Get Answers to all your Questions

header-bg qa

Let \mathrm{f(x)=x^3-x^2+x+1, g(x)=\left\{\begin{array}{cc}\max \cdot\{f(t), 0 \leq t \leq x\}, & 0 \leq x \leq 1 \\ 3-x, & 1<x \leq 2\end{array}\right.}
Then in  [0,2] the points where \mathrm{g(x)}  is not differentiable is/are

Option: 1

1


Option: 2

2


Option: 3

 1 and 2


Option: 4

none of these


Answers (1)

best_answer

\mathrm{f(t)=t^3-t^2+t+1}

\mathrm{\therefore \quad f^{\prime}(t)=3 t^2-2 t+1>0}
 

\mathrm{\therefore \quad f(t) } is an increasing function.
Since \mathrm{0 \leq t \leq x \therefore \quad \max . f(t)=f(x)=x^3-x^2+x+1 }

Thus \mathrm{g(x)=x^3-x^2+x+1,0 \leq x \leq 1\: and \: g(x)=3-x, 1<x \leq 2 }
The only doubtful point for differentiability of \mathrm{g(x) } in \mathrm{[0,2]} is \mathrm{x=1}.

Clearly, \mathrm{\lim _{x \rightarrow 1^{+} 0} g(x)=\lim _{x \rightarrow 1^{+}}(3-x)=2\: and \: g(1)=2}

\mathrm{\therefore \quad g(x)} is continuous at \mathrm{x=1}

Also g^{\prime}(x)=3 x^2-2 x+1,0 \leq x<1\: and\: g^{\prime}(x)=-1,1<x \leq 2
\therefore \quad g^{\prime}(1-0)=3 \cdot 1^2-2 \cdot 1+1=2$ and $g^{\prime}(1+0)=-1

Hence g(x) is not differentiable at x=1.

Posted by

Riya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE