Question

If the function,

$f\left(x\right)=\left\{\begin{array}{l}{k}_1{(x-\mathrm{\pi })}^2-1,\mathrm{x}\le \mathrm{\pi }\\ k_2\cos \left(x\right),x>\mathrm{\pi }\end{array}\right.$ is twice differentiable, then the ordered pair $\left(k_1,k_2\right)$ is equal to:

• A. $(1,1)$
• B. $(1,0)$
• C. $\left(\frac{1}{2},-1\right)$
• D. $\left(\frac{1}{2},1\right)$

Answer 1

The correct option is D

$\left(\frac{1}{2},1\right)$

Explanation for the correct option:

Given function, $f\left(x\right)=\left\{\begin{array}{l}{k}_1{(x-\mathrm{\pi })}^2-1,\mathrm{x}\le \mathrm{\pi }\\ k_2\cos \left(x\right),x>\mathrm{\pi }\end{array}\right.$

It is given that the function is continuous and differentiable,

so, $f\left({\mathrm{\pi }}^{-}\right)=f\left({\mathrm{\pi }}^{+}\right)$

$$\begin{gathered} \Rightarrow k_1{(x-\mathrm{\pi })}^2-1={\mathrm{k}}_2\cos \left(x\right) \\ \Rightarrow {\mathrm{k}}_1{(\mathrm{\pi }-\mathrm{\pi })}^2-1={\mathrm{k}}_2\cos \left(\pi \right) \\ \Rightarrow -1={\mathrm{k}}_2(-1)[\because \cos \left(\pi \right)=-1] \end{gathered}$$

$\Rightarrow 1=k_2$

Now differentiating $f\left(x\right)$, we get,

$f\text{'}\left(x\right)=\left\{\begin{array}{l}2k_1(x-\mathrm{\pi }),\mathrm{x}\le \mathrm{\pi }\\ -k_2\sin \left(x\right),x>\mathrm{\pi }\end{array}\right.$

Again, $f\text{'}\left({\mathrm{\pi }}^{-}\right)=f\text{'}\left({\mathrm{\pi }}^{+}\right)$

$$\begin{gathered} \Rightarrow 2k_1(x-\mathrm{\pi })=-{\mathrm{k}}_2\sin \left(x\right) \\ \Rightarrow {\mathrm{k}}_1(\mathrm{\pi }-\mathrm{\pi })=-{\mathrm{k}}_2\sin \left(\mathrm{\pi }\right) \\ \Rightarrow 0=0[\because \sin \left(\pi \right)=0] \end{gathered}$$

Now, evaluate the double derivative,

$f\text{'}\text{'}\left(x\right)=\left\{\begin{array}{l}2k_1,\mathrm{x}\le \mathrm{\pi }\\ -k_2\cos \left(x\right),x>\mathrm{\pi }\end{array}\right.$

So, $f\text{'}\text{'}\left({\mathrm{\pi }}^{-}\right)=f\text{'}\text{'}\left({\mathrm{\pi }}^{+}\right)$

$$\begin{gathered} \Rightarrow 2k_1=-k_2\cos \left(x\right) \\ \Rightarrow 2k_1=1[\because {\mathrm{k}}_2=1;\cos \left(\mathrm{\pi }\right)=-1] \\ \Rightarrow k_1=\frac{1}{2} \end{gathered}$$

Hence, option (D), i.e. $\left(\frac{1}{2},1\right)$ is the correct answer.