Question

Consider a function f: `[0, pi/2] ->` R, given by f(x) = sinx and `g[0, pi/2] ->` R given by g(x) = cosx then f and g are

पर्याय

• one – one

• onto

• bijective

• injective

Answer 1

one – one

Explanation:

Since for any two distinct elements x₁ and x₂ in `[0, pi/2], sinx_1 ≠ sinx_2` and `cosx_1^1 cosx_2` both f and g must be one – one