Question

Let ƒ(x) = $\left\{ \begin{matrix} x^{3} - x^{2} + 10x - 5, & x \leq 1 \\ –2x + \log_{2}(b^{2} - 2), & x > 1 \end{matrix} \right.\$ the set of values of b for which ƒ(x) have greatest value at x = 1 is given by –

• A. 1 £ b £ 2
• B. b = {1, 2}
• C. b Ī (–, –1)
• D. None

Answer 1

Answer: (D)

None

Answer 2

For x £ 1

ƒ¢(x) = 3x² – 2x + 10 = 3 $\left\lbrack \left( x - \frac{1}{3} \right)^{2} + \frac{29}{3} \right\rbrack$ > 0

So ƒ(x) is an increasing function for x £ 1

For x > 1, ƒ¢(x) = –2

So, ƒ(x) is decreasing function for x > 1.

Now, ƒ(x) will have greatest value at x = 1. If

$\lim_{x \rightarrow 1 +}$ƒ(x) £ ƒ(1) Ž $\lim_{h \rightarrow 0}$ƒ(1 + h) £ 5

Ž $\lim_{h \rightarrow 0}$ – 2(1 + h) + log₂ (b² – 2) £ 5

Ž –2 + log₂ (b² – 2) £ 5 Ž log₂ (b² – 2) £ 7

Ž b² £ 130 but b² > 2

Ž 2 < b² £ 130

\ b Ī [–$\sqrt{130}$, –$\sqrt{2}$) Č ($\sqrt{2}$,$\sqrt{130}$]