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) = \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

k

so that

f(x)

becomes differentiable at

x = b-k

and

x = b+k

\dfrac32

\dfrac12

correct

1

2

Solution

Step 1: Unfold the moduli so the three pieces are explicit. The condition

|x-b| \le k

means

b-k \le x \le b+k

, and outside that

|x-b|

b-x

on the left and

x-b

on the right:

f(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) = \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+k

2(x-b)\Big|_{x=b+k} = 2k \qquad \text{must equal} \qquad +1 .

Step 4: Solve.

2k = 1 \implies k = \frac12 .

Step 5: Check the left-hand join

x = b-k

gives the same value.

2(x-b)\Big|_{x=b-k} = -2k \qquad \text{must equal} \qquad -1 \implies k = \frac12 .

Step 6: (Consistency.) Continuity at

x = b+k

then forces

a+k^{2} = c+k

, which fixes a relation between

a

and

c

but does not affect

k

. So the answer depends only on the slopes. Answer: (2).

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