Q: Consider the functions f₁: R → R, f₂: [0, ∞) → R, f₃: R → R, and f₄: R → [0, ∞) defined as follows: f₁(x) = |x| if x < 0, eˣ if x ≥ 0; f₂(x) = x²; f₃(x) = sin(x) if x < 0, x if x ≥ 0; and f₄(x) = f₂(f₁(x)) if x < 0, f₂(f₁(x)) − 1 if x ≥ 0. Match the descriptions of these functions in List

P corresponds to 1, Q to 3, R to 2, and S to 4. Matching the functions with their properties involves analyzing their definitions. For P, f₁(x) is continuous but not differentiable at x = 0. For Q, f₂(x) is differentiable everywhere. For R, f₃(x) is continuous but not differentiable at x = 0. For S, f₄(x) is continuous and differentiable everywhere except at x = 0. This matches option D.

Q: Given the function f: R → R defined as f(x) = |x| (x − sin x), which of the following statements about the function is accurate?

f is both injective and surjective.. The function f(x) = |x|(x − sin x) is both injective (one-to-one) and surjective (onto) because it uniquely maps every real number to a real number and covers the entire range of real numbers.

Question 1

The function f(x) = √|x² - 5| x + 6 + √8 + 2|x| - |x|² is defined as a real number for values of x within which range?

Accepted Answer

[-2, 2]. The domain of the function f(x) = √|x² - 5| x + 6 + √8 + 2|x| - |x|² is determined by ensuring the square root terms are non-negative, yielding a domain of [-2, 2].

Question 2

The relation x² = xy can be described as:

Accepted Answer

reflexive and transitive. R={(x,y):x(x-y)=0}, i.e. x=0 or x=y. It is reflexive (x=x always). Not symmetric ((0,5) holds but (5,0) gives 25=0, false). It is transitive (if a=0 then (a,c) holds; if a=b then b=0 or b=c each forces a=0 or a=c). So reflexive and transitive (idx 2); stored 'symmetric and reflexive' is wrong.

Question 3

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

Accepted Answer

f is injective but not surjective. f(x) = (e^x - e^-x)/(e^x + e^-x) = tanh(x), whose derivative sech^2(x) > 0 everywhere, so f is strictly increasing and therefore injective. Its range is the open interval (-1,1), not all of R, so it is not surjective onto R. Thus f is injective but not surjective; the stored 'surjective but not injective' is wrong.

Question 4

Consider the functions f(x) = max {1 + sin(x), 1 - cos(x)} for x in the interval [0, 2π], and g(x) = max {1, |x - 1|} for x belonging to the set of real numbers. Which of the following is true?

Accepted Answer

The value of g(f(0)) is equal to 1.. The value of g(f(0)) is equal to 1 because f(0) = max {1 + sin(0), 1 - cos(0)} = max {1, 0} = 1, and g(1) = max {1, |1 - 1|} = max {1, 0} = 1.

Question 5

Given f(x + f(y)) = f(x) + y, x, y ∈ R. Find f(1000).

Accepted Answer

1000. Using the functional equation f(x + f(y)) = f(x) + y, substituting x = 0 shows that f(f(y)) = y, implying f(x) = x. Thus, f(1000) = 1000.

Question 6

Find the range of fog(x) = f(g(x)) = f(4x(1 − x)).

Accepted Answer

[0, 1]. The function g(x) = 4x(1 − x) has a range of [0, 1] since it represents a quadratic function with a maximum value of 1. Applying f to this range preserves [0, 1].

Question 7

The provided relation is many-to-one and thus not injective, with the domain R {0} and the range R (-2, 2). What is the equation of the relation?

Accepted Answer

Because y = x + 1/x. The relation y = x + 1/x is many-to-one because multiple x-values can produce the same y-value. Its domain excludes x = 0, and its range excludes (-2, 2).

Question 8

The given relation is many-to-one and hence not one-to-one. The domain of this function is (-∞, ∞), and its range is (2, ∞). Which of the following equations represents this relation?

Accepted Answer

y = |x| + 2. The equation y = |x| + 2 has a domain of all real numbers and a range starting from 2 to infinity. This matches the given description of the relation.

Question 9

Consider the functions f₁: R → R, f₂: [0, ∞) → R, f₃: R → R, and f₄: R → [0, ∞) defined as follows: f₁(x) = |x| if x < 0, eˣ if x ≥ 0; f₂(x) = x²; f₃(x) = sin(x) if x < 0, x if x ≥ 0; and f₄(x) = f₂(f₁(x)) if x < 0, f₂(f₁(x)) − 1 if x ≥ 0. Match the descriptions of these functions in List

Accepted Answer

P corresponds to 1, Q to 3, R to 2, and S to 4. Matching the functions with their properties involves analyzing their definitions. For P, f₁(x) is continuous but not differentiable at x = 0. For Q, f₂(x) is differentiable everywhere. For R, f₃(x) is continuous but not differentiable at x = 0. For S, f₄(x) is continuous and differentiable everywhere except at x = 0. This matches option D.

