Question

The function $f\left(x\right)$ is defined as follows $f\left(x\right)=\{\begin{array}{cc}{x}^3;& x\le 1\\ a{x}^2+bx+c;& x>1\end{array}$. What must be the values of $a,b,c$ so that $f^{\prime \prime }\left(x\right)$ is continuous everywhere ?

• A. $a=3,b=-3,c=1$
• B. $a=3,b=3,c=1$
• C. $a=3,b=3,c=2$
• D. can not be determined

Answer 1

The correct option is A $a=3,b=-3,c=1$

$f\left(x\right)=x^{3}forx\le 1a{x}^2+bx+cforx>1$

$f^{\prime }\left(x\right)=3x^{2}forx\le 12ax+bforx>1$

$f^{\prime \prime }\left(x\right)=6xforx\le 12aforx>1$
For $f^{\prime \prime }$ to be continuous at $1$

$2a=6\Rightarrow a=3f^{\prime }shouldalsobecontinuousSo2\times 3\times 1+b=3\Rightarrow b=-3fshouldalsobecontinuous3\times 1\times 1-3\times 1+c=1\Rightarrow c=1$
Hence, answer is A