If the function y = a log x + bx^2 + x has ex | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $y = a \log x + bx^2 + x$. Find the derivative: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the function has extreme values at $x = 1$ an

Vedclass · Accepted Answer

$\left( -\frac{2}{3}, -\frac{1}{6} \right)$

Vedclass · Answer

(D) Given $y = a \log x + bx^2 + x$. Find the derivative: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the function has extreme values at $x = 1$ and $x = 2$,the derivative must be zero at these points. At $x = 1$: $\frac{a}{1} + 2b(1) + 1 = 0 \implies a + 2b = -1$ (Equation $1$). At $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \implies \frac{a}{2} + 4b = -1 \implies a + 8b = -2$ (Equation $2$). Subtracting Equation $1$ from Equation $2$: $(a + 8b) - (a + 2b) = -2 - (-1) \implies 6b = -1 \implies b = -\frac{1}{6}$. Substitute $b = -\frac{1}{6}$ into Equation $1$: $a + 2(-\frac{1}{6}) = -1 \implies a - \frac{1}{3} = -1 \implies a = -1 + \frac{1}{3} = -\frac{2}{3}$. Thus,$(a, b) = \left( -\frac{2}{3}, -\frac{1}{6} \right)$.

If the function $y = a \log x + bx^2 + x$ has extreme values at $x = 1$ and $x = 2$,then $(a, b) = \dots$

Similar Questions

What is the local maximum value of $\frac{\log x}{x}$?

If $x$ and $y$ are two variables such that $x > 0$ and $xy = 1$,what is the minimum value of $x + y$?

The minimum value of $2x^2+x-1$ is:

$A$ triangular garden is fenced on two sides,and the third side is a straight river bank. The two fenced sides have an equal length of $x$. What is the maximum area that can be enclosed by the garden?

$A$ closed cylinder of given volume will have the least surface area when the ratio of its height to its base radius is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module