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Question: If \(y = a\log x + bx^{2} + x\) has its extremum value at \(x = 1\) and \(x = 2\), then \((a,b) =\)...

If y=alog⁡x+bx2+xy = a\log x + bx^{2} + x has its extremum value at x=1x = 1 and

x=2x = 2, then (a,b)=(a,b) =

A

(1,12)\left( 1,\frac{1}{2} \right)

B

(12,2)\left( \frac{1}{2},2 \right)

C

(2,−12)\left( 2, - \frac{1}{2} \right)

D

) (−23,−16)\left( - \frac{2}{3}, - \frac{1}{6} \right)

Explanation

Solution

)

Sol. dydx=ax+2bx+1\frac{dy}{dx} = \frac{a}{x} + 2bx + 1 ⇒ (dydx)x=1=a+2b+1=0\left( \frac{dy}{dx} \right)_{x = 1} = a + 2b + 1 = 0

⇒ a=−2b−1a = - 2b - 1

and (dydx)x=2=a2+4b+1=0\left( \frac{dy}{dx} \right)_{x = 2} = \frac{a}{2} + 4b + 1 = 0 ⇒ −2b−12+4b+1=0\frac{- 2b - 1}{2} + 4b + 1 = 0

⇒ −b+4b+12=0- b + 4b + \frac{1}{2} = 0 ⇒ 3b=−123b = \frac{- 1}{2} ⇒ b=−16b = \frac{- 1}{6}

and a=13−1=−23a = \frac{1}{3} - 1 = \frac{- 2}{3}.