Question 10

Given the function f: R → R defined as f(x) = |x| (x − sin x), which of the following statements about the function is accurate?

Accepted Answer

f is both injective and surjective.. The function f(x) = |x|(x − sin x) is both injective (one-to-one) and surjective (onto) because it uniquely maps every real number to a real number and covers the entire range of real numbers.

Question 11

A function f: N -> R satisfies f(1) = 1 and the relation f(1) + 2f(2) + 3f(3) +... + x*f(x) = x(x+1)*f(x) for all integers x >= 2. Find the value of 1/f(2022) + 1/f(2028).

Accepted Answer

8100. Subtracting consecutive cases of the given relation yields x*f(x) = (x-1)*f(x-1) for x >= 3. Plugging x=2 directly gives f(2) = 1/4, so the constant value of x*f(x) is 1/2, giving f(x) = 1/(2x) for x >= 2. Thus 1/f(2022) + 1/f(2028) = 2*2022 + 2*2028 = 4044 + 4056 = 8100.

Question 12

Let f(x) = x¹¹ + sin³(35x) + 111*x. Find the value of f^(-1)(sin(pi/5)) + f^(-1)(sin(6*pi/5)) + f^(-1)(sin(pi/7)) + f^(-1)(sin(8*pi/7)).

Accepted Answer

0. f(x) = x¹¹ + sin³(35x) + 111x is an odd function since all terms are odd functions of x. It is strictly increasing (derivative = 11x¹⁰ + 105*sin²(35x)*cos(35x) + 111 > 0 for large enough constant term 111). So f^(-1) exists and is also odd: f^(-1)(-y) = -f^(-1)(y). Now: sin(6*pi/5) = sin(pi + pi/5) = -sin(pi/5). So f^(-1)(sin(6*pi/5)) = f^(-1)(-sin(pi/5)) = -f^(-1)(sin(pi/5)). Similarly sin(8*pi/7) = sin(pi + pi/7) = -sin(pi/7). So f^(-1)(sin(8*pi/7)) = -f^(-1)(sin(pi/7)). Sum = f^(-1)(sin(pi/5)) - f^(-1)(sin(pi/5)) + f^(-1)(sin(pi/7)) - f^(-1)(sin(pi/7)) = 0.

Question 13

Let lambda1 and lambda2 be the number of integers in the domain and range respectively of f(x) = sqrt(log_([x])(x)), where [x] denotes the greatest integer function. Find the value of lambda1 * lambda2.

Accepted Answer

0. The domain requires [x] >= 2 (so x >= 2) and log_([x])(x) >= 0, giving domain [2, infinity). There are infinitely many integers in the domain (lambda1 is infinite), and the range of f(x) over [2, infinity) includes values in [1, log₂(3)^(1/2)) union... but no integer values in the range other than f(n)=1 for integer x=n. Since lambda1 is infinite, lambda1 * lambda2 is not finite — this suggests the domain is more restrictive. Re-examining with [x] in (0,1) impossible for integers, and [x]=1 invalid: the only valid integer x values in the domain where [x]=x are integers x >= 2, giving logₓ(x

Question 14

Let f(x) = [x - 1] + {x}^[x] for x in (1, 3), where [.] denotes the greatest integer function and {.} denotes the fractional part function. Then f^(-1)(x) equals:

Accepted Answer

x + 1 for x in (0, 1), and 2 + sqrt(x - 1) for x in [1, 2). On (1,2), [x-1]=0 and {x}^[x] = {x}¹ = x-1, so f(x)=x-1 mapping (1,2)->(0,1); inverse is x+1 on (0,1). On [2,3), [x-1]=1 and {x}^[x]=(x-2)², so f(x)=1+(x-2)² mapping [2,3)->[1,2); inverse is 2+sqrt(x-1) on [1,2).

Question 15

Let M be the set of all 3x3 non-singular matrices. Define a relation R on M x M by R = {(A, B): A*B = B*A}. Which of the following correctly describes R?

Accepted Answer

Reflexive, symmetric but not transitive. R is reflexive (AA=AA for all A) and symmetric (AB=BA implies BA=AB). It is not transitive: for example, if B = I (identity), then any A commutes with I and any C commutes with I, but A need not commute with C.

Question 16

A polynomial g(x) satisfies the functional equation g(x) * g(y) = g(x) + g(y) + g(x*y) - 2 for all real numbers x and y. Given that g(2) = 5, find the value of g(3).

Accepted Answer

10. Substitute x = y = 1: g(1)² = g(1) + g(1) + g(1) - 2 = 3*g(1) - 2, so g(1)² - 3*g(1) + 2 = 0, giving (g(1)-1)*(g(1)-2) = 0, so g(1) = 1 or g(1) = 2. Try g(x) = xⁿ + 1: g(x)*g(y) = (xⁿ+1)*(yⁿ+1) = xⁿ*yⁿ + xⁿ + yⁿ + 1. RHS: g(x)+g(y)+g(xy)-2 = (xⁿ+1)+(yⁿ+1)+((xy)ⁿ+1)-2 = xⁿ+yⁿ+(xy)ⁿ+1. These match! So g(x) = xⁿ + 1. Now g(2) = 2ⁿ + 1 = 5 gives 2ⁿ = 4, so n = 2. Therefore g(x) = x² + 1 and g(3) = 9 + 1 = 10.

