Prime Number for ISI BStat | TOMATO Objective 70

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Prime number | ISI BStat Entrance | Problem no. 70

The number of integers $n>1$, such that n, n+2, n+4 are all prime numbers is ......

Zero
One
Infinite
More than one,but finite

Key Concepts

Number theory

Algebra

Prime

Check the Answer

Answer: One

TOMATO, Problem 70

Challenges and Thrills in Pre College Mathematics

Try with Hints

taking n=3, 5, 7, 11, 13, 17....prime numbers we will get

Case of n=3

n= 3

n+2=5

n+4=7

Case of n=5

then $n$=5

n+2=7

n+4=9 which is not prime....

Case of n=7,

then n=7

n+2=9 which is not prime ...

n+4=11

Can you now finish the problem ..........

We observe that when n=3 then n,n+2,n+4 gives the prime numbers.....other cases all are not prime.Therefore any no can be expressed in anyone of the form 3k, 3k+1 and 3k+2.

can you finish the problem........

If n is divisible by 3 , we are done. If the remainder after the division by 3 is 1, the number n+2 is divisible by 3. If the remainder is 2, the number n+4 is divisible by 3

The three numbers must be primes! The only case n=3 and gives$(3,5,7)$

Hundred Integers | ISI-B.Stat Entrance | TOMATO 82

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Hundred Integers | ISI B.Stat Entrance | Problem-82

Let $x_1,x_2,......,x_100$ be hundred integers such that the sum of any five of them is 20. Then..

the largest $x_i$ equals 5
the smallest $x_i$ equals to 3
$x_{17} = x_{83}$
none of the foregoing statements is true

Key Concepts

Number theory

Divisor

integer

Check the Answer

Answer:$x_{17} = x_{83}$

TOMATO, Problem 82

Challenges and Thrills in Pre College Mathematics

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Let us take the numbers be $x_i , x_j , x_k ,x_l , x_m $

Now $x_i + x_j + x_k + x_l + x_m = 20$ and again $x_i + x_j + x_k + x_l + x_n = 20$

Can you now finish the problem ..........

From the above relation there are three case arise that....

1)$x_m = x_n$

2)All the integers are equal.

3)$x_{17} =x_{83}$

So the correct answer is $x_{17} =x_{83}$

Prime number Problem | ISI BStat | TOMATO Objective 96

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Prime number | ISI B.Stat Entrance | Problem no. 96

The number of different prime factors of 3003 is.....

Key Concepts

Number theory

Algebra

Prime numbers

Check the Answer

Answer: 16

TOMATO, Problem 96

Challenges and Thrills in Pre College Mathematics

Try with Hints

At first, we have to find out the prime factors. Now $3003$=$3 \times 7 \times 11 \times 13$. but now it can be expressed as another prime number also such as $3003=3 \times 1001$. So we have to find different prime factors.

Can you now finish the problem ..........

Now, if you have a number and its prime factorisation, $n={p_1}^{m_1} {p_2}^{m_2}⋯{p_r}^{m_r}$ you can make divisors of the number by taking up to $m_1$ lots of $p_1$, up to $m_2$ lots of $p_2$ and so on. The number of ways of doing this is going to be$ (m_1+1)(m_2+1)⋯(m_r+1)$.

can you finish the problem........

for the given case $3003$ has $2^4=16 $divisors.

Discontinuity Problem | ISI B.Stat Objective | TOMATO 734

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Discontinuity Problem | ISI B.Stat TOMATO 734

$\quad$ Let $f(x)=|| x-1|-1|$ if $x<1$ and $f(x)=[x]$ if $x \geq 1,$ where, for any real number $x,[x]$ denotes the largest integer $\leq x$ and $|y|$ denotes absolute value of $y .$ Then, the set of discontinuity-points of the function f consists of

all integers $\geq 0$
all integers $\geq 1$
all integers $> 1$
the integer 1

Key Concepts

Limit

Calculas

Continuous

Check the Answer

Answer: $C$

TOMATO, Problem 734

Challenges and Thrills in Pre College Mathematics

Try with Hints

Let us first check at $x=1$
$\lim f(x)$ as $x\to1-=\lim || x-1|-1|$ as $x\to 1-=1$
$\lim f(x)$ as $x\to1+=\lim [x]$ as $x\to1+=1$
So, continuous at $x=1$
Let us now check at $x=0$
$\lim f(x)$ as $x\to 0-=\lim || x-1|-1|$ as $x\to 0-=0$
$\lim f(x)$ as $x\to 0+=\lim || x-1|-1|$ as $x\to 0+=0$
So continuous at $x=0$

Can you now finish the problem ..........

Let us now check at $x=2$
$\lim f(x)$ as $x\to 2-=\lim [x]$ as $x\to 2-=1$
$\lim f(x)$ as $x\to 2+=\lim [x]$ as $x \rightarrow 2+=2$
Discontinuous at $x=2$
Option (c) is correct.

