Question

If $f(x)=\{\begin{array}{cl}a+{\tan }^{-1}(x-b)& ;x\ge 1\\ \frac{x}{2}& ;x<1\end{array}$ is differentiable at $x=1,$ then $4a-b$ can be

• A. $0$
• B. $1$
• C. $-1$
• D. $\pi$

Answer 1

Answer: (D)

Also $f\left(x\right)$ must be continuous at $x=1$
$\therefore f\left(1^{-}\right)=f\left(1^{+}\right)=f\left(1\right)$
$\frac{1}{2}=a+{\left(tan\right)}^{-1}\left(1-b\right)\dots \left(i\right)$
$f^{\prime }(x)=\{\begin{array}{ccc}\frac{1}{1+(x-b{)}^2}& ;& x\ge 1\\ \frac{1}{2}& ;& x<1\end{array}$
Now for $f\left(x\right)$ to be differentiable at $x=1$
$f^{{}^{\prime }}\left(1^{-}\right)=f^{{}^{\prime }}\left(1^{+}\right)$
$\frac{1}{1+{\left(1-b\right)}^2}=\frac{1}{2}\Rightarrow 1+{\left(1-b\right)}^2=2$
$\Rightarrow b=2$ or $0\dots \left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right)$ , we get
$\frac{1}{2}=a\pm \frac{\pi }{4}\Rightarrow a=\frac{1}{2}-\frac{\pi }{4}$ or $\frac{1}{2}+\frac{\pi }{4}$
$\left(a,b\right)\equiv \left(\frac{1}{2}-\frac{\pi }{4},0\right)$ or $\left(\frac{1}{2}+\frac{\pi }{4},2\right)$
Hence, $4a-b=2-\pi ,\pi$