RBSE Solutions Class 12 Maths Chapter 1 Composite Functions Exercise 1.1

Question 1. If f : R → R and g : R → R be two functions defined as below, then find (fog) (x) and (gof)(x):
(i) \( f(x) = 2x + 3, g(x) = x^2 + 5 \)
(ii) \( f(x)= x^2 +8, g(x) = 3x^3 +1 \)
(iii) \( f(x) = x, g(x) = |x| \)
(iv) \( f(x) = x^2 + 2x + 3, g(x) = 3x - 4 \)
Answer:
(i) Given functions are \( f(x) = 2x + 3 \) and \( g(x) = x^2 + 5 \).
To find \( (fog)(x) \), we use the definition \( (fog)(x) = f(g(x)) \).
So, we substitute \( g(x) \) into \( f(x) \):
\( (fog)(x) = f(x^2 + 5) \)
Now, replace \( x \) in \( f(x) \) with \( (x^2 + 5) \):
\( (fog)(x) = 2(x^2 + 5) + 3 \)
\( = 2x^2 + 10 + 3 \)
\( = 2x^2 + 13 \)

(ii) Given functions are \( f(x) = x^2 + 8 \) and \( g(x) = 3x^3 + 1 \).
To find \( (fog)(x) \), we use \( (fog)(x) = f(g(x)) \).
\( (fog)(x) = f(3x^3 + 1) \)
Substitute \( (3x^3 + 1) \) into \( f(x) \):
\( (fog)(x) = (3x^3 + 1)^2 + 8 \)
Expand the square: \( (a+b)^2 = a^2 + 2ab + b^2 \)
\( = (3x^3)^2 + 2(3x^3)(1) + (1)^2 + 8 \)
\( = 9x^6 + 6x^3 + 1 + 8 \)
\( = 9x^6 + 6x^3 + 9 \)
Now, to find \( (gof)(x) \), we use \( (gof)(x) = g(f(x)) \).
\( (gof)(x) = g(x^2 + 8) \)
Substitute \( (x^2 + 8) \) into \( g(x) \):
\( (gof)(x) = 3(x^2 + 8)^3 + 1 \)

(iii) Given functions are \( f(x) = x \) and \( g(x) = |x| \).
To find \( (fog)(x) \), we use \( (fog)(x) = f(g(x)) \).
\( (fog)(x) = f(|x|) \)
Substitute \( |x| \) into \( f(x) \):
\( (fog)(x) = |x| \)
To find \( (gof)(x) \), we use \( (gof)(x) = g(f(x)) \).
\( (gof)(x) = g(x) \)
Substitute \( x \) into \( g(x) \):
\( (gof)(x) = |x| \)

(iv) Given functions are \( f(x) = x^2 + 2x + 3 \) and \( g(x) = 3x - 4 \).
To find \( (fog)(x) \), we use \( (fog)(x) = f(g(x)) \).
\( (fog)(x) = f(3x - 4) \)
Substitute \( (3x - 4) \) into \( f(x) \):
\( (fog)(x) = (3x - 4)^2 + 2(3x - 4) + 3 \)
Expand the square and distribute:
\( = (9x^2 - 24x + 16) + (6x - 8) + 3 \)
\( = 9x^2 - 24x + 6x + 16 - 8 + 3 \)
\( = 9x^2 - 18x + 11 \)
To find \( (gof)(x) \), we use \( (gof)(x) = g(f(x)) \).
\( (gof)(x) = g(x^2 + 2x + 3) \)
Substitute \( (x^2 + 2x + 3) \) into \( g(x) \):
\( (gof)(x) = 3(x^2 + 2x + 3) - 4 \)
Distribute and simplify:
\( = 3x^2 + 6x + 9 - 4 \)
\( = 3x^2 + 6x + 5 \)
In simple words: To find \( (fog)(x) \), you take the function \( g(x) \) and put it inside \( f(x) \). For \( (gof)(x) \), you put \( f(x) \) inside \( g(x) \). This helps you create a new function by combining two existing ones.

🎯 Exam Tip: Remember the order of composition: \( (fog)(x) \) means "f of g of x," so you apply g first, then f. Conversely, \( (gof)(x) \) means "g of f of x," so you apply f first, then g. Pay close attention to parentheses and algebraic expansion.

