If C 3 9 + C 5 9 = C r 10 , then find r .

Given C 3 9 + C 5 9 = C r 10 We know that C r n = C n - r n When, C 3 9 = C 6 9 Therefore, C 6 9 + C 5 9 = C r 10 We know that C r n + C r - 1 n = C r ( n + 1 ) ∴ r = 6 When C 5 9 = C 4 9 ⇒ C 3 9 + C 4 9 = C r 10 ∴ r = 4 Hence, r = 4 , r = 6

Resolve x 2 + 1 x 2 + x + 1 2 into partial fraction.

Given, x 2 + 1 x 2 + x + 1 2 Let, x 2 + 1 x 2 + x + 1 2 = A x + B x 2 + x + 1 + C x + D x 2 + x + 1 2 ⇒ x 2 + 1 = ( A x + B ) x 2 + x + 1 + ( C x + D ) ...(i) Equating the coefficients of x 3 in (i), we get A = 0 Equating the coefficients of x 2 in (i), we get A + B = 1 ⇒ B = 1 Equating the coefficients of x in (i), we get A + B + C = 0 ⇒ 1 + C = 0 ⇒ C = - 1 Equating the constant in (i), we get B + D = 1 ⇒ D = 1 - B ⇒ D = 1 - 1 ⇒ D = 0 ∴ A x + B = 1 , C x + D = - x ∴ x 2 + 1 x 2 + x + 1 2 = 1 x 2 + x + 1 - x x 2 + x + 1 2

Resolve x 3 + x 2 + 1 ( x - 1 ) x 3 - 1 into partial fraction.

Given, x 3 + x 2 + 1 ( x - 1 ) x 3 - 1 = x 3 + x 2 + 1 ( x - 1 ) ( x - 1 ) x 2 + x + 1 Let, x 3 + x 2 + 1 ( x - 1 ) ( x - 1 ) x 2 + x + 1 = A x - 1 + B ( x - 1 ) 2 + C x + D x 2 + x + 1 ⇒ x 3 + x 2 + 1 = A ( x - 1 ) x 2 + x + 1 + B x 2 + x + 1 + ( C x + D ) ( x - 1 ) 2 ...(i) Put x = 1 in (i), 1 + 1 + 1 = A ( 0 ) + B ( 1 + 1 + 1 ) + ( C ( 1 ) + D ) ( 0 ) ⇒ 3 B = 3 ⇒ B = 1 Equating the coefficients of x 3 in (i), we get 1 = A + C ...(ii) Equating the coefficients of x 2 in (i), we get 1 = A ( 1 - 1 ) + B ( 1 ) + C ( - 2 ) + D ( 1 ) ⇒ 1 = B - 2 C + D ⇒ 1 = 1 - 2 C + D ⇒ 2 C = D ...(iii) Put x = 0 in (i), we get 1 = A ( - 1 ) ( 1 ) + B ( 1 ) + D ( - 1 ) 2 ⇒ - A + B + D = 1 ⇒ - A + 1 + D = 1 ⇒ A = D ...(iv) From (ii), (iii), (iv), 1 = D + D 2 ⇒ 3 D 2 = 1 ⇒ D = 2 3 From (iii), A = 2 3 From (iv), C = D 2 = 2 3 2 = 1 3 Then, x 3 + x 2 + 1 ( x - 1 ) x 3 - 1 = 2 3 x - 1 + 1 ( x - 1 ) 2 + 1 3 x + 2 3 x 2 + x + 1 ∴ x 3 + x 2 + 1 ( x - 1 ) x 3 - 1 = 2 3 ( x - 1 ) + 1 ( x - 1 ) 2 + x + 2 3 x 2 + x + 1

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Solved Mathematics Paper>Chapter 10 - Solved Mathematics Paper-2>Exercise

Embibe Experts Solutions for Exercise 1: Exercise

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Exercise 1: Exercise

Attempt the practice questions from Exercise 1: Exercise with hints and solutions to strengthen your understanding. Mathematics Crash Course (Based on Revised Syllabus-2023) solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Exercise 1: Exercise with Hints & Solutions

MEDIUM

12th Telangana Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 10 - Permutations and Combinations>Exercise>Q 24

If the letters of the word $B R I N G$ are permuted in all possible ways and the words thus formed are arranged in the dictionary order, then find the $59^{t h}$ word.