If y = a x + bx^2 + x has extreme values at x = 1 and x = 2, then Option 1: a=2/3, b=-1/6 Option 2: a=-2/3, b=1/6 Option 3: a=-2/3, b=-1/6 Option 4: a=2/3, b=1/6

Answer: a=-2/3, b=-1/6. Solution: y' = a/x+2bx+1. Extreme values need y'=0. At x=1: a+2b+1=0. At x=2: a/2+4b+1=0 a+8b+2=0. Subtracting: 6b+1=0 b=-1/6, then a=-2b-1=-2/3.

Calculus

Application of Derivatives

Easy

common

If $y = a\log x + bx^2 + x$ has extreme values at $x = 1$ and $x = 2$ , then

$a=\frac{2}{3}, b=-\frac{1}{6}$

$a=-\frac{2}{3}, b=\frac{1}{6}$

$a=-\frac{2}{3}, b=-\frac{1}{6}$

$a=\frac{2}{3}, b=\frac{1}{6}$

✅ Correct Option: 3

$y' = \frac{a}{x}+2bx+1$ . Extreme values need $y'=0$ .

At $x=1$ : $a+2b+1=0$ .

At $x=2$ : $\frac{a}{2}+4b+1=0 \Rightarrow a+8b+2=0$ .

Subtracting: $6b+1=0 \Rightarrow b=-\frac{1}{6}$ , then $a=-2b-1=-\frac{2}{3}$ .

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Match List-I with List-II

List-I	List-II
Function	Increasing on the interval
(A) $f(x) = -x^2 - 2x + 1$	(I) $(-\infty, -1)$
(B) $f(x) = x^2 + 1$	(II) $(1, \infty)$
(C) $f(x) = x^2 - 2x + 3$	(III) $(-\infty, 0)$
(D) $f(x) = -x^2$	(IV) $(0, \infty)$

Choose the correct answer from the options given below:

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If $y = a\log x + bx^2 + x$ has extreme values at $x = 1$ and $x = 2$ , then