n+1}}\]

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

Let \( f: \mathbb{R} \to \mathbb{R} \) be a function that is differentiable $n+1$ times for some positive integer $n$. The $ith$ derivative of $f$ is denoted by $f^{(i)}$. Suppose \[ f(1)=f(0)=f^{(1)}(0)=\dots=f^{(n)}(0)=0.\] Prove that $f^{(n+1)}=0$ for some $x \in (0,1)$. $$$$ Since $f$ is $n+1$ times differentiable(everywhere in $\mathbb{R}$).\( f,f^{(1)},\dots,f^{(n)} \) are all differentiable and continuous in \( (0,1) \quad and \quad [0,1] \) respectively.$\dots (A)$ $$$$ Since \(f(0)=f(1)=0 \) By Rolles' Theorem $\exists \quad c_1$ in $(0,1)$ suct that $f^{(1)}(c_1)=0$ $$$$ Now consider the interval $[0,c_1]$. Note that \( f^{(1)}(0)=f^{(1)}(c_1)=0 \) and $f^{(1)}$ is continuous and differentiable in $[0,c_1]$ and $(0,c_1)$ respectively. (Using $\dots (A)$). Thus by Rolle's Theorem $\exists \quad c_2$ in $(0,c_1)$ suct that $f^{(2)}(c_2)=0$ $$$$ Now consider the interval $[0,c_2]$. Note that \( f^{(2)}(0)=f^{(2)}(c_2)=0 \) and $f^{(2)}$ is continuous and differentiable in $[0,c_2]$ and $(0,c_2)$ respectively. (Using $\dots (A)$). Thus by Rolle's Theorem $\exists \quad c_3$ in $(0,c_2)$ suct that $f^{(3)}(c_3)=0$ $$$$ Continuing like this, we get a point $c_{n}$ such that $f^{(n)}(c_{n})=0$ where $c_{n} \in (0,c_{n-1})$. $$$$ Now consider the interval $[0,c_n]$. Note that \( f^{(n)}(0)=f^{(n)}(c_n)=0 \) and $f^{(n)}$ is continuous and differentiable in $[0,c_n]$ and $(0,c_n)$ respectively. (Using $\dots (A)$). Thus by Rolle's Theorem $\exists \quad c_{n+1}$ in $(0,c_n)$ suct that $f^{(n+1)}(c_{n+1})=0$. Since \( 0 < c_{n+1} < c_{n} < \dots < c_1 < 1 \) thus $c_{n+1} \in (0,1)$ $$$$

Indian Statistical Institute B.Math & B.Stat: Integration and Continuity

If \( c \int_{0}^{1} xf(2x) dx = \int_{0}^{2} tf(t) dt,\) where $f$ is a positive continuous functions, then find then value of $c$. $$$$ In the R.H.S put $2x=t$, this gives \( c \int_{0}^{1} xf(2x) dx = 4\int_{0}^{1} xf(2x) dx = (c-4) \int_{0}^{1} xf(2x) dx =0 \implies c =4\) $$$$ Since $f$ is given to be a positive continuous function, $xf(x)$ is continuous and $>0 \in (0,1)$. Therefore \( \int_{0}^{1} xf(2x) dx > 0 \). This follows from the fact that if a continuous functions is positive at a point in its domain, the $\exists$ a open neighborhood containing the point and contained in the domain throughout which the function is positive.

