Continuity & DifferentiabilityhardFree

Value of k Making a Piecewise Quadratic-Modulus Function Differentiable

JEE Maths question with a full step-by-step solution.

Question
Let
f(x)={a+(x−b)2for ∣x−b∣≤kc+∣x−b∣for ∣x−b∣>kf(x) = \begin{cases} a+(x-b)^{2} & \text{for } |x-b| \le k \\ c+|x-b| & \text{for } |x-b| > k \end{cases}
 then the positive value of kk so that f(x)f(x) becomes differentiable at x=b−kx = b-k and x=b+kx = b+k is
A32\dfrac32
B12\dfrac12correct
C11
D22
Solution
Step 1: Unfold the moduli so the three pieces are explicit. The condition ∣x−b∣≤k|x-b| \le k means b−k≤x≤b+kb-k \le x \le b+k, and outside that ∣x−b∣|x-b| is b−xb-x on the left and x−bx-b on the right:
f(x)={c+b−xx<b−ka+(x−b)2b−k≤x≤b+kc+x−bx>b+kf(x) = \begin{cases} c+b-x & x < b-k \\ a+(x-b)^{2} & b-k \le x \le b+k \\ c+x-b & x > b+k \end{cases}
 Step 2: Differentiate each piece.
f′(x)={−1x<b−k2(x−b)b−k<x<b+k+1x>b+kf'(x) = \begin{cases} -1 & x < b-k \\ 2(x-b) & b-k < x < b+k \\ +1 & x > b+k \end{cases}
 Step 3: Match the derivatives at the right-hand join x=b+kx = b+k.
2(x−b)∣x=b+k=2kmust equal+1.2(x-b)\Big|_{x=b+k} = 2k \qquad \text{must equal} \qquad +1 .
 Step 4: Solve.
2k=1  ⟹  k=12.2k = 1 \implies k = \frac12 .
 Step 5: Check the left-hand join x=b−kx = b-k gives the same value.
2(x−b)∣x=b−k=−2kmust equal−1  ⟹  k=12.2(x-b)\Big|_{x=b-k} = -2k \qquad \text{must equal} \qquad -1 \implies k = \frac12 .
 Step 6: (Consistency.) Continuity at x=b+kx = b+k then forces a+k2=c+ka+k^{2} = c+k, which fixes a relation between aa and cc but does not affect kk. So the answer depends only on the slopes. Answer: (2).
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