Let a function f: \mathbbR arrow \mathbbR be defined as : f(x)= \begincases\int\limits_0^x(5-|t-3|) d t, & x\gt 4 x^2+b x & , x \leq 4\endcases where b \in \mathbbR . If f is continuous at x=4 , then which of the following statements is NOT true?

f is increasing in (-\infty, \frac18) \cup(8, \infty)

f^\prime(3)+f^\prime(5)=\frac354

f has a local minima at x=\frac18

Definite Integration PYQ — JEE Main 2022 · 27 Jul · Shift 1 · Q38 with Solution

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Correct answer: C

Use continuity at $x=4$ to find $b$

For $x>4$ ,

f(x)=\int_0^x (5-|t-3|)\,dt

So,

f(4^+)=\int_0^4 (5-|t-3|)\,dt

Now evaluate the integrand piecewise:

For $0\le t\le 3$ , $|t-3|=3-t$ , so $5-|t-3|=5-(3-t)=t+2$
For $3\le t\le 4$ , $|t-3|=t-3$ , so $5-|t-3|=5-(t-3)=8-t$

Hence,

\int_0^4 (5-|t-3|)dt=\int_0^3 (t+2)dt+\int_3^4 (8-t)dt

Compute:

\int_0^3 (t+2)dt=\left[\frac{t^2}{2}+2t\right]_0^3=\frac{9}{2}+6=\frac{21}{2}

\int_3^4 (8-t)dt=\left[8t-\frac{t^2}{2}\right]_3^4=(32-8)-\left(24-\frac{9}{2}\right)=24-\frac{39}{2}=\frac{9}{2}

Thus,

f(4^+)=\frac{21}{2}+\frac{9}{2}=15

For $x\le 4$ ,

f(x)=x^2+bx

So,

f(4)=16+4b

Continuity at $x=4$ gives

16+4b=15\implies 4b=-1\implies b=-\frac14

Therefore,

f(x)=x^2-\frac{x}{4},\quad x\le 4

Check differentiability at $x=4$

For $x<4$ ,

f'(x)=2x-\frac14

So left derivative at $4$ is

f'_-(4)=8-\frac14=\frac{31}{4}

For $x>4$ , by Fundamental Theorem of Calculus,

f'(x)=5-|x-3|

So right derivative at $4$ is

f'_+(4)=5-|4-3|=4

Since

\frac{31}{4}\ne 4,

$f$ is not differentiable at $x=4$ .

So A is true.

Check option B

First, $3\le 4$ , so use the left formula:

f'(3)=2(3)-\frac14=6-\frac14=\frac{23}{4}

Next, $5>4$ , so use the right formula:

f'(5)=5-|5-3|=5-2=3

Thus,

f'(3)+f'(5)=\frac{23}{4}+3=\frac{23}{4}+\frac{12}{4}=\frac{35}{4}

So B is true.

Check option C: increasing intervals

For $x\le 4$ ,

f'(x)=2x-\frac14

Hence,

f'(x)>0 \iff 2x-\frac14>0 \iff x>\frac18

So on $(-\infty,4]$ , $f$ is increasing on $(\tfrac18,4]$ , not on $(-\infty,\tfrac18)$ .

For $x>4$ ,

f'(x)=5-|x-3|

Now for $x>4$ , $|x-3|=x-3$ , so

f'(x)=5-(x-3)=8-x

Thus,

f'(x)>0 \iff x<8

So on $(4,\infty)$ , $f$ is increasing on $(4,8)$ and decreasing on $(8,\infty)$ .

Therefore overall, $f$ is increasing on

\left(\frac18,8\right)

(except not differentiable at $4$ , but monotonic increase continues across it by derivative signs).

So the statement

\text{"$f$ is increasing in }(-\infty,\tfrac18)\cup(8,\infty)\text{"}

is false.

So C is NOT true.