Indian Statistical Institute B.Math & B.Stat : Continuity and Bijections

$Problem-1.$ Let \(f: \mathbb{R} \to \mathbb{R} \) be given by \(f(x) = x(x-1)(x+1) \). Then show that $f$ is onto but not 1-1. $$$$ The injectivity part is very much obvious. See that \( f(0)=f(1)=f(-1) = 0 \). Now for the surjectivity, observe that $f$ is an odd degree polynomial. Hence it must be surjective! (Think if not, then $\exists$ a real number which has no pre-image (say $r$ ). Now consider the polynomial \( g(x) = f(x)-r \)!! what else g(x) is also an odd degree polynomial with no real roots!!! ) $$$$ $Problem-2.$ Another Problem with a kick of continuity. Let \(f: \mathbb{R} \to \mathbb{R} \) be given by \(f(x) = x^3-3x^2+6x-5 \). Then show that $f$ is both onto and 1-1. $$$$ Again note that $f$ is an odd degree polynomial so it is surjective. Since $f$ is a polynomial it is continuous on $\mathbb{R}$. Also \(f'(x)=3x^2-6x+6 = 3(x^2-2x+2) = 3\{(x-1)^2 + 1 \} > 0 \) \( \forall x \in \mathbb{R}\). This shows that $f$ is strictly increasing. Thus $f$ is continuous and strictly increasing hence must be 1-1. $$$$ $Problem-3.$ Let \(f: \mathbb{R} \to \mathbb{R} \) be given by \(f(x) = x^2 - \frac{x^2}{1+x^2} \). Then show that $f$ is neither onto nor 1-1. $$$$ Clearly $f$ is not injective since \(f(x)=f(-x) \). Also \( f(x) = \frac{x^4}{1+x^2} \) which shows that $f$ assume non-negative values. Thus the \(Range_f \subset \mathbb{R} = Co-domain_f\). Thus $f$ is neither surjective. $$$$ $Problem-4.$ Let \( \phi :[0,1] \to [0,1] \) be a continuous and 1-1 function. Let \( \phi(0) = 0, \phi(1) = 1, \phi \big(\frac{1}{2}\big) = c, \phi \big(\frac{1}{4}\big) = d.\) Then show that $ c > d $. $$$$ This follows form the property of continuous functions. On an interval I ( not necessarily compact ), if a function is continuous and 1-1 it is strictly increasing. To prove this you can use the Intermediate Value Theorem.

Problems : Limits

Problems for Indian Statistical Institute, Chennai Mathematical Institute, JEE Main and Advanced. $$$$ \[1.Evaluate: \lim_{x \to \infty} \frac{20+2\sqrt{x}+3\sqrt[3]{x}}{2+\sqrt{4x-3}+\sqrt[3]{8x-4}}\] \[2.Evaluate: \lim_{x \to \infty} \big( x \sqrt{x^2+a^2}-\sqrt{x^4+a^4}\big)\] \[3.Evaluate: \lim_{x \to \infty} x^3 \big\{ \sqrt{x^2+\sqrt{x^4+1}}-\sqrt{2}x \big \}\] \[4.Evaluate: \lim_{x \to \infty} \sqrt{\frac{x-\cos^2 x}{x+\sin x}}\] \[5.Evaluate: \lim_{x \to \infty} [2\log(3x)-\log(x^2+1) ]\] 6. Let \( R_n =2+\sqrt{2+\sqrt{2+\dots+\sqrt{2}}}\) (n square roots signs). Then evaluate \(\lim_{n \to \infty} R_n \) $$$$ 7. If \(a_n = \bigg( 1+\frac{1}{n^2}\bigg)\bigg( 1+\frac{2^2}{n^2}\bigg)^2 \bigg( 1+\frac{3^2}{n^2}\bigg)^3 \dots \bigg( 1+\frac{n^2}{n^2}\bigg)^n \), then evaluate \[ \lim_{n \to \infty} a_n^{-\frac{1}{n^2}} \] $$$$ \[8.Evaluate \lim_{x \to \infty} \sqrt{x^2+x}-\sqrt{x^2+1}\] \[9. \lim_{x \to \frac{\pi}{2}} (\sin x)^{\tan x}\] \[10. \lim_{x \to 0} \frac{\cos x -1}{\sin^2 x}\] For PDF click here