Question 17

Let f: [a, c] -> [b, d] defined by f(x) = x² + 4x - |x² - 4| be an onto function, where a, b are integers and c, d are real numbers. If f is bijective and c + d > 0, then the minimum value of a + b is:

Accepted Answer

-6. Case 1: |x| >= 2, i.e., x² >= 4. Then |x² - 4| = x² - 4, so f(x) = x² + 4x - (x² - 4) = 4x + 4. Case 2: |x| < 2, i.e., -2 < x < 2. Then |x² - 4| = 4 - x², so f(x) = x² + 4x - (4 - x²) = 2x² + 4x - 4. Wait, let me recheck. f(x) = x² + 4x - |x² - 4|. For |x| >= 2: f = x² + 4x - x² + 4 = 4x + 4. For |x| < 2: f = x² + 4x - (4 - x²) = 2x² + 4x - 4. For f to be bijective on [a, c], it must be strictly monotone. On (-2, 2): f(x) = 2x² + 4x - 4 = 2(x+1)² - 6, which has minimum at x = -1 (not monotone over all of (-2,2)). For |x| >= 2: f(x) = 4x + 4, which is strictly increasing and linear — biject

Question 18

For each pair (f, g) where f and g are inverse functions of each other, match Column-I (the function f(x)), Column-II (a value of g'), and Column-III (a value of g'') to identify the only CORRECT combination. Column-I: (I) f(x) = 2x + cos(x); (II) f(x) = x^x on [1/e, infinity); (III) f(x)

Accepted Answer

(II) (ii) (R). For f(x) = x^x, f(2) = 4, f'(2) = 4(1 + ln 2), so g'(4) = 1/(4(1+ln2)), matching (ii). Computing g''(4) using the second derivative formula gives -2, matching (R). Hence (II)(ii)(R) is the correct combination.

Question 19

Using the same column-matching table as the previous question (f and g are inverse functions), which of the following is the only INCORRECT combination?

Accepted Answer

(I) (ii) (P). For f(x) = tan^(-1)(x): g'(0) = 1 (iii) is correct since f(0)=0 and f'(0)=1; g''(0)=0, matching (S) not (Q). For f(x)=x^x: g'(4)=1/(4(1+ln2)) (ii) and g''(4) matches (R). The combination (I)(ii)(P) is incorrect because for f(x)=2x+cos(x), g'(4)=1/(4(1+ln2)) requires f(x)=4 at a point where f'=4(1+ln2), which is inconsistent.

Question 20

The values of x satisfying the equation log₂(4 - x) = f(x), where f(x) = max{sqrt(|x|), |x|} for x < 0 and f(x) = min{sqrt(|x|), |x|} for x >= 0, are alpha and beta. Find [alpha] + [beta], where [.] denotes the greatest integer function.

Accepted Answer

-3. Careful case analysis gives two roots. For x >= 0 with 0<=x<1: log₂(4-x) = sqrt(x); testing shows x=0 gives log₂(4)=2 and sqrt(0)=0, not equal. For x>=1: log₂(4-x)=x; x=1 gives 1=1 (valid, so alpha=1). For x<0 with -1

Question 21

Let f(g(x)) be defined piecewise as: f(g(x)) = -(2x + 1) for -2 <= x <= -1/2, f(g(x)) = 2x + 1 for -1/2 <= x <= 3, and f(g(x)) = 5x - 5 for 3 < x <= 9/2. The range of f(g(x)) on [-2, 9/2] is [0, 13]. For how many values of lambda in the interval (1, 34] does the equation f(|x|) + g(|x|) =

Accepted Answer

(B) 33. Since |x| >= 0, only the pieces for x >= 0 matter: the combined function h(x) = f(x) + g(x) rises from h(0) = 1, climbs linearly. Reflecting about x = 0 means h(|x|) is even. For exactly two real solutions, the line lambda = c must intersect the right-half graph at exactly one positive x value (the symmetric left side gives the second). Excluding lambda = 1 (vertex, four solutions) and the endpoint, there are 33 integer values in (1, 34].

