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

Options

• a = 0

• b = 0

• c = 0

• none of these

Answer 1

(b) b = 0 

We have, 
\[f\left( x \right) = a + b\left| x \right| + c \left| x \right|^4 \]

`f(x) = {(a +bx +c|x|^4 ,xge 0),(a-bx +cx^4,x<0):}`
\[\text { Here }, f\left( x \right)\text {  is differentiable at x = 0 }\]
\[ \therefore \left(\text {  LHD at x } = 0 \right) = \left(\text {  RHD at x }= 0 \right)\]
\[ \Rightarrow \lim_{x \to 0^-} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0} = \lim_{x \to 0^+} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0}\]
\[ \Rightarrow \lim_{x \to 0^-} \frac{a - bx + c x^4 - a}{x} = \lim_{x \to 0^+} \frac{a + bx + c x^4 - a}{x}\]
\[ \Rightarrow \lim_{h \to 0} \frac{a - b\left( 0 - h \right) + c \left( 0 - h \right)^4 - a}{0 - h} = \lim_{h \to 0} \frac{a + b\left( 0 + h \right) + c \left( 0 + h \right)^4 - a}{0 + h}\]
\[ \Rightarrow \lim_{h \to 0} \frac{a + bh + c h^4 - a}{- h} = \lim_{h \to 0} \frac{a + bh + c h^4 - a}{h}\]
\[ \Rightarrow \lim_{h \to 0} \frac{bh + c h^4}{- h} = \lim_{h \to 0} \frac{bh + c h^4}{h}\]
\[ \Rightarrow \lim_{h \to 0} \left( - b - c h^3 \right) = \lim_{h \to 0} \left( b + c h^3 \right)\]
\[ \Rightarrow - b = b\]
\[ \Rightarrow 2b = 0\]
\[ \Rightarrow b = 0\]