Let x = 1 2 and x = 1 are the extreme points of f x = a l o g x + b x x + 2 c o s π x π , then

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Question: Answered & Verified by Expert

Let

x = \frac{1}{2}

and

x = 1

are the extreme points of

f (x) = a l o g |x| + b x |x| + 2 \frac{c o s |π x|}{π}

, then

MathematicsApplication of DerivativesJEE Main

Options:

A $a - b = 2$
B $a + b = 1$
C $\frac{a}{b} = \frac{1}{2}$
D $\frac{a}{b} = 2$

Solution:

1518 Upvotes Verified Answer

The correct answer is:

a - b = 2

f^{'} (x) = \frac{a}{x} + 2 b |x| - 2 s i n π x

Now

f^{'} (\frac{1}{2}) = 0 \Rightarrow 2 a + b - 2 = 0

&

f^{'} (1) = 0 \Rightarrow a + 2 b = 0

\Rightarrow b = \frac{- 2}{3}

and

a = \frac{4}{3}

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