Q: For the function f(x)=√(x-√(1-x²)), what is its domain?

[1/√(2),1]. Require 1-x^2 >= 0 so x in [-1,1], and x - sqrt(1-x^2) >= 0 means x >= sqrt(1-x^2), which forces x >= 0 and x^2 >= 1-x^2, i.e. x >= 1/sqrt(2). Combining gives the domain [1/sqrt(2), 1].

Question 1

Given f(x)=(x(x-p))/(q-p)+(x(x-q))/(p-q), where p≠ q, determine the value of f(p)+f(q).

Accepted Answer

f(p+q). Combining the terms, f(x) = x(x-p)/(q-p) + x(x-q)/(p-q) simplifies to f(x) = x. Hence f(p)+f(q) = p+q = f(p+q).

Question 2

A real-valued function f(x) obeys the relation f(x-y)=f(x)f(y)-f(a-x)f(a+y), where a is a fixed constant, and f(0)=1. Then the value of f(2a-x) is

Accepted Answer

-f(x). The given functional equation can be manipulated by substituting specific values for x and y, revealing that f(2a-x) is indeed the negative of f(x). This relationship holds due to the symmetry and properties of the function as defined by the equation.

Question 3

The set of all real values of x for which the function f(x)=(3)/(4-x²)+log₁₀(x³-x) is defined is:

Accepted Answer

(-1,0)∪(1,2)∪(2,∞). The term 3/(4-x^2) requires x != +/-2, and log10(x^3-x) requires x^3-x = x(x-1)(x+1) > 0, i.e. x in (-1,0) U (1,inf). Removing x = 2 gives the domain (-1,0) U (1,2) U (2,inf).

Question 4

A relation R is defined on the set of integers Z by the condition that (x,y) belongs to R exactly when x² + y² = 9. Which statement below is false?

Accepted Answer

None of these. Integer solutions of x^2+y^2=9 are (0,+/-3) and (+/-3,0). The relation set, its domain {-3,0,3}, and its range {-3,0,3} are all stated correctly. Since none of statements (a)-(c) is false, the false-statement answer is 'None of these'.

Question 5

If f(x)=√(1+x²), then which of the following statements is true?

Accepted Answer

f(xy)≤ f(x) f(y). f(x)f(y)=sqrt((1+x^2)(1+y^2))=sqrt(1+x^2+y^2+x^2y^2), while f(xy)=sqrt(1+x^2y^2). Since x^2+y^2>=0, we always have f(xy) <= f(x)f(y).

Question 6

For the function f(x)=√(x-√(1-x²)), what is its domain?

Accepted Answer

[1/√(2),1]. Require 1-x^2 >= 0 so x in [-1,1], and x - sqrt(1-x^2) >= 0 means x >= sqrt(1-x^2), which forces x >= 0 and x^2 >= 1-x^2, i.e. x >= 1/sqrt(2). Combining gives the domain [1/sqrt(2), 1].

Question 7

Let f(x)=x/(1-x), where a is a real number. Define the sequence by x0=a, x1=f(x0), x2=f(x1), x3=f(x2), and so on. If x2009=1, then what is the value of a?

Accepted Answer

1/2010. With f(x)=x/(1-x), iterating gives x_n = a/(1 - n*a). Setting x_2009 = a/(1 - 2009a) = 1 gives a = 1 - 2009a, so 2010a = 1 and a = 1/2010.

Question 8

Find the domain of the function f(x)=log₂ (-log_(1/2)(1+1/x^(1/4))-1).

Accepted Answer

(0, 1). The function involves a logarithm, which requires its argument to be positive. The inner logarithm, -log_(1/2)(1+1/x^(1/4))-1, must also be positive, leading to constraints on the values of x. Analyzing these conditions reveals that the valid domain is (0, 1), where the expression remains defined.

Question 9

Find the domain of the function f(x)=1/√(x²−3x+2).

Accepted Answer

(-∞, 1) ∪ (2, ∞). The function is undefined where the expression under the square root is non-positive, which occurs at the roots of the quadratic equation x²−3x+2=0. The roots are x=1 and x=2, leading to the intervals where the function is defined: (-∞, 1) and (2, ∞).

Question 10

For the function f(x) = ln((x² + e)/(x² + 1)), what is the set of all possible values of f(x)?

Accepted Answer

(0, 1]. The function f(x) is the natural logarithm of a fraction that approaches 1 as x approaches infinity, resulting in f(x) approaching 0, but never reaching it. The logarithm of 1 is 0, and since the fraction is always greater than 1 for all x, the function can take values just above 0 up to and including 1.

