JEE Main 2022 · 26 Jul · Shift 1 · Mathematics Q9

← JEE Main 2022 · 26 Jul · Shift 1

MathematicsSingle correct+4 / -1Question 9 of 90

Let $f(x)=\begin{cases}x^3-x^2+10x-7, & x\le1\\-2x+\log_2\left(b^2-4\right), & x>1\end{cases}$ . Then the set of all values of b, for which $f(x)$ has maximum value at $x=1$ , is :

(-6,-2)

(2,6)

[-6,-2)\cup(2,6]

\left[-\sqrt6,-2\right)\cup\left(2,\sqrt6\right]

Answer and worked solution

Official answer: C

Left piece increases, reaching $f(1)=3$ . Need $-2+\log_2(b^2-4)\le3$ and $b^2>4$ : $4<b^2\le36$ .

(C)

Topics:Maxima and minima Logarithms

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JEE Main 2022 · 26 Jul · Shift 1, Mathematics question 9