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

for replacement of both the Test Booklet and the Answer Sheet.

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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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