Question

If $0<a<b<1,$ then which of the following is/are correct

• A. $\frac{b-a}{1+b^2}<{\tan }^{-1}b-{\tan }^{-1}a<\frac{b-a}{1+a^2}$
• B. $\frac{8}{25}<{\tan }^{-1}\frac{8}{19}<\frac{8}{17}$
• C. If $\log (1+x)=\frac{x}{1+ax},$ then $\frac{x}{1+x}<\log (1+x)<x,\forall x>0$
• D. If $\log (1+x)=\frac{x}{1+ax},$ then $\frac{x}{1+x}>\log (1+x)>x,\forall x>0$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\frac{b-a}{1+b^2}<{\tan }^{-1}b-{\tan }^{-1}a<\frac{b-a}{1+a^2}$
B $\frac{8}{25}<{\tan }^{-1}\frac{8}{19}<\frac{8}{17}$
C If $\log (1+x)=\frac{x}{1+ax},$ then $\frac{x}{1+x}<\log (1+x)<x,\forall x>0$

Let $f\left(x\right)={\tan }^{-1}x\Rightarrow f^{\prime }\left(x\right)=\frac{1}{1+x^2}$
By using mean value theorem
$\frac{{\tan }^{-1}b-{\tan }^{-1}a}{b-a}=\frac{1}{1+c^2};(a<c<b)$
$\because a<c<b$
$\therefore 1+a^2<1+c^2<1+b^2$
$\Rightarrow \frac{1}{1+b^2}<\frac{1}{1+c^2}<\frac{1}{1+a^2}$
$\Rightarrow \frac{1}{1+b^2}<\frac{{\tan }^{-1}b-{\tan }^{-1}a}{b-a}<\frac{1}{1+a^2}$
$\Rightarrow \frac{b-a}{1+b^2}<{\tan }^{-1}b-{\tan }^{-1}a<\frac{b-a}{1+a^2}$
Put $a=\frac{1}{4},b=\frac{3}{4},$ we get
$\Rightarrow \frac{\frac{2}{4}}{1+\frac{9}{16}}<{\tan }^{-1}\left(\frac{\frac{3}{4}-\frac{1}{4}}{1+\frac{3}{4}\times \frac{1}{4}}\right)<\frac{\frac{2}{4}}{1+\frac{1}{16}}$
$\Rightarrow \frac{8}{25}<{\tan }^{-1}\frac{8}{19}<\frac{8}{17}$

If $\log (1+x)=\frac{x}{1+ax}$
$0<a<1\therefore 0<ax<x\forall x>0$
$\Rightarrow 1<1+ax<1+x$
$\Rightarrow 1>\frac{1}{1+ax}>\frac{1}{1+x}$
$\Rightarrow x>\frac{x}{1+ax}>\frac{x}{1+x}$
$\Rightarrow \frac{x}{1+x}<\log (1+x)<x,\forall x>0$