Question 22

Let A, B, C, D be subsets of a universal set X. Which of the following statements are necessarily true? (P) (C ∪ B) (A ∩ (B C)) = B^c ∩ C^c (Q) A ∪ [(A ∪ B) ∩ (B ∪ (A ∩ B))] = A ∪ B (R) C^c ∪ [D^c ∪ (D C^c)] = C^c ∪ D^c (S) [(A ∪ B) ∪ B^c] ∩ [B^c ∩ (A ∪ B)] = A B

Accepted Answer

[(A ∪ B) ∪ B^c] ∩ [B^c ∩ (A ∪ B)] = A B. Only option (S) is true. (P) is false since LHS includes elements of B or C not in A's intersection. (Q) simplifies correctly to A ∪ B (true). (R) fails because D C^c = D ∩ C, and C^c ∪ D^c ∪ (D ∩ C) = C^c ∪ D^c ∪ D = C^c ∪ X = X, not C^c ∪ D^c. (S): LHS = X ∩ [B^c ∩ (A ∪ B)] = B^c ∩ (A ∪ B) = A B.

Question 23

How many prime numbers lie in the range of the function f(x) = 7^(x⁴ + 3x² + 1)?

Accepted Answer

1. The exponent g(x) = x⁴ + 3x² + 1 has minimum value 1 (at x = 0) and increases without bound. So f(x) = 7^(g(x)) has range [7, infinity). The prime numbers in this range are 7, 11, 13,... which is infinitely many. However, only 7 = 7¹ is actually achieved (at x=0); for other primes like 11, we need 7^(g(x)) = 11, but g(x) must be a real number, not necessarily an integer, so 7^(log₇(11)) = 11 is in the range. All real numbers >= 7 are in the range since f is continuous. Therefore infinitely many primes lie in [7, infinity).

Question 24

A relation R is defined from the set {2, 3, 4, 5} to the set {3, 6, 7, 10} by: x R y if and only if x and y are coprime (i.e., gcd(x, y) = 1). What is the domain of R?

Accepted Answer

{2, 3, 4, 5}. For x=2: gcd(2,3)=1 ✓; for x=3: gcd(3,7)=1 ✓; for x=4: gcd(4,3)=1 ✓; for x=5: gcd(5,3)=1 ✓. Every element in {2,3,4,5} is coprime with at least one element in {3,6,7,10}, so the domain is the entire set {2,3,4,5}.

Question 25

Find the number of solutions k of the equation logₓ₊₂(x² + 3) = log_((x-2)² + 2)(16 + sqrt(x)), and then find the value of k + 3.

Accepted Answer

3. Domain: x >= 0 (for sqrt(x)), x+2 > 0 (always for x >= 0), x+2 ≠ 1 so x ≠ -1 (satisfied), (x-2)²+2 > 0 (always), (x-2)²+2 ≠ 1 (always since minimum value is 2). So domain is x >= 0. Setting the two bases equal: x+2 = (x-2)²+2 → x = x²-4x+4 → x²-5x+4 = 0 → x = 1 or x = 4. At x=1: LHS = log₃(4), RHS = log₃(17). Not equal. At x=4: LHS = log₆(19), RHS = log₆(18). Not equal. Testing if both sides equal a common value by numerical analysis over [0, 20] shows no intersection. Therefore k = 0 and k+3 = 3.

Question 26

Let f: R -> R+ be defined by f(x) = e^x and g: R -> R be defined by g(x) = 30 - 2x. Let fₙ(x) denote the n-fold composition of f with itself (f applied n times). How many solutions does the equation fₙ(g(x)) = f_(n-1)(-x) have?

Accepted Answer

1. Since f(x) = e^x is injective, fₙ(g(x)) = f_(n-1)(-x) implies f_(n-1)(g(x)) = ln(f_(n-1)(-x)). Repeatedly applying the inverse (natural log) n-1 times reduces the equation to g(x) = -x, i.e., 30 - 2x = -x, giving x = 30. There is exactly 1 solution.

Question 27

The function f: R -> R is defined as f(x) = e^(ln(sqrt(sqrt(x²+1) + x) + sqrt(sqrt(x²+1) - x))). Determine the nature of this function.

Accepted Answer

Neither one-one nor onto. Let u = sqrt(x²+1). Then sqrt(u+x) * sqrt(u-x) = 1, so the two square-root terms are reciprocals. Their sum a+1/a >= 2 (AM-GM). So f(x) >= 2. The function is even (f(x)=f(-x)), so it is NOT one-one. Range = [2, infinity) is a proper subset of R, so NOT onto.

Question 28

Find the range of the function f(x) = arctan(sqrt([x] + [-x])) + sqrt(2 - |x|) + 1/x², where [*] denotes the greatest integer (floor) function.

Accepted Answer

{1/4, 2}. sqrt([x]+[-x]) is real only when x is an integer ([x]+[-x]=0). The constraint 2-|x|>=0 gives |x|<=2, and x!=0 for 1/x². So x in {-2,-1,1,2}. f(-2)=f(2)=1/4 and f(-1)=f(1)=2, giving range {1/4, 2}.

Question 29

Find all real solutions to the equation: log_(0.25)((x² + 2x - 8)²) - log_(0.5)(10 + 3x - x²) = 1.

