# RBSE Solutions for Class 12 Maths Chapter 1 Composite Functions Miscellaneous Exercise

## Rajasthan Board RBSE Class 12 Maths Chapter 1 Composite Functions Miscellaneous Exercise

Question 1.
If f : R → R, f(x) = 2x – 3; g : R → R, g(x) = x3 + 5, then the value of (fog)-1 (x):

Solution:
Given,
f : R → R, f(x) = 2x – 3
g : R → R, g(x)= x3 + 5
∴ (fog)(x) = f(g(x)]
= f(x3 +5)
= 2(x3 + 5) – 3 = 2x3 + 10 – 3
= 2x3 + 7
Let y = (fog)(x) = 2x3 + 7
∴ (fog)-1(y) = x

So, option (d) is correct.

Question 2.
If
, then the value of f(y):
(a) x
(b) x – 1
(c) x + 1
(d) 1 – x
Solution:

So, no any option is correct.

Question 3.
If
, then f{f{f(x)}] is equal to :
(a) x
(b) 1/x
(c) – x
(d) – 1/x
Solution:

So, option (a) is correct.

Question 4.
If f(x) = cos (log x) then,

is equal to:
(a) -1
(b) 0
(c) 1/2
(d) – 2
Solution:
Given, f(x) = cos (log x)

So, option (b) is correct.

Question 5.
If f : R → R, f(x) = 2x + 1 and g : R → R, g(x) = x3, then (gof)-1(27) is equal to :
(a) 2
(b) 1
(c) -1
(d) 0
Solution:
Let (gof)-1(27) = x
∴ (gof)(x) = 27
g{f(x)} = 27
g{2x + 1} = 27
(2x + 1)3 = 27
2x + 1 = 271/3
2x + 1 = 33 x 1/3
2x + 1 = 3
∴ 2x = 2
x = 1
So, option (b) is correct.

Question 6.
If f: R → R and g: R+ R, where f(x) = 2x + 3 and g(x) = x2 + 1, then (gof)(2):
(a) 38
(b) 42
(c) 46
(d) 50
Solution:
Let (gof)(2) = y
∴ y = g{{{2}}
= g{2 x 2 + 3}
= g(7)
= (7)2 + 1
= 49 + 1
= 50
So, option (d) is correct.

Question 7.
If * is an operation on Q0, defined by a*b = ab/2, ∀ a, b ∈ Q0, then identity element of this operation is :
(a) 1
(b) 0
(c) 2
(d) 3
Solution:
If e is identity element
For a ∈ Q0
a*e = a
⇒ $$\frac { ae }{ 2 }$$ = a
e = 2
So, option (c) is correct.

Question 8.
A binary operation on R is defined by a*b = 1 + ab, ∀ a, b ∈ R, then identity element of this operation is :
(a) commutative but not associative
(b) associative but not commutative
(c) neither commutative nor associative
(d) commutative and associative both
Solution:
Given, a*b = 1 + ab, ∀ a, b ∈ R
Commutativity a*b = 1 + ab
= 1 + b.a
= b*a
∵ Set of real numbers is commutative.
So, a*b = b*a
∴ is commutative Associativity (a*b)*c = (1 + ab)*c
= (1 + ab)c
= 2 + abc
a*(b*c) = a*(1 + bc) = 1 + a*(1 + bc)
= 1 + a + abc
It is clear that (a*b)*c ≠ a*(b*c)
So, * operation is not associative.
So, option (a) is correct.

Question 9.
Subtraction is an operation on Z, which is
(a) commutative and associative
(b) associative but not commutative
(c) neither commutative and nor associative
(d) commutative but not associative
Solution:
Let a, b ∈ z
Commutativity : As we know,
a-b ≠ b – a
Associativity : (a – b) – c ≠ a – (b – c)
So, option (c) is correct.

Question 10.
* be an operation on Z definied by a*b = a + b – ab, ∀ a, b ∈ Z, for any a ∈ Z, a(≠ 1), inverse of a with respect to * is :

Solution:
Given, a*b = a + b – ab, ∀ a, b ∈ z
Let a ∈ Z is possible and x is inverse of a, then According to question
a*x = 0 (∵ 0 is identity)
⇒ a + x – ax = 0
⇒ x(1 – a) = – a

Hence, option (a) is correct.