Question 2. If A = {a,b,c}, B = {u, v, w}. If f: A→B and g : B → A, defined as f = {(a, v), (b, u), (c, w)} and g = {(u, b), (v, a), (w, c)}, then find (fog) and (gof).
Answer: Given are sets A = {a,b,c} and B = {u, v, w}.
Functions are defined as:
\( f = \{(a, v), (b, u), (c, w)\} \)
\( g = \{(u, b), (v, a), (w, c)\} \)

To find \( (fog)(x) \), we apply \( g \) first and then \( f \). We do this for each element in the domain of \( g \) (which is B).
\( (fog)(u) = f(g(u)) = f(b) = u \)
\( (fog)(v) = f(g(v)) = f(a) = v \)
\( (fog)(w) = f(g(w)) = f(c) = w \)
So, \( (fog) = \{(u, u), (v, v), (w, w)\} \).

To find \( (gof)(x) \), we apply \( f \) first and then \( g \). We do this for each element in the domain of \( f \) (which is A).
\( (gof)(a) = g(f(a)) = g(v) = a \)
\( (gof)(b) = g(f(b)) = g(u) = b \)
\( (gof)(c) = g(f(c)) = g(w) = c \)
So, \( (gof) = \{(a, a), (b, b), (c, c)\} \).
In simple words: We are combining two functions. For \( (fog) \), we use what \( g \) gives us and then apply \( f \) to that result. For \( (gof) \), we use what \( f \) gives us and then apply \( g \) to that result. This tells us how elements from one set map to another after two steps.

🎯 Exam Tip: When dealing with functions defined by sets of ordered pairs, always remember that the output of the inner function must be a valid input for the outer function. List out each element's transformation clearly to avoid errors.

Question 3. If f: R\( ^+ \) → R\( ^+ \) and g: R\( ^+ \) → R\( ^+ \), defined as f(x) = x\( ^2 \) and g(x)= \( \sqrt{x} \), then find gof and fog. Are they identity functions?
Answer: Given functions are \( f: R^+ \rightarrow R^+ \) with \( f(x) = x^2 \) and \( g: R^+ \rightarrow R^+ \) with \( g(x) = \sqrt{x} \).

To find \( (gof)(x) \), we use \( (gof)(x) = g(f(x)) \).
Substitute \( f(x) \) into \( g(x) \):
\( (gof)(x) = g(x^2) \)
Now, replace \( x \) in \( g(x) \) with \( x^2 \):
\( = \sqrt{x^2} \)
Since the domain is \( R^+ \) (positive real numbers), \( \sqrt{x^2} = x \).
So, \( (gof)(x) = x \).

To find \( (fog)(x) \), we use \( (fog)(x) = f(g(x)) \).
Substitute \( g(x) \) into \( f(x) \):
\( (fog)(x) = f(\sqrt{x}) \)
Now, replace \( x \) in \( f(x) \) with \( \sqrt{x} \):
\( = (\sqrt{x})^2 \)
\( = x \)
So, \( (fog)(x) = x \).

Since both \( (fog)(x) = x \) and \( (gof)(x) = x \) for all \( x \in R^+ \), both functions \( (fog) \) and \( (gof) \) are identity functions. This means they return the same value that was put into them. Identity functions are important because they effectively "undo" the operation of another function if they are inverses.
In simple words: When we combine these two functions, \( f \) and \( g \), in any order, the result is always just the original input number. This means they are "identity functions" because they don't change the number you start with.

🎯 Exam Tip: An identity function always maps an element to itself, i.e., \( I(x) = x \). If \( (fog)(x) = x \) and \( (gof)(x) = x \), then f and g are inverses of each other.

Question 4. If f : R → R and g : R → R be such two functions that defined as f(x) = 3x +4 and g(x) = \( \frac{1}{3} \) (x - 4), then find (fog)(x) and (gof)(x), also find (gog)(1).
Answer: Given functions are \( f: R \rightarrow R \) with \( f(x) = 3x + 4 \) and \( g: R \rightarrow R \) with \( g(x) = \frac{1}{3} (x - 4) \).

