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Calculus Level 4

If y = a log ⁡ ∣ x ∣ + b x 2 + x y=a \log{|x| }+bx^{2}+x has extreme values at x = − 1 a n d x = 2 x=-1~and~x=2 .

Then find a b ab .


The answer is -1.00.

Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Rahul Singh
Jun 23, 2015

We have,

y=alogx +b x 2 x^{2} +x , clearly it is defined for all x>0

Therefore, d y d x \frac{dy}{dx} = a x \frac{a}{x} +2bx+1 ........(i)

Since, the function has extreme values at x=-1 & x=2

Therefore, ( d y d x ) a t x = − 1 , 2 (\frac{dy}{dx})_{at x= -1,2} = 0

On substituting the values x= -1,2 respectively in equation (i) we get,

-a-2b+1=0 .........(ii)

and, a 2 \frac{a}{2} +4b+1=0 ...........(iii)

On solving equations (ii) & (iii) we get a = 2 \boxed{a=2} & b = − 1 / 2 \boxed{b= -1/2}

Therefore, ab = 2 × ( − 1 2 ) 2 \times\ ( \frac{-1}{2})

or, a b = − 1 \boxed{ab = -1}

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