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JEE Main Mathematics Sets Relations and Functions Previously Asked Questions
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Question
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Answer
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Question 1:
If A = [(x, y) : x2 + y2 = 25] and B = [(x, y) : x2 + 9y2 = 144], then A ∩ B contains _______ points.
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A = Set of all values (x, y) : x2 + y2 = 25 = 52
B = [x2 / 144] + [y2 / 16] = 1
i.e., [x2 / (12)2] + [y2 / (4)2] = 1.
Clearly, A ∩ B consists of four points.
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Question 2:
In a college of 300 students, every student reads 5 newspapers, and every newspaper is read by 60 students. The number of newspapers is ________.
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Solution:
Let the number of newspapers be x.
If every student reads one newspaper, the number of students would be x (60) = 60x
Since every student reads 5 newspapers, the number of students = [x * 60] / [5] = 300
x = 25
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Question 3:
Let R be the relation on the set R of all real numbers defined by a R b if and only if |a − b| ≤ 1. Then R is __________.
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a − a| = 0 < 1
Therefore, a R a
∀
a
∈
R
Therefore, R is reflexive.
Again a R b, |a − b| ≤ 1
⇒
|b − a| ≤ 1
⇒
b R a
Therefore, R is symmetric.
Again 1 R [½] and [½] R1 but [½] ≠ 1
Therefore, R is not anti-symmetric.
Further, 1 R 2 and 2 R 3, but 1 R 3 is not possible, [Because, |1 − 3| = 2 > 1]
Hence, R is not transitive.
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Question 4:
Let a relation R be defined by R = {(4, 5); (1, 4); (4, 6); (7, 6); (3, 7)} then R−1 o R is ________.
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First, find R−1.
R−1 = {(5, 4) ; (4, 1) ; (6, 4) ; (6, 7) ; (7, 3)}.
Obtain the elements of R−1 o R.
Pick the element of R and then of R−1.
Since (4, 5)
∈
R and (5, 4)
∈
R−1, we have (4, 4)
∈
R−1 o R
Similarly, (1, 4)
∈
R, (4, 1)
∈
R−1
⇒
(1, 1)
∈
R−1 o R
(4, 6)
∈
R, (6, 4)
∈
R−1
⇒
(4, 4)
∈
R−1 o R,
(4, 6)
∈
R, (6, 7)
∈
R−1
⇒
(4, 7)
∈
R−1 o R
(7, 6)
∈
R, (6, 4)
∈
R−1
⇒
(7, 4)
∈
R−1 o R,
(7, 6)
∈
R, (6, 7)
∈
R−1
⇒
(7, 7)
∈
R−1 o R
(3, 7)
∈
R, (7, 3)
∈
R−1
⇒
(3, 3)
∈
R−1 o R,
Hence, R−1 o R = {(1, 1); (4, 4); (4, 7); (7, 4), (7, 7); (3, 3)}.
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Question 5:
If f (x) = cos (log x), then find the value of f (x) * f (4) − [1 / 2] * [f (x / 4) + f (4x)].
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f (x) = cos (log x)
Let y = f (x) * f (4) − [1 / 2] * [f (x / 4) + f (4x)]
y = cos (log x) * cos (log 4) − [1 / 2] * [cos log (x / 4) + cos (log 4x)]
y = cos (log x) cos (log 4) − [1 / 2] * [cos (log x −log 4) + cos (log x + log 4)]
y = cos (log x) cos (log 4) − [1 / 2] * [2 cos (log x) cos (log 4)]
y = 0
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Question 6: Let f : R → R be defined by f (x) = 2x + |x|, then f (2x) + f (−x) − f (x) = _______.
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f(x) = 2x + |x|
f(2x) = 2(2x) + |2x| = 4x + 2|x|
f(-x) = -2x + |-x| = -2x + |x|
-f(x) = -2x + (-|x|) = -2x – |x|
Hence, f(2x) + f(-x) – f(x) = 4x + 2|x| – 2x + |x| – 2x – |x|
= 2|x|
= 2x; x ≥ 0 and -2x; x < 0
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Question 7: If f (x) = cos [π2] x + cos[−π2] x, where [x] stands for the greatest integer function, then find the function of the right angle.
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f (x) = cos [π2] x + cos[−π2] x
f (x) = cos (9x) + cos (−10x) {since π = 3.14}
= cos (9x) + cos (10x)
= 2 cos (19x / 2) cos (x / 2)
Now, right angle = π/2
So, f (π / 2) = 2 cos (19π / 4) cos (π / 4)
f (π / 2) = 2 *(−1 / √2) * (1/ √2)
= −1
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Question 8: The function f : R → R defined by f (x) = ex is ________.
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Function f: R → R is defined by f (x) = ex.
Let x1, x2 ∈ R and f (x1) = f (x2) or ex1 = ex2 or x1 = x2.
Therefore, f is one-one.
Let f (x) = ex = y.
Taking log on both sides, we get x = log y.
As we know, negative real numbers have no pre-image, or the function is not onto, and zero is not the image of any real number. Therefore, function f is one-one and into.
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Question 9:
If f: R → S defined by f (x) = sin x − √3 cos x + 1 is onto, then what is the interval of S?
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Given,
f (x) = sin x − √3 cos x + 1
As we know, the range of the function f(x) = a cos x + b sin x + c is given by:
c – √(a2 + b2) ≤ f(x) ≤ c + √(a2 + b2)
− √[1 + (√−3)2] ≤ (sin x − √3 cos x) ≤ √[1 + (√−3)2]
−2 ≤ (sin x − √3 cos x) ≤ 2
−2 + 1 ≤ (sin x − √3 cos x + 1) ≤ 2 + 1
−1 ≤ (sin x − √3 cos x + 1) ≤ 3
i.e., range = [−1, 3]
For f to be onto, the interval of S = [−1, 3].
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Question 10: If f (x) = a cos (bx + c) + d, then what is the range of f (x)?
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f (x) = a cos (bx + c) + d ..(i)
As we know, -1 ≤ cos θ ≤ 1
For minimum, cos (bx + c) = −1
From (i), f (x) = −a + d = (d − a)
For maximum, cos (bx + c) = 1
From (i), f (x) = a + d = (d + a)
Range of f (x) = [d − a, d + a]
Alternatively,
-1 ≤ cos(bx + c) ≤1
-a ≤ a cos(bx + c) ≤ a
-a + d ≤ a cos(bx + c) + d ≤ a + d
Range of f(x) = [d – a, a + d]
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Question 11: The function f: R → R is defined by f (x) = cos2 x + sin4x for x ∈ R, then what is f (R)?
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f (x) = cos2 x + sin4x
y = f (x) = cos2 x + sin2x (1 − cos2x)
y = cos2 x + sin2x − sin2x cos2x
y = 1 − sin2x cos2x
y = 1 − [1 / 4] * [sin22x]
3 / 4 ≤ f (x) ≤ 1, (Because 0 ≤ sin22x ≤ 1)
f (R) ∈ [3/4, 1]
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