Question

Let $f\left(x\right)=\{\begin{array}{cc}{x}^3& x<1\\ a{x}^2+bx+c& x\ge 1\end{array}$. If $f^{\prime }\left(1\right)$ exists the value of $(a^2+b^2+c^2)$ is

• A. $20$
• B. $21$
• C. $19$
• D. $17$

Answer 1

The correct option is C $19$

$f^{\prime }\left(x\right)=\{\begin{array}{c}3x^{2}x<1\\ 2ax+bx\ge 1\end{array}$

For $\underset{x\to 1}{lim}{f}^{\prime }\left(x\right)$ to exist

$\underset{x\to 1^{-}}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 1^{+}}{lim}{f}^{\prime }\left(x\right)$

$\Rightarrow 3=2a+b$

Also, $\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)\Rightarrow 1=a+b+c$

$\underset{x\to 1^{-}}{lim}{f}^{\prime \prime }\left(x\right)=\underset{x\to 1^{+}}{lim}{f}^{\prime \prime }\left(x\right)$

$\Rightarrow \underset{x\to 1^{-}}{lim}6x=\underset{x\to 1^{+}}{lim}2a\Rightarrow 2a=6\Rightarrow a=3$

And $b=-3c=1$

$\Rightarrow a^2+b^2+c^2=19$

Hence, correct answer is $19$