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

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 $\frac{1}{p}+\frac{1}{q}+\frac{1}{r} \geq 9$

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

A point $P$ with coordinates $(x,y)$ is said to be good if both $x$ and $y$ are positive integers. Find the number of good points on the curve $xy=27027.$ First note that, $27027 = 3^3 \times 13 \times 11 \times 7$. Now for any good point on the curve $xy=27027$, $x$ must be of the form $3^p \times 13^q \times 11^r \times 7^s$ and $y$ must be of the form $3^{p'} \times 13^{q'} \times 11^{r'} \times 7^{s'}.$ Where $p+p'=3,q+q'=1,r+r'=1$ and $s+s'=1$ Now considers the coordinates $(3^3 \times ?, 3^0 \times ?),(3^2 \times ?, 3^1 \times ?),(3^1 \times ?, 3^2 \times ?),(3^0 \times ?, 3^3 \times ?)$ where in the place of $?$ in the $x-$ coordinate we can have any combinations of the products $13^q \times 11^r \times 7^s$ and in the place of $?$ in the $y-$ coordinate we can have any combinations of the products $13^{q'} \times 11^{r'} \times 7^{s'}$ such that $q+q'=1,r+r'=1$ and $s+s'=1.$ Note that for the equation $q+q'=1$ the possible non-negative solutions are $(1,0), (0,1)$. Similarly for the other two equations $r+r'=1$ and $s+s'=1$. Counting the combinations for $?$ is the $x-$coordinate for the coordinate $(3^3 \times ?, 3^0 \times ?)$ we can have CASE I : All the three numbers $13,11,7$ appears which can be done in $\binom {3}{3}$ ways. $?$ in the $y-$ coordinate the must contain $13^0,11^0,7^0$. CASE II : Any two of the three numbers $13,11,7$ appears which can be done in $\binom {3}{2}$ ways. $?$ in the $y-$ coordinate the must contain $13$ or $11$ or $7$. CASE III : Any one of the three numbers $13,11,7$ appears which can be done in $\binom {3}{1}$ ways. $?$ in the $y-$ coordinate the must contain $13,11$ or $11,7$ or $7,13$. CASE IV : None of the three numbers $13,11,7$ appears which can be done in $\binom {3}{0}$ ways. $?$ in the $y-$ coordinate the must contain $13,11,7$. Giving a total of $\binom {3}{3}+ \binom {3}{2}+ \binom {3}{1}+ \binom {3}{0}=8$ possibilities. Similarly we have $8$ possibilities for each of the coordinates of the form $(3^2 \times ?, 3^1 \times ?),(3^1 \times ?, 3^2 \times ?),(3^0 \times ?, 3^3 \times ?)$, giving in total $4\times8=32$ good coordinates Another problem of same flavor form $Indian- Statistical- Institute$ What is the number of ordered triplets $(a, b, c)$, where $a, b, c$ are positive integers (not necessarily distinct), such that $abc = 1000$? First note that, $1000 = 2^3 \times 5^3$. Any ordered triplet with the given condition must be of the form $(2^l\times5^m,2^p\times5^q,2^r\times5^s)$ where $l+p+r=3$ and $m+q+s=3$. Number of non-negative solutions of the above equations is $\binom {3+3-1}{2} = 10$ in both cases. Now consider one ordered triplet $(2^1\times5^m,2^0\times5^q,2^2\times5^s)$ ( 10 such triplet are possible, varying powers of 2 subjected to $l+p+r=3$ ), for this triplet now we can vary $m,q,s$ subjected to $m+q+s=3$ in 10 ways.Thus in total $10\times10=100$ ordered triplets are possible. Another good problem with a kick of $Derangement!!$ There are $8$ balls numbered $1,2,\dots,8$ and $8$ boxes numbered $1,2,\dots,8$. Find the number of ways one can put these balls in the boxes so that each box gets one ball and exactly $4$ balls go in their corresponding numbered boxes. $4$ balls can be selected in $\binom{8}{4}$ ways and for each for such selection, say $\{Ball_2,Ball_5,Ball_3,Ball_7\}$ the balls can go in their corresponding boxes in exactly one way (why?). Now the remaining $4$ balls $\{Ball_3,Ball_4,Ball_1,Ball_8\}$ has to be put in the boxes in such a way none of the balls goes to the box of same number. Since, exactly $4$ balls go in their corresponding numbered boxes. Which is nothing but $D_4$, so total number of ways is $\binom{8}{4} \times D_4$ where $D_4 = 4!\big( 1 - \frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}\big) = 9$ In general $D_n$ is Derangement of $n$ objects.

