Question

Let $f(x)= \begin{cases} x^3-x^2+10 x-7, & x \leq 1 \\ -2 x+\log _2\left(b^2-4\right), & x>1\end{cases}$ Then the set of all values of $b$, for which $f(x)$ has maximum value at $x=1$, is :

• A. $(-6,-2)$
• B. $(2,6)$
• C. $[-6,-2) \cup(2,6]$
• D. $[-\sqrt{6},-2) \cup(2, \sqrt{6}]$

Answer 1

The Correct Option is C

 To determine the set of all values of \(b\), for which the function \(f(x)\) has its maximum value at \(x=1\), we need to analyze the behavior of \(f(x)\) at that point and the surrounding interval.

1. The function is defined piecewise:
   • \(f(x) = x^3 - x^2 + 10x - 7\) for \(x \leq 1\)
   • \(f(x) = -2x + \log_2(b^2 - 4)\) for \(x > 1\)
2. First, we find \(f(x)\) at \(x=1\):
   • For \(x \leq 1\), \(f(1) = 1^3 - 1^2 + 10 \times 1 - 7 = 3\)
   • For \(x > 1\), to satisfy the condition that \(f(x)\) has a maximum value at \(x=1\), \(-2x + \log_2(b^2 - 4)\) for \(x > 1\) must be less than or equal to 3 when \(x=1\).
3. Calculate the derivative of both parts to check behavior around \(x = 1\):
   • For \(x \leq 1\), the derivative \(f'(x) = 3x^2 - 2x + 10\). At \(x = 1\), \(f'(1) = 3(1)^2 - 2(1) + 10 = 11\), which indicates that the left side has a positive slope, meaning it increases to \(x=1\).
   • For \(x > 1\), the derivative \(f'(x) = -2\), constant and negative, hence decreasing for \(x > 1\).
4. To ensure a maximum at \(x=1\), the function must have its peak at the transition, hence we must adjust:
   • Set \(-2x + \log_2(b^2-4) \leq 3\).
   • At \(x = 1\), \(-2(1) + \log_2(b^2-4) = 3\) simplifies to \(\log_2(b^2-4) = 5\).
   • Solve for \(b^2 - 4 = 2^5 = 32\) so \(b^2 = 36\).
   • This gives possible \(b = \pm 6\).
     • When \(b = 6\), \(-2(1) + \log_2(32) = 3\), maintaining maximum at \(x=1\).
     • When \(b = -6\), same logic applies.
5. Therefore, based on \(\log_2(b^2 - 4)\), \(-\sqrt{36}=-6\) to \(\sqrt{36}=6\).
   • Ensure \(b^2 > 4\), so exclude ends of intervals where equation is \((b^2=4)\).

The correct set where maximum occurs at \(x = 1\) is \([-6,-2) \cup (2,6]\).