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

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.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

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)} $

Test of Mathematics Solution Subjective 42- Polynomial with Integer Coefficients

This is a Test of Mathematics Solution of Subjective 42 (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 Visit: I.S.I & CMI Entrance Course of Cheenta

Problem

Let f(x) be a polynomial with integer coefficients. Suppose that there exist distinct integers $ a_1 , a_2 , a_3 , a_4 $ , such that $ f( a_1 ) = f( a_2 ) = f( a_3 ) = f( a_4 ) = 3 $ . Show that there does not exist any integer b with f(b) = 14.

Solution:

Consider the auxiliary polynomial g(x) = f(x) - 3. Clearly, according to the problem, g(x) has four distinct integer roots $ a_1 , a_2 , a_3 , a_4 $. Hence we may write $ g(x) = (x - a_1)(x- a_2)(x- a_3)(x- a_4) Q(x)$ where Q(x) is an integer coefficient polynomial. (Since by factor theorem if 't' is a root of a polynomial f(x) then (x-t) is it's factor).

Suppose there exists an integer b such that f(b) = 14; then g(b) = 14-3 =11. Hence $ g(b) = 11 = (b- a_1 )(b- a_2 )(b- a_3 )(b- a_4 ) Q(b)$. $ a_1 , a_2 , a_3 , a_4 $ are distinct so are $ (b- a_1 ) , (b- a_2 ) , (b- a_3 ) , (b- a_4 ) $. Therefore the equation $ g(b) = 11 = (b- a_1 ) (b- a_2 )(b- a_3 )(b- a_4 ) Q(b)$ indicates that 11 can be written as the product of at least 4 different integers which is impossible (as 11 is a prime). Indeed even if we consider -1, 1 and 11 , we have only three integers whose product is 11.

Thus we have a contradiction and proving that there does not exist an integer b such that f(b) 14.

