Let 3f (x) - 2 f (1/x) = x, then f '(2) is equal to

Question

Let 3f (x) - 2 f (1/x) = x, then f '(2) is equal to (A) (2/7) (B) (1/2) (C) 2 (D) (7/2)

Tardigrade Admin · Accepted Answer

3f (x) - 2 f (1/x) = x, .....(1) Put x = (1/x) , then 3 f( (1/x) ) - 2 f(x) = (1/x) ....(2) Solving (1) and (2) , we get 5f(x) = 3x + (2/x) ⇒ f'(x) = (3/5) - (2/5x2) ∴ : :: f'(2) = (3/5) - (2/20) = (1/2)

Q. Let $3 f (x) - 2 f (1/ x) = x,$ then $f^{'} (2)$ is equal to

$\frac{2}{7}$

$\frac{1}{2}$

2

$\frac{7}{2}$

$3 f (x) - 2 f (1/ x) = x,$ .....(1)
Put $x = \frac{1}{x},$ then $3 f (\frac{1}{x}) - 2 f (x) = \frac{1}{x}$ ....(2)
Solving (1) and (2) , we get
$5 f (x) = 3 x + \frac{2}{x} \Rightarrow f^{'} (x) = \frac{3}{5} - \frac{2}{5 x ^{2}}$
$∴ f^{'} (2) = \frac{3}{5} - \frac{2}{20} = \frac{1}{2}$

Q. Let 3f(x)−2f(1/x)=x, then f′(2) is equal to

72​

21​

2

27​

3f(x)−2f(1/x)=x, .....(1) Put x=x1​, then 3f(x1​)−2f(x)=x1​ ....(2) Solving (1) and (2) , we get 5f(x)=3x+x2​⇒f′(x)=53​−5x22​ ∴f′(2)=53​−202​=21​

Q. Let $3 f (x) - 2 f (1/ x) = x,$ then $f^{'} (2)$ is equal to

$\frac{2}{7}$

$\frac{1}{2}$

$\frac{7}{2}$

$3 f (x) - 2 f (1/ x) = x,$ .....(1)
Put $x = \frac{1}{x},$ then $3 f (\frac{1}{x}) - 2 f (x) = \frac{1}{x}$ ....(2)
Solving (1) and (2) , we get
$5 f (x) = 3 x + \frac{2}{x} \Rightarrow f^{'} (x) = \frac{3}{5} - \frac{2}{5 x ^{2}}$
$∴ f^{'} (2) = \frac{3}{5} - \frac{2}{20} = \frac{1}{2}$