JEE Main 2022 (Online) 26th July Morning ShiftQuestion 1
Let $f(x)=\begin{cases}x^3-x^2+10x-7, & x\le1\\ -2x+\log_2(b^2-4), & x>1\end{cases}$. Then the set of all values of $b$, for which $f(x)$ has maximum value at $x=1$, is :
Differentiation and its Application
Solution
Concept. On the left $f$ increases up to $f(1)=3$; on the right $f$ decreases from just below its limit at $1^+$. So we need the right-hand limit at $1$ to be at most $3$. Formulas used. (1) $3x^2-2x+10>0$ (discriminant $<0$)Step 1 - Left side. By (1), $f$ increases on $(-\infty,1]$, so its largest value there is $f(1)=1-1+10-7=3$. Step 2 - Right side. $f=-2x+\log_2(b^2-4)$ (needs $b^2>4$) decreases, with values just below $-2+\log_2(b^2-4)$. Step 3 - Condition. $-2+\log_2(b^2-4)\le3$ gives $b^2-4\le32$, i.e. $b^2\le36$. Step 4 - Set. $4<b^2\le36$: $b\in[-6,-2)\cup(2,6]$. Answer: (C)