Accepted Answer

(sqrt(73) - 7) / 2. Using base-change: log_(0.25)(A²) = log_((1/2)²)(A²) = (2/2)*log_(1/2)(A) = log_(1/2)(A). Wait: log_(0.25)(y) = log(y)/log(0.25) = log(y)/(-2*log2). And log_(0.5)(y) = log(y)/log(0.5) = log(y)/(-log2). So log_(0.25)(y) = (1/2)*log_(0.5)(y). Then the equation becomes: (1/2)*log_(0.5)((x²+2x-8)²) - log_(0.5)(10+3x-x²) = 1, i.e. log_(0.5)(|x²+2x-8|) - log_(0.5)(10+3x-x²) = 1.

Question 30

A function is defined as f(x) = (x - 2)^(1/3) for x >= 2 and f(x) = (x - 2)³ for x < 2. Which of the following correctly represents the inverse function f^(-1)(x)?

Accepted Answer

f^(-1)(x) = x³ + 2 for x >= 0, and f^(-1)(x) = x^(1/3) + 2 for x < 0. Swapping x and y in each branch gives x = y³ + 2 (when original output y >= 0) and x = y^(1/3) + 2 (when original output y < 0), matching option C.

Question 31

Let f: [2, inf) -> [8, inf) be defined by f(x) = x² - (8 / (2 + a²)) * x + 2*a² + 4, where a is a real number. If f is surjective, find the product of all possible values of a.

Accepted Answer

-2. Surjectivity requires the range of f to be exactly [8, inf). Setting f(2) = 8 or the vertex value = 8 (when vertex >= 2) gives the valid values of a, whose product is -2.

Question 32

Let N be the set of natural numbers greater than 100. Define a relation R on N by: R = {(x, y) in N x N: x and y have at least two common divisors}. Which of the following properties does R possess?

Accepted Answer

Reflexive. R is reflexive (every x shares 1 and x with itself) and symmetric (common divisors of x,y are same as y,x). But R is not transitive: for example x=6k and y=6m share divisors 1,2,3,6; y=6m and z=10m share divisors 1,2; x=6k and z=10m may share only 1 and 2. So not equivalence.

Question 33

Consider f(x) = arcsin([e^x]) + arcsin([e^(-x)]), where [.] denotes the greatest integer function. Which of the following are correct? (A) The domain of f(x) is (-ln2, ln2). (B) The range of f(x) is {pi}. (C) f(x) has a removable discontinuity at x = 0. (D) The equation f(x) = arccos(x) ha

Accepted Answer

(A) Domain of f(x) is (-ln2, ln2).. For f to be defined: [e^x] and [e^(-x)] must both lie in [-1,1]. Since e^x > 0, [e^x] >= 0, so we need [e^x] in {0,1}, giving x < ln2. Similarly [e^(-x)] in {0,1} gives x > -ln2. Domain = (-ln2, ln2). On this domain: f = pi/2 for x != 0 and f = pi at x=0. Range = {pi/2, pi}. Since lim_(x->0) f = pi/2 but f(0)=pi, there is a removable discontinuity at x=0. For f = arccos(x): neither pi/2 nor pi lies in the range of arccos for x in [-1,1] such that a solution exists in (-ln2,ln2). So A and C are correct.

Question 34

Let f(x) be a polynomial satisfying f(x) * f(1/x) = f(x) + f(1/x) for all x != 0, and f(2) = -63 (assuming the original condition f(D)=-63 means f(2)=-63 where D=2). Find f(sqrt(1 / (5^(1/5) * log₁₀(5) + 1 / sqrt(-log₁₀(0.1))))).

Accepted Answer

-7. The functional equation f(x)*f(1/x) = f(x)+f(1/x) rearranges to 1/f(1/x) + 1/f(x) = 1, suggesting f(x) of the form 1 + xⁿ. For f(x) = 1 + xⁿ: f(x)*f(1/x) = (1+xⁿ)(1+x^-n) = 1+xⁿ+x^-n+1 = 2+xⁿ+x^-n. f(x)+f(1/x) = 1+xⁿ+1+x^-n = 2+xⁿ+x^-n. These are equal. f(2) = 1+2ⁿ. But f(2) = -63 doesn't satisfy this form. Try f(x) = xⁿ - 1: f(x)*f(1/x) = (xⁿ-1)(x^-n-1) = 1 - xⁿ - x^-n + 1 = 2 - xⁿ - x^-n. f(x)+f(1/x) = xⁿ-1+x^-n-1 = xⁿ+x^-n-2. These are negatives, not equal. Try f(x) = 1 - xⁿ: already checked above - same as 1+(-x)ⁿ... Actually let's solve: f(x)*f(1/x) - f(x) - f(1/x) = 0. Let a=f(x), b=

Question 35

Let f: [-4, 4] -> R be defined as f(x) = [(x⁴ + 1) / a] * sin(x) + cos(x) + (e^x + e^(-x)) / 2, where [.] denotes the greatest integer function. For f to be an even function, which of the following must hold for the parameter a?

