If x≠0 and fx satisfies 8fx+6f1/x=x+5, then d/dxx^2fx at x=1 is

Correct answer: (A) (-1)/14. We have, 8fx+6f1/x=x+5⇒f-1=2/7 Differentiating on both sides 8f^(')x-(6x^2f^(')1)/x=1 Put x=-1⇒f^(')-1=1/2 Now differentiating y=x^2fx we get, dy/dx=2xfx+x^2f^(')x at x=-1,dy/dx=2-1×2/7+-1^21/2=-1/14

Q102 / 160 · AP EAMCET 2022 (04 Jul Shift 2) · Mathematics

If $x \neq 0$ and $f (x)$ satisfies $8 f (x) + 6 f (\frac{1}{x}) = x + 5$ , then $\frac{d}{d x} (x^{2} f (x))$ at $x = 1$ is

Correct answer
A
$\frac{- 1}{14}$
B
$\frac{25}{14}$
C
$\frac{9}{14}$
D
$\frac{19}{14}$

Solution

We have,
$8 f (x) + 6 f (\frac{1}{x}) = x + 5 \Rightarrow f (- 1) = \frac{2}{7}$
Differentiating on both sides
$8 f^{'} (x) - \frac{6}{x^{2}} f^{'} (\frac{1}{x}) = 1$
Put $x = - 1 \Rightarrow f^{'} (- 1) = \frac{1}{2}$
Now differentiating $y = x^{2} f (x)$ we get,
$\frac{d y}{d x} = 2 x f (x) + x^{2} f^{'} (x)$
at $x = - 1, \frac{d y}{d x} = 2 (- 1) \times \frac{2}{7} + {(- 1)}^{2} \frac{1}{2} = - \frac{1}{14}$

Hence, the correct answer is option: 'A'.

If x≠0 and fx satisfies 8fx+6f1x=x+5, then ddxx2fx at x=1 is

Solution

Relevant Questions

If $x \neq 0$ and $f (x)$ satisfies $8 f (x) + 6 f (\frac{1}{x}) = x + 5$ , then $\frac{d}{d x} (x^{2} f (x))$ at $x = 1$ is