Check option D: local minima at $x=\frac18$

For $x<\frac18$ ,

f'(x)=2x-\frac14<0

For $x>\frac18$ (and near it),

f'(x)=2x-\frac14>0

Thus derivative changes from negative to positive at $x=\frac18$ , so $f$ has a local minimum there.

So D is true.

Conclusion

The only statement which is NOT true is

\boxed{\text{C}}

More from Definite Integration

View written solutionFree

Correct answer: C

Use continuity at $x=4$ to find $b$

For $x>4$ ,

f(x)=\int_0^x (5-|t-3|)\,dt

So,

f(4^+)=\int_0^4 (5-|t-3|)\,dt

Now evaluate the integrand piecewise:

For $0\le t\le 3$ , $|t-3|=3-t$ , so $5-|t-3|=5-(3-t)=t+2$
For $3\le t\le 4$ , $|t-3|=t-3$ , so $5-|t-3|=5-(t-3)=8-t$

Hence,

\int_0^4 (5-|t-3|)dt=\int_0^3 (t+2)dt+\int_3^4 (8-t)dt

Compute:

\int_0^3 (t+2)dt=\left[\frac{t^2}{2}+2t\right]_0^3=\frac{9}{2}+6=\frac{21}{2}

\int_3^4 (8-t)dt=\left[8t-\frac{t^2}{2}\right]_3^4=(32-8)-\left(24-\frac{9}{2}\right)=24-\frac{39}{2}=\frac{9}{2}

Thus,

f(4^+)=\frac{21}{2}+\frac{9}{2}=15

For $x\le 4$ ,

f(x)=x^2+bx

So,

f(4)=16+4b

Continuity at $x=4$ gives

16+4b=15\implies 4b=-1\implies b=-\frac14

Therefore,

f(x)=x^2-\frac{x}{4},\quad x\le 4

Check differentiability at $x=4$

For $x<4$ ,

f'(x)=2x-\frac14

So left derivative at $4$ is

f'_-(4)=8-\frac14=\frac{31}{4}

For $x>4$ , by Fundamental Theorem of Calculus,

f'(x)=5-|x-3|

So right derivative at $4$ is

f'_+(4)=5-|4-3|=4

Since

\frac{31}{4}\ne 4,

$f$ is not differentiable at $x=4$ .

So A is true.

Check option B

First, $3\le 4$ , so use the left formula:

f'(3)=2(3)-\frac14=6-\frac14=\frac{23}{4}

Next, $5>4$ , so use the right formula:

f'(5)=5-|5-3|=5-2=3

Thus,

f'(3)+f'(5)=\frac{23}{4}+3=\frac{23}{4}+\frac{12}{4}=\frac{35}{4}

So B is true.

Check option C: increasing intervals

For $x\le 4$ ,

f'(x)=2x-\frac14

Hence,

f'(x)>0 \iff 2x-\frac14>0 \iff x>\frac18

So on $(-\infty,4]$ , $f$ is increasing on $(\tfrac18,4]$ , not on $(-\infty,\tfrac18)$ .

For $x>4$ ,

f'(x)=5-|x-3|

Now for $x>4$ , $|x-3|=x-3$ , so

f'(x)=5-(x-3)=8-x

Thus,

f'(x)>0 \iff x<8

So on $(4,\infty)$ , $f$ is increasing on $(4,8)$ and decreasing on $(8,\infty)$ .

Therefore overall, $f$ is increasing on

\left(\frac18,8\right)

(except not differentiable at $4$ , but monotonic increase continues across it by derivative signs).

So the statement

\text{"$f$ is increasing in }(-\infty,\tfrac18)\cup(8,\infty)\text{"}

is false.

So C is NOT true.

Check option D: local minima at $x=\frac18$

For $x<\frac18$ ,

f'(x)=2x-\frac14<0

For $x>\frac18$ (and near it),

f'(x)=2x-\frac14>0

Thus derivative changes from negative to positive at $x=\frac18$ , so $f$ has a local minimum there.

So D is true.

Conclusion

The only statement which is NOT true is

\boxed{\text{C}}

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