Question

If $0<a<b<1,$ then which of the following is true

• A. $\frac{1}{\sqrt{1-a^2}}<\frac{{\sin }^{-1}b-{\sin }^{-1}a}{b-a}<\frac{1}{\sqrt{1-b^2}}$
• B. $\frac{1}{\sqrt{1-b^2}}<\frac{{\sin }^{-1}b-{\sin }^{-1}a}{b-a}<\frac{1}{\sqrt{1-a^2}}$
• C. $\frac{{\sin }^{-1}b-{\sin }^{-1}a}{b-a}=\frac{1}{\sqrt{1-a^2}}$
• D. $\frac{{\sin }^{-1}b-{\sin }^{-1}a}{b-a}=\frac{1}{\sqrt{1-b^2}}$

Answer 1

The correct option is A $\frac{1}{\sqrt{1-a^2}}<\frac{{\sin }^{-1}b-{\sin }^{-1}a}{b-a}<\frac{1}{\sqrt{1-b^2}}$

Let $f\left(x\right)={\sin }^{-1}x,x\in [a,b]$
Let $f\left(x\right)$ be continuous on $[a,b]$ and differentiable on $(a,b).$
By $LMVT,$ there exist at least one $c\in (a,b)$
$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
$\Rightarrow \frac{1}{\sqrt{1-c^2}}=\frac{{\sin }^{-1}b-{\sin }^{-1}a}{b-a},c\in (a,b)$

$\Rightarrow a<c<b$
$\Rightarrow a^2<c^2<b^2$
$\Rightarrow -b^2<-c^2<-a^2$
$\Rightarrow 1-b^2<1-c^2<1-a^2$
$\Rightarrow \sqrt{1-b^2}<\sqrt{1-c^2}<\sqrt{1-a^2}$
$\Rightarrow \frac{1}{\sqrt{1-a^2}}<\frac{1}{\sqrt{1-c^2}}<\frac{1}{\sqrt{1-b^2}}$
$\therefore \frac{1}{\sqrt{1-a^2}}<\frac{{\sin }^{-1}b-{\sin }^{-1}a}{b-a}<\frac{1}{\sqrt{1-b^2}}$