Now, put y = -x in eq. (1)
f(x) + f(-x) = f(0) { f(0) = 0 }
⇒ f(-x) = -f(x)
⇒ f is an odd function
From eq. (2)
f(-nx) = n f(-x)
⇒ f(-nx) = -n f(x)
⇒ f(mx) = m f(x) ∀ m ∈ ℤ⁻ … (3)
From eq. (2) and eq. (3)
f(nx) = n f(x) ∀ n ∈ ℤ … (4)
Now, put x = pq where p, q ∈ ℤ, q ≠ 0
f (npq) = n f (pq) ∀ n ∈ ℤ
Put n = q
f(p) = q f pq
⇒ p f(1) = q f (pq) {from eq. (4)}
Let f(1) = a
Then, p a = q f (pq)
⇒ f (pq) = apq
⇒ f(x) = a x ∀ x ∈ ℚ
Now, f (-35) = a (-35) = 12 ⇒ a = -20
⇒ f(x) = -20x ∀ x ∈ ℚ … (5)
From the given functional equation, it is not possible to find a unique function for irrational values of ‘x’. There are infinitely many such functions satisfying the given functional equation for irrational values of x, but in this problem, we finally need the function at rational values of ‘x’ only. So, for rational values of x, we get a unique function mentioned in (5).
Now, g(x + y) = g(x) ⋅ g(y)
⇒ ln(g(x + y)) = ln(g(x)) + ln(g(y))
Let ln(g(x)) = h(x)
⇒ h(x + y) = h(x) + h(y)
⇒ h(x) = k x ∀ x ∈ ℚ
⇒ g(x) = ekx ∀ x ∈ ℚ … (6)
And g (-13) = e -k3 = 2 ⇒ K = -3 ln(2)
⇒ K = ln (18)
⇒ g(x) = e(ln(1/8) . x) = 18x = 2-3x∀ x ∈ ℚ
Now, f (14) = -5, g(-2) = 2⁶ = 64
g(0) = 1
So, (f 14 + g(-2) - 8 g(0))
= (-5 + 64 - 8)(1) = 51
Q2: Let the function f : ℝ → ℝ be defined by
.
Then the number of solutions of f(x) = 0 in ℝ is ______. [JEE Advanced 2024 Paper 2]
Ans: 1
f(x) m= x2023 + 2024x + 2025eπx (x² - x + 3) (sin x + 2)
∴ (sin x + 2) is never zero.
∴ For x2023 + 2024x + 20251 = 0
Let ϕ(x) = x2023 + 2024x + 2025.
ϕ'(x) = 2023x2022 + 2024 > 0 ∀ x ∈ ℝ
∴ ϕ(x) is a strictly increasing function.
∴ ϕ(x) = 0 for exactly one value of x.
∴ f(x) = 0 has one solution.
Let domain and co-domain of a function y = f(x) are S and T respectively.
(A) There are infinitely many elements in domain and four elements in co-domain.
⇒ There are infinitely many functions from S to T.
⇒ Option (A) is correct
(B) If number of elements in domain is greater than number of elements in co-domain, then number of strictly increasing function is zero.
⇒ Option (B) is incorrect
(C) Maximum number of continuous functions = 4 × 4 × 4 = 64
(Every subset (0, 1),(1, 2),(3, 4) has four choices)
∵ 64<120⇒ option (C) is correct.
(D) For every point at which f(x) is continuous, f(x) = 0
⇒ Every continuous function from S to T is differentiable.
Option (D) is correct.
Q2: Let f : [0, 1] → [0, 1] be the function defined by . Consider the square region S = [0, 1] × [0, 1]. Let
be called the green region and
be called the red region. Let
be the horizontal line drawn at a height ℎ ∈ [0, 1]. Then which of the following statements is(are) true?
(a) There exists an ℎ ∈ [1/4, 2/3] such that the area of the green region above the line Lℎ equals the area of the green region below the line Lℎ
(b) There exists an ℎ ∈ [1/4, 2/3] such that the area of the red region above the line Lℎ equals the area of the red region below the line Lℎ
(c) There exists an ℎ ∈ [1/4, 2/3] such that the area of the green region above the line Lℎ equals the area of the red region below the line Lℎ
(d) There exists an ℎ ∈ [1/4, 2/3] such that the area of the red region above the line Lℎ equals the area of the green region below the line [JEE Advanced 2023 Paper 1]
Ans: (b), (c) & (d)
Given,
option (A) is incorrect
option (B) is correct.
