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The set of triplets (a, b, c) of real numbers with a \mathrm{\neq 0} , for which the function 

\mathrm{f(x)=\left\{\begin{array}{c} x \text { for } x \leq 1 \\ a x^2+b x+c \text { otherwise, is differentiable, is } \end{array}\right.}

Option: 1

\mathrm{\{(a, 1-2 a, a) / a \in R ; a \neq 0\}}


Option: 2

\mathrm{\{(a, 1-2 a, c) / a, c \in R ; a \neq 0\}}


Option: 3

\mathrm{\{(a, b, c) / a, b, c \in R ; a+b+c=1\}}


Option: 4

\mathrm{\{(a, 1-2 a, 0) / a \in R ; a \neq 0\}}


Answers (1)

best_answer

Given f is differentiable for all real x

\mathrm{\Rightarrow f} is continuous for all real x.

so, \mathrm{\lim _{x \rightarrow 1} f(x)=f(1) \Rightarrow a+b+c=1 \ldots (1)}

\mathrm{\text { Also } f^{\prime}(x)=\left\{\begin{array}{cl} 1 \quad \text { for } x \leq 1 \\ 2 a x+b & \text { otherwise }(\text { since } f \text { is continuous) } \end{array}\right.}

\mathrm{f^{\prime}(1+)=f^{\prime}(1-) \Rightarrow 1=2 a+b \Rightarrow b=-2 a+1}                               ......(2)

\mathrm{\text { since } a, b, c \in R \text { and } a \neq 0 \text {, using (1) and (2) }}

\mathrm{\Rightarrow c=a}

Posted by

vinayak

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