Question

Function \(f(x)=\begin{cases} k\left(x^2-6x+13\right) & : x<1 \\ |x-3| & : x\ge 1 \end{cases}\) is differentiable at \(x=1\), then the value of \(k\) is

• A. \(4\)
• B. \(\frac{1}{4}\)
• C. \(\frac{-1}{4}\)
• D. \(-4\)

Hint:

Use continuity and equal left and right derivatives at \(x=1\).

Answer 1

The Correct Option is B

Step 1: Rewrite the second piece.
Near $x=1$ (for $x$ slightly above 1), $x-3$ is negative, so $|x-3| = 3-x$.

Step 2: Use the derivative test first.
Left slope is $k(2x-6)$ at $x=1$, which is $-4k$. Right slope of $3-x$ is $-1$. Setting them equal gives $-4k=-1$.

Step 3: Solve for k.
\[ k = \frac{1}{4} \]

Step 4: Verify by continuity.
With $k=\frac14$ the left value is $\frac14(1-6+13) = \frac{8}{4} = 2$, and the right value is $|1-3|=2$. They match, so $f$ is continuous, and the equal slopes make it differentiable.

Final Answer:
So $k=\frac14$. \[ \boxed{\frac{1}{4}} \]