Key Idea: Factor Theorem

I.S.I. Entrance Solution Sequence of isosceles triangles -2018 Problem 6

[et_pb_section bb_built="1" admin_label="Blog Hero" _builder_version="3.0.82" use_background_color_gradient="on" background_color_gradient_start="rgba(114,114,255,0.24)" background_color_gradient_end="#ffffff" background_blend="multiply" custom_padding="0|0px|0|0px|false|false" animation_style="slide" animation_direction="top" animation_intensity_slide="2%" locked="off" next_background_color="#ffffff"][et_pb_row custom_width_px="1280px" custom_padding="27px|0px|27px|0px" custom_margin="|||" _builder_version="3.0.82" background_size="initial" background_position="top_left" background_repeat="repeat"][et_pb_column type="4_4"][et_pb_text _builder_version="3.12.2" text_text_color="#474ab6" text_line_height="1.9em" background_size="initial" background_position="top_left" background_repeat="repeat" text_orientation="center" max_width="540px" module_alignment="center" locked="off"] 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. [/et_pb_text][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section bb_built="1" admin_label="Blog" _builder_version="3.0.82" custom_margin="|||" custom_padding="0px|0px|21px|0px|false|false" prev_background_color="#000000" next_background_color="#f7f8fc"][et_pb_row use_custom_width="on" custom_width_px="960px" custom_padding="0|0px|24px|0px|false|false" _builder_version="3.0.82" background_size="initial" background_position="top_left" background_repeat="repeat"][et_pb_column type="4_4"][et_pb_tabs _builder_version="3.12.2"][et_pb_tab title="Hint 1 - Triangular Inequality" _builder_version="3.12.2" use_background_color_gradient="off" background_color_gradient_start="#2b87da" background_color_gradient_end="#29c4a9" background_color_gradient_type="linear" background_color_gradient_direction="180deg" background_color_gradient_direction_radial="center" background_color_gradient_start_position="0%" background_color_gradient_end_position="100%" background_color_gradient_overlays_image="off" parallax="off" parallax_method="on" background_size="cover" background_position="center" background_repeat="no-repeat" background_blend="normal" allow_player_pause="off" background_video_pause_outside_viewport="on" tab_text_shadow_style="none" body_text_shadow_style="none" tab_text_shadow_horizontal_length="0em" tab_text_shadow_vertical_length="0em" tab_text_shadow_blur_strength="0em" body_text_shadow_horizontal_length="0em" body_text_shadow_vertical_length="0em" body_text_shadow_blur_strength="0em"] If a, b, c are sides of a triangle, triangular inequality assures that difference of two sides is lesser than the third side. Since $ a \ge b \ge c > 0 $, hence using triangular inequality we have a - b < c. Infact for all n, $ a^n - b^n < c^n $ [/et_pb_tab][et_pb_tab title="Hint 2 - Factor and estimate" _builder_version="3.12.2" use_background_color_gradient="off" background_color_gradient_start="#2b87da" background_color_gradient_end="#29c4a9" background_color_gradient_type="linear" background_color_gradient_direction="180deg" background_color_gradient_direction_radial="center" background_color_gradient_start_position="0%" background_color_gradient_end_position="100%" background_color_gradient_overlays_image="off" parallax="off" parallax_method="on" background_size="cover" background_position="center" background_repeat="no-repeat" background_blend="normal" allow_player_pause="off" background_video_pause_outside_viewport="on" tab_text_shadow_style="none" body_text_shadow_style="none" tab_text_shadow_horizontal_length="0em" tab_text_shadow_vertical_length="0em" tab_text_shadow_blur_strength="0em" body_text_shadow_horizontal_length="0em" body_text_shadow_vertical_length="0em" body_text_shadow_blur_strength="0em"] We have $ a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + b^{n-1} ) < c^n $ Replacing every a by b in the left hand side, we make the expression to the left even smaller. i.e. $(a-b)(b^{n-1} + b^{n-2}b + ... + b^{n-1} ) \le (a-b)(a^{n-1} + a^{n-2}b + ... + b^{n-1} ) < c^n $ Hence $ (a-b) \times n \times b^{n-1} < c^n $ [/et_pb_tab][et_pb_tab title="Hint 3 - Final Steps" _builder_version="3.12.2" use_background_color_gradient="off" background_color_gradient_start="#2b87da" background_color_gradient_end="#29c4a9" background_color_gradient_type="linear" background_color_gradient_direction="180deg" background_color_gradient_direction_radial="center" background_color_gradient_start_position="0%" background_color_gradient_end_position="100%" background_color_gradient_overlays_image="off" parallax="off" parallax_method="on" background_size="cover" background_position="center" background_repeat="no-repeat" background_blend="normal" allow_player_pause="off" background_video_pause_outside_viewport="on" tab_text_shadow_style="none" body_text_shadow_style="none" tab_text_shadow_horizontal_length="0em" tab_text_shadow_vertical_length="0em" tab_text_shadow_blur_strength="0em" body_text_shadow_horizontal_length="0em" body_text_shadow_vertical_length="0em" body_text_shadow_blur_strength="0em"] Now notice $ (a-b) < \frac {c^n}{n\times b^{n-1}} = \frac{c}{n} \times \frac {c^{n-1}}{b^{n-1}} = \frac {c}{n} (\frac{c}{b})^{n-1}$ Clearly $ \frac{c}{b} \le 1 $ by given hypothesis. Hence $ a-b \le \frac{c}{n} $ for all n. But letting n go to infinity, we see that a and b can be made arbitrarily close to each other. This implies a=b. Hence each triangle in the sequence is isosceles [/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section bb_built="1" admin_label="Footer" _builder_version="3.0.82" background_color="#f7f8fc" custom_padding="0px|0px|2px|0px|false|false" animation_style="zoom" animation_direction="bottom" animation_intensity_zoom="6%" animation_starting_opacity="100%" saved_tabs="all" prev_background_color="#ffffff"][et_pb_row use_custom_gutter="on" gutter_width="2" custom_padding="24px|0px|145px|0px|false|false" _builder_version="3.0.82" background_size="initial" background_position="top_left" background_repeat="repeat"][et_pb_column type="1_2"][et_pb_text _builder_version="3.12.2" text_text_color="#7272ff" header_font="|on|||" header_text_color="#7272ff" header_font_size="36px" header_line_height="1.5em" background_size="initial" background_position="top_left" background_repeat="repeat" custom_margin="||20px|" animation_style="slide" animation_direction="bottom" animation_intensity_slide="10%"]

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