Indian Statistical Institute B.Math & B.Stat : Roots of Polynomial

Let $c$ be a fixed real number. Show that a root of the equation \[x(x+1)(x+2)\dots(x+2009)=c\] can have multiplicity at most $2.$ $$$$ Let \( f(x) = x(x+1)(x+2)\dots(x+2009)-c \) $$$$ First we compute the derivative of $f(x)$ and see that \( f'(x) = (x+1)(x+2)\dots(x+2009)+x(x+2)\dots(x+2009)+\dots+x(x+1)\dots(x+r-1)(x+r+1)\dots(x+2009)+\) \(\dots+x(x+1)(x+2)\dots(x+2008) \) where $r$ is a positive integer less than $2009$. $$$$ Now \(f'(-r) = (-r)(-r+1)\dots(-1)(1)\dots(-r+2009) =(-1)^r r! (2009-r)! > 0\) $if$ $r$ is even, else $<0$.where \( r \in \{0,1,2,\dots,2008\}\) $$$$ Thus we have the following inequalities, \( f'(0) > 0, f'(-1) < 0, f'(-2) > 0, \dots, f'(2008) > 0, f'(2009) < 0 \) $$$$ This shows that $f'(x)=0$ has one real root in each of the intervals \( (-1,0),(-2,-1),\dots,(-2009,-2008) \). Since $degree$ of $f'(x)$ is $2009$, all the roots of $f'(x)=0$ is real and simple. Thus a root of $f'(x)=0$ cannot be a root of the equation $f''(x)=0$. So a root of $f(x)=0$ can have $multiplicity$ at most $2$. $$$$

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

Let $n$ be a positive integer Define \[f(x) = min \{|x-1|,|x-2|,\dots,|x-n|\} \] $$$$ Then evaluate \[ \int_{0}^{n+1} f(x) dx \] $$$$ When \(0 < x < 1+ \frac{1}{2} \), $|x-1|=min \{|x-1|,|x-2|,\dots,|x-n|\}$. Now $|x-1|= 1-x$ for \( 0 < x < 1\) and $|x-1|= x-1 $ for $1 < x < 1+ \frac{1}{2}$ $$$$ So, \(\int_{0}^{1+\frac{1}{2}} f(x) dx = \int_{0}^{1} (1-x) dx + \int_{1}^{1+ \frac{1}{2}} (x-1) dx = \frac{1}{2}+ \frac{1}{2} \times \frac{1}{4} \dots (A)\) $$$$ When \(n+ \frac{1}{2} < x < n+1 \), $|x-n|=min \{|x-1|,|x-2|,\dots,|x-n|\}$. Now $|x-n|= x-n$ for \( n+ \frac{1}{2} < x < n+1 \)$$$$ So, \(\int_{n+\frac{1}{2}}^{n+1} f(x) dx = \int_{n+\frac{1}{2}}^{n+1} (x-n) dx = \frac{1}{2}- \frac{1}{2} \times \frac{1}{4} \dots (B)\) $$$$ Consider the diagram given below where $1 < k \leq n$,. When \( x \in \big(k-\frac{1}{2},k+\frac{1}{2} \big) \), $|x-k|$ is minimum among $|x-i|$ where $i=1,2,3,\dots,k-1,k+1,\dots,n$ $$$$ $$$$ So, \( \int_{k-\frac{1}{2}}^{k+\frac{1}{2}} f(x) dx = \int_{k-\frac{1}{2}}^{k+\frac{1}{2}} |x-k| dx = \int_{k-\frac{1}{2}}^{k} (k-x) dx + \int_{k}^{k+\frac{1}{2}} (x-k) dx = \frac{1}{4}\) for $k=2,3,4\dots,n$.$$$$ Summing for $k=2,3,4\dots,n$ we get \[\int_{1+\frac{1}{2}}^{n+\frac{1}{2}} f(x) dx = \frac{(n-1)}{4} \dots (C)\] $$$$ Adding equations $A,C$ and $B$ we get \[ \int_{0}^{n+1} f(x) dx = \int_{0}^{1+\frac{1}{2}} f(x) dx+ \int_{1+\frac{1}{2}}^{n+\frac{1}{2}} f(x) dx+\int_{n+\frac{1}{2}}^{n+1} f(x) dx = \frac{1}{2}+ \frac{1}{2} \times \frac{1}{4}+ \frac{(n-1)}{4}+ \frac{1}{2}- \frac{1}{2} \times \frac{1}{4} \] $$$$ \[= \frac{(n-1)}{4}+1 = \frac{n+3}{4} \]

