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

Problem 1: Points on a Curve

Question: 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 \( xy=27027 \).

Solution:

First, note the prime factorization of the number: \( 27027 = 3^3 \times 13^1 \times 11^1 \times 7^1 \).

Since \( x \) and \( y \) must be positive integers, for every chosen positive integer \( x \) that divides 27027, there is exactly one corresponding integer \( y \) (where \( y = \frac{27027}{x} \)). Therefore, the number of good points is simply equal to the total number of positive divisors of 27027.

Using the divisor function formula \( \tau(n) \), if \( n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k} \), the number of divisors is \( (a_1 + 1)(a_2 + 1) \dots (a_k + 1) \).

Applying this to our factorization:

\[ \tau(27027) = (3+1)(1+1)(1+1)(1+1) = 4 \times 2 \times 2 \times 2 = 32 \]

There are 32 good points on the curve.

Problem 2: Ordered Triplets

Question: What is the number of ordered triplets \( (a, b, c) \), where \( a, b, c \) are positive integers (not necessarily distinct), such that \( abc = 1000 \)?

Solution:

First, find the prime factorization of 1000, which is \( 2^3 \times 5^3 \).

Any ordered triplet \( (a, b, c) \) satisfying the condition must be of the form:

• \( a = 2^{l_1} \times 5^{m_1} \)
• \( b = 2^{l_2} \times 5^{m_2} \)
• \( c = 2^{l_3} \times 5^{m_3} \)

For \( abc = 2^3 \times 5^3 \) to hold true, the powers of each prime base must add up to 3. This gives us two independent equations:

1. \( l_1 + l_2 + l_3 = 3 \)
2. \( m_1 + m_2 + m_3 = 3 \)

We need to find the number of non-negative integer solutions for each equation. Using the "Stars and Bars" combinatorics method, the number of solutions to \( x_1 + x_2 + x_3 = n \) is given by \( \binom{n+r-1}{r-1} \), where \( r \) is the number of variables.

For the powers of 2: \( \binom{3+3-1}{3-1} = \binom{5}{2} = 10 \) solutions.

For the powers of 5: \( \binom{3+3-1}{3-1} = \binom{5}{2} = 10 \) solutions.

Since the distributions of the powers of 2 and 5 are independent, we multiply the possibilities:

\[ 10 \times 10 = 100 \]

Thus, in total 100 ordered triplets are possible.

Problem 3: Derangements and Boxes

Question: 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 into their corresponding numbered boxes.

Solution:

First, we must choose which 4 balls will go into their correct boxes. This can be done in \( \binom{8}{4} \) ways. For any such selection—say we choose \( \{Ball_2, Ball_5, Ball_3, Ball_7\} \)—there is exactly 1 way for them to go into their corresponding boxes.

Now, the remaining 4 balls—in this example, \( \{Ball_1, Ball_4, Ball_6, Ball_8\} \)—must be placed into the remaining 4 boxes such that none of them end up in a box matching their own number. This is the classic definition of a derangement.

The number of ways to derange \( n \) objects is denoted as \( D_n \). The formula for a derangement is:

\[ D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + (-1)^n\frac{1}{n!} \right) \]

For our remaining 4 balls, we calculate \( D_4 \):

\[ D_4 = 4! \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right) = 24 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right) = 12 - 4 + 1 = 9 \]

The total number of valid configurations is the product of our choices:

\[ \text{Total Ways} = \binom{8}{4} \times D_4 = 70 \times 9 = 630 \]

```

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 \(S = \{(1+\omega+\omega^2+\dots+\omega^n)^m \mid m,n = 1,2,3,\dots\}\)

Note that the sum \(1+\omega+\omega^2+\dots+\omega^n\) contains \(n+1\) terms. The power \(n\) must be of the form \(3k\), \(3k+1\) or \(3k+2\) where \(k \in \mathbb{N} \cup \{0\}\).

When \(n\) is of the form \(3k+2\), there are \(3k+3\) terms (a multiple of 3). Because \(1+\omega+\omega^2 = 0\), the sum evaluates to \(0\). Whence \((1+\omega+\omega^2+\dots+\omega^n)^m = 0^m = 0\) \(\forall m \in \mathbb{N}\).

When \(n\) is a multiple of 3 (i.e., \(3k\)), there are \(3k+1\) terms. The sum evaluates to \(1\). Whence \((1+\omega+\omega^2+\dots+\omega^n)^m = 1^m = 1\) \(\forall m \in \mathbb{N}\).

When \(n\) is of the form \(3k+1\), there are \(3k+2\) terms. The sum evaluates to \(1+\omega = -\omega^2\). Whence \((1+\omega+\omega^2+\dots+\omega^n)^m = (-\omega^2)^m = (-1)^m \omega^{2m}\) \(\forall m \in \mathbb{N}\).

In this final case, as \(m\) varies over the natural numbers, the expression \((-1)^m \omega^{2m}\) generates 6 distinct values: \(-\omega^2, \omega, -1, \omega^2, -\omega, 1\).

Combining the three cases, we see that \( S = \{0, -1, 1, \omega, -\omega, \omega^2, -\omega^2\} \), therefore \(|S|=7\).

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  

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