TS EAMCET · Maths · Application of Derivatives
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
- A \([-1,1]\)
- B \([0,1]\)
- C \([0,2]\)
- D \([-1,0]\)
Answer & Solution
Correct Answer
(A) \([-1,1]\)
Step-by-step Solution
Detailed explanation
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 \(f(x)=x+\log _2\left(b^2+7\right)\)…
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