If x=-1 and x=2 are extreme points of f(x)=al | vedclass

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

Question

(C) Given $f(x)=\alpha \log |x|+\beta x^2+x$.The derivative is $f^{\prime}(x)=\frac{\alpha}{x}+2\beta x+1$.Since $x=-1$ and $x=2$ are extreme points,$

Vedclass · Accepted Answer

$\alpha=2, \beta=-\frac{1}{2}$

Vedclass · Answer

(C) Given $f(x)=\alpha \log |x|+\beta x^2+x$.The derivative is $f^{\prime}(x)=\frac{\alpha}{x}+2\beta x+1$.Since $x=-1$ and $x=2$ are extreme points,$f^{\prime}(-1)=0$ and $f^{\prime}(2)=0$.For $x=-1$: $\frac{\alpha}{-1}+2\beta(-1)+1=0 \Rightarrow -\alpha-2\beta+1=0 \Rightarrow \alpha+2\beta=1$ (Equation $i$).For $x=2$: $\frac{\alpha}{2}+2\beta(2)+1=0 \Rightarrow \frac{\alpha}{2}+4\beta+1=0 \Rightarrow \alpha+8\beta=-2$ (Equation $ii$).Subtracting $(i)$ from (ii): $(\alpha+8\beta)-(\alpha+2\beta)=-2-1 \Rightarrow 6\beta=-3 \Rightarrow \beta=-\frac{1}{2}$.Substituting $\beta=-\frac{1}{2}$ into $(i)$: $\alpha+2(-\frac{1}{2})=1 \Rightarrow \alpha-1=1 \Rightarrow \alpha=2$.Thus,$\alpha=2$ and $\beta=-\frac{1}{2}$.

If $x=-1$ and $x=2$ are extreme points of $f(x)=\alpha \log |x|+\beta x^2+x$,then

Similar Questions

If the functions $f(x) = \frac{x^3}{3} + 2bx + \frac{ax^2}{2}$ and $g(x) = \frac{x^3}{3} + ax + bx^2$,where $a \neq 2b$,have a common extreme point,then $a + 2b + 7$ is equal to

The maximum value of the function $f(x) = x^{3} - 12x^{2} + 36x + 17$ in the interval $[1, 10]$ is:

If ${a^2}{x^4} + {b^2}{y^4} = {c^6}$,then the maximum value of $xy$ is

Let a function $f:[0,5] \rightarrow R$ be continuous. $f(1)=3$ and $F$ be defined as $F(x)=\int_{1}^{x} t^{2} g(t) dt$,where $g(t)=\int_{1}^{t} f(u) du$. Then for the function $F$,the point $x=1$ is

If $x$ is real,then the minimum value of $\frac{x^2-x+1}{x^2+x+1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module