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I.S.I. and C.M.I. Entrance

Sign change | ISI-B.stat | Objective Problem 709

Try this beautiful problem on Sign change, useful for ISI B.Stat Entrance.

Sign change | ISI B.Stat Entrance | Problem 709


In the interval \((-2\pi, 0)\) the function \(f(x) = sin(1/x^3)\)

  • (a) never changes sign
  • (b) changes sign only once
  • (c) changes sign more than once, but a finite number of times
  • (d) changes sign infinite number of times

Key Concepts


Calculus

Limit

Trigonometry

Check the Answer


Answer: (d)

TOMATO, Problem 709

Challenges and Thrills in Pre College Mathematics

Try with Hints


As x becomes \(\leq 1\) and tends to zero then it crosses \(\pi, 2\pi, 3\pi, ….\).can you draw the graph?

Can you now finish the problem ……….

Sign change - graph

If we draw the graph then we can see that the function \(f(x) = sin(1/x^3)\) crosses many times. Therefore number of sign changes is infinite.

Therefore option \((d)\) is correct…..

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I.S.I. and C.M.I. Entrance

Limit Problem | ISI-B.stat | Objective Problem 694

Try this beautiful problem on Limit, useful for ISI B.Stat Entrance.

Limit Problem | ISI B.Stat Entrance | Problem 694


Let \(a_1 = 1\) and \(a_n = n(a_{n-1} + 1)\) for \(n = 2, 3, ….\) Define \(P_n = (1 +1/a_1)(1 + 1/a_2)….(1 + 1/a_n)\). Then \(\lim\limits_{x \to \infty} {P_n}\)?

  • (a) \(1+e\)
  • (b) \(e\)
  • (c) \(1\)
  • (d) \(\infty\)

Key Concepts


Calculus

Limit

Trigonometry

Check the Answer


Answer: (b)\(e\)

TOMATO, Problem 709

Challenges and Thrills in Pre College Mathematics

Try with Hints


Given that \(P_n = (1 +1/a_1)(1 + 1/a_2)….(1 + 1/a_n)\)

Therefore \(P_n=\frac{a_1 +1}{a_1}.\frac{a_2 +1}{a_2}.\frac{a_3 +1}{a_3}…..\frac{a_n +1}{a_n}\)

Now \(a_n = n(a_{n-1} + 1)\)

Put \(n=2\), we will get \(a_1+1=\frac{a_2}{2}\)

\(a_2+1=\frac{a_3}{3}\)……………….

………………………..

…………………………

\(a_n+1=\frac{a_n}{n}\)

Therefore \(P_n=\frac{a_1 +1}{a_1}.\frac{a_2 +1}{a_2}.\frac{a_3 +1}{a_3}…..\frac{a_n +1}{a_n}\)

\(\Rightarrow {P_n}= \frac{a_2}{2a_1}.\frac{a_3}{3a_2}.\frac{a_4}{4a_3}……..\frac{a_{n+1}}{(n+1).{a_n}}\)

\(\Rightarrow {P_n}=\frac{a_{n+1}}{{a_1}\{2.3.4………..(n+1)\}}\)

\(\Rightarrow {P_n}=\frac{a_{n+1}}{\{1.2.3.4………..(n+1)\}}\) (as \(a_1=1\))

\(\Rightarrow {P_n}=\frac{a_{n+1}}{(n+1)!}\)

\(\Rightarrow {P_n}=\frac{(n+1)(a_n +1)}{(n+1)!}\)

\(\Rightarrow {P_n}=\frac{(a_n +1)}{n!}\)

\(\Rightarrow {P_n}=\frac{a_n}{n!} +\frac{1}{n!}\)

\(\Rightarrow {P_n}=\frac{n(a_{n-1}+1)}{n!}+\frac{1}{n!}\)

\(\Rightarrow {P_n}=\frac{a_{n-1}+1}{(n-1)!}+\frac{1}{n!}\)

\(\Rightarrow {P_n}=\frac{a_{n-1}}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!}\)

\(\Rightarrow {P_n}=\frac{a_2}{2!}+\frac{1}{2!}+\frac{1}{3!}+…….+\frac{1}{n!}\)

\(\Rightarrow {P_n}=\frac{2(1+1)}{2!}+\frac{1}{2!}+….+\frac{1}{n!}\)

\(\Rightarrow {P_n}=1+\frac{1}{1!}+…..+\frac{1}{n!}\)

Can you now finish the problem ……….

