JEE Main 2022 · 27 Jul · Shift 1 · Mathematics Q20

MathematicsSingle correct+4 / -1Question 20 of 90

Let a function $f:\mathbf R\to\mathbf R$ be defined as :

$f(x)=\begin{cases}\displaystyle\int_0^x(5-|t-3|)\,dt, & x>4\\ x^2+bx, & x\le4\end{cases}$

where $b\in\mathbf R$ . If $f$ is continuous at $x=4$ , then which of the following statements is NOT true?

f

is not differentiable at

x=4

f'(3)+f'(5)=\dfrac{35}4

f

is increasing in

\left(-\infty,\dfrac18\right)\cup(8,\infty)

f

has a local minima at

x=\dfrac18

Answer and worked solution

Official answer: C

Continuity. $\int_0^4(5-|t-3|)dt=10.5+4.5=15=16+4b$ , so $b=-\frac14$ .

Check the options.

$f'(4^-)=\frac{31}4$ and $f'(4^+)=4$ , so f is not differentiable at 4. (A) is true.
$f'(3)+f'(5)=\frac{23}4+3=\frac{35}4$ . (B) is true.
$f'=2x-\frac14$ changes sign at $\frac18$ : a local minimum, so (D) is true. Then f is decreasing on $\left(-\infty,\frac18\right)$ .

JEE Main 2022 · 27 Jul · Shift 1, Mathematics question 20