Class 12 RD Sharma Solutions - Chapter 2 Functions - Exercise 2.1 | Set 2

Question 12: Show that the exponential function f: R → R, given by f(x) = e^x is one one but not onto. What happens if the codomain is replaced by Ro+.

Solution:

We have f: R → R, given by f(x) = e^x.

Let x,y ϵ R, such that

=> f(x) = f(y)

=> e^x = e^y

=> e^(x-y) = 1 = e⁰

=> x - y = 0

=> x = y

Hence,

f is one-one.

Clearly range of f = (0, INFINITY) not equals to R.

Hence, f is not onto.

When co-domain is replaced by Ro+ i.e. (0, INFINITY) then f becomes an onto function.

Question 13: Show that the logarithmic function f: Ro+→ R given by f(x) = log(a)x, a>0 is a bijection.

Solution:

We have f: Ro+→ R given by f(x) = log(a)x, a>0.

let x,y ϵ Ro+ such that,

f(x) = f(y)

=> log(a)x = log(a)y

=> log(a)x (x/y) = 0 [log(a)x = 0]

=> x/y = 1

=> x = y

Hence, f is not one-one.

Now, let y ϵ R be arbitrary, then f(x) = y

=> log (a)x = y

=>x = a^y ϵ Ro+

Thus for all yϵR there exist x = a^y such that f(x) = y.

Hence,

f is onto.

f is one-one.

=> f is bijective.

Question 14: If A = {1,2,3}, show that a one-one function f: A → A must be onto.

Solution:

Since f is one-one, three elements of {1,2,3} must be taken to the 3 different elements of the co-domain {1,2,3} under f. Hence f has to be onto.

Question 15: If A={1, 2, 3}, show that an onto function f: A→A must be one-one.

Solution:

A={1,2,3}

Possible onto functions from A to A can be the following:

(0){(1,1),(2,2).(3,3)

(ii) (1,1),(2,3),(3,2)

(ii){(1,2),(2,2),(3,3)}

(iv){(1,2),(2,1),(3,3)}

(v){(1,3), (2,2),(3,1)]

(vi){(1,3),(2,1),(3,2)

Here, in each function, different elements of the domain have different images.

Therefore,

All the functions are one-one

Question 16: Find the number of all onto functions from the set A = { 1, 2, 3, ….n} to itself.

Solution:

We know that every onto function from A to itself is one-one.

Therefore,

The number of one-one functions=number of bijections =n!

Question 17: Give examples of two one-one functions f1 and f2. From R to R such that f1 + f2: R→ R defined by (f1 + f2)(x) = f1(x) + f2(x) is not one-one.

Solution:

We know that f1: R→ R, given by f1 (x)= x, and f2 (x)=-x are one-one.

Proving f1, is one-one:

Let f1(x)= f1(y)

Implies that x = y

Therefore,

f1 is one-one.

Proving f2 is one-one:

Let f2(x) = f2(y)

Implies that - x=-y

Implies that x = y

Therefore,

f2 is one-one.

Proving (f1 + f2) is not one-one:

Given:(f1+f2)(x)= f1(x)+ f2 (x)= x+(-x) = 0

Therefore,

For every real number x,(f1 + f2)(x)=0

Therefore,

The image of ever number in the domain is same as 0.

Thus, (f1 + f2) is not one-one.

Question 18: Given an example of two surjective functions f1 and f2 from Z to Z such that f1 + f2 is not surjective.

Solution:

We know that f1: R → R, given by f1(x) = x, and f2(x)=-x are surjective functions.

Proving f1 is surjective:

Let y be an element in the co-domain (R), such that f1(x)= y.

f1(x)= y

Implies that x = y, which is in R.

Therefore,

for every element in the co-domain, there exists some pre-image in the domain.

Therefore,

f1 is surjective

Proving f2 is surjective:

Let f2 (x)= y

x = y, which is in R.

Therefore,

for every element in the co-domain, there exists some pre-image in the domain.

Therefore,

f2 is surjective.

Proving (f1+f2) is not surjective:

Given:(f1 + f2)(x) = f1(x)+ f2(x)=x+(-x)=0

Therefore, for every real number x, (f1 + f2)(x) = 0

Therefore, the image of every number in the domain is same as 0.

Implies that Range = {0}

Co-domain = R

Therefore, both are not same.

Therefore, f1+f2 is not surjective.

Question 19: Show that if f1 and f2 are one-one maps, from R to R then the product f1 X f2: R→R defined by (f1 X f2)(x) = f1(x)f2(x) need not be one-one.

Solution:

We know that f: R→ R, given by f1(x) = x, and f2(x) = x are one-one.

