Let g(x) = 2f(frac{x}{2}) + f(2 - x) and f''( | vedclass

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

Question

(D) Given $g(x) = 2f(\frac{x}{2}) + f(2 - x)$.Taking the derivative with respect to $x$,we get $g'(x) = 2f'(\frac{x}{2}) \cdot \frac{1}{2} + f'(2 - x)

Vedclass · Accepted Answer

$(0, 4/3)$

Vedclass · Answer

(D) Given $g(x) = 2f(\frac{x}{2}) + f(2 - x)$.Taking the derivative with respect to $x$,we get $g'(x) = 2f'(\frac{x}{2}) \cdot \frac{1}{2} + f'(2 - x) \cdot (-1) = f'(\frac{x}{2}) - f'(2 - x)$.We are given $f''(x) For $g(x)$ to be increasing,we require $g'(x) > 0$,which means $f'(\frac{x}{2}) - f'(2 - x) > 0$,or $f'(\frac{x}{2}) > f'(2 - x)$.Since $f'(x)$ is strictly decreasing,$f'(a) > f'(b)$ implies $a Therefore,$\frac{x}{2} Solving for $x$: $\frac{x}{2} + x Thus,$g(x)$ increases in the interval $(0, 4/3)$.

Let $g(x) = 2f(\frac{x}{2}) + f(2 - x)$ and $f''(x) < 0$ for all $x \in (0, 2)$. Then $g(x)$ increases in

Similar Questions

$f(x) = \frac{x}{2} + \frac{2}{x}, x \neq 0$ is strictly decreasing in

If $x > 0$,then $\frac{x}{1+x} - \log(1+x)$

The complete set of values of $m$ for which the function $f(x) = e^{\sin x} + 2m\sin x + 1$ is increasing for all $x \in \left( 0, \frac{\pi}{2} \right)$ is:

If $y = 8x^3 - 60x^2 + 144x + 27$ is a decreasing function on the interval $(a, b)$,then $(a, b) = $

The function $f(x)$ is defined by $f(x)=(x+2) e^{-x}$. Which of the following statements is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module