Q.Let f(x)=x2 and g(x)=2x+1 be two real functions. Find (f+g)(x), (f−g)(x), (fg)(x), (gf)(x).
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Concept understanding — Function Operations
Function Operations: Combining Machines
Think of a function as a machine. You feed it an input (say, a number x), it does something, and out comes an output f(x). Now imagine you have two such machines, f and g. Function operations are simply ways to hook these machines together — to add, subtract, multiply, or divide their outputs, or to feed one machine's output into the other.
The core idea is simple: if you can do arithmetic with numbers, you can do arithmetic with functions. The only catch is that both functions must be "ready to work" on the same input at the same time.
The Four Arithmetic Operations
Let f and g be two functions. For any input x that belongs to both their domains (the set of numbers each can accept), we define:
Operation
Notation
What it means
Sum
(f+g)(x)
f(x)+g(x)
Difference
(f−g)(x)
f(x)−g(x)
Product
(f⋅g)(x)
f(x)⋅g(x)
Quotient
(gf)(x)
g(x)f(x), provided g(x)=0
Note
The domain of the new function is the intersection of the domains of f and g — the numbers both machines can handle. For the quotient, you must also exclude any x where g(x)=0, because division by zero is undefined.
Example. Let f(x)=x (domain: x≥0) and g(x)=x−1 (domain: all real numbers). Then:
(f+g)(x)=x+x−1, domain: x≥0.
(gf)(x)=x−1x, domain: x≥0andx=1.
Composition: Feeding One Machine into Another
This is the most powerful operation. Instead of adding outputs side by side, you take the output of one function and feed it as the input to the other.
(f∘g)(x)=f(g(x))
Read "f composed with g". You do g first, then f on the result.
Intuition. Suppose g is a machine that converts Celsius to Fahrenheit, and f is a machine that converts Fahrenheit to Kelvin. Then f∘g converts Celsius directly to Kelvin — one combined machine.
Domain trap. For f(g(x)) to make sense, two conditions must hold:
x must be in the domain of g (so g(x) exists).
g(x) must be in the domain of f (so f can accept it).
So the domain of f∘g is: all x in the domain of g such that g(x) is in the domain of f.
Watch out
Composition is not commutative. f∘g is almost never the same as g∘f. For example, if f(x)=x2 and g(x)=x+1, then:
(f∘g)(x)=(x+1)2=x2+2x+1
(g∘f)(x)=x2+1
These are different functions.
Why This Matters
Function operations let you build complex behaviour from simple pieces. A polynomial like 3x2+2x−5 is just a sum of products of simpler functions. A rational function like x−2x+1 is a quotient. And composition is the engine behind everything from transformations of graphs (shifting, stretching) to the chain rule in calculus.
One final exam tip. When asked to find (f+g)(x) or (f∘g)(x), always write down the domains explicitly. Many marks are lost by forgetting that the quotient excludes zeros of the denominator, or that composition requires the inner output to be valid for the outer function.
Operations on functions, including sum, difference, product, quotient, and composition, are covered in the NCERT Class 11-12 Mathematics curriculum on Relations and Functions, and "composition of functions formula and domain" is a frequently searched topic for CBSE board and JEE Main preparation. Correctly tracking the domain restrictions in these operations is also a common source of lost marks flagged in "functions important questions" for competitive exams.
Concept: Function Operations — adding, subtracting, multiplying, and dividing two functions pointwise.
The results are (f+g)(x)=x2+2x+1, (f−g)(x)=x2−2x−1, (fg)(x)=2x3+x2, and (gf)(x)=2x+1x2 for x=−21.
Function operations combine two functions pointwise: add, subtract, multiply, or divide their outputs for the same input. For f(x)=x2 and g(x)=2x+1, we get (f+g)(x)=x2+2x+1, (f−g)(x)=x2−2x−1, (fg)(x)=2x3+x2, and (gf)(x)=2x+1x2 (with x=−21).
When you have two functions f and g, you can combine them using ordinary arithmetic — but applied pointwise. That means for each x in the domain, you evaluate f(x) and g(x) separately, then perform the operation. The result is a new function whose rule is that combination.
This is exactly like adding or multiplying numbers, except the numbers come from plugging x into each function. The only catch is division: you must exclude any x that makes the denominator zero, because division by zero is undefined.
Let’s work through each operation step by step.
Addition: (f+g)(x)
By definition, (f+g)(x)=f(x)+g(x).
So:
(f+g)(x)=x2+(2x+1)=x2+2x+1.
Notice that x2+2x+1 factors as (x+1)2, but the simplified form is perfectly fine.
Subtraction: (f−g)(x)
Here (f−g)(x)=f(x)−g(x).
(f−g)(x)=x2−(2x+1)=x2−2x−1.
No further simplification is needed — it’s a quadratic expression.
Multiplication: (fg)(x)
The product is (fg)(x)=f(x)⋅g(x).
(fg)(x)=x2⋅(2x+1)=2x3+x2.
Just distribute x2 across the binomial.
Division: (gf)(x)
For division, (gf)(x)=g(x)f(x), provided g(x)=0.
(gf)(x)=2x+1x2.
Now, g(x)=2x+1=0 when x=−21. So the domain of this new function is all real numbers except −21.
Watch out
A common mistake is to forget the domain restriction for division. The expression 2x+1x2 is not defined at x=−21, so you must explicitly state that x=−21 when writing the quotient function.
Tip
Notice that (f+g)(x)=x2+2x+1=(x+1)2. This is a perfect square — a neat observation, but not required for the answer. It can help you check your work quickly.
✓Final answer
The results are (f+g)(x)=x2+2x+1, (f−g)(x)=x2−2x−1, (fg)(x)=2x3+x2, and (gf)(x)=2x+1x2 for x=−21.