Now we have to find out \(\lim\limits_{x \to \infty} {P_n}\)

we know that \(e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+……….+\infty\)

So,\(e^1=1+\frac{1}{1!}+\frac{1^2}{2!}+……….+\infty\)

\(\lim\limits_{x \to \infty} {P_n}\)=\(1+\frac{1}{1!}+\frac{1^2}{2!}+……….+\infty\)

\(\lim\limits_{x \to \infty} {P_n}\)=\(e\)

Therefore option (b) is correct…..

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I.S.I. and C.M.I. Entrance

Graph in Calculus | ISI-B.stat | Objective Problem 699

Try this beautiful problem on Graph in Calculus, useful for ISI B.Stat Entrance.

Graph in Calculus | ISI B.Stat Entrance | Problem 699


The adjoining figure is the graph of

graph
  • (a) \(y = 2e^x\)
  • (b) \(y = 2e^-x\)
  • (c) \(y = e^x + e^-x\)
  • (d) \(y = e^x – e^-x + 2\)

Key Concepts


Calculus

Graph

Functions

Check the Answer


Answer: (C)

TOMATO, Problem 699

Challenges and Thrills in Pre College Mathematics

Try with Hints


Griaph in calculus

At first we have to check all the options…..

  1. For \(y = 2e^x\), this is not possibe because \(y = 2e^x\) expression represent only positive side’s of x axis
  2. For \(y = 2e^-x\),This expression also represents one side of x-axis
  3. For \(y = e^x + e^-x\).This expression represents both sides of the x-axise.as we put \(x=0\) then \(y=2\) i.e the curve not starts from the origin
  4. \(y = e^x – e^-x + 2\), this expression also represents both sides of the x-axis

Can you now finish the problem ……….

Therefore ,the correct ans is (c)

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I.S.I. and C.M.I. Entrance

Negative & Positive Roots | ISI-B.stat | Objective Problem 708

Try this beautiful problem on Negative & Positive Roots, useful for ISI B.Stat Entrance.

Negative & Positive Roots | ISI B.Stat Entrance | Problem 708


If n stands for the number of negative roots and p for the number of positive roots of the equation \(e^x = x\), then

  • (a) \(n = 1\), \(p = 0\)
  • (b) \(n = 0\), \(p = 1\)
  • (c) \(n = 0\), \(p > 1\)
  • (d) \(n = 0\), \(p = 0\)

Key Concepts


Calculus

Limit

Trigonometry

Check the Answer


Answer: (d)

TOMATO, Problem 708

Challenges and Thrills in Pre College Mathematics

Try with Hints


Negative & Positive Roots- graph

Draw the figure of given equation \(e^x = x\) and check all the points that are in the options…..

Can you now finish the problem ……….

Negative & Positive Roots - graph

The given equation is \(e^x = x\)

as \(x \)increases exponential function increases more rapidly than any polynomial function say \(x\) . Hence from graph we can say they never intersect . So, there is no solution for the equation . Therefore \(n=0\) and \(p=0.\)

Therefore option \((d)\) is correct…..

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I.S.I. and C.M.I. Entrance

Graphs in Calculus | ISI-B.stat | Objective Problem 698

Try this beautiful problem on Graphs in Calculus, useful for ISI B.Stat Entrance.

Graphs in Calculus | ISI B.Stat Entrance | Problem 698


Four graphs marked G1, G2, G3 and G4 are given in the figure which are graphs of the four functions \(f_1(x) = |x – 1| – 1, f_2(x) = ||x –1| – 1|, f_3(x) = |x| – 1, f_4(x) = 1 – |x|\), not necessarily in the correct order.The correct order is

graph in calculus 1
graph in calculus 2
graph 3
graph 4
  • (a) \(G_2, G_1, G_3, G_4\)
  • (b) \(G_3, G_4, G_1, G_2\)
  • (c) \(G_2, G_3, G_1, G_4\)
  • (d) \(G_4, G_3, G_1, G_2\)

Key Concepts


Calculus

Graph

Functions

Check the Answer


Answer: (C)

TOMATO, Problem 698

Challenges and Thrills in Pre College Mathematics

Try with Hints


We take the each functions and express it in intercept form.we expand the mod i,e take the value once positive and once negetive .so we will get two equations and solve them,we will get the intersecting point also and draw the graph……..

