Question

Let $f(x)=a{x}^2+1$ 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)

$L{f}^{\prime }\left(1\right)=L{t}_{x\to 1-}\frac{f\left(x\right)-f\left(1\right)}{x-1}$
$=L{t}_{x\to 1}\frac{x+a-1-a}{x-1}=1$
$R{f}^{\prime }\left(1\right)=L{t}_{x\to 1+}\frac{f\left(x\right)-f\left(1\right)}{x-1}$
$=L{t}_{x\to 1}\frac{a{x}^2+1-1-a}{x-1}=L{t}_{x\to 1}\frac{a\left(x^2-1\right)}{x-1}$
$=L{t}_{x\to 1}a\left(x+1\right)=2a$
Since $f^{\prime }\left(1\right)$ exists $\therefore 1=2a\Rightarrow a=\frac{1}{2}$.