Indian Statistical Institute B.Math & B.Stat : Square of a real is non-negative

Show that the following system of inequalities has exactly one solution $a-b^2 \geq \frac{1}{4},$ $b-c^2 \geq \frac{1}{4},$ $c-d^2 \geq \frac{1}{4}$ and $d-a^2 \geq \frac{1}{4}.$ $$$$ Adding up all the inequalities we get \( a-b^2 + b-c^2 + c-d^2 + d-a^2 \geq \frac{1}{4} + \frac{1}{4} + \frac{1}{4} +\frac{1}{4} \) $$$$ \( \implies a-a^2 - \frac{1}{4} + b-b^2- \frac{1}{4} + c-c^2 - \frac{1}{4} + d-d^2 - \frac{1}{4} \geq 0 \) $$$$ \( \implies -\big(a- \frac{1}{2} \big)^2 -\big(b- \frac{1}{2} \big)^2 - \big(c- \frac{1}{2} \big)^2 - \big(d- \frac{1}{2} \big)^2 \geq 0 \) $$$$ which is possible only when R.H.S is zero i.e., \( a=b=c=d= \frac{1}{2} \), since the R.H.S is always non-positive.

Indian Statistical Institute B.Math & B.Stat : Limits at Infinity

Let \( a_1 > a_2 > \dots > a_r \) be positive real numbers. Compute \(\lim_{n \to \infty} \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}}\). $$$$ Since \( a_1 > a_2 > \dots > a_r \) and each of them is positive we have \(a_1^n>a_2^n>\dots>a_r^n \) $$$$ \( \implies a_1^n+a_2^n+\dots+a_r^n < a_1^n+a_1^n+\dots+a_1^n = ra_1^n \) $$$$ Letting \( n \to \infty \) we have, \(\lim_{n \to \infty} \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}} < \lim_{n \to \infty}(ra_1^n)^{\frac{1}{n}} \) \( = a_1\lim_{n \to \infty}r^{\frac{1}{n}}= a_1 \) Note $r>0$ $$$$ Now, \( \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}} = \bigg( a_1^n \big(1+\frac{a_2^n}{a_1^n}+\dots+\frac{a_r}{a_1}\big) \bigg)^{\frac{1}{n}} = a_1\bigg( 1+\frac{a_2^n}{a_1^n}+\dots+\frac{a_r}{a_1}\big) \bigg)^{\frac{1}{n}} > a_1 \) Since \(\bigg( 1+\frac{a_2^n}{a_1^n}+\dots+\frac{a_r}{a_1} \bigg)^{\frac{1}{n}} > 1\) $$$$ Letting \( n \to \infty \) we have, \(\lim_{n \to \infty} \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}} > \lim_{n \to \infty}a_1 =a_1 \) $$$$ Thus by $Sandwhich-theorem$ \(\lim_{n \to \infty} \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}} = a_1\)