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

Using only the digits $2$, $3$ and $9$, how many six digit numbers can be formed which are divisible by $6$? First note that a number is divisible by $6$, $iff$ it is divisible by $2$ and by $3.$ ($2$ and $3$ being co-prime) The above argument clearly shows that the number $222222$ is divisible by $6$. Use divisibility test of $2$ and $3$. First note that a number will be divisible by $2$, $iff$ the digit at the unit place is $2.$ Now observe that, if the number which is divisible by $2$ has to be divisible by $3$, the sum of the digits must be divisible by $3.$ So the 5 places of the number (except the digit at the unit place, which is $2$) is a combination of the digits $2$, $3$ and $9$ such that the sum including the digit at the unit place is divided by $3$. Now observe that among the $5$ places to be filled, exactly $two$ places can be the digit $2$. (Convince yourself why?! thin if not...) And the remaining $three$ places can be filled by either $3$ or $9$. Thus giving $\binom {5} {2} \times 2\times2\times2$ possibilities with desired condition. therefore, total number of numbers $= \binom {5} {2} \times 2\times2\times2+1 = 80+1=81$

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

Evaluate: $$\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x}{x} dx$$ Let $I=\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x}{x} dx$, Put $x=\frac{1}{t}$ $\implies I = -\int_{2014}^\frac{1}{2014} \frac{tan^{-1}\frac{1}{t}}{t} dt =\int_\frac{1}{2014}^{2014} \frac{tan^{-1}\frac{1}{t}}{t} dt = \int_\frac{1}{2014}^{2014} \frac{tan^{-1}\frac{1}{x}}{x} dx$ $\mathbb{Therefore,}$ $$2I=\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x}{x} dx+\int_\frac{1}{2014}^{2014} \frac{tan^{-1}\frac{1}{x}}{x} dx$$ $$=\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x+tan^{-1}\frac{1}{x}}{x} dx$$ $$=\int_\frac{1}{2014}^{2014} \frac{\frac{\pi}{2}}{x} dx$$ $$=\frac{\pi}{2}\int_\frac{1}{2014}^{2014} \frac{dx}{x}$$ $$=\frac{\pi}{2}[\log x]_{\frac{1}{2014}}^{2104}$$ $$=\frac{\pi}{2}\big[\log 2014-\log \frac{1}{2014}\big]$$ $$=\frac{\pi}{2}\log 2014^2$$ $$=\pi\log 2014$$

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

Let $P: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that $P(x)=x$ has no real solution. Prove that $P(P(x))=x$ has no real solution. If possible let, $P(P(x))=x$ has a real solution for $x=x_0$. Then $P(P(x_0))=x_0\dots(1)$ Now let, $P(x_0)=y_0$ $\implies P(y_0)=x_0$ using $(1)$ Note that $x_0 \neq y_0$, otherwise we will have a solution to the equation $P(x)=x$! A contradiction to the hypothesis. Without loss of generality assume that $x_0 < y_0$ Construct a function $Q:[x_0,y_0]\rightarrow \mathbb{R}$ where $Q(x)=P(x)-x$,since $P$ is given to be continuous on $\mathbb{R}$, $Q$ is continuous on $[x_0,y_0].$ Observe that, $Q(x_0)=P(x_0)-x_0=y_0-x_0 > 0$ and $Q(y_0)=P(y_0)-y_0=x_0-y_0 < 0$ $\implies Q(x_0)Q(y_0) < 0 \implies$ there exists a point $c\in$ $[x_0,y_0]$ such that $Q(c)=0$ using $Intermediate-Value-Theorem$ $\implies P(c)-c=0 \implies P(c)=c$ for a real value, which contradicts the hypothesis, thus the assumption $P(P(x))=x$ has a real solution is not tenable.

