If f(x) + 2f(1/x) = 3x for x neq 0 and S = {x | vedclass

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

Question

(B) Given the equation: $f(x) + 2f(1/x) = 3x$  ........$(1)$Replacing $x$ with $1/x$,we get: $f(1/x) + 2f(x) = 3/x$  ........$(2)$From equation $(2)$,

Vedclass · Accepted Answer

contains exactly two elements.

Vedclass · Answer

(B) Given the equation: $f(x) + 2f(1/x) = 3x$  ........$(1)$Replacing $x$ with $1/x$,we get: $f(1/x) + 2f(x) = 3/x$  ........$(2)$From equation $(2)$,$f(1/x) = 3/x - 2f(x)$.Substituting this into equation $(1)$:$f(x) + 2(3/x - 2f(x)) = 3x$$f(x) + 6/x - 4f(x) = 3x$$-3f(x) = 3x - 6/x$$f(x) = 2/x - x$Now,for the set $S$,we solve $f(x) = f(-x)$:$2/x - x = 2/(-x) - (-x)$$2/x - x = -2/x + x$$4/x = 2x$$2/x = x$$x^2 = 2$$x = \pm \sqrt{2}$Thus,$S = \{\sqrt{2}, -\sqrt{2}\}$.Therefore,$S$ contains exactly two elements.

If $f(x) + 2f(1/x) = 3x$ for $x \neq 0$ and $S = \{x \in R : f(x) = f(-x)\}$,then $S$:

Similar Questions

If $f(x)$ is a polynomial function satisfying the condition $f(x) \cdot f(1/x) = f(x) + f(1/x)$ and $f(2) = 9$,then:

Let $f: R \rightarrow R$ be defined by $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$ for all $x$ and $y$. If $f^{\prime}(0)$ exists and equals $-1$ and $f(0)=1$, then $f(2)=$

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:

Let $f: N \times N \rightarrow N$ be a function such that $f(1,1)=2$,$f(m+1, n)=f(m, n)+2(m+n)$,and $f(m, n+1)=f(m, n)+2(m+n-1)$ for all $m, n \in N$. Find the value of $f(2,2)$.

Let $f(x + y) = f(x) + f(y)$ for all $x, y \in R.$ Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module