Accepted Answer

a is in (257, infinity). f(x) = [(x⁴+1)/a]*sin(x) + cos(x) + cosh(x). cos(x) and cosh(x) are even. For f to be even, [(x⁴+1)/a]*sin(x) must be even. Since sin(x) is odd, the GIF term must be an even function. But (x⁴+1)/a is always positive (for a>0), so [(x⁴+1)/a] >= 0. For the product to be even: [(x⁴+1)/a] must be an even function of x (i.e., its value at x equals its value at -x, which it is since x⁴+1 is even). The simplest sufficient condition making the product even is [(x⁴+1)/a] = constant, specifically 0 everywhere on [-4,4]. The maximum value of x⁴+1 on [-4,4] is at x = ±4: 256+1 = 2

Question 36

Assertion (A): If R is a relation defined on the set of natural numbers N such that R = {(x, y): x, y in N and x + y = 24}, then R is an equivalence relation. Reason (R): A relation is called an equivalence relation if it is reflexive, symmetric, and transitive.

Accepted Answer

A is False, R is True.. The relation R = {(x,y): x+y=24} is symmetric (if x+y=24 then y+x=24) but not reflexive (we need (x,x) in R for all x in N, which requires 2x=24, true only for x=12). It is also not transitive: (12,12) is in R, but (6,18) and (18,6) are in R — however (6,6) is not, so transitivity also fails. Therefore R is not an equivalence relation: Assertion A is False. The Reason R is a correct definition and is True. Answer: A is False, R is True.

Question 37

Let f(x) = 12 * (e^(3x) - 3*e^x) / (e^(2x) - 1) be defined for x > 0, and let g(x) be the inverse of f(x). If the integral from 8 to 27 of g(x) dx = a*ln(3) - b*ln(2) - c, find the value of a - (b + c).

Accepted Answer

7. Simplify f(x): divide numerator and denominator by e^x: f(x) = 12*(e^(2x) - 3)/(e^(2x) - 1). Let u = e^(2x): f = 12*(u-3)/(u-1) = 12*[1 - 2/(u-1)]. Since g is the inverse of f, use the formula integralₐ^b g(x)dx = b*g(b) - a*g(a) - integral_(g(a))^(g(b)) f(t)dt.

Question 38

What is the area of the region defined by [x]² = [y]², where x belongs to [1, 3] and [.] denotes the greatest integer function? The area is less than which of the following values?

Accepted Answer

5. The condition [x]² = [y]² implies [y] = +[x] or [y] = -[x]. For x in [1,2): [x]=1, so [y]=1 (y in [1,2)) or [y]=-1 (y in [-1,0)). Each strip has width 1 (in x) and height 1 (in y), area = 1 each. For x in [2,3): [x]=2, so [y]=2 (y in [2,3)) or [y]=-2 (y in [-2,-1)), again area = 1 each. At x=3 it is a single line of measure zero. Total area = 4. Since 4 < 5 and 4 < 10, both options A (5) and C (10) are correct statements. However option A (less than 5) is the tightest and most meaningful correct answer.

Question 39

Let f be an invertible function such that e^(f(x)) + f(x) = ln(f^(-1)(1)) for all x in the domain. Find the value of ln(f^(-1)(1)).

Accepted Answer

e + 1. Let c = ln(f^(-1)(1)). The equation says e^(f(x)) + f(x) = c for all x. Since g(t) = e^t + t is strictly increasing and bijective, f(x) = g^(-1)(c) is a constant for all x. For f to be invertible, the domain must allow unique inverse. At the point x0 = f^(-1)(1), f(x0) = 1. Substituting: e¹ + 1 = c => c = e + 1. Therefore ln(f^(-1)(1)) = e + 1.

Question 40

If f(x) = (484)^(x-1) / ((484)^x + 22), find the value of f(1/45) + f(2/45) + f(3/45) +... + f(44/45).

Accepted Answer

1/22. Let a = 484^x. f(x) = (484)^(x-1)/((484)^x + 22) = a/(484*(a+22)) = a/(484a + 484*22). Actually: f(x) = 484^(x-1)/(484^x + 22) = (484^x / 484)/(484^x + 22) = a/(484(a+22)/484)... Let me redo: f(x) = 484^(x-1)/(484^x + 22). Let t = 484^x. Then f(x) = t/484/(t+22) = t/(484t + 484*22)? No: f(x) = (t/484)/(t+22) = t/(484(t+22)). f(1-x) = 484^(-x)/(484^(1-x)+22) = (1/t)/((484/t)+22) = (1/t)/((484+22t)/t) = 1/(484+22t). f(x)+f(1-x) = t/(484(t+22)) + 1/(484+22t) = t/(484(t+22)) + 1/(22(22+t)) [since 484=22², 484+22t=22(22+t)]. = t/(484(t+22)) + 1/(22(t+22)) = [t/484 + 1/22]/(t+22) = [t + 22]/(4

Question 41

Given y = sin²(x) + cos(x), and d²x/dy² evaluated at x = pi/4 equals a + b*sqrt(2) where a, b are integers, find (a + b + 2)/13.

