Question

If y = a log x + bx2 + x has its extreme values at x = 1 and x = 2, then (a, b) = ____________________.

Solution

It is given that, $y=a\mathrm{log}x+b{x}^{2}+x$ has its extreme values at x = 1 and x = 2. $\therefore \frac{dy}{dx}=0$ at x = 1 and x = 2 $y=a\mathrm{log}x+b{x}^{2}+x$ Differentiating both sides with respect to x, we get $\frac{dy}{dx}=\frac{a}{x}+2bx+1$ Now, ${\left(\frac{dy}{dx}\right)}_{x=1}=0$ $⇒a+2b+1=0$ Also, ${\left(\frac{dy}{dx}\right)}_{x=2}=0$ $⇒\frac{a}{2}+4b+1=0$ Subtracting (1) from (2), we get 6b = −1 $⇒b=-\frac{1}{6}$ Putting $b=-\frac{1}{6}$ in (1), we get $a+2×\left(-\frac{1}{6}\right)=-1$ $⇒a=-1+\frac{1}{3}=-\frac{2}{3}$ Thus, the values of a and b are $-\frac{2}{3}$ and $-\frac{1}{6}$, respectively. Hence, the ordered pair (a, b) is $\left(-\frac{2}{3},-\frac{1}{6}\right)$. If y = alogx + bx2 + x has its extreme values at x = 1 and x = 2, then (a, b) = .MathematicsRD Sharma XII Vol 1 (2019)All

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