用泰勒公式解決2021年高考數(shù)學(xué)全國(guó)乙卷比大小
原題:

泰勒公式是高等數(shù)學(xué)中的一種 對(duì)函數(shù)近似 的工具(具體描述見(jiàn)此動(dòng)態(tài))。構(gòu)造函數(shù)f(x)=ln(x)和g(x)=√(x)。則:
f'(x)=1/x
f''(x)=-1/(x^2)
f'''(x)=2/(x^3)
g'(x)=1/(2√x)
g''(x)=-1/4*x^(-3/2)
g'''(x)=3/8*x^(-5/2)
其中^符號(hào)為乘方符號(hào)
則根據(jù)泰勒公式,在x=x0處分別展開(kāi)f(x)與g(x):
f(x)
≈ln(x0)+f'(x0)(x-x0)+1/(2!)*f''(x0)(x-x0)^2+1/(3!)*f'''(x0)(x-x0)^3
=ln(x0)+1/x0*(x-x0)-1/2*1/(x0^2)*(x-x0)^2+1/6*2/(x0^3)(x-x0)^3
g(x)
≈√(x0)+g'(x0)(x-x0)+1/(2!)*g''(x0)(x-x0)^2+1/(3!)*g'''(x0)(x-x0)^3
=√(x0)+1/(2√x0)*(x-x0)-1/2*1/4*x0^(-3/2)*(x-x0)^2+1/6*3/8*x0^(-5/2)*(x-x0)^3
觀察到a b c選項(xiàng)的x值都接近1(距離1的距離小于1),且f(1)與g(1)非常容易求值,故取x0=1:
f(x)
≈ln(1)+1/1*(x-1)-1/2*1/(1^2)*(x-1)^2+1/6*2/(1^3)*(x-1)^3
=(x-1)-1/2*(x-1)^2+1/6*2*(x-1)^3
g(x)
≈√(1)+1/(2√1)*(x-1)-1/2*1/4*1^(-3/2)*(x-1)^2+1/6*3/8*1^(-5/2)*(x-1)^3
=1+1/2*(x-1)-1/8*(x-1)^2+1/16*(x-1)^3
則:
ln(1.01)
≈(1.01-1)-1/2*(1.01-1)^2+1/6*2*(1.01-1)^3
≈0.01-0.5*0.0001+0.00000033
=0.00995033
ln(1.02)
≈(1.02-1)-1/2*(1.02-1)^2+1/6*2*(1.02-1)^3
=0.02-1/2*0.0004+0.0000027
=0.0198027
√(1.04)
≈1+1/2*(1.04-1)-1/8*(1.04-1)^2+1/16*(1.04-1)^3
=1+0.5*0.04-1/8*0.0016+1/16*0.000064
=1.019804
所以
a=2ln(1.01)≈2*0.00995033=0.01990066
b=ln(1.02)≈0.0198027
c=√(1.04)-1≈0.019804
所以a>c>b,故選B