A real valued function y = f(x) satisfies the | vedclass

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

Question

(B) Given the inequality: $f\left( x - \frac{4}{9} \right) + 2x \le \frac{9}{4}x^2 + \frac{8}{9} \le f\left( x + \frac{4}{9} \right) - 2x$. Let $g(x)

Vedclass · Accepted Answer

$\frac{9}{2}$

Vedclass · Answer

(B) Given the inequality: $f\left( x - \frac{4}{9} \right) + 2x \le \frac{9}{4}x^2 + \frac{8}{9} \le f\left( x + \frac{4}{9} \right) - 2x$. Let $g(x) = \frac{9}{4}x^2 + \frac{8}{9}$. We can rewrite the inequality as $f\left( x - \frac{4}{9} \right) \le g(x) - 2x$ and $g(x) + 2x \le f\left( x + \frac{4}{9} \right)$. Let $x - \frac{4}{9} = t$,then $x = t + \frac{4}{9}$. Substituting this,we get $f(t) \le \frac{9}{4}(t + \frac{4}{9})^2 + \frac{8}{9} - 2(t + \frac{4}{9}) = \frac{9}{4}t^2 + 2t + \frac{4}{9} + \frac{8}{9} - 2t - \frac{8}{9} = \frac{9}{4}t^2 + \frac{4}{9}$. Similarly,for the second part,let $x + \frac{4}{9} = t$,then $x = t - \frac{4}{9}$. Substituting this,we get $f(t) \ge \frac{9}{4}(t - \frac{4}{9})^2 + \frac{8}{9} + 2(t - \frac{4}{9}) = \frac{9}{4}t^2 - 2t + \frac{4}{9} + \frac{8}{9} + 2t - \frac{8}{9} = \frac{9}{4}t^2 + \frac{4}{9}$. Since $f(t) \le \frac{9}{4}t^2 + \frac{4}{9}$ and $f(t) \ge \frac{9}{4}t^2 + \frac{4}{9}$,it must be that $f(x) = \frac{9}{4}x^2 + \frac{4}{9}$. Now,find the second derivative: $f'(x) = \frac{d}{dx}\left( \frac{9}{4}x^2 + \frac{4}{9} \right) = \frac{9}{2}x$. $f''(x) = \frac{d}{dx}\left( \frac{9}{2}x \right) = \frac{9}{2}$. Therefore,$f''(2) = \frac{9}{2}$.

$A$ real valued function $y = f(x)$ satisfies the relation $f\left( x - \frac{4}{9} \right) + 2x \le \frac{9}{4}x^2 + \frac{8}{9} \le f\left( x + \frac{4}{9} \right) - 2x$. The value of $f''(2)$ is

Similar Questions

Let $u+v+w=3$, where $u, v, w \in \mathbb{R}$ and $f(x)=u x^2+v x+w$ be such that $f(x+y)=f(x)+f(y)+x y$ for all $x, y \in \mathbb{R}$. Then $f(1)$ is equal to:

Suppose $f : R \rightarrow (0, \infty)$ be a differentiable function such that $5f(x + y) = f(x) \cdot f(y), \forall x, y \in R$. If $f(3) = 320$,then $\sum_{n=0}^5 f(n)$ is equal to:

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:

Let $f: R \rightarrow R$ be a function defined by $f(x)=(2+3a)x^2 + \left(\frac{a+2}{a-1}\right)x + b$,where $a \neq 1$. If $f(x+y) = f(x) + f(y) + 1 - \frac{2}{7}xy$,then the value of $28 \sum_{i=1}^3 |f(i)|$ is:

The function $t$ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t(C) = \frac{9C}{5} + 32$. Find $t(28)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module