Let a, b \in R be such that the function f given by f(x) = ln | x | + bx2 + ax, x \neq 0 has extreme values at x = – 1 and x = 2. Statement 1 : f has local maximum at x = – 1 and at x = 2. Statement 2 : a = \frac{1}{2} \textrm{ }\textrm{ } and \textrm{ }\textrm{ } b = \frac{- 1}{4}

Engineering

Mathematics

Maxima and Minima

Question

Let a, b $\in$ R be such that the function f given by

f(x) = ln | x | + bx² + ax, x $\neq$ 0 has extreme values at x = – 1 and x = 2.

Statement 1 : f has local maximum at x = – 1 and at x = 2.

Statement 2 : $a = \frac{1}{2} and b = \frac{- 1}{4}$

Statement 1 is true, Statement 2 is false.

Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

Statement 1 is false, Statement 2 is true.

Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 2.

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f(x) = ln | x | + bx² + ax, x $\neq$ 0

$f' (x) = \frac{1}{x} + 2 bx + a$

extreme values at x = – 1, 2

$\Rightarrow$ – 1 – 2b + a = 0

$\Rightarrow$ a – 2b = 1 ...(1)

and $\frac{1}{2} + 4 b + a = 0 \Rightarrow a + 4 b = \frac{- 1}{2}$ ...(2)

From (A) and (B) $a = \frac{1}{2}, b = \frac{- 1}{4}$ ,

again $f " (x) = 2 b - \frac{- 1}{x^{2}} = \frac{- 1}{2} - \frac{1}{x^{2}}$

$\Rightarrow$ f "(–1) < 0 and f "(B) < 0

$\Rightarrow$ f has local maximum at x = – 1 and x = 2.