function value

VITEEE 2015 Papers › Calculus questions › function value

\hspace{-16}$If $\bf{f(x)=\frac{x^2}{1+x^2}.}$ Then Determine value of the following expression\\\\\\ $\bf{f\left(\frac{1}{2000}\right)+f\left(\frac{2}{2000}\right)+...+f\left(\frac{1999}{2000}\right)+f\left(\frac{2000}{2000}\right)+f\left(\frac{2000}{1999}\right)+...+f\left(\frac{2000}{1}\right)}$

#Mathematics #Calculus

UP 0 DOWN 0 0 1
Post on FacebookPost anonymously?

1 Answers

2305

Shaswata Roy ·2013-02-28 06:48:34

\inline \dpi{200} f(x)+f(\frac{1}{x}) = \frac{x^{2}}{1+x^{2}}+\frac{\frac{1}{x^{2}}}{1+\frac{1}{x^{2}}} = \frac{x^{2}+1}{x^{2}+1} = 1

Required sum = \inline \dpi{200} [f(\frac{1}{2000})+f(\frac{2000}{1})]+[f(\frac{2}{2000})+f(\frac{2000}{2})]......+f(\frac{2000}{2000})

= 1999 + 1/2
=1999.5

UP 1 DOWN 1
man111 singh Yes .
Upvote·0· Reply ·2013-02-28 07:09:32
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