Can you now finish the problem ……….

\(f_1(x) = |x – 1| – 1\)

\((x-1)-1=y\)

\(\Rightarrow x-y=2\)……………..(1)

\(\Rightarrow \frac{x}{2} +\frac{y}{-2}=0\)\(\Rightarrow (2,0) ,(0,-2)\)

And

\(-(x-1)-1=y\)

\(\Rightarrow x+y=0\)……..(2)

\(\frac{x}{1}+\frac{y}{1}=0\)\(\Rightarrow (1,0),(0,1)\)

Now if we draw the graph of (1) & (2) we will get the figure \(G_2\) and the intersecting point is \((1,-1)\)

Similarly we can draw the graphs for other functions………….

The second function is \(f_2(x) = ||x –1| – 1|\) i.e \(x-y=2\),\(x=y\),\(x+y=1\)…which represents the two figure as given in \(G_3\).

The third function \(f_3(x) = |x| – 1\) which gives \(x-y=1\) & \(x+y=-1\)..if we solve this two equations as first function then we will get \(G_1\)

The third function will gives the \(G_4\) graph

Similarly we will draw the graph for all given functions….

Therefore ,the correct ans is (c)

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I.S.I. and C.M.I. Entrance

Problem on Calculus | ISI-B.stat | Objective Problem 696

Try this beautiful problem on Calculus, useful for ISI B.Stat Entrance.

Problem on Calculus | ISI B.Stat Entrance | Problem 696


If k is an integer such that lim \(\{{cos}^n(k\pi/4) – {cos}^n(k\pi/6)\} = 0\),
then

  • (a) k is divisible neither by 4 nor by 6
  • (b) k must be divisible by 12, but not necessarily by 24
  • (c) k must be divisible by 24
  • (d) either k is divisible by 24 or k is divisible neither by 4 not by 6

Key Concepts


Calculus

Limit

Trigonometry

Check the Answer


Answer: (d)

TOMATO, Problem 694

Challenges and Thrills in Pre College Mathematics

Try with Hints


There are four options ,at first we have to check each options…..

If k is divisible by 24 then cos(kπ/4) = cos(kπ/6) = 1
\(\Rightarrow\) The limit exists and equal to RHS i.e. 0
If k is not divisible by 4 or 6 then cos(kπ/4), cos(kπ/6) both <1

Can you now finish the problem ……….

Therefore ,

lim cosn(kπ/4), cosn(kπ/6) = 0. so we may say that
\(\Rightarrow \)The equation holds.

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I.S.I. and C.M.I. Entrance Math Olympiad

2016 ISI Objective Solution Problem 1

Problem

The polynomial \(x^7+x^2+1\) is divisible by

    • (A) \(x^5-x^4+x^2-x+1\)             (B) \(x^5-x^4+x^2+1\)
    • (C)   \(x^5+x^4+x^2+x+1\)          (D)   \(x^5-x^4+x^2+x+1\)

.
Also Visit: I.S.I. & C.M.I Entrance Program


Understanding the Problem:

The problem is easy to understand right?

STOP

Before scrolling down, your first task is to try the problem YOURSELF.

All the best

We will guide you along a short path to the solution in a step by step approach.

Hint 1: 
Try to factorize \(x^7 + x^2 + 1\).

Hint 2: 
Observe that  \(\omega\) and \(\omega^2 \) are the roots of \(x^7 + x^2 + 1\).

Hint 3:
\(x^7 + x^2 + 1\) =  (\(x^2 + x + 1\)).(\(x^5 – x^4 + x^2 -x + 1\))

A shorter solution or approach can always exist. Think about it. If you find an alternative solution or approach, mention it in the comments. We would love to hear something different from you.

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I.S.I. and C.M.I. Entrance

Parity of the terms of a sequence | Tomato Problem 7

Try this problem from TOMATO Problem 7 based on the Parity of the terms of a sequence.

Problem: Parity of the terms of a sequence

If \( a_0 = 1 , a_1 = 1 \) and \( a_n = a_{n – 1} a_{n – 2} + 1 \) for \( n > 1 \), then:

(A) \(a_{465} \) is odd and \(a_{466} \) is even;
(B) \(a_{465} \) is odd and \(a_{466} \) is odd;
(C) \(a_{465} \) is even and \(a_{466} \) is even;
(A) \(a_{465} \) is even and \(a_{466} \) is odd;

Discussion:

First we note a pattern and then we prove that the pattern actually holds.

