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

Have you ever looked at a polynomial and wondered if it has a root in a specific interval, without actually having to solve it?

Calculus gives us an incredibly elegant workaround for this exact scenario: Rolle's Theorem. Let's walk through a classic problem that demonstrates how we can find hidden roots by stepping up one level of abstraction.

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The Problem

Suppose we have four real numbers \(a_0, a_1, a_2,\) and \(a_3\) that satisfy the following condition:

$$a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \frac{a_3}{4} = 0$$

The Goal: Prove that the polynomial \(f(x) = a_0 + a_1x + a_2x^2 + a_3x^3\) has at least one root in the interval \((0, 1)\).

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The Intuition: Where do we start?

When asked to prove that a function \(f(x)\) equals zero at some point, a calculus student's first instinct should be to look for an antiderivative.

Why? Because Rolle's Theorem tells us that if a continuous, differentiable function starts and ends at the exact same height on a graph (for instance, if it equals \(0\) at both ends of an interval), it must flatten out somewhere in between. At that flat point, its derivative is exactly zero.

So, if we can find a parent function \(g(x)\) such that \(g'(x) = f(x)\), and we can show that \(g(x)\) is tied down at \(x=0\) and \(x=1\), the theorem will do the heavy lifting for us.

---

The Proof

Step 1: Construct the parent function, \(g(x)\).
To find our function \(g(x)\), we simply integrate \(f(x)\) with respect to \(x\):

$$g(x) = \int (a_0 + a_1x + a_2x^2 + a_3x^3) \,dx$$
$$g(x) = a_0x + \frac{a_1}{2}x^2 + \frac{a_2}{3}x^3 + \frac{a_3}{4}x^4$$

(Note: We can ignore the constant of integration \(+C\) here by simply setting it to \(0\), as we just need one valid parent function).

Step 2: Evaluate the endpoints of our interval \([0, 1]\).
Let's plug our bounds into \(g(x)\) to see how the function behaves at the edges.

At \(x = 0\):

$$g(0) = a_0(0) + \frac{a_1}{2}(0)^2 + \frac{a_2}{3}(0)^3 + \frac{a_3}{4}(0)^4 = 0$$

At \(x = 1\):

$$g(1) = a_0(1) + \frac{a_1}{2}(1)^2 + \frac{a_2}{3}(1)^3 + \frac{a_3}{4}(1)^4$$
$$g(1) = a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \frac{a_3}{4}$$

Notice something familiar? The problem explicitly states that \(a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \frac{a_3}{4} = 0\). Therefore:

$$g(1) = 0$$

Step 3: Apply Rolle's Theorem.
We now have a function \(g(x)\) where:

1. \(g(x)\) is a polynomial, meaning it is perfectly continuous on the closed interval \([0,1]\) and differentiable on the open interval \((0,1)\).
2. \(g(0) = g(1) = 0\).

Because both endpoints are at the same \(y\)-value, the graph of \(g(x)\) must turn around at least once between \(x=0\) and \(x=1\).

By Rolle's Theorem, there must exist at least one point \(c\) in the interval \((0, 1)\) where the derivative is zero:

$$g'(c) = 0$$

Since \(g'(x)\) is just our original polynomial \(f(x)\), this means:

$$f(c) = a_0 + a_1c + a_2c^2 + a_3c^3 = 0$$

Conclusion:

We have successfully proven that \(f(x)\) has at least one root exactly where we wanted it: inside the interval \((0, 1)\).

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

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

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

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

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

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JEE MAIN MATHEMATICS SOLVED PAPER 2016

Counting

If the integers \(m\) and \(n\) are chosen at random from \(1\) to \(100\), then find the probability that the number of the form \(7^n+7^m\) is divisible by \(5\). $$ .$$ Number of numbers of the form \(7^n+7^m\) is 100×100= \(100^2\) , since m,n ∈{1 ,…………,100}. Now \(7^1=7,7^2=49,7^3=343\) and \(7^4=2401\) , the digits at the unit places are 7,9,3 and 1. $$ .$$ For n ≥5 the digit at the unit place for the number \(7^n\) will be one of among the numbers 1,3,7 and 9 $$ .$$ A number is divisible by 5 iif the digit at the unit’s place is either 0 or 5. Therefore a number of the form \(7^n+7^m\) will be divisible by 5 iff the digits at the unit place of them add up to 10. The possible cases being {1,9} and {3,7} in any order. $$ .$$ Since there are 4 distinct digits at the unit place for the number \(7^n\) and they are repeated modulo 4, we have 100/4=25 distinct number of the form \(7^n\) having the digit 1 or 3 or 7 or 9 at the unit place. $$ .$$ When \(7^n\) has the digit 1 at the unit place, \(7^m\) need to have the digit 9 at the unit place giving 25 such choices. In total 2 x 25 x 25 choices ( n and m can be interchanged!) $$ .$$ Similarly for the pair {3,7} we have \( 2 \times 25 \times 25 \) choices, in total giving \( 4 \times 25 \times 25 \) choices!! $$ .$$ Required probability $$ = {4 \times 25 \times 25 \over 100 \times 100} = {1 \over 4} .$$

ISC Mathematics Question Paper Solved 2015

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Integration

Integration Special Methods