JEE Maths›Sets, Relations & Functions›Visual Solution
Visual SolutionHard
Question
Let f:[1,∞)→[2,∞)f: [1, \infty) \to [2, \infty) be defined by f(x)=x+1xf(x) = x + \frac{1}{x}. Then f−1(x)f^{-1}(x) equals:
(A)x+x2−42\frac{x + \sqrt{x^2 - 4}}{2}
(B)x−x2−42\frac{x - \sqrt{x^2 - 4}}{2}
(C)x−1xx - \frac{1}{x}
(D)x+x2+42\frac{x + \sqrt{x^2 + 4}}{2}
Solution Path
Verify bijectivity via f'(x) >= 0 on [1,inf). Solve y = x + 1/x as quadratic x^2 - yx + 1 = 0. Take + root since x >= 1. f^{-1}(x) = (x + sqrt(x^2 - 4))/2.
01Question Setup
1/4
Find f−1(x)f^{-1}(x) for f:[1,∞)→[2,∞)f: [1, \infty) \to [2, \infty) defined by f(x)=x+1xf(x) = x + \frac{1}{x}.
f(x)=x+1xf(x) = x + \frac{1}{x}
02Verify Bijectivity
2/4
f′(x)=1−1x2≥0f'(x) = 1 - \frac{1}{x^2} \geq 0 for x≥1x \geq 1, so ff is increasing. f(1)=2f(1) = 2 and f(x)→∞f(x) \to \infty. So ff is bijective.
f′(x)≥0 on [1,∞)f'(x) \geq 0 \text{ on } [1, \infty)
03Solve for InverseKEY INSIGHT
3/4
Set y=x+1/xy = x + 1/x, multiply by xx: x2−yx+1=0x^2 - yx + 1 = 0. Quadratic formula gives x=y±y2−42x = \frac{y \pm \sqrt{y^2 - 4}}{2}. Since x≥1x \geq 1, take the ++ sign.
x=y+y2−42x = \frac{y + \sqrt{y^2 - 4}}{2}
04Final Answer
4/4
Quick check: f(2)=5/2f(2) = 5/2, f−1(5/2)=5/2+3/22=2f^{-1}(5/2) = \frac{5/2 + 3/2}{2} = 2.
f−1(x)=x+x2−42f^{-1}(x) = \dfrac{x + \sqrt{x^2 - 4}}{2}
Concepts from this question2 concepts unlocked
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Inverse Function Existence and Computation

STANDARD

A function f: A to B has an inverse f^(-1): B to A if and only if f is bijective. To find f^(-1): write y = f(x), solve for x in terms of y, then swap x and y. The composition f^(-1)(f(x)) = x for all x in A and f(f^(-1)(y)) = y for all y in B. If f is one-one but not onto, restrict the codomain to range(f) to obtain a bijection.

f−1 exists  ⟺  f is bijective,f−1(f(x))=xf^{-1} \text{ exists} \iff f \text{ is bijective}, \quad f^{-1}(f(x)) = x

Inverse function questions test both the existence condition (bijectivity check) and the computation. JEE often gives a function and asks for its inverse, or asks for which values of a parameter the inverse exists.

Find inverse functionVerify invertibilityParametric inverse existenceComposition with inverse
Practice (7 Qs) →

Domain, Range, and Composite Functions

STANDARD

The domain of a function is the set of all valid inputs. For composite functions g(f(x)), the domain consists of all x in dom(f) such that f(x) is in dom(g). The range of a composite function is a subset of the range of the outer function. Key domain rules: square root requires nonneg argument, log requires positive argument, denominator must be nonzero.

dom(g∘f)={x∈dom(f):f(x)∈dom(g)}\text{dom}(g \circ f) = \{x \in \text{dom}(f) : f(x) \in \text{dom}(g)\}

Domain and range problems are guaranteed in JEE. The domain of composite functions is a common trap where students forget the intermediate domain restriction. Floor function (greatest integer function) domain/range questions are also frequently tested.

Find domain of compositeFind range of functionFloor/ceiling function problemsDomain restrictions
Practice (8 Qs) →
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