If f(x) is a real-valued function defined by | vedclass

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

Question

(A) Given $f(x) = \frac{a x^{10} + b x^8 + c x^6 + d x^4 + e x^2 + 12 x + 15}{x}$.Dividing each term by $x$, we get:$f(x) = a x^9 + b x^7 + c x^5 + d

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(A) Given $f(x) = \frac{a x^{10} + b x^8 + c x^6 + d x^4 + e x^2 + 12 x + 15}{x}$.Dividing each term by $x$, we get:$f(x) = a x^9 + b x^7 + c x^5 + d x^3 + e x + 12 + \frac{15}{x}$.Now, consider $f(-x)$:$f(-x) = a(-x)^9 + b(-x)^7 + c(-x)^5 + d(-x)^3 + e(-x) + 12 + \frac{15}{-x}$$f(-x) = -(a x^9 + b x^7 + c x^5 + d x^3 + e x) + 12 - \frac{15}{x}$.Adding $f(x)$ and $f(-x)$:$f(x) + f(-x) = (a x^9 + b x^7 + c x^5 + d x^3 + e x + 12 + \frac{15}{x}) + (-(a x^9 + b x^7 + c x^5 + d x^3 + e x) + 12 - \frac{15}{x})$$f(x) + f(-x) = 12 + 12 = 24$.For $x = 4$:$f(4) + f(-4) = 24$.Given $f(4) = -4$, we have:$-4 + f(-4) = 24$$f(-4) = 24 + 4 = 28$.

If $f(x)$ is a real-valued function defined by $f(x) = \frac{a x^{10} + b x^8 + c x^6 + d x^4 + e x^2 + 12 x + 15}{x}$ for $x \neq 0$ and $f(4) = -4$, then find the value of $f(-4)$.

Similar Questions

Let $S = \{1, 2, 3, 4\}$. Then the number of elements in the set $\{f: S \times S \rightarrow S : f \text{ is onto and } f(a, b) = f(b, a) \geq a; \forall (a, b) \in S \times S\}$ is

If $f(x) = \cos (\log x)$,then $f(x)f(y) - \frac{1}{2}[f(x/y) + f(xy)] = $

Let $f(x) = a^x$ $(a > 0)$ be written as $f(x) = f_1(x) + f_2(x)$,where $f_1(x)$ is an even function and $f_2(x)$ is an odd function. Then $f_1(x + y) + f_1(x - y)$ equals

The total number of functions $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5, 6\}$ such that $f(1) + f(2) = f(3)$ is equal to:

If $R \subset A \times B$ and $S \subset B \times C$ be two relations,then $(S \circ R)^{-1} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module