Test of Mathematics Solution Objective 398 - Complex Number and Binomial Theorem

This is a Test of Mathematics Solution (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

Problem

If $a_0, a_1, \cdots, a_n $ are real numbers such that $$ (1+z)^n = a_0 + a_1 z + a_2 z^2 + \cdots + a_n z^n $$ for all complex numbers z, then the value of $$ (a_0 - a_2 + a_4 - a_6 + \cdots )^2 + (a_1 - a_3 + a_5 - a_7 + \cdots )^2 $$ equals

(A) $ 2^n $ ; (B) $ a_0^2 + a_1^2 + \cdots + a_n^2 $ ; (C) $ 2^{n^2} $ ; (D) $ 2n^2 $ ;

Sequential Hints

(How to use this discussion: Do not read the entire solution at one go. First, read more on the Key Idea, then give the problem a try. Next, look into Step 1 and give it another try and so on.)

Key Idea

This is the generic use case of Complex Number $ \iota =\sqrt {-1} $ and binomial theorem.

Step 1

Note that $ i^2 = -1 $. Also, geometrically speaking, i = (0,1). Hence adding (1,0) to i (=(0,1)) gives us the point (1, 1). Polar coordinate of this point is $ ( \sqrt 2, \frac{pi}{4} ) $. Here is a picture:

Try the problem with this hint before looking into step 2. Remember, no one learnt mathematics by looking at solutions.

At Cheenta we are busy with Complex Number and Geometry module. Additionally I.S.I. Entrance Mock Test 1 is also active now.

Replace $ z $ by $ i $. We have $ (1+z)^n = (\sqrt 2 , \frac {\pi}{4} )^n = (2^{n/2}, \frac{n \cdot \pi }{4} ) $ on the left hand side.

Now, replace $ z $ by $ i $ on the right hand side.

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Replacing z by $ i $ on the right hand side we have $$(2^{n/2}, \frac{n \cdot \pi }{4} ) = a_0 + a_1 i + a_2 i^2 + a_3 I^3 \cdots + a_n I^n $$. This implies $$ (2^{n/2}, \frac{n \cdot \pi }{4} ) = a_0 - a_2 + a_4 - \cdots + i (a_1 - a_3 + a_5 - \cdots ) $$

Think now, what the following expression represents: $$ (a_0 - a_2 + a_4 - a_6 + \cdots )^2 + (a_1 - a_3 + a_5 - a_7 + \cdots )^2 $$

It represents the square of the length of point $ (2^{n/2}, \frac{n \cdot \pi }{4} ) $. That is simply $ (2^{n/2})^2 = 2^n $

Hence the answer is option A.

More Resources

Look into the following notes on Complex Number and Geometry module.

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 ..........

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

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.....

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

(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

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

For $y = 2e^x$, this is not possibe because $y = 2e^x$ expression represent only positive side's of x axis
For $y = 2e^-x$,This expression also represents one side of x-axis
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
$y = e^x – e^-x + 2$, this expression also represents both sides of the x-axis

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

Therefore ,the correct ans is (c)

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

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 ..........

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.....

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.

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

(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)

2016 ISI Objective Solution Problem 1

Understand the 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) $x^5-x^4+x^2+x+1$

Solution

The problem is easy to understand. Your first task is to try the problem yourself.

Try to factorize $x^7 + x^2 + 1$.

Observe that $\omega$ and $\omega^2 $ are the roots of $x^7 + x^2 + 1$.

$x^7 + x^2 + 1$ = ($x^2 + x + 1$).($x^5 - x^4 + x^2 -x + 1$)

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.

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

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. course, Math Olympiad Program discusses these topics in the ‘Induction’ module.
Book Suggestions: Elementary number theory by David Burton

Paper folding Geometry - Video

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) $\frac {d}{m+r}$ ; (D) $\frac {md}{m+r}$;

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 $\frac {md}{m+r}$ 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

Problem

Sequential Hints

Key Idea

Step 1

More Resources

Sign change | ISI B.Stat Entrance | Problem 709

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Limit Problem | ISI B.Stat Entrance | Problem 694

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

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

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

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

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

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

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

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

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Understand the Problem:

Solution

Chatuspathi:

Chatuspathi: