If f(x)=a log |x|+b x^2+x has extreme values | vedclass

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

Question

(B) Given,$f(x)=a \log |x|+b x^2+x$.The derivative is $f^{\prime}(x) = \frac{a}{x} + 2bx + 1$.Since the function has extreme values at $x=-1$ and $x=2

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given,$f(x)=a \log |x|+b x^2+x$.The derivative is $f^{\prime}(x) = \frac{a}{x} + 2bx + 1$.Since the function has extreme values at $x=-1$ and $x=2$,we have $f^{\prime}(-1)=0$ and $f^{\prime}(2)=0$.For $x=-1$: $f^{\prime}(-1) = \frac{a}{-1} + 2b(-1) + 1 = 0 \Rightarrow -a - 2b + 1 = 0 \Rightarrow a + 2b = 1$ $(i)$.For $x=2$: $f^{\prime}(2) = \frac{a}{2} + 2b(2) + 1 = 0 \Rightarrow \frac{a}{2} + 4b + 1 = 0 \Rightarrow a + 8b = -2$ (ii).Subtracting $(i)$ from (ii): $(a + 8b) - (a + 2b) = -2 - 1 \Rightarrow 6b = -3 \Rightarrow b = -\frac{1}{2}$.Substituting $b = -\frac{1}{2}$ into $(i)$: $a + 2(-\frac{1}{2}) = 1 \Rightarrow a - 1 = 1 \Rightarrow a = 2$.Thus,the ordered pair $(a, b) = \left(2, -\frac{1}{2}\right)$.

If $f(x)=a \log |x|+b x^2+x$ has extreme values at $x=-1$ and $x=2$,then the ordered pair $(a, b)=$

Similar Questions

For the function $f(x)=x^3-6x^2+12x-3$,the point $x=2$ is

If $f(x)=x^5-5 x^4+5 x^3-10$ has its local maxima and minima at $x=a$ and $x=b$ respectively,then $2 a+b$ is equal to

Find all the points of local maxima and local minima of the function $f$ given by $f(x) = 2x^3 - 6x^2 + 6x + 5$.

Let $f : (-\infty, \infty) \to (-\infty, \infty)$ be defined by $f(x) = x^3 + 1$.
Statement-$1$: The function has a local extremum at $x = 0$.
Statement-$2$: The function $f(x)$ is continuous and differentiable on $(-\infty, \infty)$ and $f'(0) = 0$.

Divide $10$ into two parts such that the sum of double of the first and the square of the second is minimum.

Vedclass Test Series

Exam Paper Generator

Online Exam Module