Find the range of the expression x + 2 2 x 2 + 3 x + 6 .

Let, y = x + 2 2 x 2 + 3 x + 6 ⇒ 2 x 2 y + 3 x y + 6 y - x - 2 = 0 ⇒ x 2 ( 2 y ) + x ( 3 y - 1 ) + 6 y - 2 = 0 Discriminant ≥ 0 We know that, for a x 2 + b x + c = 0 , Discriminant = b 2 - 4 a c . ⇒ ( 3 y - 1 ) 2 - 4 ( 2 y ) ( 6 y - 2 ) ≥ 0 ⇒ 9 y 2 - 6 y + 1 - 48 y 2 + 16 y ≥ 0 ⇒ - 39 y 2 + 10 y + 1 ≥ 0 ⇒ 39 y 2 - 10 y - 1 ≤ 0 ⇒ 39 y 2 - 13 y + 3 y - 1 ≤ 0 ⇒ 13 y ( 3 y - 1 ) + ( 3 y - 1 ) ≤ 0 ⇒ ( 13 y + 1 ) ( 3 y - 1 ) ≤ 0 ⇒ - 1 13 ≤ y ≤ 1 3 Hence, range of y = x + 2 2 x 2 + 3 x + 6 is - 1 13 , 1 3 .

Evaluate: ∫ 1 x + 3 x + 2 d x

Let I = ∫ 1 x + 3 x + 2 d x Let x + 2 = t 2 Differentiate it, ⇒ d x = 2 t d t Now, I = ∫ 1 t 2 - 2 + 3 t 2 2 t d t ⇒ I = ∫ 2 t 2 + 1 d t ⇒ I = 2 tan - 1 t + c ⇒ I = 2 tan - 1 x + 2 + c Hence, ∫ 1 x + 3 x + 2 d x = 2 tan - 1 x + 2 + c

Evaluate ∫ 0 1 log ( 1 + x ) 1 + x 2 d x

Let I = ∫ 0 1 log ( 1 + x ) 1 + x 2 d x Put x = tan θ ⇒ θ = tan - 1 x ⇒ d x = sec 2 θ d θ ⇒ θ = tan - 1 ( 1 ) = π / 4 ⇒ θ = tan - 1 ( 0 ) = 0 ⇒ I = ∫ 0 π / 4 log ( 1 + tan θ ) 1 + tan 2 θ · sec 2 θ d θ = ∫ 0 π 4 log 1 + tan θ d θ ...... i Now, we know that ∫ 0 a f ( x ) d x = ∫ f ( a - x ) d x ∴ I = ∫ 0 π / 4 log ( 1 + tan ( π / 4 - θ ) ) d θ Now, tan π 4 - θ = tan π 4 - tan θ 1 + tan π 4 tan θ = 1 - tan θ 1 + tan θ ∴ I = ∫ 0 π / 4 log 1 + 1 - tan θ 1 + tan θ d θ I = ∫ 0 π / 4 log 2 1 + tan θ d θ = ∫ 0 π / 4 [ log 2 - log ( 1 + tan θ ) ] d θ I = ∫ 0 π / 4 log 2 d θ - I [Using i ] ⇒ 2 I = log 2 ∫ 0 π / 4 d θ ⇒ I = 1 2 log 2 [ θ ] 0 π / 4 ⇒ I = 1 2 log 2 π 4 - 0 Hence, ∫ 0 1 log ( 1 + x ) 1 + x 2 d x = π 8 log 2

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

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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.