Test of Mathematics Solution Subjective 176 - Value of a Polynomial at x = n+1

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 176 (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.


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Problem

Suppose that P(x) is a polynomial of degree n such that $ P(k) = \frac {k}{k+1} $ for k = 0, 1, 2, ..., n . Find the value of P(n+1).


Solution

Consider an auxiliary polynomial g(x) = (x+1)P(x) - x . g(x) is an n+1 degree polynomial (as P(x) is n degree and we multiply (x+1) with it). We note that g(0) = g(1) = ... = g(n) = 0  (as the given condition allows (k+1) P(k) - k = 0 for all k from 0 to n). Hence 0, 1, 2, ... , n are the n+1 roots of g(x).

Therefore we may write g(x) = (x+1)P(x) - x = C(x)(x-1)(x-2)...(x-n) where C is a constant. Put x = -1. We get g(-1) = (-1+1)P(-1) - (-1) = C(-1)(-1-1)(-1-2)...(-1-n).

Thus 1 = C $ (-1)^{(n+1) } (n+1)! $ gives us the value of C. We put the value of C in the equation (x+1)P(x) - x = C(x)(x-1)(x-2)...(x-n) and replace x by n+1 to get the value of P(n+1).

$ (n+2)P(n+1) - (n+1) = \frac { (-1)^{(n+1)}}{(n+1)!} (n+1)(n)(n-1) ... (1) $ implying $ P(n+1) = \frac { (-1)^{(n+1)} + (n+1)}{(n+2)} $

2016 ISI Objective Solution Problem 1

Problem

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


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Understanding the Problem:

The problem is easy to understand right?

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

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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.

ISI Entrance Paper 2018 - B.Stat, B.Math Subjective

Here, you will find all the questions of ISI Entrance Paper 2018 from Indian Statistical Institute's B.Stat Entrance. You will also get the solutions soon of all the previous year problems.

Problem 1:

Find all pairs $(x,y)$ with $x,y$ real, satisfying the equations:

$\sin(\frac{x+y}{2})=0,\vert x\vert+\vert y\vert=1$

Problem 2:

Suppose that $PQ$ and $RS$ are two chords of a circle intersecting at a point $O$. It is given that $PO=3 \mathrm{cm}$ and $SO=4 \mathrm{cm}$. Moreover, the area of the triangle $POR$ is $7 \mathrm{cm}^2$. Find the area of the triangle $QOS$.

Problem 3:

Let $f:\mathbb{R} \to \mathbb{R}$ be a continuous function such that for all $x \in \mathbb{R}$ and for all $t \geq 0$, $f(x)=f(e^{t}x)$. Show that $f$ is a constant function.

Problem 4:

Let $f:(0,\infty)\to \mathbb{R}$ be a continuous function such that for all $x \in(0,\infty)$, $f(2x)=f(x)$. Show that the function $g$ defined by the equation $g(x)=\int_{x}^{2x} f(t)\frac{dt}{t}$ for $x>0$ is a constant function.

Problem 5:

Let $f:\mathbb{R}\to \mathbb{R}$ be a differentiable function such that its derivative $f'$ is a continuous function. Moreover, assume that for all $x \in\mathbb{R}$, $0 \leq \vert f'(x)\vert\leq \frac{1}{2}$. Define a sequence of real numbers $ \{a_n\}_{n\in\mathbb{N}}$ by : $a_1=1$ and $a_{n+1}=f(a_n)$ for all $n\in\mathbb{N}$. Prove that there exists a positive real number $M$ such that for all $n\in\mathbb{N}$,

|an|M

Problem 6:

Let, $a\geq b\geq c >0$ be real numbers such that for all natural number $n$, there exist triangles of side lengths $a^{n} , b^{n} ,c^{n}$. Prove that the triangles are isosceles.

Problem 7:

Let $a, b, c$ are natural numbers such that $a^{2}+b^{2}=c^{2}$ and $c-b=1$

Prove that,

(i) $a$ is odd,

(ii) $b$ is divisible by $4$,

(iii) $a^{b}+b^{a}$ is divisible by $c$.

Problem 8:

Let $n\geq 3$. Let $A=((a_{ij}))_{1\leq i,j\leq n}$ be an $n\times n$ matrix such that $a_{ij}\in\{-1,1\}$ for all $1\leq i,j\leq n$. Suppose that $a_{k1}=1$ for all $1\leq k\leq n$ and $\sum_{k=1}^n a_{ki}a_{kj}=0$ for all $i\neq j$. Show that $n$ is a multiple of $4$.

Some useful Links:

Rotation of triangle (B.Stat 2006, Problem 4 solution)

Problem:

In the figure below, $E$ is the midpoint of the arc $ABEC$ and the segment $ED$ is perpendicular to the chord $BC$ at $D$. If the length of the chord $AB$ is $\mathbf{\ell_1} $, and that of the segment $BD$ is $\mathbf{\ell_2} $, determine the length of $DC$ in terms of $\mathbf{\ell_1, \ell_2} $.

B.Stat Entrance 2006 Geometry Problem

Discussion:

Teacher: Here is a clue: rotate ($ \Delta EDC$ ) about point $E$ such that $EC$ falls along $EA$. Can you draw the diagram after rotation?

