Question

If $f(x)=\left\{\begin{array}{cl}a+\tan ^{-1}(x-b) & ; \quad x \geq 1 \\ \frac{x}{2} & ; \quad x<1\end{array}\right.$ 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 $x=1$
$\therefore f\left(1^{-}\right)=f\left(1^{+}\right)=f\left(1\right)$
$\frac{1}{2}=a+\left(tan\right)^{- 1} \left(1 - b\right)\ldots \left(i\right)$
$f^{\prime}(x)=\left\{\begin{array}{ccc}\frac{1}{1+(x-b)^{2}} & ; & x \geq 1 \\ \frac{1}{2} & ; & x<1\end{array}\right.$
Now for $f\left(x\right)$ to be differentiable at $x=1$
$f^{'} \left(1^{-}\right) = f^{'} \left(1^{+}\right)$
$\frac{1}{1 + \left(1 - b\right)^{2}}=\frac{1}{2}\Rightarrow 1+\left(1 - b\right)^{2}=2$
$\Rightarrow b=2$ or $0\ldots \left(i i\right)$
From equation $\left(i\right)$ and $\left(i i\right)$ , 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 $