設(shè)S(x) = ln[x + √(x² + 1)]
S(- x) = ln[- x + √(x² + 1)]
= ln{[√(x² + 1) - x] * [√(x² + 1) + x]/[√(x² + 1) + x]}
= ln{[(x² + 1) - x²]/[x + √(x² + 1)]}
= ln{1/[x + √(x² + 1)]}
= - ln[x + √(x² + 1)]
= - S(x)
∴S(x)是奇函數(shù)
∴∫(- 2→2) ln[√(x² + 1) + x] dx = 0
y = tanx、- π/2 ≤ x ≤ π/2
令tanx = 0 ==> x = kπ、k∈Z
- π/2 ≤ x ≤ π/2
- π/2 ≤ kπ ≤ π/2
- 1/2 ≤ k ≤ 1/2
k的實(shí)數(shù)整數(shù)解只有k = 0
所以只有一個(gè)實(shí)數(shù)解.第二題答案是3，所以你確定答案是錯(cuò)的嗎？沒錯(cuò)，因?yàn)樵? π/2 ≤ x ≤ π/2內(nèi)tanx也只有x = 0這個(gè)答案而y = x與y = tanx也只有x = 0，y = 0這個(gè)交點(diǎn)