To find \( (fog)(x) \), we use \( (fog)(x) = f(g(x)) \).
Substitute \( g(x) \) into \( f(x) \):
\( (fog)(x) = f\left(\frac{1}{3}(x - 4)\right) \)
Now, replace \( x \) in \( f(x) \) with \( \frac{1}{3}(x - 4) \):
\( = 3\left(\frac{1}{3}(x - 4)\right) + 4 \)
\( = (x - 4) + 4 \)
\( = x \)
So, \( (fog)(x) = x \).

To find \( (gof)(x) \), we use \( (gof)(x) = g(f(x)) \).
Substitute \( f(x) \) into \( g(x) \):
\( (gof)(x) = g(3x + 4) \)
Now, replace \( x \) in \( g(x) \) with \( (3x + 4) \):
\( = \frac{1}{3}((3x + 4) - 4) \)
\( = \frac{1}{3}(3x) \)
\( = x \)
So, \( (gof)(x) = x \).

Next, we need to find \( (gog)(1) \). First, let's find the general form for \( (gog)(x) \).
\( (gog)(x) = g(g(x)) \)
\( = g\left(\frac{1}{3}(x - 4)\right) \)
Substitute \( \frac{1}{3}(x - 4) \) into \( g(x) \):
\( = \frac{1}{3}\left(\left(\frac{1}{3}(x - 4)\right) - 4\right) \)
Now, substitute \( x = 1 \) into the expression for \( (gog)(x) \):
\( (gog)(1) = g(g(1)) \)
First, find \( g(1) \):
\( g(1) = \frac{1}{3}(1 - 4) = \frac{1}{3}(-3) = -1 \)
Then, substitute this result back into \( g \):
\( (gog)(1) = g(-1) \)
\( = \frac{1}{3}(-1 - 4) \)
\( = \frac{1}{3}(-5) \)
\( = -\frac{5}{3} \)
Functions \( f \) and \( g \) here are inverses of each other, which is why their composition results in the identity function \( x \). This property is fundamental in function theory.
In simple words: We combine the functions \( f \) and \( g \) in both orders and find that they cancel each other out, giving us back the original input \( x \). Then, we combine \( g \) with itself and find the value when \( x \) is 1.

🎯 Exam Tip: When composing functions and evaluating at a specific point, it's often easier to first calculate the inner function's value, then use that result as the input for the outer function. This avoids complex algebraic manipulation if only a single value is needed.

Question 5. If f, g, h be three functions from R to R, defined as f(x) = x\( ^2 \), g(x) = cos x and h(x) = 2x + 3, then find {ho(gof)}(√2π).
Answer: Given functions are \( f(x) = x^2 \), \( g(x) = \cos x \), and \( h(x) = 2x + 3 \). All functions map from R to R.

We need to find \( \{ho(gof)\}(\sqrt{2}\pi) \). This involves three functions composed together. We work from the innermost function outwards.
First, find \( (gof)(x) \):
\( (gof)(x) = g(f(x)) \)
\( = g(x^2) \)
Substitute \( x^2 \) into \( g(x) \):
\( = \cos(x^2) \)

Next, find \( \{ho(gof)\}(x) \):
\( \{ho(gof)\}(x) = h((gof)(x)) \)
\( = h(\cos(x^2)) \)
Substitute \( \cos(x^2) \) into \( h(x) \):
\( = 2\cos(x^2) + 3 \)

Finally, evaluate this expression at \( x = \sqrt{2}\pi \):
\( \{ho(gof)\}(\sqrt{2}\pi) = 2\cos((\sqrt{2}\pi)^2) + 3 \)
\( = 2\cos(2\pi) + 3 \)
We know that \( \cos(2\pi) = 1 \) because \( 2\pi \) represents one full rotation on the unit circle, bringing us back to the starting point where cosine is 1.
\( = 2(1) + 3 \)
\( = 2 + 3 \)
\( = 5 \)
Thus, the value of the composite function at \( \sqrt{2}\pi \) is 5.
In simple words: We have three functions linked together. We first apply \( f \), then \( g \) to that result, and finally \( h \) to that. When we put \( \sqrt{2}\pi \) into this chain, we find the final answer is 5.