Q.Let f(x)=7−x and g(x)=x−5. Then the domain of the function h(x)=f(x)g(x) is
(A) [5,7]
(B) (−∞,7]
(C) [5,∞)
(D) [−7,−5]
(E) (−∞,∞)
›Reveal solutionSolution
Both radicals must be defined: x≤7 and x≥5, giving [5,7].
f(x)=7−x requires 7−x≥0⇒x≤7.
g(x)=x−5 requires x−5≥0⇒x≥5.
The product h(x)=f(x)g(x) is defined where both hold:
[5,∞)∩(−∞,7]=[5,7].
✓Final answer
The correct option is (A).
KEAM 2024Set eng-2024-06074 marksMCQ
Q.Let f(x)=4−x2, g(x)=x2−1. Then the domain of the function h(x)=f(x)+g(x) is equal to
(A) (−∞,−1]∪[1,∞)
(B) (−∞,−2]∪[2,∞)
(C) [−2,−1]
(D) [−2,−1]∪[1,2]
(E) [1,2]
›Reveal solutionSolution
f needs 4−x2≥0⇒−2≤x≤2; g needs x2−1≥0⇒x≤−1 or x≥1. The domain of h=f+g is the intersection, [−2,−1]∪[1,2].
For f(x)=4−x2 to be real, 4−x2≥0, i.e. −2≤x≤2. For g(x)=x2−1 to be real, x2−1≥0, i.e. x≤−1 or x≥1. Since h(x)=f(x)+g(x) requires both, take the intersection of [−2,2] with (−∞,−1]∪[1,∞), giving
[−2,−1]∪[1,2].
✓Final answer
The correct option is (D).
KEAM 2024Set eng-2024-06074 marksMCQ
Q.Let f(x)=6x2+9x+10 and g(x)=x2−9x−9. Then the value of (f∘g)(10) is equal to
(A) 10
(B) 15
(C) 25
(D) 35
(E) 45
›Reveal solutionSolution
Compute the inner function first, then feed it into f.
(f∘g)(10)=f(g(10)). First g(10)=102−9(10)−9=100−90−9=1. Then f(1)=6(1)2+9(1)+10=6+9+10=25.
✓Final answer
The correct option is (C).
KEAM 2024Set eng-2024-06074 marksMCQ
Q.Let g(x)=4x+3 and f(g(x))=x2+9. Then the value of f(7) is equal to
(A) 7
(B) 9
(C) 10
(D) 12
(E) 14
›Reveal solutionSolution
Choose x so that g(x)=7, then read off f(g(x)).
We need the input to f to be 7, so set g(x)=7:
4x+3=7⇒x=1.
Then
f(7)=f(g(1))=12+9=10.
✓Final answer
The correct option is (C).
KEAM 2024Set eng-2024-06084 marksMCQ
Q.If f(x)=x+8, and g(x)=2x2, then (g∘f)(x) is equal to
(A) (2x+8)2
(B) 2(x+8)2
(C) 2x2+8
(D) 2x2+64
(E) 2x3+8x
›Reveal solutionSolution
Compose by substituting f(x)=x+8 into g(x)=2x2: (g∘f)(x)=2(x+8)2.
By definition of composition:
(g∘f)(x)=g(f(x))
Substitute f(x)=x+8 into g(x)=2x2 (replace the variable of g by x+8):
g(x+8)=2(x+8)2
Hence (g∘f)(x)=2(x+8)2.
✓Final answer
The correct option is (B).
KEAM 2022Set eng-2022-P2-B14 marksMCQ
Q.Let ⊙ be a binary operation on Q−{0} defined by a⊙b=ba. Then 1⊙(2⊙(3⊙4)) is equal to
(A) 23
(B) 38
(C) 34
(D) 43
(E) 83
›Reveal solutionSolution
The result is 83.
Concept and Intuition
The operation a⊙b=a/b is non-associative, so we must respect the given parenthesization from the inside out.
Step-by-Step Solution
3⊙4=43.
2⊙43=3/42=38.
1⊙38=8/31=83.
Common Mistakes
Evaluating left to right instead of honoring the inner brackets.
Inverting the division direction (a/b not b/a).
✓Final answer
The correct option is (E) — 83.
ANSWER: E
KEAM 2021Set eng-2021-P2-B14 marksMCQ
Q.Let f(x)=x2 and g(x)=9+x. Then the value of (f∘g−g∘f)(4) is equal to
(A) 6
(B) 6
(C) 8
(D) 8
(E) 5
›Reveal solutionSolution
(f∘g−g∘f)(4)=8.
Concept and Intuition
Composition applies the inner function first, then the outer. We evaluate the two composite functions separately at x=4 and subtract.
Step-by-Step Solution
g(4)=9+4=13, so (f∘g)(4)=f(13)=(13)2=13.
f(4)=42=16, so (g∘f)(4)=g(16)=9+16=25=5.
Difference: 13−5=8.
Common Mistakes
Reversing the order of composition, or squaring/rooting in the wrong sequence.
✓Final answer
The correct option is (D) — 8.
ANSWER: D
KEAM 2021Set eng-2021-P2-B14 marksMCQ
Q.For any two positive rational numbers m and n, a binary operation ∗ is defined by m∗n=3m+n, then 27∗25 is equal to
(A) 4
(B) 6
(C) 2
(D) 8
(E) 9
›Reveal solutionSolution
27∗25=2.
Concept and Intuition
A binary operation is just a rule; here we substitute the two operands into 3m+n and simplify.
Step-by-Step Solution
27+25=212=6.
Divide by 3: 36=2.
Common Mistakes
Adding the fractions incorrectly or forgetting to divide by 3.