Question

For what triplets of real numbers (a, b, c) with $a\ne 0$ the function
$f\left(x\right)=\{\begin{array}{cc}x& x\le 1\\ a{x}^2+bx+c& otherwise\end{array}$ is differentiable for all real x?

• A. $\left\{\right(a,1-2a,a)/a\in R,a\ne 0\}$
• B. $\left\{\right(a,1-2a,c)/a,c\in R,a\ne 0\}$
• C. $\left\{\right(a,b,c)/a,b,c\in R,a+b+c=1\}$
• D. $\left\{\right(a,1=2a,0)/a\in R,a\ne 0\}$

Answer 1

The correct option is A $\left\{\right(a,1-2a,a)/a\in R,a\ne 0\}$

graph of the given fraction is

$\therefore$ From graph differentiality should be checked at point $x=1$, alse friction is differentiable.

To check differentiality at $x=1$ we must firstly check continuity at $x=1$

$\therefore$ For fraction to be continuous

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

$\therefore 1=1=a(1{)}^2+b(1)+c$

$\therefore a+b+c=\mathrm{1.......}\left(1\right)$

Now checking differentiality

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

$\therefore 1=2a\left(1\right)+b$

$\therefore 2a+b=\mathrm{1.......}\left(2\right)$

from equation $\left(1\right)$ and $$(2)$

$b=1-2a$

$\therefore$ triplet is $(a,1-2a,a)$

such that $a\in R$ also $a\ne 0$