If f is an even function defined on the inter | vedclass

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

Question

(A) Since $f$ is an even function,$f(-x) = f(x)$ for all $x \in (-5, 5)$.Given $f(x) = f\left( \frac{x + 1}{x + 2} \right)$.Since $f(x) = f(-x)$,we ha

Vedclass · Accepted Answer

$\frac{-3 - \sqrt{5}}{2}, \frac{-3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2}$

Vedclass · Answer

(A) Since $f$ is an even function,$f(-x) = f(x)$ for all $x \in (-5, 5)$.Given $f(x) = f\left( \frac{x + 1}{x + 2} \right)$.Since $f(x) = f(-x)$,we have $f(x) = f\left( \frac{-x + 1}{-x + 2} \right)$.Case $1$: $x = \frac{-x + 1}{-x + 2} \Rightarrow -x^2 + 2x = -x + 1 \Rightarrow x^2 - 3x + 1 = 0$.Solving for $x$,we get $x = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$.Case $2$: $-x = \frac{x + 1}{x + 2} \Rightarrow -x^2 - 2x = x + 1 \Rightarrow x^2 + 3x + 1 = 0$.Solving for $x$,we get $x = \frac{-3 \pm \sqrt{9 - 4}}{2} = \frac{-3 \pm \sqrt{5}}{2}$.Thus,the four values are $\frac{-3 - \sqrt{5}}{2}, \frac{-3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2}$.

If $f$ is an even function defined on the interval $(-5, 5)$,then four real values of $x$ satisfying the equation $f(x) = f\left( \frac{x + 1}{x + 2} \right)$ are

Similar Questions

At room temperature,a $P-$ type semiconductor has

An atomic power nuclear reactor can deliver $300 \, MW$. The energy released due to the fission of each nucleus of uranium atom $U^{238}$ is $170 \, MeV$. The number of uranium atoms fissioned per hour will be:

The diode used in the circuit shown in the figure has a constant voltage drop of $0.5 \, V$ at all currents and a maximum power rating of $100 \, mW$. What should be the value of the resistor $R$,connected in series with the diode,for obtaining maximum current?

Element having highest $I.P.$ value is

When $n-p-n$ transistor is used as an amplifier:

Vedclass Test Series

Exam Paper Generator

Online Exam Module