Let f be a composite function of x defined by f(u)=1u3−6u2+11u−6 where u(x)=1x. Then the number of points x where f is discontinuous is :

f(u)=1u3−6u2+11u−6 =1(u−1)(u−2)(u−3)' as u(x)=1x,So, f=11x−11x−21x−3∴ f is discontinuous at x = 1,12,13,0

Let f be a composite function of x defined by f(u)=1u3−6u2+11u−6 where u(x)=1x. Then the number of points x where f is discontinuous is :

f(u)=1u3−6u2+11u−6 =1(u−1)(u−2)(u−3)' as u(x)=1x,So, f=11x−11x−21x−3∴ f is discontinuous at x = 1,12,13,0

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Let f be a composite function of x defined by $f (u) = \frac{1}{u^{3} - 6 u^{2} + 11 u - 6} w h e r e u (x) = \frac{1}{x} .$ Then the number of points x where f is discontinuous is :

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$f (u) = \frac{1}{u^{3} - 6 u^{2} + 11 u - 6}$

$= \frac{1}{(u - 1) (u - 2) (u - 3)'} a s u (x) = \frac{1}{x},$

So, $f = \frac{1}{(\frac{1}{x} - 1) (\frac{1}{x} - 2) (\frac{1}{x} - 3)}$

$∴$ f is discontinuous at x = $1, \frac{1}{2}, \frac{1}{3}, 0$

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Let f be a composite function of x defined by f(u)=1u3−6u2+11u−6 where u(x)=1x. Then the number of points x where f is discontinuous is :

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Let f be a composite function of x defined by $f (u) = \frac{1}{u^{3} - 6 u^{2} + 11 u - 6} w h e r e u (x) = \frac{1}{x} .$ Then the number of points x where f is discontinuous is :