Q112 / 160 · TS EAMCET 2022 (19 Jul Shift 2) · Mathematics

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

  • Correct answer
    A
    $[-1,1]$
  • B
    $[0,1]$
  • C
    $[0,2]$
  • D
    $[-1,0]$

Solution

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 $f(x)=x+\log _2\left(b^2+7\right)$ $$ \begin{aligned} & f(1)=1+\log _2\left(b^2+7\right) \\ & \Rightarrow 1+\log \left(b^2+7\right)=4 \\ & \Rightarrow b^2=1 \Rightarrow b \varepsilon[-1,1] \end{aligned} $$

Hence, the correct answer is option: 'A'.

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