TextBook Solutions
- If sin-1(sin\(\frac { 2π }{ 3}\)) sin⁻¹(sin 2π / 3 ) = sin⁻¹[ sin(π – π / 3 ) ] = sin⁻¹[sin π / 3 ] = π / 3 . Please note : sin⁻¹(sin 2π / 3 ) ≠ sin 2π / 3 , since the range... 1
- tan-1(tan\(\frac { 3π }{ 4 }\)) tan⁻¹(tan 3π / 4 ) = tan⁻¹ tan(π – π / 4 ) = tan⁻¹(- tan π / 4 ) = tan⁻¹[tan( -π / 4 ) ] = – π / 4 Again note that tan⁻¹ tan 3π / 4 ≠ 3π / 4... 1
- cos-1(cos \(\frac { 7π }{ 6 }\)) is equal to (A) \(\frac { 7π }{ 6 }\) (B) \(\frac { 5π }{ 6 }\) (B) 5π / 6 cos⁻¹(cos 7π / 6 ) = cos⁻¹cos ( π + π / 6 ) = cos⁻¹(- cos π / 6 ) = cos⁻¹[cos ( π – π / 6 ) = cos⁻¹cos 5π / 6 = 5π / 6 . Note tha... 1
- sin[ \(\frac { π }{ 3 }\)-sin-1(-\(\frac { 1 }{ 2 }\)) ] is equal to (A) \(\frac { 1 }{ 2 }\) Answer: (D) 1 The principal value branch of sin⁻¹ is [-π/2, π/2] , and sin(-π/6) = -1/2 . So sin⁻¹(-1/2) = -π/6. Then sin[π/3 + π/6] = sinπ/... 1
- tan-1\(\sqrt{3}\) – cot-1(-\(\sqrt{3}\)) is equal to (A) π (B) \(\frac { π }{ 2 }\) (C) 0 (D) 2\(\sqrt{3}\) (B) π / 2 tan⁻¹ √(3) = tan⁻¹(tan π / 3 ) = π / 3 and cot⁻¹ √(3) = cot⁻¹(-cot π / 6 ) = cot⁻¹cot ( π – π / 6 ) = cot⁻¹cot 5π / 6 = 5π / 6 . s... 1
- sin-1(-\(\frac { 1 }{ 2 }\)) Let sin⁻¹(- 1 / 2 ) = y ∴ sin y = 1 / 2 = – sin π / 6 = sin(- π / 6 ) The range of principal value branch of sin⁻¹ is [- π / 2 , π / 2 ]. He... 1
- cos-1\(\left(-\frac{\sqrt{3}}{2}\right)\) Answer: cos⁻¹(-√(3)/2) = 5π/6 The principal value branch of cos⁻¹ is [0, π] . We know cosπ/6 = √(3)/2 . Using cos(π - θ) = -cosθ : cos(π - π... 1
- cos-1(-\(\frac { 1 }{ 2 }\)) Let cos⁻¹(- 1 / 2 ) = y ∴ cos y = – 1 / 2 = – cos π / 3 = cos(π – π / 3 = cos 2π / 3 The range of principal value branch is (0, π) ∴ princip... 1
- Find the principal value of tan-1(-1) Answer: tan⁻¹(-1) = -π/4 The principal value branch of tan⁻¹ is (-π/2, π/2) . Since tanπ/4 = 1 and tan is an odd function, tan(-π/4) = -1. -... 1
- sec-1\(\left(-\frac{2}{\sqrt{3}}\right)\) Answer: sec⁻¹(-2/√(3)) = 5π/6 The principal value branch of sec⁻¹ is [0, π] - \π/2\ . We know secπ/6 = 2/√(3) . For a negative value we use... 1
- cot-1 (\(\sqrt{3}\)) Let cot⁻¹( √(3) ) = y ∴ cot y = √(3) = cot π / 6 The range of principal value branch of cot⁻¹ is [0, π] ∴ principal value of cot⁻¹ √(3) is π... 1
- cos-1\(\left(-\frac{1}{\sqrt{2}}\right)\) Let cos⁻¹ (-1/√(2)) = y ∴ cos y = (-1/√(2)) = cos( π / 4 ) = cos(π – π / 4 ) The range of principal value branch of cos⁻¹ is [0, π]. ∴ princ... 1
- cosec-1 (-\(\sqrt{2}\)) Let cosec-1(- √(2) ) = y ∴ cosec y = – √(2) = – cosec π / 4 = cosec (- π / 4 ). The range of principal value branch of cosec-1 is (- π / 2 ,... 1
