If g:[-2, 2] to R where g(x) = x^3 + tan x + | vedclass

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

Question

(C) For $g(x)$ to be an odd function,it must satisfy $g(-x) = -g(x)$,which implies $g(x) + g(-x) = 0$.Given $g(x) = x^3 + 	an x + \left[ \frac{x^2 +

Vedclass · Accepted Answer

$P > 5$

Vedclass · Answer

(C) For $g(x)$ to be an odd function,it must satisfy $g(-x) = -g(x)$,which implies $g(x) + g(-x) = 0$.Given $g(x) = x^3 + 	an x + \left[ \frac{x^2 + 1}{P} ight]$.Then $g(-x) = (-x)^3 + 	an(-x) + \left[ \frac{(-x)^2 + 1}{P} ight] = -x^3 - 	an x + \left[ \frac{x^2 + 1}{P} ight]$.Adding these,$g(x) + g(-x) = 2 \left[ \frac{x^2 + 1}{P} ight] = 0$.This implies $\left[ \frac{x^2 + 1}{P} ight] = 0$ for all $x \in [-2, 2]$.For the greatest integer function to be zero,we must have $0 \le \frac{x^2 + 1}{P} Since $x \in [-2, 2]$,the maximum value of $x^2 + 1$ is $2^2 + 1 = 5$.Thus,we require $\frac{5}{P}  5$ (assuming $P > 0$ for the function to be well-defined in this context).

If $g:[-2, 2] \to R$ where $g(x) = x^3 + \tan x + \left[ \frac{x^2 + 1}{P} \right]$ is an odd function,then the value of the parameter $P$ is:

Similar Questions

If $f$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \in N$ such that $f(1)=3$ and $\sum\limits_{x = 1}^n {f(x) = 120}$,find the value of $n$.

If $f(x)$ is a polynomial function satisfying $f(x) \cdot f(\frac{1}{x}) = f(x) + f(\frac{1}{x})$ and $f(4) = 65$,then the value of $f(6)$ is:

Let $f: R \rightarrow R$ be a differentiable function that satisfies the relation $f(x + y) = f(x) + f(y) - 1$ for all $x, y \in R$. If $f'(0) = 2$,then $|f(-2)|$ is equal to:

If $f(f(0)) = 0$,where $f(x) = x^2 + ax + b$ and $b \neq 0$,then $a + b =$

If $f(x + y) = f(x)f(y)$ for all $x$ and $y$ and $f(5) = 2$,$f'(0) = 3$,then $f'(5)$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module