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: State the condition.
A function is differentiable at a point only if it is continuous there and the left and right derivatives are equal.

Step 2: Apply continuity at x = 1.
Left limit: \(k(1-6+13) = 8k\). Right value: \(|1-3| = 2\).
\[ 8k = 2 \Rightarrow k = \frac14 \]

Step 3: Check the derivatives.
For \(x<1\), \(f'(x) = k(2x-6)\), so at \(x=1\) the left derivative is \(-4k\). For \(x\ge 1\) near 1, \(x-3<0\), so \(f(x)=3-x\) and \(f'(x)=-1\).
\[ -4k = -1 \Rightarrow k = \frac14 \]

Step 4: Compare.
Both conditions give the same \(k=\frac14\), so the function is differentiable at 1.

Step 5: Check the options.
Option 1 (\(4\)) and option 4 (\(-4\)) give \(8k=\pm32\), not 2. Option 3 (\(-\frac14\)) gives \(8k=-2\), not 2. Only option 2 works.

Final Answer:
The value of \(k\) is \(\frac14\), option 2. \[ \boxed{\frac{1}{4}} \]