The function f(x)=ae-bx2,x 0 is differentiable. Then which of the following is not true?

f(0−)=f(0)=f(0+)⇒a=1=cf1(0−)=f1(0+)⇒ae−bx2(−2bx)x=0=−sinx at x-0⇒0=0

The function f(x)=ae-bx2,x 0 is differentiable. Then which of the following is not true?

f(0−)=f(0)=f(0+)⇒a=1=cf1(0−)=f1(0+)⇒ae−bx2(−2bx)x=0=−sinx at x-0⇒0=0

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$The function f (x) = \{\begin{array}{cl} {ae}^{- {bx}^{2}}, & x < 0 \\ 1, & x = 0 \\ c \cdot cosx, & x > 0 \end{array} is differentiable. Then which of the following$ is not true?

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$b \in R$

c<0

a=c

a=1

answer is C.

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Detailed Solution

$\begin{matrix} f (0 -) = f (0) = f (0 +) \\ \Rightarrow a = 1 = c \\ f^{1} (0 -) = f^{1} (0 +) \\ \Rightarrow a e^{- b x^{2} (- 2 b x)] x = 0} = - \sin x a t x - 0 \\ \Rightarrow 0 = 0 \end{matrix}$

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