- tan-1(1) + cos-1(-\(\frac { 1 }{ 2 }\)) + sin-1(\(\frac { -1 }{ 2 }\)) Answer: 3π/4 Use the principal value branches: tan⁻¹(1) = π/4 , since tanπ/4 = 1 and π/4in(-π/2, π/2) . cos⁻¹(-1/2) = π - cos⁻¹1/2 = π - π/3... 1
- cos-1(\(\frac { 1 }{ 2 }\)) + 2 sin-1(\(\frac { 1 }{ 2 }\)) cos-1( 1 / 2 ) + 2 sin-1( 1 / 2 ) Now, cos-1( 1 / 2 ) = π / 3 and sin-1( 1 / 2 ) = π / 6 ∴ cos-1( 1 / 2 ) + 2 sin-1( 1 / 2 ) = π / 3 + 2 x π... 1
- If sin-1 = y, then (A) 0 ≤ y ≤ π (B) – \(\frac { π }{ 2 }\) ≤ y ≤ \(\frac { π }{ 2 }\) (C) 0 < y < π optin B The range of principal value branch of sin-1 is [- π / 2 , π / 2 ]∴ if sin-1 x = y, then – π / 2 ≤ y ≤ π / 2 1
- tan-1\(\sqrt{3}\) – sec-1(-2) is equal to (A) π (B) – \(\frac { π }{ 3 }\) (C) \(\frac { π }{ 3 }\) option B tan-1 √(3) = π / 3 and sec-1(-2) = π – π / 3 = 2π / 3 , since principal value branch of sec-1 is [0, π] – π / 2 ∴ tan-1 √(3) – sec-... 1
- Let R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that gof = fog = IR. Let f : X → Y, where X ⊆ R, Y ⊆ R. Consider an arbitrary element y of Y. By definition, y = 10x + 7 for some x ∈ X. ⇒ x = y-7 / 10 . Let us... 1
- Let f : W → W be defined as f(n) = n -1, if n is odd and f(n) = n + 1, if n is even. We are given, f : W → W, defined as \beginarrayll n-1, & n is odd \\ n+1, & n is even endarray. Let f(n1) =f(n2). If n1 is odd and n2 is eve... 1
- If f : R → R is defined by f(x) = x² – 3x + 2, find f[f(x)]. f(x) = x² – 3x + 2 f[f(x)] = [f(x)]² – 3[f(x)] + 2 = (x² – 3x + 2)² – 3(x² -3x + 2) + 2 = (x4 + 9x² + 4 – 6x³ – 12x + 4x²) – 3x² + 9x – 6 +... 1
- Show that the function f : R → { x ∈ R : – 1 – x < 1} defined by f(x) = \(\frac{x}{1+|x|}\) x ∈ R is one-one and onto. (i) Let x >0, f(x) = x / 1+x So, f(x1) = f(x2) ⇒ fracx₁1+x₁ = fracx₂1+x₂ or x1(1 + x2) = x2(1 + x1) or x1 + x1x2 = x2 + x1x2 ⇒ x1 = x2 When... 2
- Show that the function f : R → R given by f(x) = x³ is injective. f(x) = x³ ⇒ f(x1) = f(x2) ⇒ x³1 = x³1 ⇒ x1 = x2 ⇒ f is one-one, i.e., f is injective. 1
- Give examples of two functions f : N → Z and g : Z → Z such that gof is injective but g is not injective. Example: f(x) = x for f:mathbbN→mathbbZ , and g(x) = |x| for g:mathbbZ→mathbbZ . g is not injective: g(-1) = |-1| = 1 and g(1) = 1 , so two... 1
- Give examples of two functions f : N → N and g : N → N such that gof is onto but f is not onto. Example: f(x) = x + 1 and g(x) = begincases x - 1, & x > 1 \\ 1, & x = 1 endcases f is not onto: f(x) = x + 1 ≥ 2 for every x in mathbbN , s... 1