Let $f(x) = \left\{ {\begin{array}{*{20}{c}}
{\int\limits_0^x {\left\{ {1 + \left| {1 - t} \right|} \right\}dt{\text{ if }}x > 2} } \\
{5x - 7{\text{ if }}x \leqslant 2}
\end{array}} \right.$ then
A. f is not continuous at x = 2
B. f is continuous but not differentiable at x = 2
C. f is differentiable everywhere
D. f’(2+) does not exist
Answer
645k+ views
Hint: Here, we are given two parts. One part contains integral function, so simplify the function by applying integration formulae. By applying limits properly you will get a simplified value of function. Now differentiate the function with respect to x to check whether the given function is continuous at x = 2 or not. After finding differential function you can also check whether function is differentiable or not at x = 2.
Complete step-by-step answer:
$f(x) = \left\{ {\begin{array}{*{20}{c}}
{\int\limits_0^x {\left\{ {1 + \left| {1 - t} \right|} \right\}dt{\text{ if }}x > 2} } \\
{5x - 7{\text{ if }}x \leqslant 2}
\end{array}} \right.$
Let y = f(x) and simplifying the function given
$y = \int\limits_0^x {\left\{ {1 + \left| {1 - t} \right|} \right\}dt{\text{ if }}x > 2} $
Dividing integral limits in two parts, we have
\[y = \int\limits_0^1 {(1 + 1 - t)dt + } \int\limits_1^x {tdt} \]
Integrating with respect to dt
$y = 2 - \dfrac{1}{2} + \dfrac{{\left( {{x^2} - 1} \right)}}{2}$
On simplifying, we have
$y = 1 + \dfrac{{{x^2}}}{2}$
Now, given function can be rewritten as
Now, $f(x) = \left\{ {\begin{array}{*{20}{c}}
{1 + \dfrac{{{x^2}}}{2}{\text{ if }}x > 2} \\
{5x - 7{\text{ if }}x \leqslant 2}
\end{array}} \right.$
Finding values of y both left side and right side
$f({2^ + }) = 1 + \dfrac{4}{2} = 3$
And $f({2^ - }) = 10 - 7 = 3$
LHS and RHS values of function at x =2 are equal hence function is continuous at x = 2.
Differentiating with respect to x we get
\[f'(x) = \left\{ {\begin{array}{*{20}{c}}
{{\text{2}}x{\text{ if }}x > 2} \\
{5{\text{ if }}x \leqslant 2}
\end{array}} \right.\]
Here RHS and LHS value at x = 2 is not equal hence differential function at x = 2 is not continuous at x = 2. So, we can say that the function is not differentiable at x = 2. If a function is differentiable at a point then it must be continuous at that point.
We can see that it is not differentiable at x = 2 but continuous at x = 2.
So, the correct answer is “Option B”.
Note: In these types of questions, find the simplified function containing one variable only. Here, use the fact of continuity and differentiability of a function at a point and relation between them.As we know that if a function is continuous at a point then it must be differentiable at that point but if a function is continuous at a point it is necessary that it will be differentiable at that point.
Complete step-by-step answer:
$f(x) = \left\{ {\begin{array}{*{20}{c}}
{\int\limits_0^x {\left\{ {1 + \left| {1 - t} \right|} \right\}dt{\text{ if }}x > 2} } \\
{5x - 7{\text{ if }}x \leqslant 2}
\end{array}} \right.$
Let y = f(x) and simplifying the function given
$y = \int\limits_0^x {\left\{ {1 + \left| {1 - t} \right|} \right\}dt{\text{ if }}x > 2} $
Dividing integral limits in two parts, we have
\[y = \int\limits_0^1 {(1 + 1 - t)dt + } \int\limits_1^x {tdt} \]
Integrating with respect to dt
$y = 2 - \dfrac{1}{2} + \dfrac{{\left( {{x^2} - 1} \right)}}{2}$
On simplifying, we have
$y = 1 + \dfrac{{{x^2}}}{2}$
Now, given function can be rewritten as
Now, $f(x) = \left\{ {\begin{array}{*{20}{c}}
{1 + \dfrac{{{x^2}}}{2}{\text{ if }}x > 2} \\
{5x - 7{\text{ if }}x \leqslant 2}
\end{array}} \right.$
Finding values of y both left side and right side
$f({2^ + }) = 1 + \dfrac{4}{2} = 3$
And $f({2^ - }) = 10 - 7 = 3$
LHS and RHS values of function at x =2 are equal hence function is continuous at x = 2.
Differentiating with respect to x we get
\[f'(x) = \left\{ {\begin{array}{*{20}{c}}
{{\text{2}}x{\text{ if }}x > 2} \\
{5{\text{ if }}x \leqslant 2}
\end{array}} \right.\]
Here RHS and LHS value at x = 2 is not equal hence differential function at x = 2 is not continuous at x = 2. So, we can say that the function is not differentiable at x = 2. If a function is differentiable at a point then it must be continuous at that point.
We can see that it is not differentiable at x = 2 but continuous at x = 2.
So, the correct answer is “Option B”.
Note: In these types of questions, find the simplified function containing one variable only. Here, use the fact of continuity and differentiability of a function at a point and relation between them.As we know that if a function is continuous at a point then it must be differentiable at that point but if a function is continuous at a point it is necessary that it will be differentiable at that point.
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