ISI2017-DCG-3

1 vote
59 views

If  $2f(x)-3f(\frac{1}{x})=x^2 \: (x \neq0)$,  then $f(2)$ is

1. $\frac{2}{3}$
2. $– \frac{3}{2}$
3. $– \frac{7}{4}$
4. $\frac{5}{4}$
in Calculus
recategorized

Answer: $\mathbf C$

Explanation:

$2f(x) - 3f(\frac{1}{x}) = x^2\tag{1}\;\;\text{[Given]}$

Substitute, $x = \frac{1}{x}$, we get:

$2f(\frac{1}{x}) - 3f(x) = \frac{1}{x^2}\tag{2}$

Multiplying Equation $(1)$ by $3$, we get:

$6f(x) - 9f(\frac{1}{x}) = 3x^2\tag{3}$

Multiplying Equation $(2)$ by $2$, we get:

$4f(\frac{1}{x}) - 6f(x) = \frac{2}{x^2}\implies -6f(x) + 4f(\frac{1}{x}) = \frac{2}{x^2}\tag{4}$

Adding Equations $(3)$ and $(4)$, we get:

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

Now, on putting $x = \frac{1}{2}$ in above equation, we get:

$x = -\frac{7}{4}$

$\therefore \mathbf C$ is the correct option.

edited by

Related questions

1
79 views
Let $f(x) = \dfrac{x-1}{x+1}, \: f^{k+1}(x)=f\left(f^k(x)\right)$ for all $k=1, 2, 3, \dots , 99$. Then $f^{100}(10)$ is $1$ $10$ $100$ $101$
2
62 views
If $f(x)=e^{5x}$ and $h(x)=f’’(x)+2f’(x)+f(x)+2$ then $h(0)$ equals $38$ $8$ $4$ $0$
3
44 views
The value of $\underset{n \to \infty}{\lim} \bigg( \dfrac{1}{1-n^2} + \dfrac{2}{1-n^2} + \dots + \dfrac{n}{1-n^2} \bigg)$ is $0$ $– \frac{1}{2}$ $\frac{1}{2}$ none of these
The limit of the sequence $\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$ is $1$ $2$ $2\sqrt{2}$ $\infty$