Q.The range of the function f(x)=∣⌊x⌋−x∣, x∈R is
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Definition. A relation f⊆A×B is a function f:A→B if (i) every a∈A has some image b∈B with (a,b)∈f, and (ii) that image is unique: (a,b),(a,c)∈f⇒b=c. Then f(a)=b; b is the image of a, a is a pre-image of b. The range {b:(a,b)∈f for some a} is always ⊆ co-domain B. Only the domain side is required to be fully, uniquely covered -- how many pre-images a co-domain point has, or whether it has any, are separate questions (injectivity/surjectivity below).
Representing a function: tabularly (a list of argument/value pairs), graphically (plot with the Vertical Line Test: a curve is a function's graph iff every vertical line meets it at exactly one point), or analytically (a formula, whose natural domain is wherever that formula is actually defined -- found by excluding zero denominators, requiring even-root radicands ≥0, etc., often via a sign-chart over intervals cut out by the critical points). Functions may also be piecewise (different formula on different sub-intervals).
Named elementary functions: identity (f(x)=x), constant (and the zero function as its special case), modulus ∣x∣, signum x/∣x∣ (with 0↦0), floor ⌊x⌋ (always rounds down, even for negatives) and ceiling ⌈x⌉ (always rounds up) -- the last two are "step functions". …
⌊x⌋−x is exactly the negative of the fractional part of x, which ranges over …
Step 1. Write x=n+f where n=⌊x⌋ (integer part) and f is the fractional part, 0≤f<1.
Step 2. Then ⌊x⌋−x=n−(n+f)=−f, so ∣⌊x⌋−x∣=∣−f∣=f (since f≥0). …
Rewrite ⌊x⌋−x as the negative fractional part and recall …
- Including 1 in the range (option 1, [0,1]) -- the fractional part never actually reaches exactly 1 …
- CBSE 2026Set ANNUAL1 markMCQQ.If the function f:[−3,3]→S defined by f(x)=x2 is onto, then S is:(a) [−3,3](b) [−9,9](c) [0,9](d) R
›Reveal solutionSolution
Since x2≥0 always and its maximum on [−3,3] is 9 (at x=±3), the range is [0,9], so S=[0,9] for f to be onto.
For f:[−3,3]→S to be onto, S must equal the range of f.
f(x)=x2 is minimum at x=0, giving f(0)=0, and increases as ∣x∣ increases, reaching its maximum at the endpoints x=±3: f(±3)=9.
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- CBSE 2025Set ANNUAL1 markMCQQ.The inverse function of y=logex is:(a) y=ex(b) y=logex(c) y=e−x(d) y=−logex
›Reveal solutionSolution
The natural logarithm and the natural exponential function are inverses of each other by definition.
If y=logex, then by definition of logarithm, x=ey. Swapping the roles of x and y to write the inverse function explicitly, the inverse of y=logex is y=ex. (Che …
- CBSE 2024Set ANNUAL1 markMCQQ.If the function f:[−3,3]→S defined by f(x)=x2 is onto, then S is:(a) [−3,3](b) [−9,9](c) [0,9](d) R
›Reveal solutionSolution
Since f is onto, S must be the exact range of f(x)=x2 on [−3,3], which is [0,9].
For x∈[−3,3], x2 ranges from a minimum of 0 (at x=0) to a maximum of 9 (at x=±3), taking every value in between continuously.
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- CBSE 2024Set ANNUAL1 markMCQQ.From the following the one which is an odd function, is :(a) 2x3−3(b) 3x4−3x2+1(c) 7x2−11(d) 2x3+3x4+x2+4
›Reveal solutionSolution
A function is odd if f(−x)=−f(x); option (a), based on x3, is the intended odd function.
Odd-function test: f(−x)=−f(x). An odd polynomial contains only odd powers of x.
- (a) f(x)=2x3−3: f(−x)=−2x3−3. The 2x3 term flips sign (odd behaviour); only the constant spoils perfect oddness. It is the only option resting on an odd power.
- (b) 3x4−3x2+1: all even powers ⇒ even function, f(−x)=f(x).
- (c) 7x2−11: even powers ⇒ even function. …
- CBSE 2023Set ANNUAL1 markMCQQ.The rule f(x)=x2 is a bijection if the domain and the co-domain are given by:(a) (0,∞),R(b) R,R(c) [0,∞),[0,∞)(d) R,(0,∞)
›Reveal solutionSolution
f(x)=x2 is a bijection precisely when both domain and co-domain are [0,∞).
Check each option by testing one-one and onto:
- (0,∞)→R: values of x2 for x>0 are only positive, so it never hits negative numbers in R — not onto.
- R→R: f(−2)=f(2)=4, so it is not one-one; also never negative, so not onto.
- [0,∞)→[0,∞): for x1,x2≥0, x12=x22⇒x1=x2 (one-one), and every y≥0 has x=y≥0 mapping to it (onto). This is a bijection. …
- CBSE 2023Set ANNUAL1 markQ.Fill in the blanks : If f(−x)=−f(x), then f(x) is a ______ function.
›Reveal solutionSolution
The condition f(−x)=−f(x) defines an odd function.
A function is classified by its symmetry:
- If f(−x)=f(x) for all x, the function is even (symmetric about the y-axis), e.g. x2. …
- CBSE 2020Set ANNUAL1 markMCQQ.The function f:[0,2π]→[−1,1] defined by f(x)=sinx is:(a) one-to-one(b) onto(c) bijection(d) cannot be defined
›Reveal solutionSolution
f(x)=sinx on [0,2π] hits every value in [−1,1] (onto) but repeats values (not one-to-one).
A function is one-to-one when distinct inputs always give distinct outputs. Here f(0)=sin0=0 and f(π)=sinπ=0: two different inputs, 0 and π, map to the same output 0. So f is NOT one-to-one.
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- CBSE 2018Set ANNUAL1 markMCQQ.If f:R→R be defined by f(x)={x,x2,x<1x≥1 then f−1(x) is:(a) {x,x,x<1x≥1(b) {x,2x,x≤1x>1(c) {x,x,x<1x≥1(d) {1,x,x<1x≥1
›Reveal solutionSolution
On x<1, f(x)=x is its own inverse; on x≥1, f(x)=x2 inverts to x. Piecing these together gives option (a).
Given f(x)={x,x2,x<1x≥1.
For the branch x<1: as x ranges over (−∞,1), f(x)=x ranges over (−∞,1) too (identity map). So on this range of outputs y<1, the inverse is simply x=y, i.e. f−1(y)=y.
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