WBJEE · Maths · Continuity and Differentiability
Let \(f(x)=a_0+a_1|x|+a_2|x|^2+a_3|x|^3\), where \(a_0, a_1, a_2, a_3\) are real constants. Then \(f(x)\) is differentiable at \(x=0\)
- A whatever be \(a_0, a_1, a_2, a_3\)
- B for no values of \(a_0, a_1, a_2, a_3\)
- C only if \(a_1=0\)
- D only if \(a_1=0, a_3=0\)
Answer & Solution
Correct Answer
(C) only if \(a_1=0\)
Step-by-step Solution
Detailed explanation
\(f^{\prime}\left(0^{+}\right)=\lim _{x \rightarrow 0}\left(a_1+2 a_1 x+3 a_3 x^2\right)=a_1\) \(f^{\prime}\left(0^{-}\right)=\lim _{x \rightarrow 0}\left(-a_1+2 a_1 x-3 a_3 x^2\right)=-a_1\) \(f^{\prime}\left(0^{+}\right)=f^{\prime}\left(0^{-}\right) \quad \Rightarrow a_1=0\)
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