Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find all points of discontinuity of the function where f(t)=1/t^2+t-2, where t=1/x-1.

Find all points of discontinuity of the function   f(t)=\frac{1}{t^{2}+t-2}, \quad t=\frac{1}{x-1}.

Answers (1)

f(t)=\frac{1}{t^{2}+t-2}

We have to find: Points discontinuity of function f(t) where \mathrm{t}=\frac{1}{\mathrm{x}-1}

As t is not defined at x = 1 as denominator becomes 0, at x = 1.

\therefore x = 1 is a point of discontinuity

\because f(t) = f\left(\frac{1}{x-1}\right)=\frac{1}{\left(\frac{1}{x-1}\right)^{2}+\frac{1}{x-1}-2}=\frac{(x-1)^{2}}{1+x-1-2(x-1)^{2}}

\Rightarrow f(t)=\frac{(x-1)^{2}}{1+x-1-2\left(x^{2}-2 x+1\right)}=\frac{(x-1)^{2}}{-2 x^{2}+5 x-2}

The f(t) is not going to be defined whenever denominator is 0 and thus will give a point of discontinuity.


∴ Solution of the following equation gives other points of discontinuities.

 \\ -2x\textsuperscript{2} + 5x - 2 = 0 \\ \Rightarrow 2x\textsuperscript{2} - 5x + 2 = 0 \\ \Rightarrow 2x\textsuperscript{2} - 4x - x + 2 = 0 \\ \Rightarrow 2x(x - 2) - (x - 2) = 0 \\ \Rightarrow (2x - 1)(x - 2) = 0 \\ \therefore x = 2 or x = 1/2 \\

Hence,

f(t) is discontinuous at x = 1, x = 2 and x = 1/2

Posted by

infoexpert21

View full answer

Similar Questions

Latest Question