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  

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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

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

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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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