Question

Let f (x) = a + b |x| + c |x|⁴, where a, b, and c are real constants. Then, f (x) is differentiable at x = 0, if
(a) a = 0
(b) b = 0
(c) c = 0
(d) none of these

Answer 1

(b) b = 0

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ f\left(x\right)=a+b\left|x\right|+c{\left|x\right|}^4 \\ f\left(x\right)=\left\{\begin{array}{l}a+bx+c{x}^{4}x\ge 0\\ a-bx+c{x}^{4}x<0\end{array}\right. \\ \mathrm{Here},f\left(x\right)\mathrm{is}\mathrm{differentiable}\mathrm{at}x=0 \\ \therefore \left(\mathrm{LHD}\mathrm{at}x=0\right)=\left(\mathrm{RHD}\mathrm{at}x=0\right) \\ \Rightarrow \underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ \Rightarrow \underset{x\to 0^{-}}{\lim }\frac{a-bx+c{x}^4-a}{x}=\underset{x\to 0^{+}}{\lim }\frac{a+bx+c{x}^4-a}{x} \\ \Rightarrow \underset{h\to 0}{\lim }\frac{a-b\left(0-h\right)+c{\left(0-h\right)}^4-a}{0-h}=\underset{h\to 0}{\lim }\frac{a+b\left(0+h\right)+c{\left(0+h\right)}^4-a}{0+h} \\ \Rightarrow \underset{h\to 0}{\lim }\frac{a+bh+c{h}^4-a}{-h}=\underset{h\to 0}{\lim }\frac{a+bh+c{h}^4-a}{h} \\ \Rightarrow \underset{h\to 0}{\lim }\frac{bh+c{h}^4}{-h}=\underset{h\to 0}{\lim }\frac{bh+c{h}^4}{h} \\ \Rightarrow \underset{h\to 0}{\lim }\left(-b-c{h}^3\right)=\underset{h\to 0}{\lim }\left(b+c{h}^3\right) \\ \Rightarrow -b=b \\ \Rightarrow 2b=0 \\ \Rightarrow b=0 \end{gathered}$$