μ(x)=x−11, which is discontinous at x=1 f(u)=u2+u−21=(u+2)(u−1)1, which is discontinous at u=−2,1 when u=−2, then x−11=−2⇒x=21 when u=1, then x−11=1⇒x=2 Hence given composite function is discontinous at three points, x=1,21 and 2 .
Let f be a composite function of x defined by f(u)=u2+u−21,u(x)=x−11. Then the number of points x where f is discontinuous is :
Held on 23 Apr 2013 · Verified 6 Jul 2026.
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