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

How many $three-digits$ numbers of distinct digits can be formed by using the digits $1,2,3,4,5,9$ such that the sum of digits is at least 12? First note that since the digits must be distinct there must be no repetition and further note that the maximum sum of such three digits is $9+5+4=18.$ We will find all such possible by considering their sum of the digits. Case I: Sum of the digits is 12. In this case the selection of digits can be $\{1,2,9\},\{3,4,5\}$ So, a total of $3!+3!=12$ $three-digits$ numbers possible. Case II: Sum of the digits is 13. In this case the selection of digits can be $\{1,3,9\}$ So, a total of $3!=6$ $three-digits$ numbers possible. Case III: Sum of the digits is 14. In this case the selection of digits can be $\{1,4,9\},\{2,3,9\}$ So, a total of $3!+3!=12$ $three-digits$ numbers possible. case IV: Sum of the digits is 15. In this case the selection of digits can be $\{1,5,9\},\{2,4,9\}$ So, a total of $3!+3!=12$ $three-digits$ numbers possible. Case V: Sum of the digits is 16. In this case the selection of digits can be $\{2,5,9\},\{2,5,9\}$ So, a total of $3!+3!=12$ $three-digits$ numbers possible. Case VI: Sum of the digits is 17. In this case the selection of digits can be $\{3,5,9\}$ So, a total of $3!=6$ $three-digits$ numbers possible. Case VII: Sum of the digits is 18. In this case the selection of digits can be $\{4,5,9\}$ So, a total of $3!=6$ $three-digits$ numbers possible. Adding all the cases we have $12+6+12+12+12+6+6=66$, $three-digits$ numbers of distinct digits can be formed.

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

Let $\omega$ be the complex cube root of unity. Find the cardinality of the set $S$ where $S = \{(1+\omega+\omega^2+\dots+\omega^n)^m | m,n = 1,2,3,......\}$ $n$ must be of the form $3k, 3k+1$ or $3k+2$ where $k \in N$ when $n$ is multiple of $3$,$1+\omega+\omega^2+\dots+\omega^n = 0$ whence $(1+\omega+\omega^2+\dots+\omega^n)^m = 0$ $\forall$ $n,m \in N$ when $n$ is of the form $3k+1$,$1+\omega+\omega^2+\dots+\omega^n = 1$ whence $(1+\omega+\omega^2+\dots+\omega^n)^m = 1^m=1$ $\forall$ $n,m \in N$ when $n$ is of the form $3k+2$,$1+\omega+\omega^2+\dots+\omega^n = 1+\omega(**)$ whence $(1+\omega+\omega^2+\dots+\omega^n)^m = (1+\omega)^m =(-\omega)^{2m}$ $\forall$ $n,m \in N$ In this case the possible values are possible values of $(-\omega)^{2m}$ which are $-1,1,\omega,-\omega,\omega^2,-\omega^2$, note m varies over the set of Natural numbers Combining the three cases we see that $S = \{0,-1,1,\omega,-\omega,\omega^2,-\omega^2\}$, therefore $|S|=7$. $(**)$ To understand these first observe that $\omega^n=1,\omega,\omega^2$ for any natural number $n$. So when $n$ is of the for $3k+2$ we can couple the three consecutive recurring terms $1,\omega,\omega^2$ to get the sum as $0$. After that we are left with two more terms $1$ and $\omega$, since the series $1+\omega+\omega^2+\omega^3+\omega^4+\omega^5\dots+\omega^n$ has repeated terms after every 3 terms (similar to the first three terms). Similarly for the othere cases the same argument follows. This concludes the result .

Common terms of two A.P Series : Indian Statistical Institute B.Math & B.Stat

Consider the two arithmetic progressions $3,7,11,\dots,407$ and $2,9,16,\dots,709$. Find the number of common terms of these two progressions. Let $a_n$ and $a_m$ be the last terms of the progressions respectively $\Rightarrow 407 = 3+(n-1)4$ and $709 = 2+(m-1)7$ Solving we get, $n,m = 102$. To find the common terms, assume that the $n^{th}$ term of the first progression is equal to the $m^{th}$ term of the second progression. $\Rightarrow 3+(n-1)4=2+(m-1)7 \Rightarrow 3+4n-4-2+7=7m \Rightarrow 4(n+1)=7m,$ where $n,m \in \{1,2,3,\dots,102\}$ R.H.S is a multiple of 7, while L.H.S is 4(n+1). Since $g.c.d(4,7)=1$ L.H.S will be multiple of $7$, $iff$ $n+1$ is a multiple of $7$. $\Rightarrow n = 6,13,20,\dots$ Again since $n$ is bounded by $102$. The largest possible value of $n$ is $97$. So, $n \in \{6,13,20,\dots\,97}$ Which has $14$ terms. Thus the number of common terms of the progression is $14$

