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  • <img alt="\text { If the function } f(x)=\left\{\begin{array}{cc} A x-B, & x \leq 1 \\ 3 x, & 1<x<2 \\ B x^2-A, & x \geq 2 \end{array}\right." src="/latex-image/?%5Ctext%20%7B%20I

\text { If the function } f(x)=\left\{\begin{array}{cc} A x-B, & x \leq 1 \\ 3 x, & 1<x<2 \\ B x^2-A, & x \geq 2 \end{array}\right.

 be continuous at  x=1  and discontinuous at x=2  then 

Option: 1

\mathrm{A=3+B, B \neq 3}


Option: 2

\mathrm{A=3+B, B=3}


Option: 3

\mathrm{A=3+B}


Option: 4

none of these


Answers (1)

best_answer

Since  f(x)  is continuous at  x=1 , therefore 

\mathrm{\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \Rightarrow A-B=3 \Rightarrow A=3+ }
If f(x) is continuous at x=2, then
\mathrm{ \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x) \Rightarrow 6=4 B-A . }
Solving (i) and (ii) we get B=3
As f(x) is not continuous at x=2, therefore  \mathrm{ B \neq 3 }
Hence, A=3+B and   \mathrm{ B \neq 3 }

Posted by

manish painkra

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