functions....4!!!!

IIT JEE 2015 Preparation › Calculus questions › functions....4!!!!

If f(x) = 2x + |x|, g(x) = 1/3 (2x - |x|) and h(x) = f(g(x)), then domain of sin^-1(h(h(h(h........h.h(x)........)))) { in brackets.. h(x) is n times} is:
a. [-1, 1]
b. [-1, -1/2] U [1/2. 1]
c. [-1, -1/2]
d. [1/2, 1]

#Mathematics #Calculus

UP 0 DOWN 0 0 2
Post on FacebookPost anonymously?

2 Answers

Mani Pal Singh ·2009-04-23 02:52:19

let us assume that x>0
so

f(x)=3x and g(x)=x/3

f(g(x))=x=h(x)

sin^-1(x.x.x.x.x.....x)(n times)
we get

sin^-1xⁿ
and we know that the domain of sin^-1x is [0 ,1](x>0)

if x<0

we will get f(g(x))=x

so again

sin^-1xⁿ
and we know that the domain of sin^-1x is [-1 ,0](x<0)

so A is correct

UP 0 DOWN 0
Post on FacebookPost anonymously?

Surbhi Agrawal ·2009-04-23 02:57:49

ohk!!!!

UP 0 DOWN 0
Post on FacebookPost anonymously?

Your Answer

Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

Close [X]

functions....4!!!!

2 Answers

Your Answer

Add Image

Similar Questions