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

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

Click here to download  2015

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

Get the solved ISC Mathematics Question paper for the year 2015. The solutions are short need not to be considered as the way of answering in board exams. Click Below

ISC MATHEMATICS PAPER 2015

ISC MATHEMATICS PAPER 2016 (SOLVED)

Solved CBSE Mathematics Question paper 2015. 

JEE MAIN 2016 MATHEMATICS SOLVED PAPER

JEE MAIN 2015 MATHEMATICS SOLVED PAPER

Integration

Integration Special Methods

Bayes' Theorem: Solved Problems for CBSE and ISC Students

Solved Problems on Bayes' Theorem for CBSE and ISC Students

These problems will help you prepare for Board Examination and several other competitive exams. All possible variations of Bayes' theorem are present  in these problems. If you have any query don't forget to comment or whatsapp me on +91-9038126497

Problem 1. Prasad is known to speak the truth 3 times out of 5. He throws a die and reports that it is 1. Find the probability that it is actually 1.

Problem 2. Pawan speaks the truth 8 times out of 10 times. A die is tossed. He reports that it was 5. What is the probability that it was actually 5?

Problem 3. A bag contains 1 white and 6 red balls, and a second bag contains 4 white and 3 red balls. One of the bags is picked up at random and a ball is randomly drawn from it, and is found to be white in colour. Find the probability that the drawn ball was from the first bag.

Problem 4. There are two bags I and II. Bag I contains 3 white and 4 black balls and Bag II contains 5 white and 6 black balls. One ball is drawn at random from one of the bags and is found to be white. Find the probability that it was drawn from bag I.

Problem 5. There are two bags I and II. Bag I contains 3 white and 2 red balls and Bag II contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag II.

Problem 6. A company has two plants to manufacture motor cycle. 70% motor cycles are manufactured at the first plant, while 30% are manufactured at the second plant. At the first plant, 80% motor cycles are rated of the standard quality while at the second plant, 90% are rated of standard quality. A motor cycle , randomly picked up, is found to be of standard quality. Find the probability that it has come out from the second plant.

Problem 7. A company has two plants to manufacture bicycles. 60% bicycles are manufactured at the first plant, while 40% are manufactured at the second plant. At the first plant, 80% motor cycles are rated of the standard quality while at the second plant, 90% are rated of standard quality. A bicycle , randomly picked up, is found to be of standard quality. Find the probability that it has come out from the second plant.

Problem 8. A insurance company insured 2000 scooters and 3000 motor cycles. The probability of an accident involving a scooter is 0.01 and that of a motor cycle is 0.02. An insured vehicle met with an accident. Find the probability that the accident vehicle was a motor cycle.

Problem 9. Three bags contain balls as shown in the following table: A bag is chosen at random and two balls are drawn. They happen to be white and red. What is the probability that they came from the third bag?

Bag	White Balls	Black Balls	Red Balls
I	1	2	3
II	2	1	1
III	4	3	2

 

Problem 10. Three urns  A,B and C contains 6 red and 4 white balls; 2 red and 6 white balls; and 1 red and 5 white balls respectively. An urn is chosen at random and a ball is drawn. If the ball drawn is found to be red, find the probability that the ball was drawn from urn A.

Problem 11. A factory has three machines A,B and C, which produce 100,200 and 300 items of a particular type daily. The machines produce 2%, 3% and 5% defective items respectively. One day when the production was over, an item was picked up randomly and found to be defective. Find the probability that it is manufactured by machine A.

Problem 12. In a bolt factory, machines A,B and C manufacture 25%,35% and 40% respectively of total bolts. Of their outputs, 5%, 4% and 2% respectively are defective bolts. A bolt is drawn at random and is found to be defective. Find the probability that it is manufactured by machine B.

Problem 13. A bag contains 4 red and 4 black balls, and a second bag contains 2 red and 6 black balls. One of the bags is picked up at random and a ball is randomly drawn from it, and is found to be red in colour. Find the probability that the drawn ball was from the first bag.

Problem 14. Probability that Ravi speaks truth is 4/5 . A coin is tossed, Ravi reports that a head appears. Find the probability that actually there was head.

Problem 15. There are three coins. One is a two headed coin, another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Problem 16. Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

Problem 17. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars ( not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their examination. One student is chosen at random and he has an A grade, what is the probability that the student is from hostel?

Problem 18. A laboratory COVID-19 test is 99% effective in detecting when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e., if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1% of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Problem 19. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Problem 20. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1,2,3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1,2,3 or 4 with the die?

Problem 21. Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second group will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Problem 22. A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on for 30% and C is on for 20% of the time. A defective item is produced, what is the probability it was produced by A?

Problem 23. In answering a question on a multiple choice test, a student either knows the answer or guesses. Let ¾ be the probability that he knows the answer and ¼ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability ¼ . What is the probability that the student knows the answer given that he answered it correctly?

Problem 24. There are two bags I and II. Bag I contains 3 white and 4 red balls and Bag II contains 5 white and 6 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag II.

Problem 25. There are two bags I and II. Bag I contains 2 white and 4 red balls and Bag II contains 5 white and 3 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag II

Problem 26. There are two bags I and II. Bag I contains 3 red and 4 black balls and Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag II.

Problem 27. Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also gold?

Problem 28. A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, and scooter or by other means of transport are respectively. The probability that he will be late are if he comes by train bus and scooter respectively. If he comes by other means of transport, then he will not be late. On a certain day, he arrived late. What is the probability that he came by train?

Problem 29. Suppose the reliability of a HIV test is specified as follows: Of people having HIV, 90% of the test detect the disease but 10% goes undetected (false negative). Of people free of HIV, 99% of the test are judged HIV negative but 1% are diagnosed as showing HIV positive (false positive). From a large population of which only 0.1% have HIV, one person is selected at random, given the HIV test, and the pathologist reports him/her as HIV positive. What is the probability that the person actually has HIV?

