Function \( f(x)=\begin{cases}k(x^2-6x+13) & :x\lt 1\\|x-3| & :x\ge 1\end{cases} \) is differentiable at \( x=1 \), then the value of \( k \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{1}{4} \)
Near x = 1 (for x < 3), \( x-3\lt 0 \), so \( |x-3|=3-x \).
A differentiable function is continuous, so the values at x = 1 must match:
\( \displaystyle \lim_{x\to 1^-}f(x)=k(1-6+13)=8k \)
\( \displaystyle f(1)=|1-3|=2 \)
\( \displaystyle 8k=2\ \Rightarrow\ k=\frac{1}{4} \)
The one-sided derivatives also agree for this k:
\( \displaystyle Lf'(1)=k(2x-6)\big|_{x=1} \)
\( \displaystyle =-4k=-1 \)
\( \displaystyle Rf'(1)=\frac{d}{dx}(3-x)=-1 \)
Both are −1, so f is differentiable at x = 1 when \( k=\frac{1}{4} \).