Inequality

$Problem$ #3 $$ $$ Find all real numbers $x$ for which \(\sqrt{3-x}-\sqrt{x+1} > \frac{1}{2}\) $$ $$ Let $f(x)$ \(=\sqrt{3-x}-\sqrt{x+1}\). First note that $f(x)$ is defined for \( -1 \leq x \leq 3 \) $$ $$ \( f'(x) = \frac{-1}{2\sqrt{3-x}} - \frac{1}{2\sqrt{x+1}} = - \big(\frac{1}{2\sqrt{3-x}} + \frac{1}{2\sqrt{x+1}}\big) < 0 \Rightarrow f(x)\) is strictly decreasing $$ $$ Now \( f(-1) = 2 > \frac{1}{2}\) and \( f(3) = -2 < \frac{1}{2}\) Since $f(x)$ is continuous, $\exists$ at least one x $\in$ ${(-1,3)}$ suct that $f(x) = \frac{1}{2}$ $$ $$ \( f(x) = \frac{1}{2} \Rightarrow \sqrt{3-x}-\sqrt{x+1} = \frac{1}{2} \Rightarrow 64x^2-128x+33 = 0 \Rightarrow x = 1 \pm \frac{\sqrt{31}}{8} \)$$ $$ but \(x = 1 + \frac{\sqrt{31}}{8}\) does not satisfy \(\sqrt{3-x}-\sqrt{x+1} = \frac{1}{2}\) Check yourself! So the only solution is \(x = 1 - \frac{\sqrt{31}}{8}\) $$ $$ Since $f(x)$ is strictly decreasing, the given inequality is true for \( x \in {[-1,1 - \frac{\sqrt{31}}{8}\big)}\) $$ $$

Matrices & Determinants

$Problem$ #1 $$ $$ If $A$ and $B$ are different matrices satisfying \( A^3 = B^3 \) and \(A^2B = B^2A\), find \(det(A^2+B^2)\) $$ $$ Since $A$ and $B$ are different matrices \( A-B \neq O \), Now \((A^2+B^2)(A-B) = A^3-A^2B+B^2A-B^3\) $$ $$ =$O$ since \(A^3 = B^3\) and \(A^2B = B^2A\) $$ $$ This shows that \((A^2+B^2)\) has a zero divisor, so it is not invertible hence \(det(A^2+B^2) = 0\)

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

Let $k$ be any odd integer greater that 1. Then show that \( 1^k+2^k+3^k+\dots \dots+2006^k \) is divisible by 2013021. $$$$ We will prove the general case. Let \(S= 1^k+2^k+3^k+\dots \dots+n^k\) where $n \geq 2$ and \(n\in \mathbb{N}\). $$$$ \(2S = 1^k+2^k+3^k+\dots \dots+n^k + 1^k+2^k+3^k+\dots \dots+n^k = (1^k+n^k)+(2^k+(n-1)^k)+\dots \dots+(n^k+1^k)\) $$$$ Using the result, $n^k+m^k$ is always divisible by $n+m$ if $k$ is odd we see that $2S$ is divisible by $(n+1)$ $$$$ Again \(2S = 1^k+2^k+3^k+\dots \dots+n^k + 1^k+2^k+3^k+\dots \dots+n^k\) $$$$ \( = (1^k+(n-1)^k)+(2^k+(n-2)^k)+\dots \dots+((n-1)^k+1^k)+2n^k\) $$$$ Using the same result and noting that $2n^k$ is divisible by $n$ we see that $2S$ is divisible by $n$. Now both $n$ an $n+1$ divides $2S$ and $g.c.d(n.n+1)=1$ we see that $n(n+1)$ divides $2S$ this implies $\frac{n(n+1)}{2}$ divides $S$. ( $n(n+1)$ is always even) $$$$ In the given problem $n=2006$, thus \( 1^k+2^k+3^k+\dots \dots+2006^k \) is divisible by $\frac{2006(2006+1)}{2}=2013021$

