Question

Let f(x) = (1 + b²) x² + 2bx + 1 and let m(2) be the minimum value of f(x). As b varies, the range of m(2) is-

• A. [0, 1]
• B. $\left\lbrack 0,\frac{1}{2} \right\rbrack$
• C. $\left\lbrack \frac{1}{2},1 \right\rbrack$
• D. (0, 1]

Answer 1

Answer: (D)

(0, 1]

Answer 2

f(x) = (1 + b²) x² + 2bx + 1

min.of f(x) = m(2) =$\frac{- D}{4a}$ = $\frac{- 4\left\lbrack b^{2} - (1 + b^{2}) \right\rbrack}{4(1 + b^{2})}$

m(2) = $\frac{1}{1 + b^{2}}$

m¢(2) = – $\left( \frac{2b}{(1 + b^{2})^{2}} \right)$= 0

b = 0 is critical point, $\begin{matrix} \begin{matrix} 0 < m(b) \leq 1 \end{matrix} \end{matrix}$