⇒ h= 1/2 ⇒ option (C) is correct.
(D) ∵ Option (C) is correct ⇒ option (D) is also correct.
Q1: Let |M| denote the determinant of a square matrix M. Let be the function defined by
where
Let p(x) be a quadratic polynomial whose roots are the maximum and minimum values of the function g(θ), and p(2) =2 − √2. Then, which of the following is/are TRUE ?
(a)
(b)
(c)
(d) [JEE Advanced 2022 Paper 1]Ans: (a) & (c)
Given,
Here,
and
and
Also,
(skew symmetric)
For option (A) Correct.
For option (B) Incorrect.
For option (C) Correct.
For option (D) Incorrect.
Then the value of
is ..........
Ans: 19
The given function f : [0, 1] → R be define by
So, f(x) + f(1 − x) = 1 .....(i)
=
= (19 times) {from Eq. (i)}
= 19.
Q2: Let the function be defined by
Suppose the function f has a local minimum at θ precisely when
, where
. Then the value of
is ............. [JEE Advanced 2020 Paper 2]
Ans: 0.5
The given function f : R → R be defined by
The local minimum of function 'f' occurs when
but
Where,
So, = 0.50
Q3: Let f : [0, 2] → R be the function defined by
If are such that
, then the value of β - α is .......... [JEE Advanced 2020 Paper 1]
Ans: 1
The given function f : [0, 2] → R defined by
As, so,
Therefore the value of (β - α) = 1
Q3: For a polynomial g(x) with real coefficients, let mg denote the number of distinct real roots of g(x). Suppose S is the set of polynomials with real coefficients defined byFor a polynomial f, let f' and f'' denote its first and second order derivatives, respectively. Then the minimum possible value of (mf' + mf''), where f ∈ S, is .............. [JEE Advanced 2020 Paper 1]
Ans: 5
Given set S of polynomials with real coefficients
and for a polynomial f ∈ S, Let
it have −1 and 1 as repeated roots twice, so graph of f(x) touches the X-axis at x = −1 and x = 1, so f'(x) having at least three roots x = −1, 1 and α. Where α ∈ (−1, 1) and f''(x) having at least two roots in interval (−1, 1)
So, mf' = 3 and mf'' = 2
∴ Minimum possible value of (mf' + mf'') = 5
Q4: If the function f : R → R is defined by f(x) = |x| (x − sin x), then which of the following statements is TRUE?
(a) f is one-one, but NOT onto
(b) f is onto, but NOT one-one
(c) f is BOTH one-one and onto
(d) f is NEITHER one-one NOR onto [JEE Advanced 2020 Paper 1]
Ans: (c)
The given function f : R → R is
The function 'f' is a odd and continuous function and as , so range is R, therefore, 'f' is a onto function.
⇒ f is strictly increasing function. ∀x ∈(0, ∞).
Similarly, for x < 0, −x + sin x > 0 and (− x) (1 − cos x) > 0, therefore, f′(x)> 0∀ x ∈(−∞, 0)
⇒ f is strictly increasing function, ∀x ∈ (0, ∞)
Therefore 'f' is a strictly increasing function for x ∈ R and it implies that f is one-one function.
Q1: Let X be a set with exactly 5 elements and Y be a set with exactly 7 elements. If α is the number of one-one functions from X to Y and β is the number of onto functions from Y to X, then the value of 1 / 5!(β − α) is .................. [JEE Advanced 2018 Paper 2]
Ans: 119
Given, X has exactly 5 elements and Y has exactly 7 elements.
∴ n(X) = 5
and n(Y) = 7
Now, number of one-one functions from X to Y isNumber of onto functions from Y to X is β
1, 1, 1, 1, 3 or 1, 1, 1, 2, 2
= 4 x 35 - 21= 140 - 21
= 119
Q2: Let and
(Here, the inverse trigonometric function sin−1 x assumes values in [−π/2, π/2].).
Let f : E1 → R be the function defined by and g : E2 → R be the function defined by
. [JEE Advanced 2018 Paper 2]
and
So,
The domain of f and g are
and Range of
Range of f is R − {0} or (−∞, 0) ∪ (0, ∞)
Range of g is
Now, P → 4, Q → 2, R → 1, S → 1
Hence, option (a) is correct answer.
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