Question

Let $f(x) = ax^2 + 1 $ for $x > 1 $
$= x + a $ for $x \le 1$, then $f$ is derivable at x = 1 if

• A. a = 0
• B. a = $\frac{1}{2}$
• C. a = 1
• D. a = 2

Answer 1

Answer: (B)

$Lf'\left(1\right) = Lt_{x\to1-} \frac{f\left(x\right) -f\left(1\right)}{x-1}$
$ = Lt_{x\to1} \frac{x+a-1-a}{x-1} =1 $
$Rf'\left(1\right) = Lt_{x\to1+} \frac{f\left(x\right)-f\left(1\right)}{x-1} $
$=Lt_{x\to1} \frac{ax^{2}+1-1-a}{x-1} = Lt_{x \to1} \frac{a\left(x^{2}-1\right)}{x-1} $
$=Lt_{x\to1} a\left(x+1\right)=2a$
Since $ f'\left(1\right)$ exists $ \therefore 1=2a \Rightarrow a = \frac{1}{2}$.