Problem 30. If a machine is correctly set up, it produces 90% acceptable items. If it is incorrectly set up, it produces only 40% acceptable items. Past experience shows that 80% of the setups are correctly done. If after a certain set up, the machine produces 2 acceptable items , find the probability that the machine is correctly set up.

Problem 31. Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%.  At a time a patient can choose any one of the two options with equal probabilities.  It is given that after going through one of the two options the patient selected at random suffers a heart attack.  Find the probability that the patient followed a course of meditation and yoga?

Problem 32. Vartika has an alarm which will ring at the appointed time with probability 0.9. If the alarm rings, it will awake her and she will reach the examination hall in time with probability 0.8. If the alarm doesn't ring , she will get up on her own to reach the examination hall in time, with probability 0.3. Knowing that Vartika reached the hall in time, find the probability that the alarm rang.

Problem 33. By examining the chest X−ray probability that T.B is detected when a person is actually suffering is 0.99. The probability that the doctor diagnoses incorrectly that a person has T.B on the bases of X−ray is 0.001. In a certain city 1 in 1000 person suffers from T.B. A person is selected at random is diagnoses to have T.B. What is the chance that the actually has T.B? 

Problem 34. Shoes are produced by two machines A and B. 50% of the shoes are produced by machine A with an estimate of 10% of them being defective. On machine B, 20% of the shoes produced are defective, if a shoe taken at random is found to be defective , what is the probability that shoe was produced by machine A?

Problem 35. In a large company, 15% of the employees are graduates, and of these, 80% work in administrative posts. Of the non-graduates employees of the company, 10% work in administrative posts. Find the probability that an employee of this company selected at random from those working in administrative posts will be graduate.

Problem 36. A pack or cards is counted with face downwards. It is found that one card is missing. One card is drawn and is found to be red. Find the probability that the missing card is red.

Problem 37. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both spades. Find the probability of the lost card being a spade.

Problem 38. For A,B and C the chances of being selected as the manager of a firm are in the ratio 4:1:2 respectively. The respective probabilities for them to introduce a radical change in marketing strategy are 0.3, 0.8 and 0.5. If the change does take place, find the probability that it is due to the appointment of B or C.

Problem 39. Two urns I and II contain respectively 3 white and 2 black balls, 2 white and 4 black balls. One ball is transferred from urn I to urn II and then one is drawn from the latter. It happens to be white. What is the probability that the transferred ball was white.

Problem 40. A bin contains 3 different types of disposable flashlights. The probability that a type 1 flashlight will give over 100 hours of use is 0.7, with the corresponding probabilities for type 2 and type 3 flashlights being 0.4 and 0.3 respectively. Suppose that 20% of the flashlights in the bin are type 1, 30% are type 2, and 50% are type 3.

(a)    What is the probability that a randomly chosen flashlight will give more than 100 hours of use?

(b)   Given the flashlight lasted over 100 hours, what is the conditional probability that it was a type 2 flashlight?

 Here is the solution to all the problems above:

PDF FILES

SOLVED PROBLEMS PROBABILITY DISTRIBUTION

SOLVED PROBLEMS TOTAL THEOREM OF PROBABILITY

SOLVED PROBLEMS BAYES' THEOREM

Geometry Inequality

Post by Maths.

Prove that in any quadrilateral, the sum of the diagonals is greater than the half of its perimeter.

Consider the quad. In the above diagram. Let E be the point of the intersection.

Now, AE+EB > AB

EB+EC > BC

AE+ED > AD

EC+ED > DC (Using Triangle Inequality)

Adding the above four inequalities we get

2(AE+EC+EB+ED) > AB+BC+AD+DC

=> AC + BD > ½(AB+BC+AD+DC)

Thus sum of the diagonals is greater than the half of its perimeter Q.E.D

In any triangle four times the sum of its medians is greater than 3 times its perimeter.

We know that difference of any two sides of a triangle is less than the third side (prove it)

In triangle ABE,

AE > AB-BE

In triangle ACE,

AE > AC-CE

Adding above two inequalities we get,

2AE > AB + AC -(BE+CE)

=> AE > ½(AB+AC-BC)

=> 4AE > 2(AB+AC-BC).........(1)

Similarly,

4BD > 2(AB+BC-AC).............(2) and 4CF > 2(AC+BC-AB)........(3)

Adding (1),(2) and (3) we have,

4(AE+BD+CF) > 2(AB+AC-BC+AB+BC-AC+AC+BC-AB)

=> 4(AE+BD+CF) > 2(AC+AB+BC)

=> sum of the lengths of the medians is greater than half the perimeter

We can strengthen the inequality by using the fact that the point 'O' divides the medians AE,BD,CF internally in the ration 2:1

Therefore, OD:OB = 1:2

=> (OB+OD):OB = (1+2):2

=> BD:OB=3:2

=> OB = 2/3 BD........(a)

Similarly, OC = 2/3 CF.........(b) and OA = 2/3 AE.......(c)

Now in triangle OBC, OB+OC> BC

=>2/3(BD+CF)>BC [using (a) and (b)]

=> 2(BD+CF)>3BC

Similarly, 2(CF+AE)>3AC and 2(BD+AE)>3AB

Adding the last three inequalities we get 4(AE+BD+CF) > 3(AB+BC+CA)

In the triangle ABC, AE,BD and CF are the medians where O is the point of there intersection

Trigonometrical Identities CBSE

Solved problems for C.B.S.E and I.C.S.E on Trigonometrical Identities

Here is the link : http://kolkatamaths.yolasite.com/ix-x.php