Note that:

\( a_0 = 1 \) is odd
\( a_1 = 1 \) is odd
\( a_2 = a_0 a_1 + 1 = 1\times 1 + 1 = 2 \) is even
\( a_3 = a_1 a_2 + 1 = 1 \times 2 + 1 = 3 \) is odd
\( a_4 = a_2 a_3 + 1 = 2 \times 3 + 1 = 7 \) is odd
\( a_5 = a_3 a_4 + 1 = 3 \times 7 + 1 = 22 \) is even

So the pattern that we observe is the following order: odd, odd, even, odd, odd, even…

We show this by strong form of induction. Suppose this pattern holds true for all n upto n = 3k+2

(that is \( a_{3k+2} = even , a_{3k+1} = odd, a_{3k}= odd \) ).

Our computations show that this is true for k =1 (so for initial value it is true).

Let us show for the next three values:

\( a_{3k+3} = a_{3k+2} \times a_{3k+1} + 1 = even \times odd + 1 = odd \)
\( a_{3k+4} = a_{3k+3} \times a_{3k+2} + 1 = odd \times even + 1 = odd \)
\( a_{3k+5} = a_{3k+4} \times a_{3k+3} + 1 = odd \times odd + 1 = even \)

Thus we showed that whenever the index is of the form 3j+2, the number is even, otherwise if the index is of the form 3j or 3j+1, the term is odd.

Since 465 and 466 are respectively of the form 3j and 3j+1, hence
\( a_{465} \) and \( a_{466} \) both are odd.

Chatuspathi:

  • What is this topic: Induction
  • What are some of the associated concepts: Strong form of induction
  • Where can learn these topics: Cheenta I.S.I. & C.M.I. courseMath Olympiad Program discusses these topics in the ‘Induction’ module.
  • Book Suggestions: Elementary number theory by David Burton

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I.S.I. and C.M.I. Entrance

Men and Job Problem | Tomato Question 2 | ISI Entrance

This is a problem from TOMATO Problem number 2, useful for ISI and CMI entrance exam based on Men and Job.

Problem:

If m men can do a job in d days, then the number of days in which m+r men can do the job is
(A) d+r; (B) ( \frac{d}{m} (m+r)) ; (C) ( \dfrac {d}{m+r} ) ; (D) ( \dfrac {md}{m+r} );

Discussion:

If m men can do a job in d days,
Then m men does ( \frac {1}{d} ) job in 1 day
Then 1 man does ( \frac {1}{md} ) job in 1 day
Then m+r men does ( \frac {m+r}{md} ) job in 1 day
Then m+r men does 1 job ( \frac {md}{m+r} ) days.

Hence answer is ( \dfrac {md}{m+r} ) days (D)

Chatuspathi:

  • What is this topic: Arithmetic
  • What are some of the associated concepts: Unitary Method
  • Where can learn these topics: Cheenta I.S.I. & C.M.I. course, discusses these topics in the ‘Arithmetic’ module.
  • Book Suggestions: Arithmetic by K.C. Nag
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I.S.I. and C.M.I. Entrance

Calculating Average Speed | Tomato Problem 3

This is a problem number 3 from TOMATO based on Calculating Average Speed.

Problem: Calculating Average Speed.

A boy walks from his home to school at 6 kmph. He walks back at 2 kmph. His average speed, in kmph is
(A) 3; (B) 4; (C) 5; (D) ( \sqrt {12} );

Discussion: 

Suppose the distance from home to school is t km.
Time taken for home to school journey: ( \frac {t}{6} )
Time taken for school to home journey: ( \frac {t}{2} )

Hence average speed = ( \frac {distance}{time} = \frac {2t}{\frac{t}{6} + \frac{t}{2}} = 3 )

There fore average speed is 3 kmph.

Note that it is the harmonic mean of the two given speeds instead of arithmetic mean.

Chatuspathi:

  • What is this topic: Elementary Arithmetic
  • What are some of the associated concepst: Harmonic Mean
  • Where can learn these topics: Cheenta I.S.I. & C.M.I. course, discusses these topics in the ‘Number Theory’ module.
  • Book Suggestion: Arithmetic by K.C. Nag