Proving f1 is one-one:

Let x and y be two elements in the domain R, such that f1(x) = f1(y)

f1(x) = f1(y)

x = y

Therefore,

f1 is one-one.

Proving f2 is one-one:

Let x and y be two elements in the domain R, such that f2 (x) = f2(y)

f2(x)= f2(y)

Implies that x = y

Therefore,

f2 is one-one.

Proving f1 X f2 is not one-one:

Given:

(f1 X f2)(x) = f1(x) X f2(x) = x * x = x²

Let x and y be two elements in the domain R, such that

(f1 X f2)(x) = (f1 X f2)(y)

Implies that x² = y²

Implies that x = (+-)y

Therefore,

(f1 X f2) is not one-one.

Question 20: Suppose f1 and f2 are non-zero one-one functions from R to R. Is (f1/f2) necessarily one-one? Justify.

Solution:

We know that f1: R→R given by f1(x) = x³ and f2(x)= x are one-one.

Injectivity of f1:

Consider x and y be two elements in the domain R, such that

f1(x)= f1(y)

Implies that x³ = y

x=3√y belongs to R

Therefore,

f1 is one-one.

Injectivity of f2:

Consider x and y be two elements in the domain R, such that

f2(x)= f2(y)

Implies that x = y

x belongs to R

Therefore,

f2 is one-one.

Providing (f1 / f2) is not one-one:

Given that (f1/f2)(x)= = f1(x)/f2(x) = (x³ / x) = x²

Consider x and y be two elements in the domain R, such that

(f1/f2)(x) = (f1/f2)(y)

f2 f2

x² = y²

x= (+-)y

Therefore,

(f1/f2) is not one-one.

Question 21: Given A = {2, 3, 4}, B = {2,5,6,7}. Construct an example of each of the following.

(i) An injective map from A to B.

(ii) A mapping from A to B which is not injective.

(iii)A mapping from A to B.

Solution:

Given A={1,2,3,4}, B = {2,5,6,7}

Let f: A → Bf: A → B be a mapping from A to B f = {(2,5)(3,6)(4,7)}

f is an injective mapping.

Since for every element a € A there is an unique element b € B

Let us define a mapping: A→B given by g = {(2,2)(2,5)(3,6)(4,7)}

g is not an injective mapping.

since the element 2 € A is not uniquely mapped

Since (2,2) and (2,5) both belong to the mapping g, g is not injective

Let us define a mapping h: A→B

h: A → B given by h = {(2,2),(5,3),(7,4)}

h is a mapping from A to B

B to A since the every ordered puts {2,5,7} € B to elements in {2,3,4} € A

Question 22: Show that f: R → R, Given by f(x) = x - [x] is neither one-one nor onto.

Solution:

f:R → R, given by f (x)= x-[x]

Injectivity:

f(x)=0 for all x belongs to Z,

Therefore,

f is not one-one.

Surjectivity:

Range of f = (0,1) not equals to R.

Co-domain of f = R

Both are not same.

Therefore,

f is not onto.

Question 23: Let f:N→N defined by

f(n) = n + 1, if n is odd.

f(n) = n - 1, if n is even. Show that if f is a bijection.

Solution:

Injectivity:

Let x and y be any two elements in the domain (N).

Case-1: Let both x and y be even and

Let x, y belongs to N such that f (x) = f(y)

As, f (x) = f(y)

Implies that x – 1= x-1

Implies that x = y

Case-2: Let both x and y be odd and

Let x, y belongs to N such that f (x)= f (y)

As, f (x)= f (y)

Implies that x +1 = y +1

Implies that x = y

Case-3:Let x be even and y be odd then, x y.

Then,

x+1 is odd and y-1 is even.

Implies that x+1≠ y-1

Implies that f(x) ≠ f(y)

Therefore,

x ≠ y

= f(x) ≠ f(y)

In all the 3 cases,

Therefore,

f is one-one.

Surjectivity:

Co-domain off = {1, 2,3,4,...}

Range of f = {1+1,2 – 1,3+1,4 – 1,...} = {2,1,4, 3,...}={1, 2, 3, 4,...}

Both are same.

Implies that f is onto.

Therefore,

f is a bijection.

Summary

Exercise 2.1 | Set 2 of Chapter 2 (Functions) in RD Sharma's Class 12 Solutions builds upon the concepts introduced in Set 1. It delves deeper into function properties, exploring more complex scenarios involving domain and range, function composition, and inverse functions. This set challenges students to apply their understanding to a wider variety of function types, including piecewise functions, absolute value functions, and rational functions. It also emphasizes graphical interpretations and the relationships between different function representations.