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 $f(\frac{1}{2}) = \frac{9}{8}$