For the differentiable function f: R - {0} ri | vedclass

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

Question

(D) Given the equation: $3 f(x) + 2 f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$  (Equation $1$)Replace $x$ with $\frac{1}{x}$: $3 f\left(\frac{1}{x}

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(D) Given the equation: $3 f(x) + 2 f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$  (Equation $1$)Replace $x$ with $\frac{1}{x}$: $3 f\left(\frac{1}{x}\right) + 2 f(x) = x - 10$  (Equation $2$)Multiply Equation $1$ by $3$ and Equation $2$ by $2$:$9 f(x) + 6 f\left(\frac{1}{x}\right) = \frac{3}{x} - 30$$4 f(x) + 6 f\left(\frac{1}{x}\right) = 2x - 20$Subtracting the second from the first:$5 f(x) = \frac{3}{x} - 2x - 10$$f(x) = \frac{3}{5x} - \frac{2x}{5} - 2$Now,find $f(3)$:$f(3) = \frac{3}{5(3)} - \frac{2(3)}{5} - 2 = \frac{1}{5} - \frac{6}{5} - 2 = -1 - 2 = -3$Find $f^{\prime}(x)$:$f^{\prime}(x) = -\frac{3}{5x^2} - \frac{2}{5}$Calculate $f^{\prime}\left(\frac{1}{4}\right)$:$f^{\prime}\left(\frac{1}{4}\right) = -\frac{3}{5(1/16)} - \frac{2}{5} = -\frac{48}{5} - \frac{2}{5} = -\frac{50}{5} = -10$Finally,calculate $\left|f(3) + f^{\prime}\left(\frac{1}{4}\right)\right|$:$|-3 + (-10)| = |-13| = 13$

For the differentiable function $f: R - \{0\} \rightarrow R$,let $3 f(x) + 2 f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$. Then $\left|f(3) + f^{\prime}\left(\frac{1}{4}\right)\right|$ is equal to

Similar Questions

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{x - 2}{x^2 - 3x + 2}, & x \in R - \{1, 2\} \\ 2, & x = 1 \\ 1, & x = 2 \end{cases}$,then $\lim_{x \rightarrow 2} \frac{f(x) - f(2)}{x - 2} = $

Find the derivative: $\frac{d}{dx} \{ e^{-ax^2} \log(\sin x) \}$

If $y = \frac{1}{4}u^4$ and $u = \frac{2}{3}x^3 + 5$,then $\frac{dy}{dx} = $

Given $f(x) = 4x^3 - 6x^2 \cos 2a + 3x \sin 2a \sin 6a + \sqrt{\ln(2a - a^2)}$,then:

Let $f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10\sqrt{x - 1}}$ for $1 < x < 26$ be a real-valued function. Then $f'(x)$ for $1 < x < 26$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module