If afx-bf1x-x-5x- a- b then fx equals-

The correct option is D 1a^2 b^2[x(a 5b)+1x(5a b)] af(x)+bf( 1 x )=x+ 5 x (i) Put x= 1 x , we get #160; af( 1 x )+bf(x)= 1 x + ...

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Byju's Answer

Standard XII

Mathematics

Definition of Function

If afx+bf1x...

Question

If $a f (x) + b f (\frac{1}{x}) = x + \frac{5}{x}; (a \neq b)$ then $f (x)$ equals.

\frac{1}{a^{2} - b^{2}} (x + \frac{1}{x})

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\frac{1}{a^{2} - b^{2}} [x (5 a - b) + \frac{1}{x} (5 b - a)]

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\frac{1}{a^{2} - b^{2}} [x (a - 5 b) + \frac{1}{x} (5 a - b)]

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Solution

The correct option is D $\frac{1}{a^{2} - b^{2}} [x (a - 5 b) + \frac{1}{x} (5 a - b)]$
$a f (x) + b f (\frac{1}{x}) = x + \frac{5}{x} - (i)$
Put $x = \frac{1}{x}$ , we get
$a f (\frac{1}{x}) + b f (x) = \frac{1}{x} + 5 x - (i i)$
Multiplying $(i)$ by $^{'} a^{'}$ and $(i i)$ by $^{'} b^{'}$ and subtracting them, we get,
$a^{2} f (x) - b^{2} f (x) = a x + \frac{5 a}{x} - \frac{b}{x} - 5 b x f (x) = \frac{1}{a^{2} - b^{2}} [x (a - 5 b) + \frac{1}{x} (5 a - b)]$

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If af(x)+bf(1x)=x+5x;(a≠b) then f(x) equals.

The correct option is D 1a2−b2[x(a−5b)+1x(5a−b)]af(x)+bf(1x)=x+5x−(i)Put x=1x, we get af(1x)+bf(x)=1x+5x−(ii)Multiplying (i) by ′a′ and (ii) by ′b′ and subtracting them, we get,a2f(x)−b2f(x)=ax+5ax−bx−5bxf(x)=1a2−b2[x(a−5b)+1x(5a−b)]

If $a f (x) + b f (\frac{1}{x}) = x + \frac{5}{x}; (a \neq b)$ then $f (x)$ equals.