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  • Let <img alt="f(x)=\left\{\begin{array}{ll} \int_0^x(1+|1-t|) d t, & x>2 \\ 5 x-7 & , x \leq 2 \end{array}\right. \text { then }" src="https://entrancecorner.oncodecogs.com/gif.late

Let f(x)=\left\{\begin{array}{ll} \int_0^x(1+|1-t|) d t, & x>2 \\ 5 x-7 & , x \leq 2 \end{array}\right. \text { then }

Option: 1

f(x) is discontinuous at x=2


Option: 2

f(x) is continuous but not differentiable at x = 2

 


Option: 3

f(x) is differentiable at x=2


Option: 4

R.H.D at x=2 does not exist


Answers (1)

best_answer

f(x)= \begin{cases}\int_0^1(1+|1-t|) d t+\int_1^x(1+|1-t|) d t & x>2 \\ 5 x-7 & , x \leq 2\end{cases}\begin{aligned} & =\left\{\begin{array}{ll} \int_0^1(2-t) d t+\int_1^x t d t & , x>2 \\ 5 x-7 & , x \leq 2 \end{array}= \begin{cases}\frac{3}{2}+\frac{1}{2}\left(x^2-1\right) & , x>2 \\ 5 x-7 & , x \leq 2\end{cases} \right. \\ & = \begin{cases}\frac{x^2}{2}+1 & , x>2 \\ 5 x-7 & , x \leq 2\end{cases} \\ & \end{aligned}

\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2)=3

\text { L.H.D at } x=2 \text { is } 5

\text { R.H.D at } x=2 \text { is } 2

So, f(x)  is continuous but not differentiable at x = 2.

 

Posted by

Gautam harsolia

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