🎯 Exam Tip: When evaluating composite functions at a specific value, it's often more efficient to evaluate step-by-step: first the innermost function, then the next, and so on. This prevents algebraic errors that can occur when finding the general composite function expression first.

Question 6. If functions f and g be defined as below, then find (i) (fog)(x), (ii) (gof)(x), (iii) (fof)(x), (iv) (gog)(x): f: R→ R, f(x) = x\( ^2 \) + 3x + 1; g: R → R, g(x) = 2x - 3
Answer: Given functions are \( f: R \rightarrow R \) with \( f(x) = x^2 + 3x + 1 \) and \( g: R \rightarrow R \) with \( g(x) = 2x - 3 \).

(i) To find \( (fog)(x) \), we use \( (fog)(x) = f(g(x)) \).
Substitute \( g(x) \) into \( f(x) \):
\( (fog)(x) = f(2x - 3) \)
Now, replace \( x \) in \( f(x) \) with \( (2x - 3) \):
\( = (2x - 3)^2 + 3(2x - 3) + 1 \)
Expand \( (2x - 3)^2 \) and distribute \( 3 \):
\( = (4x^2 - 12x + 9) + (6x - 9) + 1 \)
Combine like terms:
\( = 4x^2 - 12x + 6x + 9 - 9 + 1 \)
\( = 4x^2 - 6x + 1 \)

(ii) To find \( (gof)(x) \), we use \( (gof)(x) = g(f(x)) \).
Substitute \( f(x) \) into \( g(x) \):
\( (gof)(x) = g(x^2 + 3x + 1) \)
Now, replace \( x \) in \( g(x) \) with \( (x^2 + 3x + 1) \):
\( = 2(x^2 + 3x + 1) - 3 \)
Distribute \( 2 \):
\( = 2x^2 + 6x + 2 - 3 \)
Combine constant terms:
\( = 2x^2 + 6x - 1 \)

(iii) To find \( (fof)(x) \), we use \( (fof)(x) = f(f(x)) \).
Substitute \( f(x) \) into \( f(x) \):
\( (fof)(x) = f(x^2 + 3x + 1) \)
Now, replace \( x \) in \( f(x) \) with \( (x^2 + 3x + 1) \):
\( = (x^2 + 3x + 1)^2 + 3(x^2 + 3x + 1) + 1 \)
Expand \( (x^2 + 3x + 1)^2 \) using the identity \( (a+b+c)^2 = a^2+b^2+c^2+2ab+2ac+2bc \) or by multiplying it out. Here, \( a=x^2, b=3x, c=1 \).
\( = (x^2)^2 + (3x)^2 + (1)^2 + 2(x^2)(3x) + 2(x^2)(1) + 2(3x)(1) + 3x^2 + 9x + 3 + 1 \)
\( = x^4 + 9x^2 + 1 + 6x^3 + 2x^2 + 6x + 3x^2 + 9x + 3 + 1 \)
Combine like terms by degree:
\( = x^4 + 6x^3 + (9x^2 + 2x^2 + 3x^2) + (6x + 9x) + (1 + 3 + 1) \)
\( = x^4 + 6x^3 + 14x^2 + 15x + 5 \)

(iv) To find \( (gog)(x) \), we use \( (gog)(x) = g(g(x)) \).
Substitute \( g(x) \) into \( g(x) \):
\( (gog)(x) = g(2x - 3) \)
Now, replace \( x \) in \( g(x) \) with \( (2x - 3) \):
\( = 2(2x - 3) - 3 \)
Distribute \( 2 \):
\( = 4x - 6 - 3 \)
Combine constant terms:
\( = 4x - 9 \)
In simple words: This question asks us to combine functions in different ways: \( f \) of \( g \), \( g \) of \( f \), \( f \) of \( f \), and \( g \) of \( g \). We do this by putting one function's rule inside the other and then simplifying the algebra.

🎯 Exam Tip: When computing \( (fof)(x) \) or \( (gog)(x) \), it's easy to make algebraic mistakes. Break down the expansion of terms like \( (x^2 + 3x + 1)^2 \) carefully, using distributive property or binomial/trinomial square formulas to ensure accuracy.