Limits, Continuity & Differentiability Question - None

Question

Let $f(x)=\begin{cases}x^3+2x^2 & x\in\mathbb{Q}\ -x^3+2x^2+ax & x\notin\mathbb{Q}\end{cases}$ . Integral value of $$a$$ so that $$f$$ is differentiable at $$x=1$$ :

Accepted Answer

Step 1: For continuity at x=1 (both branches must agree at every point near 1, especially along rationals and irrationals approaching 1):

At x=1: $f_\mathbb{Q}(1)=1+2=3$ , $f_{\mathbb{Q}^c}(1)=-1+2+a=1+a$ . Need $$a=2$$ .

Step 2: For differentiability, need equal derivatives: $f'_\mathbb{Q}(x)=3x^2+4x\big|_{x=1}=7$ , $f'_{\mathbb{Q}^c}(x)=-3x^2+4x+a\big|_{x=1}=-3+4+2=3 eq 7$ . Not differentiable at a=2.

Step 3: Actually, for a rational/irrational piecewise function to be differentiable at a point, BOTH pieces must be equal AND have equal derivatives there. Set equal derivatives: $3+4=-3+4+a\implies a=6$ . Check continuity: $$1+2=3$$ vs $$-1+2+6=7 eq 3$$ . Not continuous.

The only way such a function is differentiable at x=1 is if both values AND derivatives match. Setting both: $1+2+a=3\implies a=0$ and derivative check gives $$7 eq 4$$ . Conclusion: not possible (D).

Answer: (D) not possible

Trap: Trying to just set the derivatives equal without checking continuity first.

Key Concept: Rational/irrational piecewise: differentiable only if both pieces agree in value AND derivative at that point

Question:

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