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Question Number 4820 by love math last updated on 16/Mar/16

y = f (x) + g (x)

f (x) - o d d f u n c t i o n

g (x) - e v e n f u n c t i o n

f i n d f (0), i f y = 2 x^{2} + \frac{s i n x}{3} + 1

Answered by prakash jain last updated on 16/Mar/16

y (x) + y (- x) = f (x) + g (x) + f (- x) + g (- x)

= f (x) + g (x) + f (x) - g (x) = 2 f (x)

y (x) + y (- x) = 2 x^{2} + \frac{\sin x}{3} + 1 + 2 {(- x)}^{2} + \frac{\sin (- x)}{3} + 1

= 4 x^{2} + 2 = 2 f (x)

f (x) = 2 x^{2} + 1

f (0) = 1

Commented by Yozzii last updated on 16/Mar/16

f (x) i s a n o d d f u n c t i o n; 2 x^{2} + 1 i s e v e n .

Commented by Rasheed Soomro last updated on 16/Mar/16

f (x) i s a n o d d f u n c t i o n, s o f (- x) = - f (x)

∴ y (x) + y (- x) = 2 g (x)

Commented by prakash jain last updated on 16/Mar/16

Sorry . I switch f (x) and g (x) .

Took f (x) a s e v e n a n d g (x) a s o d d .

So

g (x) = 2 x^{2} + 1

f (x) = y (x) - g (x) = \frac{\sin x}{3}

f (0) = 0