No links please Q (iii) For x 0 prove that x-x22 log1+x x-x221+x - Maths - Differential Equations

Question

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Q (iii) For x > 0 prove that
x - x 2 2 < log 1 + x < x - x 2 2 1 + x

Shruti Tyagi · Accepted Answer

Dear student, If a function is increasing, that means that for x10 then, the function is increasing We'll note f(x)=x2-x1 f(x) = log(1+x)-x+x22If the first deivative of f(x) is positive, then f(x) is increasing f'(x) = ddxlog1+x-x-x22 f'(x) = 1(1+x)-1+2x2 f'(x) = 1(1+x)-(1-x)f'(x) = 1-1+x2(1+x)f'(x) = x2(1+x)>0 for any x>0So, f(x)>0But fx=log(1+x)-x+x22>0⇒log(1+x)>x-x22Similarly,Consider the function h(x)=x-x221+x-log1+xh'(x)=ddxx-log1+x-x221+xh'(x)=1-1(1+x)-4x1+x-2x241+x2 --[using quotient rule]h'(x)=1-1(1+x)-4x+4x2-2x241+x2h'(x)=1-1(1+x)--2x241+x2h'(x)=41+x2-41+x+2x241+x2h'(x)=41+x2+2x-4+4x+2x241+x2h'(x)=4+4x2+8x-4+4x+2x241+x2h'(x)=6x2+12x41+x2h'(x)=6xx+241+x2h'(x)=3xx+221+x2>0∴h(x) is increasing ⇒x-x221+x-log1+x>0⇒x-x221+x>log1+xSo, log(1+x) lies between x-x22 and x-x221+x Hence Proved Regards

No links please Q. (iii) For x > 0 prove that x - x 2 2 < log 1 + x < x - x 2 2 1 + x

No links please
Q. (iii) For x > 0 prove that
$x - \frac{x^{2}}{2} < \log (1 + x) < x - \frac{x^{2}}{2 (1 + x)}$