Question

let f ( -10,10)-> R , where f(x)= sinx + (x square/ a) be an odd function then the set of value of parameter a is /an​

Answer 1

Given : f : (-10, 10) → R , where f(x) = sinx + (x²/a) be an odd function.

To find : The set of value of parameter a.

solution : a function, y = f(x) will be an odd function only if f(-x) = - f(x)

here f(x) = sinx + (x²/a)

⇒f(-x) = sin(-x) + [(-x)²/a ] = - sinx + x²/a

⇒-f(x) = - sinx - x²/a

so, f(-x) = - f(x)

⇒-sinx + x²/a = - sinx - x²/a

⇒2x²/a = 0

⇒x²/a = 0

as domain of function, x ∈ (-10, 10) , a ≠ 0 at x = 0

Therefore, value of a ≠ 0, in the domain of function.

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