Question 11.
Which of the following operations is commutative in R:
(a) a*b = a2b
(b) a*b = ab
(c) a*b = a – b + ab
(d) a * b = a + b + a2b
Solution:
(a) ∵ a*b=ab and b*a = b2a
∴ a*b ≠ b*a
So, operation is not commutative.

(b) ∵ a*b = ab and b*a = ba
∴ a*b ≠ b*a
So, operation is not commutative.

(c) ∵ a*b = a – a + ab
and b*a = b – a + ba
∴ a*b* ≠ b*a
So, operation is not commutative.

(d) ∵ a*b= a + b + a2b
and : b*a = b + a + b2a
∴ a*b* ≠ b*a
So, operation is not commutative.
Hence, any of these operation is not commutative.

Question 12.
Verify the associative law for the composite function of following three functions :
f : N → Z0, f(x) = 2x
g : Z0 → Q, g(x) = 1/x
h : Q → R, h(x) = ex
Solution:
Given,
f : N→ Z0
g : Z0 → Q
h : Q → R
Then, ho(gof) : N → R
and (hog)of : N → R
Hence, domain and co-domain of ho(gof) and go(hof) are same because both functions are defined from N to R.
Hence, we have to prove
[ho(gof)](x) = [(hog)of)(x), ∀ X ∈ N
Now, [ho(gof)](x)= h[(gof)(x)]
= h[g{f(x)]
= h[g(2x)]

From eqs. (i) and (ii),
(hog)of = ho(gof)
Hence, associativity of f, g, h is proved. Proved.

Question 13.
If f: R+ → R+ and g : R+ → R+, defined as f(x) = x2, g(x) = √x, then find gof and fog wheather are they equivalent ?
Solution:
Given, f: R+ → R+, f(x) = x2
g : R+ → R+, g(x)= √x
Then, (fog): R+ → R+ and (gof): R+ → R+ are defined
∴ (gof)(x) = g[f(x)]
= g(x2) = √x2 = x
(fog)(x) = f(g(x)]
= f(√x)= (√5)2 = x
Based on above, (gof) and (fog) are of same domain and co-domain.
(fog)(x) = (gof)(x) ∀ x ∈ R+
So, (fog) and (gof) are equivalent functions.

Question 14.
If f : R → R, f(x) = cos (x + 2), justify whether f is inverse function or not.
Solution:
Given function,
f : R → R, f(x) = cos (x + 2)
Putting x = 2π
f(2π) = cos (2π + 2) = cos (2)
Putting x = 0
f(0) = cos (0 + 2) = cos 2
Since, only one image is obtained for 0 and 2π, so function is not one-one. Hence, function is not one-one onto.
Hence, f-1 : R → R does not exist.

Question 15.
If A = {-1,1} fand g are two functions defined on A, where f(x) = x2, g(x) = sin (πx/2), prove that g-1 exists but f-1 does not exist, also find g-1.
Solution:
Given, A = {-1, 1}
f(x)= x2, g(x) = sin πx/2

One-one/many-one : f : A → A, f(x) = x2
f(-1) = f(1) = 1
⇒ Image of – 1 and 1 is same.
So, f is not one-one.

Onto/into : Elements of co-domain unpaired with any element of domain.
So ‘f’ is not onto.
So, f is neither one-one nor onto.
Hence, ‘f-1‘ does not exist.
Thus, for g(x) = sin πx/2

One-one/many-one : Let x1, x2 ∈ A, such that

Hence, function in one-one.
Let y ∈ R (co-domain), if possible then pre image is x in R, then

Since, pre-image of each value of y exist in domain R, then g(x) is onto function.
Hence, g(x) is one-one onto function, so g-1 exist.

Question 16.
If f : R → R and g : R → R, such that f(x) = 3x + 4 and g(x) = (x – 4)/3 then (fog)(x) and (gof)(x). Also, find the value of (gog) (1).
Solution:
Given, f : R → R, f (x) = 3x + 4