If $f(x)=\left\{\begin{array}{cl}1+6 x-3 x^2, & x \leq 1 \ x+\log _2\left(b^2+7 ight), & x1\end{array} ight.$ is continuous at all real $x$, then $b=$

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If $f(x)=\left\{\begin{array}{cl}1+6 x-3 x^2, & x \leq 1 \ x+\log _2\left(b^2+7ight), & x>1\end{array}ight.$ is continuous at all real $x$, then $b=$

Aicharya · Accepted Answer

$f(x)= \begin{cases}1+6 x-3 x^2, & x \leq 1 \ x+\log _2\left(b^2+7ight), & x>1\end{cases}$ Since $f(x)$ is continuous at $x=1$, Hence $ \begin{aligned} & \Rightarrow \lim _{h ightarrow 0}(1+h)=f(1) \ & \Rightarrow \lim _{h ightarrow 0}\left[(1+h)+\log _2\left(b^2+7ight)ight]=1+6-3 \ & \Rightarrow \log _2\left(b^2+7ight)=3 \Rightarrow b= \pm 1 \end{aligned} $

If $f(x)=\left\{\begin{array}{cl}1+6 x-3 x^2, & x \leq 1 \\ x+\log _2\left(b^2+7\right), &…

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