For real constants a-b-c-d- suppose fx is a function of the form fx - ax8-bx6-cx4-dx2-15x-1x for x - 0- If f 5 - 2- then the value of f- 5 is

X f(x) = ax^8 + bx^6 + cx^4 + dx^2 + 15x + 1 ⋯(1) Replacing x by x, xf( x)=ax^8 + bx^6 + cx^4 + dx^2 15x + 1 ⇒ xf ( x)= ax^8 bx^6 cx^4 dx^2+15x 1 ⋯(2) Ad ...

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Byju's Answer

Standard XII

Mathematics

Polynomial Functions

For real cons...

Question

For real constants $a, b, c, d,$ suppose $f (x)$ is a function of the form $f (x) = \frac{a x^{8} + b x^{6} + c x^{4} + d x^{2} + 15 x + 1}{x}$ for $x \neq 0.$ If $f (5) = 2,$ then the value of $f (- 5)$ is

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Solution

$x f (x) = a x^{8} + b x^{6} + c x^{4} + d x^{2} + 15 x + 1 \dots (1)$
Replacing $x$ by $- x,$
$- x f (- x) = a x^{8} + b x^{6} + c x^{4} + d x^{2} - 15 x + 1$
$\Rightarrow x f (- x) = - a x^{8} - b x^{6} - c x^{4} - d x^{2} + 15 x - 1 \dots (2)$
Adding $(1)$ and $(2),$ we get
$x (f (x) + f (- x)) = 30 x$
$\Rightarrow f (x) + f (- x) = 30$
At $x = 5,$ we have
$f (5) + f (- 5) = 30$
$∴ f (- 5) = 28$

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For real constants a,b,c,d, suppose f(x) is a function of the form f(x)=ax8+bx6+cx4+dx2+15x+1x for x≠0. If f(5)=2, then the value of f(–5) is

xf(x)=ax8+bx6+cx4+dx2+15x+1 ⋯(1) Replacing x by −x, −xf(−x)=ax8+bx6+cx4+dx2−15x+1 ⇒xf(−x)=−ax8−bx6−cx4−dx2+15x−1 ⋯(2) Adding (1) and (2), we get x(f(x)+f(–x))=30x ⇒f(x)+f(–x)=30 At x=5, we have f(5)+f(–5)=30 ∴f(–5)=28

For real constants $a, b, c, d,$ suppose $f (x)$ is a function of the form $f (x) = \frac{a x^{8} + b x^{6} + c x^{4} + d x^{2} + 15 x + 1}{x}$ for $x \neq 0.$ If $f (5) = 2,$ then the value of $f (- 5)$ is