Q.If f and g are real functions defined by f(x)=x2+7 and g(x)=3x+5, find each of the following
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →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 |
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≥0 and x=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.
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 …
Concept: Function Operations — substitute the given input into each function, then perform the indicated arithmetic.
(a) f(3)=32+7=9+7=16
g(−5)=3(−5)+5=−15+5=−10
f(3)+g(−5)=16+(−10)=6
(b) f(21)=(21)2+7=41+7=429
g(14)=3(14)+5=42+5=47
f(21)×g(14)=429×47=41363
(c) f(−2)=(−2)2+7=4+7=11
g(−1)=3(−1)+5=−3+5=2
f(−2)+g(−1)=11+2=13
(d) f(t)=t2+7, f(−2)=11
f(t)−f(−2)=(t2+7)−11=t2−4 …
Substitute the given inputs into f(x)=x2+7 and g(x)=3x+5: (a) 6, (b) 41363, (c) 13, (d) t2−4, (e) t+5.
These are function-evaluation problems: apply each rule to its input, then perform the indicated operation.
(a) f(3)+g(−5)
f(3)=32+7=16,g(−5)=3(−5)+5=−10
f(3)+g(−5)=16+(−10)=6
(b) f(21)×g(14)
f(21)=(21)2+7=41+7=429,g(14)=3(14)+5=47
f(21)×g(14)=429×47=41363
(c) f(−2)+g(−1)
f(−2)=(−2)2+7=11,g(−1)=3(−1)+5=2 …
- KEAM 2026Set eng-2026-04174 marksMCQQ.Let f(x)=x2−10x and g(x)=ex+5 for x∈R. Then, for all x, g(2x)−(f∘g)(x)= (A) 30 (B) 20 (C) -30 (D) -20 (E) 15
›Reveal solutionSolution
Compute both pieces: g(2x)=e2x+5 and (f∘g)(x)=e2x−25, giving a constant difference of 30.
First, g(2x)=e2x+5.
Next, (f∘g)(x)=f(g(x)) with g(x)=ex+5:
f(ex+5)=(ex+5)2−10(ex+5)=e2x+10ex+25−10ex−50=e2x−25. …
- KEAM 2026Set eng-2026-04184 marksMCQQ.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. …
- KEAM 2024Set eng-2024-06074 marksMCQQ.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 …
- KEAM 2024Set eng-2024-06074 marksMCQQ.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. …
- KEAM 2024Set eng-2024-06074 marksMCQQ.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 …
- KEAM 2024Set eng-2024-06084 marksMCQQ.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)) …
- KEAM 2022Set eng-2022-P2-B14 marksMCQQ.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 …
- KEAM 2021Set eng-2021-P2-B14 marksMCQQ.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. …
- KEAM 2021Set eng-2021-P2-B14 marksMCQQ.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. …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.