Question

If $af(x + 1) + bf(\dfrac{1}{{x + 1}}) = x$and $x \ne - 1$, $a \ne b$ then f(2) is equal to
A. $\dfrac{{2a + b}}{{2({a^2} - {b^2})}}$
B. $\dfrac{a}{{{a^2} - {b^2}}}$
C. $\dfrac{{a + 2b}}{{{a^2} - {b^2}}}$
D. none of these

Answer 1

Hint: We will start solving this question by putting the value of x in the function, so that we get the value of f (2). For this, we will first put x = 1 in the given expression and after it, we will again put x = $- \dfrac{1}{2}$ in the function.

Complete step-by-step answer:
Now, we will first put x = 1, in the expression $af(x + 1) + bf(\dfrac{1}{{x + 1}}) = x$. On putting we get,
$af(1 + 1) + bf(\dfrac{1}{{1 + 1}}) = 1$
$af(2) + bf(\dfrac{1}{2}) = 1$ … (1)
Now, we will put x = $- \dfrac{1}{2}$ in the expression $af(x + 1) + bf(\dfrac{1}{{x + 1}}) = x$. So, on putting the value of x we get,
$af( - \dfrac{1}{2} + 1) + bf(\dfrac{1}{{ - \dfrac{1}{2} + 1}}) = - \dfrac{1}{2}$
$af(\dfrac{1}{2}) + bf(2) = - \dfrac{1}{2}$ … (2)
Now, we will solve both equation (1) and (2). Now, we will multiply equation (1) by a and multiply equation (2) by b. So, on multiplying equation (1) becomes,
${a^2}f(2) + abf(\dfrac{1}{2}) = a$ … (3)
On multiplication, equation (2) becomes,
$abf(\dfrac{1}{2}) + {b^2}f(2) = - \dfrac{b}{2}$ … (4)
Now, subtracting equation (4) from equation (3), we get
${a^2}f(2) - {b^2}f(2) = a + \dfrac{b}{2}$
$f(2)({a^2} - {b^2}) = \dfrac{{2a + b}}{2}$
$\Rightarrow$ $f(2) = \dfrac{{2a + b}}{{2({a^2} - {b^2})}}$
So, option (A) is correct.

Note: Whenever we come up with such types of questions, where we have to find the value of the function at a particular value and we don’t know the function, we will use the relation given in the question to determine the value of function. To find the value of function, we have to put the value of x such that it will give the value of function at that point. The value which we have to put can be found through trial and error.