Application of Rolle's Theorem

Let $a_0,a_1,a_2$ and $a_3$ be real numbers such that $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\frac{a_3}{4}=0$. Then show that the polynomial $f(x)=a_0+a_1x+a_2x^2+a_3x^3$ has at least one root in the interval $( 0 , 1 )$. Consider the polynomial $g(x) = a_0x+\frac{a_1}{2}x^2+\frac{a_2}{3}x^3+\frac{a_3}{4}x^4$ Clearly $g(0)=0$ and $g(1) = a_0+\frac{a_1}{2}+\frac{a_2}{3}+\frac{a_3}{4} = 0$ { given in the problem } Since $g(x)$ is a polynomial it is continuous in [0,1] and differentiable in (0,1) Thus by Rolle's Theorem, $g'(x) = 0$ for at least one $x \in (0,1)$ $\Rightarrow a_0+a_1x+a_2x^2+a_3x^3 = 0$ for at least one $x \in (0,1)$

Problem from Indian Statistical Institute: B.Stat. (Hons.)

Let $P(x)$ be a polynomial of degree $11$ such that $P(x) = \frac{1}{x+1}$, for $x = 0,1,2,\dots,11.$ Find the value of $P(12)$ Solution: Let $f(x) = (x+1)P(x)-1$, clearly $f(x)$ is a polynomial of degree 12. Now for $x \in \{0,1,2,\dots,11\}$, $f(x) = (x+1)P(x)-1=\frac{x+1}{x+1}-1=1-1=0$ This shows that $f(x)$ vanishes at the points $x = 0,1,2,\dots,11.$ $f(x)$ being of degree 12, the above statement assures that $x = 0,1,2,\dots,11.$ are the possible roots of $f(x)$ Therefore $f(x)=(x+1)P(x)-1=a_{0}(x-0)(x-1)(x-2)\dots(x-11)$ Letting $x=-1$ in the above equality, we have $-1=a_{0}(-1)(-2)(-3)\dots(-12) \Rightarrow a_{0} = \frac{-1}{12!}$ Therefore $f(x)=(x+1)P(x)-1=\frac{-1}{12!}(x-0)(x-1)(x-2)\dots(x-11)$ Now, letting $x=12$ we have $13P(12)-1=\frac{-1}{12!}(12)(11)(10)\dots(1)=\frac{-12!}{12!}=-1$ $\Rightarrow 13P(12)-1=-1 \Rightarrow P(12) = 0$

CBSE Mathematics Solved Paper 2015 : XII

CBSE Mathematics Solved Paper 2015 : XII  

Downlaod File
To go to the download page, click here

Remainder Theorem

The term containing the highest power of $x$ in the polynomial $f(x)$ is $2x^4$. Two of the roots of the equation $f(x)=0$ are -1 and 2. Given that $x^2-3x+1$ is a quadratic factor of $f(x)$, find the remainder when $f(x)$ is divided by $2x-1.$ Since degree of $f(x)$ is $4$ and $x^2-3x+1$ is a factor of $f(x)$, it can be written as product of two quadratics Therefore, $f(x) = ( x^2-3x+1 ) (ax^2+bx+c )$. Again since, $2x^4$ is the leading term $a$, must be equal to $2$ So, $f(x) = ( x^2-3x+1 ) (2x^2+bx+c )$. Given $-1$ and $2$ are the roots of $f(x)$ $\Rightarrow$ $f(-1)=0$ & $f(2)=0$ Note that $x^2-3x+1$ does not vanishes at $x = -1 , 2$ $\Rightarrow$ $2x^2+bx+c$ must vanishes at this two points. $\Rightarrow 2-b+c = 0 , 8+2b+c=0$ Solving the equations we get $b=-2, c=-4$ Thus $f(x) = ( x^2-3x+1 ) (2x^2-2x-4 )$. Required remaninder is $f(\frac{1}{2}) = \frac{9}{8}$

