COMEDK Exam 2016 Solved Paper

We have,
f:X→Y
X : domain and Y : codomain
In one-one function,
x1‌≠x2
f(x1)‌≠f(x2)
In onto function,
Range = domain
f(x)=sin‌x
So, if X=[0,‌

π

2

] and Y=[−1,1]
For each value of x∈X and range =[0,1], there exists a unique value for y∈Y
Hence, function is one-one but not onto.