If y = a log |x| + b x^2 + x has its extremum | vedclass

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

Question

(B) Given $y = a \log |x| + b x^2 + x$. Since $\frac{d}{dx} \log |x| = \frac{1}{x}$ for all $x 
eq 0$,we differentiate $y$ with respect to $x$: $\fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given $y = a \log |x| + b x^2 + x$. Since $\frac{d}{dx} \log |x| = \frac{1}{x}$ for all $x 
eq 0$,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the function has extremum values at $x = -1$ and $x = 2$,the derivative must be zero at these points. For $x = -1$: $\frac{a}{-1} + 2b(-1) + 1 = 0 \implies -a - 2b + 1 = 0 \implies a + 2b = 1$ (Equation $1$). For $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \implies \frac{a}{2} + 4b + 1 = 0 \implies a + 8b = -2$ (Equation $2$). Subtracting Equation $1$ from Equation $2$: $(a + 8b) - (a + 2b) = -2 - 1 \implies 6b = -3 \implies b = -\frac{1}{2}$. Substituting $b = -\frac{1}{2}$ into Equation $1$: $a + 2(-\frac{1}{2}) = 1 \implies a - 1 = 1 \implies a = 2$. Thus,$a = 2$ and $b = -\frac{1}{2}$.

If $y = a \log |x| + b x^2 + x$ has its extremum values at $x = -1$ and $x = 2$,then

Similar Questions

Let $a, b \in R$ be such that the function $f(x) = \ln|x| + bx^2 + ax, x \neq 0$ has extreme values at $x = -1$ and $x = 2$.
Statement-$1$: $f$ has a local maximum at $x = -1$ and $x = 2$.
Statement-$2$: $a = \frac{1}{2}$ and $b = -\frac{1}{4}$.

The number of values of $x$ where the function $f(x) = \cos x + \cos (\sqrt{2} x)$ attains its maximum is

$x$ and $y$ are two positive integers such that $2x + 3y = 50$. If $x^2 y^3$ is maximum for $x = \alpha$ and $y = \beta$, then $\frac{\alpha}{2} + \frac{\beta}{5} =$

There are $50$ apple trees in an orchard. Each tree produces $800$ apples. For each additional tree planted in the orchard,the output per tree drops by $10$ apples. The number of trees that should be added to the existing orchard to maximize the total output is:

Find the maximum and minimum values of the function given by $g(x) = x^{3} + 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module