JEE Main 2022 (Online) 27th July Morning ShiftQuestion 1
Let a function $f:\mathbb{R}\to\mathbb{R}$ be defined as : $f(x)=\begin{cases}\int_0^x(5-|t-3|)\,dt, & x>4\\ x^2+bx, & x\le4\end{cases}$ where $b\in\mathbb{R}$. If $f$ is continuous at $x=4$, then which of the following statements is NOT true?
Differentiation and its Application
Solution
Concept. Continuity at $4$ fixes $b$; then compare $f'$ on both sides and find the sign of $f'$. Formulas used. (1) $\dfrac{d}{dx}\int_0^xh(t)\,dt=h(x)$Step 1 - Find $b$. $$\int_0^4\left(5-|t-3|\right)dt=20-\left(\tfrac92+\tfrac12\right)=15=16+4b\ \Longrightarrow\ b=-\tfrac14.$$ Step 2 - $f'$. For $x<4$: $f'=2x-\frac14$. For $x>4$, by (1): $f'=5-|x-3|=8-x$. Step 3 - (A). $f'(4^-)=\frac{31}{4}\ne4=f'(4^+)$. Not differentiable. True. Step 4 - (B). $f'(3)=\frac{23}{4}$, $f'(5)=3$, sum $\frac{35}{4}$. True. Step 5 - (C), (D). $f'<0$ for $x<\frac18$, $f'>0$ on $\left(\frac18,8\right)$, $f'<0$ for $x>8$. So $f$ decreases on $\left(-\infty,\frac18\right)$ and on $(8,\infty)$: (C) is not true. $x=\frac18$ is a local minimum: (D) is true. Answer: (C)