Problem on Limit | ISI B.Stat Objective | TOMATO 728

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Problem on Limit | ISI B.Stat TOMATO 728

The limit lim $\int\frac {h}{(h^2 + x^2)}$dx (integration running from $x =-1$to $x = 1$) as$ h \to 0$

equals 0
equals $\pi$
equals $-\pi$
deoes not exist

Key Concepts

Limit

Calculas

trigonometry

Check the Answer

Answer: does not exist

TOMATO, Problem 728

Challenges and Thrills in Pre College Mathematics

Try with Hints

Now, $\int{h}{(h^2 + x^2)}$dx (integration running from $x = -1$ to $x = 1$)
Let, $x$ = h tany

$\Rightarrow dx = h sec^2y dy$
$\Rightarrow $ $x = -1$, $y = -tan^{-1}(1/h)$ and $x = 1$, $y = tan^{-1}(1/h)$

$\Rightarrow \int \frac{h}{(h^2 + x^2)}$dx =$\int \frac{h(hsec^2ydy)}{h^2sec^2y}$ (integration running from$ y = -tan^{-1}(1/h) $ to $y = tan^-1(1/h))$

= y (upper limit =$ tan-1(1/h)$) and lower limit = $-tan^-1(1/h)$

= $2tan^-1(1/h)$

Can you now finish the problem ..........

Now, lim $2tan^-(1/h)$ as$ h \to 0$ doesn‟t exist

Real valued function | ISI B.Stat Objective | TOMATO 690

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Real valued functions | ISI B.Stat TOMATO 690

Let $f(x)$ be a real-valued function defined for all real numbers x such that $|f(x) – f(y)|≤(1/2)|x – y|$ for all x, y. Then the number of points of intersection of the graph of $y = f(x)$ and the line $y = x$ is

0
1
2
none of these

Key Concepts

Limit

Calculas

Real valued function

Check the Answer

Answer: $1$

TOMATO, Problem 690

Challenges and Thrills in Pre College Mathematics

Try with Hints

Now,

$|f(x) – f(y)| ≤ (1/2)|x – y|$

$\Rightarrow lim |{f(x) – f(y)}/(x – y)|$( as x -> y ≤ lim (1/2)) as x - > y

$\Rightarrow |f‟(y)| ≤ ½$

$\Rightarrow -1/2 ≤ f‟(y) ≤1/2$

$\Rightarrow -y/2 ≤ f(y) ≤ y/2$ (integrating)

$\Rightarrow -x/2 ≤ f(x) ≤ x/2$

Can you now finish the problem ..........

Therefore from the picture we can say that intersection point is $1$

Expansion Problem | ISI B.Stat Objective | TOMATO 102

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Expansion Problem | ISI B.Stat TOMATO 102

The number of terms in the expansion of $[(a+3b)^2 (a-3b)^2]^2$ , when simplified, is

Key Concepts

Algebra

multiplication

Expansion

Check the Answer

Answer: $5$

TOMATO, Problem 102

Challenges and Thrills in Pre College Mathematics

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The given expression is $[(a+3b)^2 (a-3b)^2]^2$.we have to find out number of terms in the expansion of $[(a+3b)^2 (a-3b)^2]^2$.

Now , $[(a+3b)^2 (a-3b)^2]^2$ $\Rightarrow [(a^2-9b^2)^2]^2$ $\Rightarrow (a^2-9b^2)^4$

Can you now finish the problem ..........

Now in the equation $(a+b)^2=a^2+2ab+b^2$ i.e total numbers of terms are $3$

$(a+b)^3=a^3+3a^2b+3ab^2+b^3$ i.e total numbers of terms are $4$

.............

$(a + b)^n = a^n + { n \choose 1} a^{n-1}b + { n \choose 2 } a^{n-2}b^2 + … + { n \choose {n-1}}ab^{n-1} + b^n$ i.e total numbers of terms are $n+1$

Similarly In this expression $(a^2-9b^2)^4$ ,The power is $4$.Therefore we say that after the expansion total number be $5$.

Integers and Divisors | ISI-B.Stat Entrance | TOMATO 98

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Integers and Divisors | ISI B.Stat Entrance | Problem-98

The number of positive integers which divide 240 (where both 1 and 240 are considered as divisors) is

Key Concepts

Integer

Divisor

Number theory

Check the Answer

Answer: 20

TOMATO, Problem 98

Challenges and Thrills in Pre College Mathematics

Try with Hints

We have to find out the number of positive integers which divide 240.so at first we have to find out the factors of 240...

$240=2 \times 120$

$240=3 \times 80$

$240=4 \times 60$

$240=5 \times 48$

$240=6 \times 40$

$240=8 \times 30$

$240=10 \times 24$

$240=12 \times 20$

$240=15 \times 16$

$240=20 \times 12$

$240=24 \times 10$ ..............

so we notice that the divisors are repeat........

Can you now finish the problem ..........

We notice that after $240=15 \times 16$ this stape all the factors are repeats.....so we have to calculate up to $240=15 \times 16$ step only....

can you finish the problem........

Therefore the total number of positive integers are $1,2,3,4,5,6,8,10,12,15,20,24,30,40,48,60,80,120,240$ i.e $20$

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

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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 ..........

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.....

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

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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.....

Prime number | ISI BStat Entrance | Problem no. 70

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Hundred Integers | ISI B.Stat Entrance | Problem-82

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Prime number | ISI B.Stat Entrance | Problem no. 96

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Discontinuity Problem | ISI B.Stat TOMATO 734

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Problem on Limit | ISI B.Stat TOMATO 728

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Real valued functions | ISI B.Stat TOMATO 690

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Expansion Problem | ISI B.Stat TOMATO 102

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Integers and Divisors | ISI B.Stat Entrance | Problem-98

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Sign change | ISI B.Stat Entrance | Problem 709

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Limit Problem | ISI B.Stat Entrance | Problem 694

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