Q.If f(x)=x+x1, prove that [f(x)]3=f(x3)+3f(x1).
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Function Equality — When Are Two Functions the Same?
You write f(x) = x * x; your friend writes g(x) = x^2. Same function? Intuitively yes — for every input they give the same output.
But what if your friend's program only works for positive numbers? Then for x=−3 your function gives 9 but theirs crashes. Now they're not the same function, even though the formulas look alike.
That's the core idea: two functions are equal only when they agree on every input they're supposed to handle.
The Precise Definition
Two functions f and g are equal (f=g) if and only if:
- They have the same domain D.
- For every x in D, f(x)=g(x).
If either condition fails, the functions are different.
A common mistake: thinking f(x)=xx2 and g(x)=x are the same. They are not — f is undefined at x=0, while g is defined everywhere. Different domains → different functions.
Why This Matters
Two expressions can look identical after simplification yet have different domains — that's the exam trap.
Equal: f(x)=x2 and g(x)=∣x∣ — both have domain R, and x2=∣x∣ for every x. So f=g.
Not equal: f(x)=x−1x−1 and g(x)=1 — f has domain R∖{1}, g has domain R. Different domains → f=g, even though f(x)=1 wherever f is defined.
Always compare domains first. If they differ, stop — the functions are not equal. Only check values if the domains match.
A Quick Test
- f(x)=x⋅x, g(x)=x — domain of f is x≥0, domain of g is all reals. Not equal (different domains).
- f(x)=sin2x+cos2x, g(x)=1 — both domain R, equal for every x. Equal. …
[f(x)]3=(x+x1)3=x3+x31+3(x+x1). Since f(x3)=x3+x31 and f(x1)=x+x1, this equals $f(x^3)+3f!\left(\tfrac …
Expand the cube of x+x1; the result splits exactly into f(x3) and 3f(1/x).
Concept. Substituting into a defined function and using the identity (p+q)3=p3+q3+3pq(p+q) verifies the relation.
Expand the left side. With f(x)=x+x1,
[f(x)]3=(x+x1)3=x3+x31+3⋅x⋅x1(x+x1)=x3+x31+3(x+x1).
Compute the right side. …
- CBSE 2026Set ANNUAL1 markMCQQ.Let f:R→R be defined by f(x)={1−1if x∈Qif x∈/Q, then f(21)+f(π)+f(2)=(a) 0(b) 1(c) −1(d) 2
›Reveal solutionSolution
21 is rational so f=1; π and 2 are irrational so f=−1 each. Sum =−1.
The function assigns 1 to every rational number and −1 to every irrational number.
- 21∈Q (rational) ⇒f(21)=1. …
- CBSE 2022Set ANNUAL1 markMCQQ.The domain in which the functions f(x) = 3x² - 2x and g(x) = 3(3x - 2) will be equal is: OR Let A = {1, 2, 3} and R be a relation defined on A, such that R = {(1,1), (1,2), (2,1)}; then the relation R will be:(a) {1, 2/3}(b) {1, 3}(c) {2/3, 3}(d) {2/3, 0}
›Reveal solutionSolution
Set f(x)=g(x), form and solve the resulting quadratic equation — its roots are exactly the domain values where the two functions agree.
Given f(x)=3x2−2x and g(x)=3(3x−2)=9x−6. We need the set of x for which f(x)=g(x).
Set the functions equal:
3x2−2x=9x−6
3x2−2x−9x+6=0
3x2−11x+6=0
Solve using the quadratic formula x=2A−B±B2−4AC with A=3,B=−11,C=6:
Discriminant =(−11)2−4(3)(6)=121−72=49, 49=7
…
- CBSE 2021Set NC1 markMCQQ.If f(x)=2, then the value of f(2) is(a) 2(b) x(c) x2(d) 2x
›Reveal solutionSolution
f(x)=2 means f outputs 2 for every input, including x=2 -- substitute directly.
Given f(x)=2 for all x (a constant function). To find f(2), substitute x=2:
f(2)=2
(The rule "f(x)=2" does not depend on x at all, so plugging in any value, including 2, still gives 2.)
…
- CBSE 2019Set ANNUAL1 markMCQQ.The domain for which the functions f(x) = 3x^2 - 2x and g(x) = 3(3x - 2) are equal, will be(a) {1, 2/3}(b) {1, 3}(c) {2/3, 3}(d) {2/3, 0}
›Reveal solutionSolution
Set f(x)=g(x), simplify to a quadratic, and solve.
Given f(x)=3x2−2x and g(x)=3(3x−2)=9x−6. Set f(x)=g(x):
3x2−2x=9x−6
3x2−11x+6=0
Using the quadratic formula: discriminant =121−72=49, 49=7. …
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