If f x -1- x -1 x -2 and g x -1- x 2- then points of discontinuity of f g x areA- -1-2-1-0-1- 1-2B- -1-0-1- 1-2C- 0-1D- 0-1- 1-2

The correct option is A f(g(x))=1/ ( 1/x^2 1 ) ( 1/x^2 2 )=x^4/(1 x^2)(1 2x^2) ⇒ f(g(x)) is discontinuous at x= ± 1, x= ±1/√(2) and x=0 Since, g(x) is discon ...

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Byju's Answer

Standard XI

Mathematics

Continuity of Composite Functions

If f x =1/ x ...

Question

If $f (x) = \frac{1}{(x - 1) (x - 2)} a n d g (x) = \frac{1}{x^{2}}$ , then points of discontinuity of f(g(x)) are

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Solution

The correct option is B
$f (g (x)) = \frac{1}{(\frac{1}{x^{2}} - 1) (\frac{1}{x^{2}} - 2)} = \frac{x^{4}}{(1 - x^{2}) (1 - 2 x^{2})} \Rightarrow f (g (x)) i s d i s c o n t i n u o u s a t x = \pm 1, x = \pm \frac{1}{\sqrt{2}} a n d x = 0$
Since, g(x) is discontinuous at x=0

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If f(x)=1(x−1)(x−2) and g(x)=1x2, then points of discontinuity of f(g(x)) are

The correct option is B f(g(x))=1(1x2−1)(1x2−2)=x4(1−x2)(1−2x2)⇒f(g(x)) is discontinuous at x=±1,x=±1√2 and x=0 Since, g(x) is discontinuous at x=0

If $f (x) = \frac{1}{(x - 1) (x - 2)} a n d g (x) = \frac{1}{x^{2}}$ , then points of discontinuity of f(g(x)) are

The correct option is B
$f (g (x)) = \frac{1}{(\frac{1}{x^{2}} - 1) (\frac{1}{x^{2}} - 2)} = \frac{x^{4}}{(1 - x^{2}) (1 - 2 x^{2})} \Rightarrow f (g (x)) i s d i s c o n t i n u o u s a t x = \pm 1, x = \pm \frac{1}{\sqrt{2}} a n d x = 0$
Since, g(x) is discontinuous at x=0