"If $$f(x) = \left\{ {\matrix{ {\int\limits_0^x {\left( {5 + \left| {1 - t} \right|} \right)dt,} } & {x 2} \cr {5x + 1,} & {x \le 2} \cr } } \right.$$, then"

Question

"If $$f(x) = \left\{ {\matrix{ {\int\limits_0^x {\left( {5 + \left| {1 - t} \right|} \right)dt,} } & {x > 2} \cr {5x + 1,} & {x \le 2} \cr } } \right.$$, then"

Accepted Answer

"$f(x) = \int\limits_0^1 {(5 + (1 - t))dt + \int\limits_1^x {(5 + (t - 1))dt} }$

$= \left. {6 - {1 \over 2} + \left( {4t + {{{t^2}} \over 2}} \right)} \right|_1^x$

$= {{11} \over 2} + 4x + {{{x^2}} \over 2} - 4 - {1 \over 2}$

$= {{{x^2}} \over 2} + 4x - 1$

$f({2^ + }) = 2 + 8 + 1 = 11$

$f(2) = f({2^ - }) = 5 \times 2 + 1 = 11$

$\Rightarrow$ continuous at x = 2

Clearly differentiable at x = 1

Lf' (2) = 5

Rf' (2) = 6

$\Rightarrow$ Not differentiable at x = 2"

If $$f(x) = \left\{ {\matrix{ {\int\limits_0^x {\left( {5 + \left| {1 - t} \right|} \right)dt,} } & {x > 2} \cr {5x + 1,} & {x \le 2} \cr } } \right.$$, then

Solution

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If $$f(x) = \left\{ {\matrix{ {\int\limits_0^x {\left( {5 + \left| {1 - t} \right|} \right)dt,} } & {x > 2} \cr {5x + 1,} & {x \le 2} \cr } } \right.$$, then

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