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JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let a function \(g:[0,4] \rightarrow R\) be defined as \(g ( x )=\left\{\begin{array}{ll}\max _{0 \leq t \leq x }\left\{ t ^{3}-6 t ^{2}+9 t -3\right\} & , 0 \leq x \leq 3 \\ 4- x & , 3 < x \leq 4\end{array}\right.\) then the number of points in the interval \((0,4)\) where \(g(x)\) is NOT differentiable, is \(.....\)
- A \(5\)
- B \(3\)
- C \(1\)
- D \(11\)
Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
Detailed explanation
\(f(X)=x^{3}-6 x^{2}+9 x-3\) \(f(x)=3 x^{2}-12 x+9=3(x-1)(x-3)\) \(f(1)=1\) and \(f(3)=-3\) \(g(x)=f(x) \quad\quad 0 \leq x \leq 1\) \(\quad\quad\quad\quad1 \quad\quad\quad 1 \leq x \leq 3\) \(\quad\quad\quad\quad4-x \quad 3\) \(g(x)\) is continuous…
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