Question

Let $f\left(x\right)=\{\begin{array}{cc}A+si{n}^{-1}(x+B),& \forall x\ge 1\\ x,& \forall x<1\end{array}$ is differentiable then

• A. $A=-1,B=-1$
• B. $A=1,B=-1$
• C. $A=B=1$
• D. $A=0,B=1$

Answer 1

Answer: (B)

$A=1,B=-1$

For continunity
$f\left(1^{-}\right)=f\left(1^{+}\right)$
$f\left(1^{+}\right)=A+si{n}^{-1}(1+B)$
$f\left(1^{-}\right)=1$
$A+si{n}^{-1}(1+B)=1$
For differentiable
$f^{\prime }\left(1^{-}\right)=f^{\prime }\left(1^{+}\right)$
$f^{\prime }\left(1^{-}\right)=f^{\prime }\left(1^{+}\right)=\frac{1}{\sqrt{1-(x+B{)}^2}}=\frac{1}{\sqrt{1-(1+B{)}^2}}$
$1=\frac{1}{1-(1+B{)}^2}$
$1-(1+B{)}^2=1$
$B=-1,A=1$