JEE MAIN - Mathematics (2021 - 25th July Evening Shift - No. 3)
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
f(x) is not continuous at x = 2
f(x) is everywhere differentiable
f(x) is continuous but not differentiable at x = 2
f(x) is not differentiable at x = 1
Explanation
$$f(x) = \int\limits_0^1 {(5 + (1 - t))dt + \int\limits_1^x {(5 + (t - 1))dt} } $$
$$ = \left. {6 - {1 \over 2} + \left( {4t + {{{t^2}} \over 2}} \right)} \right|_1^x$$
$$ = {{11} \over 2} + 4x + {{{x^2}} \over 2} - 4 - {1 \over 2}$$
$$ = {{{x^2}} \over 2} + 4x - 1$$
$$f({2^ + }) = 2 + 8 + 1 = 11$$
$$f(2) = f({2^ - }) = 5 \times 2 + 1 = 11$$
$$\Rightarrow$$ continuous at x = 2
Clearly differentiable at x = 1
Lf' (2) = 5
Rf' (2) = 6
$$\Rightarrow$$ Not differentiable at x = 2
$$ = \left. {6 - {1 \over 2} + \left( {4t + {{{t^2}} \over 2}} \right)} \right|_1^x$$
$$ = {{11} \over 2} + 4x + {{{x^2}} \over 2} - 4 - {1 \over 2}$$
$$ = {{{x^2}} \over 2} + 4x - 1$$
$$f({2^ + }) = 2 + 8 + 1 = 11$$
$$f(2) = f({2^ - }) = 5 \times 2 + 1 = 11$$
$$\Rightarrow$$ continuous at x = 2
Clearly differentiable at x = 1
Lf' (2) = 5
Rf' (2) = 6
$$\Rightarrow$$ Not differentiable at x = 2
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