Question 11

Let f: R → R be a function such that f(x + y) = f(x) + f(y) for every real x and y, and f(1) = 7. Then the value of ∑_(r=1)ⁿ f(r) is

Accepted Answer

(7n(n+1))/(2). The additive (Cauchy) relation with f(1)=7 gives f(r)=7r. Then sum_{r=1}^n f(r) = 7(1+2+...+n) = 7n(n+1)/2.

Question 12

Given that f(x) = (x−1)/(x+1), what is the value of f(2x)?

Accepted Answer

(3f(x)+1)/(f(x)+3). Let y = f(x) = (x-1)/(x+1), so x = (1+y)/(1-y). Then f(2x) = (2x-1)/(2x+1). Substituting and simplifying gives f(2x) = (3y+1)/(y+3) = (3f(x)+1)/(f(x)+3).

Question 13

Find the set of all real x for which the function f(x)=exp(√(5x-3-2x²)) is defined.

Accepted Answer

[1,3/2]. The function f(x) is defined when the expression inside the square root, 5x - 3 - 2x², is non-negative. Solving the inequality leads to the interval [1, 3/2], where the function is valid.

Question 14

Given that f(x+y)=f(x)+2y²+kxy, and that f(a)=2 and f(b)=8, the function f(x) must be of the form

Accepted Answer

2x². Put x=0: f(y)=f(0)+2y^2, so f(x)=2x^2+c, and matching the cross term gives k=4. The leading coefficient is therefore 2, so f(x)=x^2 cannot be correct; the standard form is f(x)=2x^2.

Question 15

On the set A = {1, 2, 3, 4, 5}, consider the relation R defined by R = {(x, y): |x² - y²| < 16}. Which of the following sets represents R?

Accepted Answer

None of these. R = {(x,y): |x^2-y^2|<16} on {1..5} contains 19 ordered pairs (e.g. (1,1),(1,2),(1,3),(1,4),(2,1)...,(4,5),(5,4),(5,5)). None of the small listed sets equals R, so the answer is 'None of these'.

Question 16

Which of the following relations does NOT define a function?

Accepted Answer

h = {(n, 1/n) | n ∈ I}. The relation h does not define a function because it is not defined for all integers; specifically, it is undefined for n = 0, which leads to a division by zero. A function must have a unique output for every input in its domain.

Question 17

Let R be a relation from A = {1, 2, 3, 4} to B = {1, 3, 5}. Which of the following statements is true?

Accepted Answer

R is a subset of A × B. A relation R from set A to set B is by definition a subset of the Cartesian product A x B. Hence R is a subset of A x B.

Question 18

Let S denote the collection of all lines in a plane. Define a relation R on S by aRb if and only if line a is perpendicular to line b. Then R is:

Accepted Answer

symmetric but neither reflexive nor transitive. A line is not perpendicular to itself (not reflexive); a⊥b implies b⊥a (symmetric); a⊥b and b⊥c make a parallel to c, not perpendicular (not transitive). So R is symmetric but neither reflexive nor transitive.

Question 19

Which of the following functions has a graph that is symmetric about the y-axis?

Accepted Answer

None of these. Option (a) sin[log(x+sqrt(x^2+1))] uses the odd function sinh^-1(x), so it is odd. Option (b) has an even numerator over an odd denominator (x^3 + x^4 cot x is odd), giving an odd function. Option (c) is Cauchy's equation forcing f(0)=0 and f(-x)=-f(x), odd. None is even, so answer is None of these.

Question 20

Let R be a reflexive relation on a finite set A containing n elements, and suppose R has m ordered pairs. Which statement must be true?

Accepted Answer

m is at least n. A reflexive relation requires that every element in the set A is related to itself, which means there must be at least n ordered pairs corresponding to the n elements in A. Therefore, the total number of ordered pairs m must be at least n.

Question 21

Consider the function f: R → R defined by
f(x) = {x|x| − 4, if x is rational; x|x| − √3, if x is irrational}.
Then f(x) is:

Accepted Answer

one-one and into. The function f(x) is one-one because it assigns different outputs for different inputs, distinguishing between rational and irrational numbers. It is into because it does not cover all possible real numbers as outputs, specifically missing some values that could be produced by either case.

Question 22

Let f: B → A be given by f(x) = (3x + 4)/(5x − 7) and g: A → B be given by g(x) = (7x + 4)/(5x − 3), where A = R {3/5} and B = R {7/5}. If I_A and I_B denote the identity maps on A and B respectively, then which statement is true?

