First slide

Functions (XII)

Question

Let $f (x) + f (y) = f (x \sqrt{1 - y^{2}} + y \sqrt{1 - x^{2}}) [f (x)$ is not identically zero]. Then

Moderate

A

$f (4 x^{3} - 3 x) + 3 f (x) = 0$
B

$f (4 x^{3} - 3 x) = 3 f (x)$
C

$f (2 x \sqrt{1 - x^{2}}) + 2 f (x) = 0$
D

$f (2 x \sqrt{1 - x^{2}}) = 2 f (x)$

Solution

Given $f (x) + f (y) = (x \sqrt{1 - y^{2}} + y \sqrt{1 - x^{2}})$ (1)
Replace y by x. Then $2 f (x) = f (2 x \sqrt{1 - x^{2}})$
$\begin{array}{l} 3 f (x) = f (x) + 2 f (x) \\ = f (x) + f (2 x \sqrt{1 - x^{2}}) \\ = f (x \sqrt{1 - 4 x^{2} (1 - x^{2})} + 2 x \sqrt{1 - x^{2}} \sqrt{1 - x^{2}}) \\ = f (x \sqrt{{(2 x^{2} - 1)}^{2}} + 2 x (1 - x^{2})) \end{array}$
$\begin{array}{l} = f (x |2 x^{2} - 1| + 2 x - 2 x^{3}) \\ = f (2 x^{3} - x + 2 x - 2 x^{3}) \end{array}$
or $f (x - 2 x^{3} + 2 x - 2 x^{3}) = f (x)$
or $f (3 x - 4 x^{3})$
$∴ f (x) = 0 or 3 f (x) = f (3 x - 4 x^{3})$
Consider $3 f (x) = f (3 x - 4 x^{3})$ .
Replacing x by - x, we get
$3 f (- x) = f (4 x^{3} - 3 x)$ (2)
Also, from $(1), f (x) + f (- x) = f (0)$
Putting x = y = 0 in (1), we have $f (0) = 0 or f (x) + f (- x) = 0$
Thus, f(x) is an odd function.
Now, from $(2), - 3 f (x) = f (4 x^{3} - 3 x)$
or $f (4 x^{3} - 3 x) + 3 f (x) = 0$

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