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

In how many ways one can choose three distinct numbers from the set \( \{1,2,3,\dots \dots,19,20\}\) such that their product is divisible by 4? $$$$ We partition the set \( \{1,2,3,\dots \dots,19,20\}\) into three disjoint sets \(S_1=\{4,8,12,16,20\},S_2=\{2,6,10,14,18\},S_3=\{1,3,5,7,9,11,13,15,17,19\}\) $$$$ Three selected (distinct) numbers will not be divisible by $4$ $iff$ all the $three$ numbers are selected form $S_3$ or $two$ of them are selected from $S_3$ and $one$ of them from $S_1$. Numbers of such numbers are \( \binom{10}{3}+\binom{5}{2} \times \binom{5}{1} = 345 \) $$$$ So numbers of selection such that their product is divisible by $4$ is \(\binom{20}{3}-345= 795 \)

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

Find the digit at the unit place of \[\big(1!-2!+3!-\dots \dots +25!\big )^{\big(1!-2!+3!-\dots \dots +25!\big )}\] First note that \( k! \equiv 0 (mod\ 10) \) for all $k \geq 5 , k \in \mathbb{N}$ $$$$ So, \( 5!-6!+7!-\dots \dots +25! \equiv 0 (mod\ 10) \) and \( 1!-2!+3!-4! = -19 \equiv 1 (mod\ 10)\) (Using the property of $congruences$). $$$$ Using the above two congruences \( \big(1!-2!+3!-\dots \dots +25!\big ) \equiv 1 (mod\ 10) \) $$$$ So, \[\big(1!-2!+3!-\dots \dots +25!\big )^{\big(1!-2!+3!-\dots \dots +25!\big )} \equiv 1^{\big(1!-2!+3!-\dots \dots +25!\big )} \equiv 1 (mod 10) \] giving $1$ as the last digit. $$$$ Let \(a \equiv a' (mod\ m) \) and \(b \equiv b' (mod\ m)\), then important properties of $congruences$ include the following, where $\implies$ means "implies": $$$$ 1. Reflexivity: $a\equiv a (mod- m)$. $$$$ 2. Symmetry: \(a\equiv b (mod\ m) \implies b\equiv a (mod\ m)\).$$$$ 3. Transitivity: \(a\equiv b (mod\ m)\) and \(b \equiv c (mod\ m)\implies a\equiv c (mod\ m)\). $$$$ 4. \(a+b \equiv a'+b' (mod\ m)\)$$$$ 5. \(a-b\equiv a'-b' (mod\ m)\). $$$$ 6. \(ab\equiv a'b' (mod\ m)\). $$$$ 7. \(a\equiv b (mod\ m)\implies ka \equiv kb (mod\ m)\). $$$$ 8. \(a\equiv b (mod\ m)\implies a^n\equiv b^n (mod\ m)\). $$$$ 9. \(ak\equiv bk (mod\ m)\implies\) \(a\equiv b \big(mod\ \frac{m}{(k,m)}\big),\) where $(k,m)$ is the greatest common divisor. $$$$ 11. If $a \equiv b (mod\ m)$, then $P(a) \equiv P(b) (mod\ m)$, for $P(x)$ a polynomial with integer coefficients.

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

Find the sum of all even positive divisors of $1000$. $$$$ \( 1000 = 2^3 \times 5^3 \). Now any even divisor of $1000$ must contain a factor of the form $2^j$ where $j \in \{1,2,3\}$. We note that $2$ and $5$ are the only prime factors of $1000$ , so a even factor must be of the form $2^j\times 5^i$ where $j \in \{1,2,3\}$ and $i \in \{0,1,2,3\}$. $$$$ So the required sum is \( \sum_{j=0}^{3} \sum_{i=1}^{3} 2^i \times 5^j =\sum_{j=0}^{3} 14 \times 5^j = 14 \times \frac{5^4-1}{5-1} = 2184\)

