u-R-R-v-R-R-u-x-v-x-x-2-v-x-u-x-v-x-x-3-v-x-h-x-u-x-v-x-

Question Number 560 by 123456 last updated on 26/Jan/15

u : R \to R

v : R \to R

{\begin{cases} u (x) v (x) = x^{2} - v (- x) \\ u (- x) v (x) = x^{3} + v (- x) \end{cases}

h (x) = u (x) v (x) = ?

Answered by prakash jain last updated on 26/Jan/15

u (x) v (x) = x^{2} - v (- x) \dots (i)

u (- x) v (x) = x^{3} + v (- x) \dots (ii)

add (i) and (ii)

v (x) [u (x) + u (- x)] = x^{2} + x^{3} \Rightarrow v (x) = \frac{x^{2} + x^{3}}{u (x) + u (- x)}

u (- x) v (- x) = x^{2} - v (x) replace x by - x in (i)

u (- x) \frac{x^{2} - x^{3}}{u (x) + u (- x)} = x^{2} - \frac{x^{2} + x^{3}}{u (x) + u (- x)}

u (x) = a

u (- x) = b

\frac{b (x^{2} - x^{3})}{a + b} + \frac{x^{2} + x^{3}}{a + b} = x^{2}

{b x}^{2} - {b x}^{3} + x^{2} + x^{3} = {a x}^{2} + {b x}^{2}

{b x}^{3} = x^{2} + x^{3} - {a x}^{2}

b = \frac{x^{2} + x^{3} - {a x}^{2}}{x^{3}} = \frac{1 + x - a}{x}

1 + x - a = x b

1 + x - u (x) = x u (- x)

x u (- x) + u (x) - x - 1 = 0 \dots (iii)

- x u (x) + u (- x) + x - 1 = 0 \dots (iv) replace x by - x in (iii)

multiply (iv) by x

- x^{2} u (x) + x u (- x) + x^{2} - x = 0 \dots (v)

add (iii) and (v)

u (x) + x^{2} u (x) - x^{2} - 1 = 0

u (x) = 1, u (- x) = 1

v (x) = \frac{x^{2} + x^{3}}{2}, v (- x) = \frac{x^{2} - x^{3}}{2}

C h e c k c o n d i t i o n 1

u (x) v (x) = \frac{x^{2} + x^{3}}{2} = x^{2} - \frac{x^{2} - x^{3}}{2} = \frac{x^{2} + x^{3}}{2}

C h e c k c o n d i t i o n 1

u (- x) v (x) = \frac{x^{2} + x^{3}}{2} = x^{3} + \frac{x^{2} - x^{3}}{2} = \frac{x^{2} + x^{3}}{2}

Results

u (x) = 1

v (x) = \frac{x^{2} + x^{3}}{2}

u (x) v (x) = \frac{x^{2} + x^{3}}{2}