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  • The function f(x) defined by <img alt="\mathrm{f(x)= \begin{cases}\log _{(4 x-3)}\left(x^2-2 x+5\right), & \frac{3}{4}<x<1 \text { and } x>1 \\ 4 & , x=1\end{cases}}" s

The function f(x) defined by
\mathrm{f(x)= \begin{cases}\log _{(4 x-3)}\left(x^2-2 x+5\right), & \frac{3}{4}<x<1 \text { and } x>1 \\ 4 & , x=1\end{cases}}

Option: 1

is continuous at x=1


Option: 2

is discontinuous at x=1 since  f \mathrm{\left(1^{+}\right)} does not exist though  \mathrm{f\left(1^{-}\right)}  exists.


Option: 3

is discontiuous at x=1 since \mathrm{f\left(1^{-}\right)} does not exist though  \mathrm{f\left(1^{+}\right)} 'exists


Option: 4

is discontinous at x=1 since neither\mathrm{ f\left(1^{+}\right)nor~ f\left(1^{-}\right)} exists.


Answers (1)

\mathrm{\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0} \frac{\log \left(4+h^2\right)}{\log (1-4 h)}=-\infty}

\mathrm{\text { and, } \lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0} \frac{\log \left(4+h^2\right)}{\log (1+4 h)}=\infty \text {. }}

\mathrm{\text { So, } f\left(1^{\top}\right) \text { and } f\left(1^{+}\right) \text {do not exist. }}

Posted by

Sumit Saini

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