JEE Main · 2012 · 29 Apr 2012 · Offline · Application of Derivatives

Let

a, b \in R

be such that the function

f

given by

f (x) = ln ∣ x ∣ + b x^{2} + a x, 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}

(A)Statement 1 is false, statement 2 is true
(B)Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1✓ Correct
(C)Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1
(D)Statement 1 is true, statement 2 is false

Answer: (B)

Solution

Differentiating,

f^{'} (x) = \frac{1}{x} + 2 b x + a

. Requiring

f^{'} (- 1) = 0

gives

a - 2 b = 1

, and

f^{'} (2) = 0

gives

a + 4 b = - \frac{1}{2}

; solving the pair gives

a = \frac{1}{2}

and

b = - \frac{1}{4}

, so the second statement is true. To classify the points,

f^{''} (x) = - \frac{1}{x ^{2}} + 2 b = - \frac{1}{x ^{2}} - \frac{1}{2}

, which is negative at both

x = - 1

and

x = 2

, so each is a local maximum and the first statement is true. The values of

a

and

b

named in the second statement are exactly what place those maxima, so the second statement correctly explains the first.

JEE Main · 2012 · 29 Apr 2012 · Offline · Single correct option