Here is some solved problems from IIT-JEE previous year papers on functions

**Q. [IIT-JEE 2002]**Let function f : R→R be defined by f(x)= 2x + sinx, then f is

(a) one-to-one& onto (b) one-to-one but not
onto (c) onto but not
one-to-one (d) neither
one-to-one nor onto

__Solution :__(a) Given f(x)= 2x + sinx

On differentiation ,f ′(x)= 2
+ cosx ; as -1 ≤cosx ≤ 1

So, f ′(x) ≥ 0, so f(x) is strictly
increasing. Thus f(x) is one-to-one &
onto.

**Q. [IIT-JEE 2001]**Let E = {1,2,3,4} and F = {1,2}. Then the number of onto functions from E to F is

(a) 14 (b)
16 (c) 12
(d) 8

__Solution :__(a) Each element of E should have image as 1 or 2.

Thus four elements of E can have images = 2

^{4}.
Now, for all x ∈ E, f(x)=1 or f(x)=2, do not make it as an onto function.

Thus, the total number of onto
functions from E to F is = 2

^{4 }– 2=14.**Q. [IIT-JEE 2000]**The domain of definition of the function y(x) given by the equation 2

^{x}+ 2

^{y}= 2 is

(a) 0< x ≤1 (b)
0≤ x ≤1 (c) -∞< x ≤0 (d) -∞<
x <1

__Solution :__(d) Given 2

^{y}= 2 – 2

^{x},

Taking log both sides, ylog2 =
log(2 – 2

^{x}). Now, 2 – 2^{x}> 0 ⇒-∞< x <1.