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Let $f(x)=\left\{\begin{array}{cc}1+6 x-3 x^2, & x \leq 1 \\ x+\log _2\left(b^2+7\right), & x>1\end{array}\right.$. Then the set of all possible values of $b$ such that $f(1)$ is the maximum value of $f(x)$ is
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Verified Answer
The correct answer is:
$[-1,1]$
Given $f(x)=\left\{\begin{array}{l}1+6 x-3 x^2, x \leq 1 \\ x+\log _z\left(b^2+7\right) x>1\end{array}\right.$
$\therefore \mathrm{f}(\mathrm{x})$ is maximum at $\mathrm{x}=1$
$\Rightarrow f(1)=4$ from $1+6 x-3 x^2$
low $f(x)=x+\log _2\left(b^2+7\right)$
$$
\begin{aligned}
& f(1)=1+\log _2\left(b^2+7\right) \\
& \Rightarrow 1+\log \left(b^2+7\right)=4 \\
& \Rightarrow b^2=1 \Rightarrow b \varepsilon[-1,1]
\end{aligned}
$$
$\therefore \mathrm{f}(\mathrm{x})$ is maximum at $\mathrm{x}=1$
$\Rightarrow f(1)=4$ from $1+6 x-3 x^2$
low $f(x)=x+\log _2\left(b^2+7\right)$
$$
\begin{aligned}
& f(1)=1+\log _2\left(b^2+7\right) \\
& \Rightarrow 1+\log \left(b^2+7\right)=4 \\
& \Rightarrow b^2=1 \Rightarrow b \varepsilon[-1,1]
\end{aligned}
$$
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