Suppose f(x) is a polynomial of degree four,h | vedclass

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

Question

(C) Since $f(x)$ is a polynomial of degree $4$,its derivative $f'(x)$ is a polynomial of degree $3$.Given that the critical points are $-1, 0, 1$,we h

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Since $f(x)$ is a polynomial of degree $4$,its derivative $f'(x)$ is a polynomial of degree $3$.Given that the critical points are $-1, 0, 1$,we have $f'(x) = k(x+1)(x)(x-1) = k(x^3 - x)$ for some constant $k 
eq 0$.Integrating $f'(x)$,we get $f(x) = k(\frac{x^4}{4} - \frac{x^2}{2}) + C$.We want to find $T = \{x \in \mathbb{R} \mid f(x) = f(0)\}$.Setting $f(x) = f(0)$,we have $k(\frac{x^4}{4} - \frac{x^2}{2}) + C = C$.This simplifies to $\frac{x^4}{4} - \frac{x^2}{2} = 0$,which implies $x^2(\frac{x^2}{4} - \frac{1}{2}) = 0$.Thus,$x^2 = 0$ or $x^2 = 2$.The elements of $T$ are $0, \sqrt{2}, -\sqrt{2}$.The sum of squares of these elements is $0^2 + (\sqrt{2})^2 + (-\sqrt{2})^2 = 0 + 2 + 2 = 4$.

Suppose $f(x)$ is a polynomial of degree four,having critical points at $-1, 0, 1$. If $T = \{x \in \mathbb{R} \mid f(x) = f(0)\}$,then the sum of squares of all the elements of $T$ is

Similar Questions

Consider all rectangles lying in the region $\left\{( x , y ) \in R \times R : 0 \leq x \leq \frac{\pi}{2} \text{ and } 0 \leq y \leq 2 \sin (2 x )\right\}$ and having one side on the $x$-axis. The area of the rectangle which has the maximum perimeter among all such rectangles is

Find the absolute maximum and minimum values of the function $f$ given by $f(x) = 12x^{\frac{4}{3}} - 6x^{\frac{1}{3}}$ for $x \in [-1, 1]$.

If the absolute maximum and absolute minimum values of the function $f(x) = x^3 - 2x^2 + x - 3$ defined on $[0, 2]$ are $M$ and $m$ respectively, then $M + m =$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module