Indian Statistica Institute : Special Integration

Evaluate: $\int_{-n}^n max \{x+|x|,x-[x]\} \mathrm{d}x$ where $[x]$ is the floor function When $x > 0$ we see that $x+|x| = 2x$. Let $k \leq x < k+1, k = 0,1,2,.........,n-1$ $\Rightarrow [x] = k, therefore, x-[x] = x-k$ $So, x+|x| = 2x > x - k = x-[x] \Rightarrow max \{x+|x|,x-[x]\} = 2x$ $\int_{0}^n max \{x+|x|,x-[x]\} \mathrm{d}x = 2 \int_{0}^n x \mathrm{d}x = x^2|_{0}^{n^2} = n^2 \dots (A)$ When $x < 0$ we see that $x+|x| = x-x = 0$. Let $-(k+1) \leq x < -k, k = 0,1,2,.........,n-1$ $\Rightarrow [x] = -(k+1), therefore, x-[x] = x+(k+1)$ $So, x+|x| = 0 < x + (k+1) = x-[x] \Rightarrow max \{x+|x|,x-[x]\} = x-[x]$ $\int_{-(k+1)}^{-k} max \{x+|x|,x-[x]\} \mathrm{d}x = \int_{-(k+1)}^{-k} x+(k+1) \mathrm{d}x$ $\int_{-n}^0 max \{x+|x|,x-[x]\} \mathrm{d}x$= $\int_{-n}^{-(n-1)} x+(k+1) \mathrm{d}x + \int_{-(n-1)}^{-(n-2)} x+(k+1) \mathrm{d}x +\dots+ \int_{-1}^{0} x+(k+1) \mathrm{d}x= \sum_{k=0}^{n-1} \int_{-(k+1)}^{-k} x+(k+1) \mathrm{d}x$ $= \int_{-n}^0 x \mathrm{d}x+\sum_{k=0}^{n-1} \int_{-(k+1)}^{-k} (k+1) \mathrm{d}x = {{-n^2} \over {2}}+\sum_{k=0}^{n-1} (k+1) x|_{-(k+1)}^{-k} = {{-n^2} \over {2}}+\sum_{k=0}^{n-1} (k+1) = {{-n^2} \over {2}} + {{n(n+1) \over {2}}} \dots (B)$ Adding A and B we have, $\int_{-n}^0 max \{x+|x|,x-[x]\} \mathrm{d}x + \int_{0}^n max \{x+|x|,x-[x]\} \mathrm{d}x =$ $\int_{-n}^n max \{x+|x|,x-[x]\} \mathrm{d}x = {{-n^2} \over {2}} + {{n(n+1) \over {2}}} + n^2 = {{n^2} \over {2}} + {{n(n+1) \over {2}}}$

JEE MAIN MATHEMATICS SOLVED PAPER 2015

JEE (MAIN)-2015

MATHEMATICS

Important Instructions:

1.  The test is of 3 hours duration.

 2. The Test Booklet consists of 90 questions. The maximum marks are 360.

 3. There are three parts in the question paper A, B, C consisting of Mathematics, Physics and

Chemistry having 30 questions in each part of equal weightage. Each question is allotted 4 (four)

marks for each correct response.

 4. Candidates will be awarded marks as stated above in Instructions No. 3 for correct response of each question. ¼ (one-fourth) marks will be deducted for indicating incorrect response of each question.

No deduction from the total score will be made if no response is indicated for an item in the answer

sheet.

5. There is only one correct response for each question. Filling up more than one response in each

question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 4 above.

6. Use Blue/Black Ball Point Pen only for writing particulars/marking responses on Side-1 and Side-2

of the Answer Sheet. Use of pencil is strictly prohibited.

7. No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc. except the Admit Card inside the examination room/hall.

8. The CODE for this Booklet is B. Make sure that the CODE printed on Side-2 of the Answer Sheet

and also tally the serial number of the Test Booklet and Answer Sheet are the same as that on this

booklet. In case of discrepancy, the candidate should immediately report the matter to the Invigilator

for replacement of both the Test Booklet and the Answer Sheet.

Click here to download  2015

JEE MAIN MATHEMATICS SOLVED PAPER 2016