Accepted Answer

f∘g = I_A and g∘f = I_B. Computing f(g(x)) = x and g(f(x)) = x. Since f:B->A and g:A->B, f o g maps A->A so f o g = I_A, and g o f maps B->B so g o f = I_B.

Question 23

Let f: R → R be a real-valued function such that 0 ≤ f(x) ≤ 1/2 for every x ∈ R. If, for a fixed a > 0, the relation f(x + a) = 1/2 − √(f(x) − [f(x)]²) holds for all x ∈ R, then the period of f is

Accepted Answer

2a. The relation given for f(x + a) indicates that the function's value at x + a is determined by its value at x, suggesting a repetitive behavior. By substituting x with x + a in the original equation, it can be shown that f(x + 2a) returns to f(x), confirming that the function is periodic with a period of 2a.

Question 24

Let f(x) = [x]² + [x + 1] - 3, where [x] denotes the greatest integer less than or equal to x. Then which of the following is true?

Accepted Answer

Both (a) and (b). Let n=[x]. Then f = [x]^2 + [x+1] - 3 = n^2 + (n+1) - 3 = n^2 + n - 2 = (n+2)(n-1). This is 0 when n=-2 or n=1, i.e. for all x in [-2,-1) or [1,2): infinitely many values. The function is also many-one and into. Both (a) and (b) hold.

Question 25

Let X = {x1, x2, x3} and Y = {x1, x2, x3, x4, x5}. Which of the following relations is reflexive?

Accepted Answer

R1 = {(x1, x1), (x2, x2), (x3, x3)}. X has three elements x1,x2,x3, so a reflexive relation on X must contain (x1,x1),(x2,x2),(x3,x3) and be a subset of X×X. R={(x1,x1),(x2,x2),(x3,x3)} satisfies this. The option with (x4,x4) is not even a relation on X since x4∉X.

Question 26

Given that the composition g(f(x)) equals |sin x| and that |f(g(x))| equals (sin √x)², which of the following is consistent?

Accepted Answer

Question 27

Let f: R → R be given by f(x) = (x - m)/(x - n), where m and n are distinct real numbers. Then

Accepted Answer

f is injective but not surjective. The function f(x) = (x - m)/(x - n) is injective because it is a rational function with a unique output for each input, as long as x is not equal to n (where it is undefined). However, it is not surjective because it cannot take the value 1, which is the horizontal asymptote as x approaches infinity.

Question 28

A function f is defined for inputs in the interval (0,1). For the expression f(e^x) + f(ln|x|), determine the set of all real x for which it is defined.

Accepted Answer

(-e, -1). The function f is defined for inputs in the interval (0,1), so we need to determine where the expressions e^x and ln|x| fall within this range. The expression e^x is in (0,1) when x is in the interval (-∞, 0), and ln|x| is in (0,1) when |x| is in (1,e), which corresponds to x being in the intervals (-e, -1) and (1, e). Therefore, the only overlap that satisfies both conditions is (-e, -1).

Question 29

Let f(x) = x/(x - 1). Then the 19-fold composition (f ∘ f ∘... ∘ f)(x) is equal to:

Accepted Answer

x/(x - 1). f(f(x)) = (x/(x-1))/((x/(x-1))-1) = x, so f composed with itself is the identity. Since 19 is odd, the 19-fold composition equals f(x) = x/(x-1).

Question 30

Which of the following functions is one-one?

Accepted Answer

f: [0, ∞) → R, f(x) = x² + 4x - 5. On [0,inf) the function f(x)=x^2+4x-5 has derivative 2x+4>0, so it is strictly increasing and hence one-one. The other options each fail: x^2-4x+3 has its vertex at x=2 inside (0,inf); e^x+e^-x is even; ln(x^2+x+1) has its minimum at x=-1/2, all making them many-one.

Question 31

Find the inverse function of f(x) = (2/3) · (10^x - 10^-x)/(10^x + 10^-x).

Accepted Answer

(1/2) log10((2 + 3x)/(2 - 3x)). The correct option is right because it accurately represents the inverse of the given function, which involves manipulating the original function's expression to isolate x and express it in terms of y, leading to the logarithmic form that matches the structure of option B.

Question 32

Consider the mapping f from the natural numbers to the integers given by
f(n) = (n - 1)/2 for odd n, and f(n) = -n/2 for even n. Which of the following best describes f?

Accepted Answer

both injective and surjective. Odd n give f=(n-1)/2 = 0,1,2,... (all non-negative integers); even n give f=-n/2 = -1,-2,... (all negative integers). Together they hit every integer with no repeats, so f is a bijection (both injective and surjective).

