Questions: Relations and Functions
Relations and Functions26 min readLesson 5 of 6
Q1. If (x + 1, y - 2) = (3, 1), find x and y.
Answer: Equal ordered pairs must match in both positions. So x + 1 = 3 and y - 2 = 1. Solving gives x = 2 and y = 3. Both equations are linear and independent, so the pair is unique.
Hinglish: Ordered pair equal tab hota hai jab dono positions equal hon. x + 1 = 3 se x = 2, aur y - 2 = 1 se y = 3. Answer (2, 3).
Q2. If A has 3 elements and B = {3, 4, 5}, find the number of elements in A x B.
Answer: The size of a Cartesian product multiplies the sizes of the two sets. So n(A x B) = n(A) times n(B) = 3 times 3 = 9. The actual elements need not be listed; the count follows from the product rule.
Hinglish: Product ka size dono sets ke sizes ka product hota hai. n(A x B) = 3 x 3 = 9. Elements list karne ki zaroorat nahi, count ka formula kaafi hai.
Q3. Write the relation R = {(x, y) : y = x + 1} from A = {1, 2, 3, 4} to B = {2, 3, 4, 5} in roster form, and give its domain and range.
Answer: Substitute each x. For x = 1 the image is 2, x = 2 gives 3, x = 3 gives 4 and x = 4 gives 5, so R = {(1, 2), (2, 3), (3, 4), (4, 5)}. The domain is {1, 2, 3, 4} and the range is {2, 3, 4, 5}. Here the range equals the codomain, though that need not happen in every relation.
Hinglish: Har x ke liye y = x + 1 nikalo: 2, 3, 4, 5. Roster {(1,2), (2,3), (3,4), (4,5)}. Domain {1,2,3,4} aur range {2,3,4,5}. Is relation me range codomain ke barabar hai.
Q4. Decide whether R = {(1, 3), (1, 5), (2, 5)} is a function.
Answer: Look at the first coordinates. The value 1 appears twice, once paired with 3 and once with 5. A function allows only one image for each input, so R is not a function. The input 2 is fine, but one failure is enough to disqualify the relation.
Hinglish: First coordinates dekho. 1 do baar aata hai, ek baar 3 ke saath aur ek baar 5 ke saath. Function me har input ka ek hi image hota hai, isliye R function nahi hai.
Q5. Find the domain of f(x) = (x squared + 1) / (x squared - 9).
Answer: The expression is a fraction, so the denominator must not be zero. Solve x squared - 9 = 0 to get x = 3 and x = -3. Both must be excluded, so the domain is R minus {3, -3}. The numerator has no restriction, so no other values are removed.
Hinglish: Denominator zero nahi hona chahiye. x squared - 9 = 0 se x = 3 aur x = -3 milte hain. In dono ko hata do, to domain R minus {3, -3} hai. Numerator se koi restriction nahi.
Q6. Find the domain and range of f(x) = x squared + 2 on the reals.
Answer: A polynomial is defined for every real x, so the domain is R. Since x squared is never negative, x squared + 2 is at least 2. Every value at least 2 is reached: set x squared = y - 2 and solve for real x when y is at least 2. So the range is [2, infinity).
Hinglish: Polynomial har real x ke liye defined hai, to domain R. x squared kabhi negative nahi hota, isliye f(x) at least 2 hai. Har value 2 ya usse badi milti hai, to range [2, infinity).
Q7. Find the domain of f(x) = square root of (x - 1).
Answer: The square root is real only when the expression inside it is at least zero. So x - 1 is at least zero, giving x at least 1. The domain is the interval [1, infinity). The endpoint 1 is included because the square root of zero is defined and equals zero.
Hinglish: Square root ke andar wali value non-negative honi chahiye. x - 1 at least 0 se x at least 1 milta hai. Domain [1, infinity). 1 include hai kyunki square root of zero defined hota hai.
Q8. If f(x) = x squared, find (f(1.1) - f(1)) / (1.1 - 1).
Answer: First compute f(1.1) = 1.21 and f(1) = 1. The difference is 0.21, and the denominator is 0.1. So the quotient is 0.21/0.1 = 2.1. This is the average rate of change of the square function between x = 1 and x = 1.1.
Hinglish: f(1.1) = 1.21 aur f(1) = 1, difference 0.21. Denominator 0.1 hai, to quotient 0.21/0.1 = 2.1. Ye interval [1, 1.1] par average rate of change hai.
Q9. Find the range of f(x) = x squared / (1 + x squared) for real x.
Answer: The denominator is at least 1, so the fraction has no break. At x = 0 the value is 0. For large positive or negative x the fraction gets close to 1 but the denominator is always bigger than the numerator, so 1 is never reached. The fraction is never negative. Hence the range is [0, 1).
Hinglish: Denominator hamesha 1 ya usse bada hota hai, isliye koi break nahi. x = 0 par value 0. Bade x par fraction 1 ke paas jaata hai par 1 kabhi nahi milta, aur negative bhi nahi hota. Range [0, 1).
Q10. Let f(x) = 2x - 5. Find f(0), f(7) and f(-3).
Answer: Substitute each value into the rule. f(0) = 2(0) - 5 = -5. f(7) = 2(7) - 5 = 9. f(-3) = 2(-3) - 5 = -11. The function is linear, so each computed value is unique and the substitution is all the working needed.
Hinglish: Har value rule me daalo. f(0) = -5, f(7) = 9, f(-3) = -11. Function linear hai, isliye substitution hi poora working hai.
Common mistakes on this chapter
Four habits cost marks. Writing a product in the wrong order, so that (a, r) turns into (r, a). Calling a relation a function without scanning the domain for a missing input. Computing the range as if it were the whole codomain. Forgetting to exclude the zeros of a denominator. Read your final answer once against these four checks.
Quiz
Quiz below: five multiple choice questions on products, relations, domain, range and standard graphs. Attempt them after the notes above, then continue to the previous year page.