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

Let $\alpha$ and $\beta$ be two positive real numbers. For any integer $n>0$, define \( a_n = \int_{\beta}^{n} \frac{\alpha}{u(u^\alpha+2+u^{-\alpha})}du\). Then find \( \lim_{n \to \infty} a_n \). $$$$ Multiplying $u^{\alpha-1}$ to the numerator and denominator of the integrand, we have \( a_n = \int_{\beta}^{n} \frac{\alpha u^{\alpha-1}}{u\times u^{\alpha-1}(u^\alpha+2+u^{-\alpha})}du\) $$$$ Substituting $u^{\alpha}=t$ we get the transformed integral as \(a_n = \int_{\beta^\alpha}^{n^\alpha}\frac{dt}{(t+1)^2}dt = \frac{n^\alpha-\beta^\alpha}{(1+\beta^\alpha)(1+n^\alpha)}\) $$$$ Therefore,\( \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n^\alpha-\beta^\alpha}{(1+\beta^\alpha)(1+n^\alpha)}= \lim_{n \to \infty} \frac{1-\big({\frac{\beta}{n}}\big)^\alpha}{(1+\beta^\alpha)(1+\frac{1}{n})}=\frac{1}{1+\beta^\alpha}\)

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

Let $S$ be the set of all integers $k$, \( 1 \leq k \leq n\), such that $g.c.d(k,n)=1$. What is the arithmetic mean of the integers in $S$?. $$$$ First note that \( |S| = \phi(n) \). Now let $k \in S$ then there exists intergers $u,v$ such that $ku+nv=1\dots (A)$. $$$$ The integer \( n-k \in \{1,2,\dots,n-1\} \) because $k$ can never be equals $n$, for $g.c.d(n,n)=n$ and $k \in S$. We will now show that $g.c.d(n-k,n)=1.$ Adding $-nu$ to both sides of $(A)$ we get \( -nu+ku+nv=-nu+1 \implies -u(n-k)+(v+u)n = 1 \implies g.c.d(n-k,n)=1 \) $$$$ So, for \( k \in \{1,2,\dots,n-1\}\) if \( k \in S \implies n-k \in S \) Thus $S$ can be written in the form, \( S=\{k_1,k_2,k,.....,k_r,n-k_r,......,n-k_2,n-k_1\} \) where $|S| = \phi(n)$ $$$$ Clearly the sum of the elements of $S$ is \( (k_1+n-k_1)+(k_2+n-k_2)+\dots+(k_r+n-k_r) = \frac{n\phi(n)}{2} \) ( Pairing reduces the terms to half the original ($\phi(n)))$. $$$$ arithmetic mean \[ =\frac{\frac{n\phi(n)}{2}}{\phi(n)} = \frac{n}{2} \]

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

If \(a,b,c \in (0,1) \) satisfy $a+b+c=2$, prove that \( \frac{abc}{(1-a)(1-b)(1-c)} \geq 8. \) $$$$ Let \( p = 1-a, q = 1-b, r = 1-c \). $p+q+r= 3-(a+b+c)=1$.Clearly $p,q,r$ are positive. Substituting in the given inequality, it transforms to $$$$ \( \frac{(1-p)(1-q)(1-r)}{pqr} \geq 8 \iff {(1-p)(1-q)(1-r)} \geq 8{pqr} \) $$$$ \( \iff 1-(p+q+r)+qr+rp+pq-pqr \geq 8{pqr} \iff qr+rp+pq-pqr \geq 8{pqr} \) $$$$ \( \iff qr+rp+pq \geq 9{pqr} \iff \frac{1}{p}+\frac{1}{q}+\frac{1}{r} \geq 9 \) $$$$ Thus proving the above inequality reduces to proving \(\frac{1}{p}+\frac{1}{q}+\frac{1}{r} \geq 9\) subjected to $p+q+r=1$ $$$$ Since $p,q,r$ are positive, apllying $A.M \geq H.M$ we have \( \frac{p+q+r}{3} \geq \frac{3}{\frac{1}{p}+\frac{1}{q}+\frac{1}{r}}\) $$$$ Noting $p+q+r=1$, we have \(\frac{1}{p}+\frac{1}{q}+\frac{1}{r} \geq 9 \)