Question

If \[f\left( x \right) = a\left| \sin x \right| + b e^\left| x \right| + c \left| x \right|^3\] 

पर्याय

• \[a = b = c = 0\]

• \[a = 0, b = 0; c \in R\]

   

• \[b = c = 0, a \in R\]

• \[c = 0, a = 0, b \in R\]

   

Answer 1

(b) \[a = 0, b = 0; c \in R\]

\[\text{ We have }, \]
\[f\left( x \right) = a \left| \sin x \right| + b e^\left| x \right| + c \left| x \right|^3 \]
`= {(a sin x+be^x +cx^3 , 0<x<pi/2),(-a sin x +be^(-x) -cx^3 , -pi/2 <x <0):}`
\[\text{Here,} f\left( x \right)\text { is differentiable at x} = 0\]
\[\text{Therefore}, \left(\text {  LHD at x } = 0 \right) = \left( \text { RHD at x } = 0 \right)\]
\[ \Rightarrow \lim_{x \to 0^-} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0} = \lim_{x \to 0^+} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0}\]
\[ \Rightarrow \lim_{x \to 0^-} \frac{- a \ sinx + b e^{- x} - c x^3 - b}{x} = \lim_{x \to 0^+} \frac{a \sin x + b e^x + c x^3 - b}{x}\]
\[ \Rightarrow \lim_{h \to 0} \frac{- a \sin\left( 0 - h \right) + b e^{- \left( 0 - h \right)} - c \left( 0 - h \right)^3 - b}{0 - h} = \lim_{h \to 0} \frac{a \sin \left( 0 + h \right) + b e^\left( 0 + h \right) + c \left( 0 + h \right)^3 - b}{0 + h}\]
\[ \Rightarrow \lim_{h \to 0} \frac{a \sin h + be {}^h + c h^3 - b}{- h} = \lim_{h \to 0} \frac{a \sin h + b e^h + c h^3 - b}{h}\]
\[ \Rightarrow \lim_{h \to 0} \frac{a \cos h + b e^h + 3c h^2}{- 1} = \lim_{h \to 0} \frac{a \cos h + b e^h + 3c h^2}{1} \left( By L'Hospital rule \right)\]
\[ \Rightarrow - \left( a + b \right) = a + b\]
\[ \Rightarrow - 2\left( a + b \right) = 0\]
\[ \Rightarrow a + b = 0\]
\[\text{This is true for all value of c}\]
\[ \text{therefore c} \in R\]
\[\text{In the given options, option} \left( b \right) \text { satisfies a + b = 0 and c} \in R\]