Student:

B.Stat Entrance 2006 Geometry Problem

Obviously $EC$ will fit into $EA$ as $E$ is the midpoint of larger arc $AC$. Suppose $D$ falls on $D'$. Then $ED'A$ is the rotated form of $EDC$.

Teacher: Can you show that $D'A$ and $BA$ are the same line?

Student: Okay I can try. If we can show that $\angle D'AE = \angle BAE $ , then we have shown that $D'A$ and $BA$ is the same line. Now $\angle D'AE = \angle DCE = \angle BCE$ (due to rotation) . But $\angle BCE = \angle BAE $ as they are the angle subtended by the same segment $BE$.

So we have $\angle D'AE = \angle BAE $. Therefore $D'A$ falls on $BA$.

Teacher: Let's revise the diagram then.

Now can you finish the problem?

Student: Since $ \Delta E_1 D_1' A_1 = \Delta E_1 D_1 C_1 $ due to rotation, we have $E_1 D_1' = E_1 D_1 \Rightarrow \angle E_1 D_1' D_1 = \angle E_1 D_1 D_1' $

But $\angle E_1 D_1' B_1 = \angle E_1 D_1 B_1' = 90^{\circ} \\ \Rightarrow \angle E_1 D_1' B_1 - \angle E_1 D_1' D_1 \\ = \angle E_1 D_1 B_1' - \angle E_1 D_1 D_1' \Rightarrow \angle B_1 D_1' D_1 \\ = \angle B_1 D_1 D_1' \Rightarrow B_1 D_1 \\ = B_1 D_1 ' $

Thus $ C_1D_1 = A_1 D_1' \\ = A_1 B_1 + B_1 D_1' \\= A_1 B_1 + B_1 D_1 \\= \ell_1 + \ell_2 $

Solutions to an equation | B.Stat 2005 Subjective Problem 4

Problem:

Find all real solutions of the equation $sin^{5}x+cos^{3}x=1$ .

Discussion:

Teacher: Notice that $|\sin x| \leq 1 , |\cos x | \leq 1 $ . So if you raise $\sin x$ and $\cos x$ to higher powers you necessarily lower the value. Take for example the number $\frac{1}{2}$. If you raise that to the power of $2$, you get $\frac{1}{4}$. Raise it to the power of $3$, you get $\frac{1}{8}$ (which is smaller than $\frac{1}{4}$).
When absolute value of a number is greater than $1$, values increase, with increasing power. If the absolute value is less than $1$, the values decrease with increasing power.
Use this insight to solve this problem.

Student: I see. Then $sin ^5 x \leq |\sin x|^5 \leq \sin^2 x , \cos^3 x \leq |\cos x |^3 \leq \cos^2 x $. Adding these two inequalities we get $\sin ^5 x + \cos^2 x \leq \sin^2 x + cos^2 x = 1 $

Equality holds when simultaneously $\sin^5 x = \sin^2 x $,
$\cos^3 x = \cos^2 x $

$\sin ^{5} x = \sin ^{3} x \Rightarrow \sin^{3} x (\sin^{2} x - 1 ) = 0 \Rightarrow \sin^{3} x = 0$ or $\sin^{2} x = 1 $

This implies $x = n \pi $ or $ \displaystyle {x = (2n+1) \frac{\pi}{2} } $

Similarly $\cos ^{3} x = \cos ^{2} x \Rightarrow \cos^{2} x (\cos x - 1 ) = 0 \Rightarrow \cos^{2} x = 0 $ or $\cos x = 1 $

This implies $x = 2n \pi $ or $ \displaystyle {x = (2n+1) \frac{\pi}{2} } $

Therefore simultaneously $\sin^5 x = \sin^2 x , \cos^3 x =\cos^2 x $ holds when $x = 2n \pi $ or $\displaystyle {x = (2n+1) \frac{\pi}{2} } $ where $n$ is any natural number.

Some Useful Links:

Solving a few Diophantine Equations – Video

ISI 2015 BStat – BMath Objective Problems

ISI BStat BMath problem 14 | Objective Problems Discussion

Let's discuss this objective problem number 14 from ISI BStat BMath. Try to solve the problem and then read their solution.

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Problem 14

f(x) = tan(sinx) (x > 0)

To understand the graph of a function, easiest and the most proper method is to apply techniques from calculus. We will quickly compute, derivative and second derivative and try to understand,extreme points and convexity of the curve.

f'(x) = cos (x) sec^2 (sin (x))

Hence when cos(x) is positive (correspondingly negative), derivative is positive (is negative). Therefore from (0, \frac{\pi}{2} ) function is increasing, ( \frac{\pi}{2} , \frac{3\pi}{2} ) the function is decreasing. Also it has critical points on cos (x) is 0 (at x = \frac{\pi}{2} , \frac{3\pi}{2} )

Now we compute the second derivative.

f''(x) = - \sec^2 (\sin(x)) (\sin(x) - 2 \cos^2(x) \tan (\sin(x)))

At x = \frac{\pi}{2} second derivative is -\sec^2 1 [hence we have a maxima] and at x = \frac{3\pi}{2} second derivative is \sec^2 1 [hence we have a minima]

Finally we compute f(\frac{\pi}{2} ) = \tan 1 > 1 , f(\frac{3\pi}{2} ) = - \tan 1 < -1 .

Moreover, the function is differentiable at the points of maxima and minima. Hence answer is (B)