Let fx =x3−x2+10x−7, x≤1−2x+log2⁡b2−4, x>1 Then the set of all values of b, for which f(x) has maximum valueat x = 1, is

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Let $f (x)$ $= \{\begin{array}{cc} x^{3} - x^{2} + 10 x - 7 & , x \leq 1 \\ - 2 x + \log_{2} (b^{2} - 4) & , x > 1 \end{array}$ Then the set of all values of b, for which f(x) has maximum value
at x = 1, is

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(2,6)

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

$[- \sqrt{6}, - 2) \cup (2, \sqrt{6}]$

(–6, –2)

answer is C.

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Detailed Solution

$f (1) = 3 For x < 1, f' (x) = 3 x^{2} - 2 x + 10 > 0 \Rightarrow f (x) is increasing For x > 1, f' (x) < 0 \Rightarrow f (x) is decreasing lim_{x \to 1^{+}} f (x) = - 2 + \log_{2} (b^{2} - 4) For maximum value at x = 1$

$\begin{array}{l} \Rightarrow 3 \geq - 2 + \log_{2} (b^{2} - 4) \\ \log_{2} (b^{2} - 4) \leq 5 \\ b^{2} - 4 > 0 a n d b^{2} - 4 \leq 32 o r b^{2} - 36 \leq 0 \\ b \in [- 6, - 2) \cup (2, 6] \end{array}$