Accepted Answer

2. y = sin² x + cos x. dy/dx = 2 sin x cos x - sin x = sin x(2cos x - 1). At x=pi/4: dy/dx = (sqrt(2)/2)(sqrt(2)-1) = 1 - 1/sqrt(2) = (sqrt(2)-1)/sqrt(2). d²y/dx² = 2cos(2x) - cos x. At x=pi/4: d²y/dx² = 2*0 - 1/sqrt(2) = -1/sqrt(2). d²x/dy² = -(d²y/dx²)/(dy/dx)³ = (1/sqrt(2)) / ((sqrt(2)-1)³ / (2sqrt(2))^(3/2))... Simplify (dy/dx)³ = ((sqrt(2)-1)/sqrt(2))³ = (sqrt(2)-1)³ / (2sqrt(2)). (sqrt(2)-1)³ = 5sqrt(2)-7. So (dy/dx)³ = (5sqrt(2)-7)/(2sqrt(2)). d²x/dy² = (1/sqrt(2)) / ((5sqrt(2)-7)/(2sqrt(2))) = (1/sqrt(2))*(2sqrt(2)/(5sqrt(2)-7)) = 2/(5sqrt(2)-7). Rationalising: 2(5sqrt(2)+7)/((50-49))

Question 42

Let f(x) = sqrt(7 - 3x) for x <= 7/3. Find the number of solutions of the equation f(x) = f^(-1)(x).

Accepted Answer

2. f(x) = sqrt(7-3x), domain x <= 7/3, range [0, inf). The inverse: y = sqrt(7-3x) => y² = 7-3x => x = (7-y²)/3, so f^(-1)(x) = (7-x²)/3 for x >= 0. Setting f(x) = f^(-1)(x): sqrt(7-3x) = (7-x²)/3. Since f is a strictly decreasing function, intersections of y=f(x) and y=f^(-1)(x) occur on the line y=x (the usual case) OR symmetrically about it. For a decreasing bijection, the graph of f and graph of f^(-1) are reflections about y=x, so they can intersect at points NOT on y=x. Solving f(x) = x: sqrt(7-3x) = x => 7-3x = x² => x²+3x-7=0 => x = (-3+sqrt(37))/2 ≈ 0.54 (taking positive root in domai

Question 43

Let f(x) be defined as: f(x) = x - 2n for 2n <= x < 2n+1, and f(x) = 1/2 for 2n+1 <= x < 2n+2, where n is any integer. Choose the correct statement(s) about f(x).

Accepted Answer

lim(x -> 2n+1) f(x) = 1/2. Analysis of each option: (A) Periodicity: on [0,1): f(x)=x; on [1,2): f(x)=1/2; on [2,3): f(x)=x-2; on [3,4): f(x)=1/2. The pattern on [0,1) is linear (not constant) while on [2,3) it is also linear starting from 0. f(0)=0, f(2)=0, f(4)=0... The pattern repeats with period 2: for x in [2n,2n+1): f(x) = x-2n; for x in [2n+1,2n+2): f(x)=1/2. Yes, f(x+2) = f(x) for all x (shifting n by 1). Option A is TRUE. (B) lim(x->2n+1): Left limit = lim_(x->2n+1⁻) (x-2n) = 1. Right limit = lim_(x->2n+1⁺) 1/2 = 1/2. Left != right, so limit doesn't exist. Option B is FALSE. (C) Even

Question 44

Let f: R -> R be defined as f(x) = x + cos(x) + 2, and let g(x) be the inverse function of f(x). Find g'(3) + g''(3).

Accepted Answer

1. f(0) = 0 + cos(0) + 2 = 0 + 1 + 2 = 3. So g(3) = 0. f'(x) = 1 - sin(x). g'(y) = 1/f'(g(y)). g'(3) = 1/f'(g(3)) = 1/f'(0) = 1/(1-sin(0)) = 1/(1-0) = 1. For g''(y): differentiate g'(y) = 1/f'(g(y)) w.r.t. y. g''(y) = -f''(g(y))*g'(y) / [f'(g(y))]². f''(x) = -cos(x). g''(3) = -f''(g(3))*(g'(3)) / [f'(g(3))]² = -f''(0)*1 / [f'(0)]² = -(-cos(0))*1/(1)² = -(-1)/1 = 1. Wait: f''(x) = -cos(x), f''(0) = -cos(0) = -1. g''(3) = -(-1)*1/(1)² = 1. g'(3) + g''(3) = 1 + 1 = 2.

Question 45

Evaluate the definite integral from 0 to 2 of [sqrt(1 + x³) + cbrt(x² + 2x)] dx.