Question 33

For the function f(x) = (ax + b)/(cx + d), the composition f(f(x)) equals x when:

Accepted Answer

d = -a. For f(x)=(ax+b)/(cx+d), f is an involution (f(f(x))=x) exactly when its trace vanishes, a+d=0, i.e. d=-a. Substituting d=-a indeed gives f(f(x))=x, whereas a=b=c=d=1 gives the constant 1.

Question 34

Consider the function f: R → R given by f(x) = (e^(|x|) - e^(-x))/(e^x + e^(-x)). Which of the following is true?

Accepted Answer

f is neither injective nor surjective. The function f is neither injective nor surjective because it does not produce unique outputs for every input (indicating it is not injective) and it does not cover all possible real numbers as outputs (indicating it is not surjective).

Question 35

Assertion-1: Let f: R → R and g: R → R be defined by f(x) = sin x and g(x) = x². Then f∘g is not equal to g∘f.
Assertion-2: For any x, (f∘g)(x) = f(g(x)) = (g∘f)(x).

Accepted Answer

Assertion-1 is true, Assertion-2 is false.. f(g(x)) = sin(x^2) and g(f(x)) = (sin x)^2, which are not equal, so Assertion-1 is true. Assertion-2 claims they are always equal, which is false. So A-1 true, A-2 false.

Question 36

Suppose f: R → R satisfies f((x+y)/3) = (f(x)+f(y))/3 for all real x,y, along with f(0)=0 and f'(0)=3. Which of the following must be true?

Accepted Answer

f(x) is differentiable on R. The functional equation given implies that f is a linear function, and since it is differentiable at 0 with a defined derivative, it must be differentiable everywhere on R.

Question 37

Suppose a mapping f: R → R satisfies f(x + y) = f(x) + f(y) for every pair of real numbers x and y. If f is continuous at x = 0, which of the following must be true?

Accepted Answer

f(x) is continuous for all real x. An additive function f(x+y)=f(x)+f(y) that is continuous at a single point (x=0) is continuous on all of R (in fact f(x)=cx). So f is continuous for all real x.

Question 38

For the function f(x) = [x]² - [x²], where [y] denotes the greatest integer less than or equal to y, at which points is f discontinuous?

Accepted Answer

at every integer except 1. f(x) = [x]^2 - [x^2]. At an integer n, f(n)=0. For x slightly above/below an integer the value jumps (e.g. at x=0 the left limit is 1 but f(0)=0). Checking each integer, the left and right limits both equal f only at x=1, so f is discontinuous at every integer except 1.

Question 39

If the function g is the inverse of f and the derivative of f is f'(x) = sin x, then what is g'(x)?

Accepted Answer

cosec{g(x)}. The derivative of the inverse function g can be found using the formula g'(x) = 1/f'(g(x)). Since f'(x) = sin x, we have g'(x) = 1/sin(g(x)), which simplifies to cosec(g(x)).

Question 40

Consider the functions f: R → R given by f(x) = 2x - 1 and g: R {1} → R given by g(x) = (x - 1/2)/(x - 1). The composite function f(g(x)) is:

Accepted Answer

injective but not surjective. f(g(x)) = 2*((x-1/2)/(x-1)) - 1 = (2x-1-(x-1))/(x-1) = x/(x-1). This Mobius map is one-to-one (injective) but its range is R\{1}, so it is not surjective onto R: injective but not surjective.

Question 41

Find the set of all real x for which the function f(x) = 3/(4 - x²) + log10(x³ - x) is defined.

Accepted Answer

(-1,0) ∪ (1,2) ∪ (2,∞). The function f(x) is defined when the denominator of the rational part is non-zero and the argument of the logarithm is positive. The term 4 - x² must not equal zero, leading to x ≠ ±2, and x³ - x must be greater than zero, which is satisfied in the intervals (-1,0) and (1,2) ∪ (2,∞). Thus, the correct option includes all these intervals.

Question 42

Let f: R → R be a function such that f(x + y) = f(x) + f(y) for every real x and y, and f(1) = 7. Then the value of ∑_(r=1)ⁿ f(r) is

Accepted Answer

7n(n+1)/2. The function f is additive, meaning it can be expressed in the form f(x) = cx for some constant c. Given that f(1) = 7, we find c = 7, leading to f(x) = 7x. The sum A3_(r=1)ⁿ f(r) becomes A3_(r=1)ⁿ 7r, which simplifies to 7(n(n+1)/2) using the formula for the sum of the first n integers.

