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 $Sandwhich-theorem$ $\lim_{n \to \infty} \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}} = a_1$