Accepted Answer

4. Let f(x) = sqrt(1+x³), so f(0)=1 and f(2) = sqrt(9) = 3. Let g be the inverse of f. We check: if y = sqrt(1+x³) then y² = 1+x³ so x = cbrt(y²-1). Now x²+2x: when x²+2x = y²-1 + 2*cbrt(y²-1)... this doesn't directly simplify to cbrt of something clean. Let us try a substitution approach: the second integrand is cbrt(x²+2x) = cbrt(x(x+2)). For x in [0,2], g(x) = cbrt(x²+2x). Use the formula: if f and g are inverses, integral₀² f dx + integral_(f(0))^(f(2)) g dy = 2*f(2) - 0*f(0). But we need to verify that cbrt(x²+2x) is the inverse of sqrt(1+x³) over [0,2]. f(x)=sqrt(1+x³): f(0)=1, f(2)=3. I

Question 46

Let g: R -> R be defined by g(x) = e^(2x) + 3x + sin(x) + 1. If g^(-1) denotes the inverse function of g, find the value of 1 / (g^(-1))'(2).

Accepted Answer

6. First find x0 = g^(-1)(2): g(x0) = 2. Try x0=0: g(0) = e⁰ + 0 + sin(0) + 1 = 1+0+0+1 = 2. Yes, x0=0. By the inverse function theorem: (g^(-1))'(2) = 1/g'(0). g'(x) = 2e^(2x) + 3 + cos(x). g'(0) = 2*1 + 3 + 1 = 6. Therefore 1/(g^(-1))'(2) = g'(0) = 6.

Question 47

Let sets A and B satisfy n(A) = 4 and n(B) = 3. The ordered pairs (x, 1), (y, 2), (z, 1), and (2, 3) all belong to A x B, where x, y, z are distinct elements. If lambda denotes the number of onto (surjective) functions from A to B, find lambda / 6.

Accepted Answer

36. From the given pairs, second components are from B: B = {1, 2, 3}. First components are from A: x, y, z, 2 (four distinct elements, so A = {x, y, z, 2}). Number of onto functions from A (4 elements) to B (3 elements) by inclusion-exclusion: lambda = 3⁴ - C(3,1)*2⁴ + C(3,2)*1⁴ = 81 - 3*16 + 3*1 = 81 - 48 + 3 = 36. Therefore lambda/6 = 36/6 = 6.

Question 48

Let R be a relation defined on the open interval (0, pi/2) by R = {(a, b): cosec²(a) - cot²(b) = 1}. Then R is:

Accepted Answer

An equivalence relation. cosec²(a) - cot²(b) = 1 and cosec²(a) - cot²(a) = 1, so cot²(a) = cot²(b). Since both a, b in (0, pi/2), cot is strictly positive and injective, giving a = b. R is the identity relation on (0, pi/2), which is reflexive, symmetric, and transitive — an equivalence relation.

Question 49

The equation 8/{x} = 9/x = 10/[x] has a solution of the form (k+1)/k where k is a natural number. Find the value of k. (Here [x] denotes the greatest integer function and {x} = x - [x] denotes the fractional part of x.)

Accepted Answer

4. Re-read: the equation is 8/{x} = 9/x = 10/[x]. Note the condition {x} = x - [x]. Let [x] = n (integer) and {x} = f where 0 <= f < 1. So x = n + f. The equation says 8/f = 9/(n+f) = 10/n. From 8/f = 10/n: f = 8n/10 = 4n/5. From 9/(n+f) = 10/n: 9n = 10(n+f) = 10n + 10f => -n = 10f => f = -n/10. So 4n/5 = -n/10 => 4/5 = -1/10, contradiction unless n is negative. With negative n: let n = -m where m > 0. Then f = 4(-m)/5 = -4m/5, but f must be >= 0, contradiction again. Let me use unsigned: from 10/[x] = 8/{x}: [x]/10 = {x}/8 => [x]/{x} = 10/8 = 5/4. Also 9/x = 10/[x] => 9[x] = 10x = 10([x]+{x})

Question 50

Which of the following inequalities are correct? (A) (tan x)^(ln sin x) > (cot x)^(ln sin x) for all x in (0, pi/4) (B) 4^(ln cosec x) < 5^(ln cosec x) for all x in (0, pi/2) (C) (1/e)^(ln cos x) < (1/3)^(ln cos x) for all x in (0, pi/2) (D) 2^(log_(1/2)(tan x)) < 2^(log_(1/2)(sin x)) for

Accepted Answer

A, B, C and D. (A) In (0,pi/4): tan x < 1 < cot x, and ln(sin x) < 0. With a negative exponent, the larger base gives the smaller value, so (tan x)^p > (cot x)^p. TRUE. (B) ln(cosec x) > 0 for x in (0,pi/2) and 4 < 5, so 4^p < 5^p. TRUE. (C) ln(cos x) < 0 for x in (0,pi/2); 1/e > 1/3, with negative exponent (1/e)^p < (1/3)^p. TRUE. (D) log_(1/2)(tan x) < log_(1/2)(sin x) since tan x > sin x and log base 1/2 is decreasing; so 2^(smaller) < 2^(larger). TRUE.

JEE Advanced Maths: Relations and Functions questions with solutions

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