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differentiability of absolute value function at 0

Source: LIMIT 2019 CBS1 P6

April 28, 2021
absolute valuefunctioncalculus

Problem Statement

Let f(x)=a0+a1∣x∣+a2∣x∣2+a3∣x∣3f(x)=a_0+a_1|x|+a_2|x|^2+a_3|x|^3, where a0,a1,a2,a3a_0,a_1,a_2,a_3 are constant. Then <spanclass=′latex−bold′>(A)</span> f(x) is differentiable at x=0 if whatever be a0,a1,a2,a3<span class='latex-bold'>(A)</span>~f(x)\text{ is differentiable at }x=0\text{ if whatever be }a_0,a_1,a_2,a_3 <spanclass=′latex−bold′>(B)</span> f(x) is not differentiable at x=0 if whatever be a0,a1,a2,a3<span class='latex-bold'>(B)</span>~f(x)\text{ is not differentiable at }x=0\text{ if whatever be }a_0,a_1,a_2,a_3 <spanclass=′latex−bold′>(C)</span> f(x) is differentiable at x=0 only if a1=0<span class='latex-bold'>(C)</span>~f(x)\text{ is differentiable at }x=0\text{ only if }a_1=0 <spanclass=′latex−bold′>(D)</span> f(x) is differentiable at x=0 only if a1=0,a3=0<span class='latex-bold'>(D)</span>~f(x)\text{ is differentiable at }x=0\text{ only if }a_1=0,a_3=0
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