Question 43

Let a, b, c ∈ R. If f(x) = ax² + bx + c is such that a + b + c = 3 and f(x+y) = f(x) + f(y) + xy, ∀ x, y ∈ R, then ∑ₙ₌₁¹⁰ f(n) is equal to:

Accepted Answer

330. From f(x+y)=f(x)+f(y)+xy we get 2a = 1 so a=1/2 and c=0; then a+b+c=3 gives b=5/2, so f(n)=n^2/2 + 5n/2. Sum for n=1..10 = (1/2)(385) + (5/2)(55) = 192.5 + 137.5 = 330.

Question 44

Let f: N→Y be a function defined as f(x)=4x+3 where Y={y∈N: y=4x+3 for some x∈N}. Then, the inverse of f(x) is

Accepted Answer

g(y)=(y-3)/(4). The correct option is right because to find the inverse of the function f(x) = 4x + 3, we need to solve for x in terms of y. Rearranging the equation gives y - 3 = 4x, which simplifies to x = (y - 3)/4, thus the inverse function is g(y) = (y - 3)/4.

Question 45

Let R be the real line. Consider the following subsets of the plane R×R:
S={(x,y): y=x+1 and 0<x<2}
T={(x,y): x−y is an integer}.
Which one of the following is true?

Accepted Answer

T is an equivalence relation but S is not. S = {(x,y): y=x+1} is not reflexive (no x satisfies x=x+1), so it is not an equivalence relation. T = {(x,y): x-y is an integer} is reflexive, symmetric and transitive, so T is an equivalence relation but S is not.

Question 46

DIRECTIONS: This question contains two statements:
Statement-1 (Assertion) and Statement-2 (Reason).
This question also has four alternative choices, only one of which is the correct answer. You have to select the correct choice.
Let f(x)=(x+1)²−1, x≥−1
Statement-1: The set {x: f(x)=f⁻¹(x)

Accepted Answer

Statement-1 is true, Statement-2 is true. Statement-2 is not a correct explanation for Statement-1.. f is increasing, so f(x)=f^-1(x) reduces to f(x)=x: (x+1)^2-1=x -> x^2+x=0 -> x=0,-1, so St-1 is true. f is also a bijection onto its range [-1,inf), so St-2 is true, but being a bijection does not by itself produce the specific solution set, so it is not the correct explanation.

Question 47

For real x, let f(x)=x³+5x+1, then

Accepted Answer

f is one-one and onto R. f'(x)=3x^2+5>0 for all x, so f is strictly increasing and hence one-one. Being an odd-degree polynomial it ranges over all of R, so it is onto. Therefore f is both one-one and onto.

Question 48

Consider the following relations:
R={(x,y) | x,y are real numbers and x=wy for some rational number w};
S={(m/n,p/q) | m,n,p and q are integers such that n,q≠0 and qm=pn}. Then

Accepted Answer

S is an equivalence relation but R is not an equivalence relation. S: m/n ~ p/q iff qm=pn means m/n=p/q, i.e. ordinary equality of rationals -> reflexive, symmetric, transitive, so S is an equivalence relation. R: x=w*y for rational w fails symmetry: (0,5) is in R (0=0*5) but (5,0) is not (5=w*0 impossible), so R is not an equivalence relation. Correct: S is an equivalence relation but R is not.

Question 49

Let R be the set of real numbers.
Statement-1: A={(x,y)∈R×R: x−y is an integer} is an equivalence relation on R.
Statement-2: B={(x,y)∈R×R: x=y for some rational number α} is an equivalence relation on R.

Accepted Answer

Statement-1 is true, Statement-2 is false.. A={(x,y): x-y is an integer} is reflexive (x-x=0), symmetric and transitive, so Statement-1 is true. Statement-2 as written is not a valid equivalence relation, so Statement-1 true, Statement-2 false is the correct choice.

Question 50

Let the function f be given by f(x)=(x-1)²+1 for x≥ 1.

Statement-1: The solution set of f(x)=f⁻¹(x) is {1,2}.
Statement-2: The function f is one-to-one and onto, and its inverse is f⁻¹(x)=1+√(x-1), with x≥ 1.

Accepted Answer

Statement-1 is correct, Statement-2 is correct; Statement-2 correctly explains Statement-1.. f is increasing on x>=1, so f(x)=f^-1(x) reduces to f(x)=x: (x-1)^2+1=x -> x^2-3x+2=0 -> x=1,2, giving solution set {1,2}. The inverse is indeed f^-1(x)=1+sqrt(x-1) for x>=1, and that one-to-one/onto property is exactly why the equation collapses to f(x)=x. So Statement-1 and Statement-2 are both correct and Statement-2 correctly explains Statement-1.

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