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

The Probabilities of three events A , B , C are such that P ( A ) = 0 . 3 , P ( B ) = 0 . 4 , P ( C ) = 0 . 8 , P ( A ∩ B ) = 0 . 09 , P ( A ∩ C ) = 0 . 28 , P ( A ∩ B ∩ C ) = 0 . 08 and P ( A ∪ B ∪ C ) ≥ 0 . 75 . Show that P ( B ∩ C ) ) lies in the interval 0 . 21 , 0 . 46 .

Given: P ( A ∪ B ∪ C ) ⩾ 0 . 75 ⇒ 0 . 75 ⩽ P ( A ∪ B ∪ C ) ⩽ 1 Since, probability of sample space is one i.e P ( s ) = 1 and here, P ( A ∪ B ∪ C ) ⩽ 1 . We know that, from Addition Theorem of Probability for Three Events P A ∪ B ∪ C = P ( A ) + P ( B ) + P ( C ) - P ( A ∩ B ) - P ( B ∩ C ) - P ( C ∩ A ) + P ( A ∩ B ∩ C ) Now, 0 . 75 ⩽ P ( A ) + P ( B ) + P ( C ) - P ( A ∩ B ) - P ( B ∩ C ) - P ( C ∩ A ) + P ( A ∩ B ∩ C ) ⩽ 1 ⇒ 0 . 75 ⩽ 0 . 3 + 0 . 4 + 0 . 8 - 0 . 09 - 0 . 28 - P ( B ∩ C ) + 0 . 08 ⩽ 1 ⇒ 0 . 75 ⩽ 1 . 21 - P ( B ∩ C ) ⩽ 1 ⇒ - 0 . 46 ⩽ - P ( B ∩ C ) ⩽ - 0 . 21 ⇒ 0 . 46 ⩾ P ( B ∩ C ) ⩾ 0 . 21 ∴ P ( B ∩ C ) lies in the interval [ 0 . 21 , 0 . 46 ] . Hence, showed.

Evaluate ∫ d x ( 1 - x ) 3 - 2 x - x 2 on ( - 3 , 1 )

Let I = ∫ d x ( 1 - x ) 3 - 2 x - x 2 put 1 - x = 1 t ⇒ d x = 1 t 2 d t ⇒ I = ∫ d t t 2 1 t 3 - 2 1 - 1 t - ( 1 - 1 t ) 2 ⇒ I = ∫ d t t 3 - 2 + 2 t - 1 + 1 t 2 - 2 t ⇒ I = ∫ d t t 2 t - 1 t 2 + 2 t ⇒ I = ∫ d t t 4 t - 1 t 2 ⇒ I = ∫ d t 4 t - 1 ⇒ I = ∫ 4 t - 1 - 1 2 d x = ( 4 t - 1 ) 1 2 4 × 1 2 + C I = 1 2 4 1 - x - 1 + C Hence, I = 1 2 3 + x 1 - x + C

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Solved Mathematics Paper>Chapter 9 - Solved Mathematics Paper-1>Exercise

Embibe Experts Solutions for Exercise 1: Exercise-Choice

Author:Embibe Experts

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

Attempt the practice questions from Exercise 1: Exercise-Choice 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-Choice with Hints & Solutions

EASY

12th Andhra Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Solved Mathematics Paper-1>Exercise>Q 5

Form the polynomial equation, whose roots are $3, 2, 1 + i, 1 - i$ .

EASY

12th Andhra Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Solved Mathematics Paper-1>Exercise>Q 18

If $α, β, γ$ are the roots of the equation $x^{3} - 3 a x + b = 0$ , then prove that $Σ (α - β) (α - γ) = 9 a$ .

HARD

12th Andhra Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Solved Mathematics Paper-1>Exercise>Q 20

Resolve $\frac{x^{3} + x^{2} + 1}{(x - 1) (x^{3} - 1)}$ into partial fraction.

EASY

12th Andhra Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Solved Mathematics Paper-1>Exercise-Choice>Q 23

The Probabilities of three events $A, B, C$ are such that $P (A) = 0.3, P (B) = 0.4, P (C) = 0.8$ , $P (A \cap B) = 0.09, P (A \cap C) = 0.28, P (A \cap B \cap C) = 0.08$ and $P (A \cup B \cup C) \geq 0.75$ . Show that $P (B \cap C)$ ) lies in the interval $[0.21, 0.46]$ .

EASY

12th Andhra Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Solved Mathematics Paper-1>Exercise>Q 34

Define differential equation. Write order and degree (if defined) of the following differential equation:

$\frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} + 2 y = 0$

HARD

12th Andhra Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Solved Mathematics Paper-1>Exercise>Q 39

Find the equation(s) of a tangent to the hyperbola $3 x^{2} - 2 y^{2} = 25$ and which passes through an external point $(0, - \frac{5}{2})$ .

HARD

12th Andhra Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Solved Mathematics Paper-1>Exercise>Q 43

Find the pair tangents drawn from the origin to the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ and hence deduce a condition for these tangents to be perpendicular.