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Differentiation

Section: Mathematics

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Question : 14 of 34

Marks: +1, -0

If

5 f(x)+3 f\left(\frac{1}{x}\right)=x+2

y=x f(x)

\frac{dy}{dx}

x=1

[AP EAPCET 21 May 2025 S1]

14
$\frac{7}{8}$
1
7

Solution:

Given,

5 f(x)+3 f\left(\frac{1}{x}\right)=x+2

replacing

x

by

\frac{1}{x}

5 f\left(\frac{1}{x}\right)+3 f(x)=\frac{1}{x}+2

Multiplying Eq. (i) by 5 and Eq. (ii) by 3, we get

25 f(x)+15 f\left(\frac{1}{x}\right)=5 x+10

15 f\left(\frac{1}{x}\right)+9 f(x)=\frac{3}{x}+6

- \qquad\qquad + \qquad\qquad -

_______________________

16 f(x)=5 x-\frac{3}{x}+4

\because \;\; f(x)=\frac{1}{16}\left(5 x-\frac{3}{x}+4\right)

y=x f(x)=\frac{x}{16}\left(5 x-\frac{3}{x}+4\right)

y=\frac{1}{16}\left(5 x^{2}-3+4 x\right)

\frac{dy}{dx}=\frac{1}{16}(10 x+4)

\left(\frac{dy}{dx}\right)_{\text{at }x=1}=\frac{10+4}{16}

=\frac{14}{16}=\frac{7}{8}

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