If 5f(x) + 3f (1x) = x + 2 and y = x f(x), then (dydx)x=1 is equal to

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If 5f(x) + 3f $(\frac{1}{x})$ = x + 2 and y = x f(x), then ${(\frac{d y}{d x})}_{x = 1}$ is equal to

Options:

A .

14

B .

\frac{7}{8}

C .

1

D . None of these

Answer: Option B
:
B

5 f (x) + 3 f (\frac{1}{x}) = x + 2 . . . . . . . . (i)

R e p l a c i n g x b y \frac{1}{x}

∴ 5 f (\frac{1}{x}) + 3 f (x) = \frac{1}{x} + 2 . . . (i i)

F r o m E q . (i) .

25 f (x) + 15 f (\frac{1}{x}) = 5 x + 10 . . . . (i i i)

a n d f r o m E q . (i i)

9 f (x) + 15 f (\frac{1}{x}) = \frac{3}{x} + 6 . . . . . (i v)

S u b t r a c t i n g E q . (i v) f r o m (i i i), w e g e t

∴ 16 f (x) = 5 x - \frac{3}{x} + 4

∴ x f (x) = \frac{5 x^{2} - 3 + 4 x}{16} = y

∴ \frac{d y}{d x} = \frac{10 x + 4}{16}

∴ \frac{d y}{d x} |_{x = 1} = \frac{10 + 4}{16}

= \frac{7}{8}

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