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Applications of Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsApplications of Derivatives2 Marks
MCQ
Let $f(x)=\left\{\begin{array}{ll}x^3-x^2+10 x-7 & ,x \leq 1 \\ -2 x+\log _2\left(b^2-4\right) & , \quad x>1\end{array}\right.$ Then the set of all values of $b$, for which $f(x)$ has maximum value at $x=1$, is
  • A
    $(-6,-2)$
  • B
    $(2,6)$
  • ✓
    $[-6,-2) \cup(2,6]$
  • D
    $[-\sqrt{6},-2) \cup(2, \sqrt{6}]$
✓

Answer

Correct option: C.
$[-6,-2) \cup(2,6]$
$f(x)=\left\{\begin{array}{ll}x^3-x^2+10 x-7, & x \leq 1 \\ -2 x+\log _2\left(b^2-4\right), & x>1\end{array}\right.$
Since $f(x)$ has maximum value at $x=1$, therefore
$\text{L.H.L.} \geq \text{R.H.L.}$ at $x=1$
$\Rightarrow \lim _{x \rightarrow 1}\left(x^3-x^2+10 x-7\right) \geq \lim _{x \rightarrow 1}\left[-2 x+\log _2\left(b^2-4\right)\right]$
$\Rightarrow 3 \geq-2+\log _2\left(b^2-4\right) $
$\Rightarrow 2^5 \geq b^2-4$
$\Rightarrow b^2 \leq 36 $
$\Rightarrow(b-6)(b+6) \leq 0 $
$\Rightarrow b \in[-6,6]$
$\text { but } b^2-4 \neq 0 $
$\Rightarrow b \neq \pm 2 $
$\Rightarrow b \in[-6,-2) \cup(2,6]$

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