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JEE Maths Quiz on Sets Relations And Functions
JEE Maths Quiz on Sets Relations And Functions : In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. The Quiz on Sets Relations And Functions is free of cost and will useful in performance and inculcating knowledge.
Quiz on Sets Relations And Functions
Q1. If f (x) = cos (log x), then find the value of f (x) * f (4) − [1 / 2] * [f (x / 4) + f (4x)].
a) 16
b) 4
c) 0
d) 5
c) 0
Solution:
f (x) = cos (log x)
Now 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
Q2. If A = [(x, y) : x^{2} + y^{2} = 25] and B = [(x, y) : x^{2} + 9y^{2} = 144], then A ∩ B contains _______ points.
a) Two
b) Three
c) Four
d) Six
c) Four
Solution:
A = Set of all values (x, y) : x^{2} + y^{2} = 25 = 5^{2}
B = [x^{2} / 144] + [y^{2} / 16] = 1
i.e., [x^{2 }/ (12)^{2}] + [y^{2} / (4)^{2}] = 1.
Clearly, A ∩ B consists of four points.
Q3. Let f : R → R be defined by f (x) = 2x + x, then f (2x) + f (−x) − f (x) = _______.
a) 4x
b) 8x
c) 16x
d) 2x
d) 2x
Solution:
f (2x) = 2 (2x) + 2x = 4x + 2 x,
y = x^{2} + 1,
f (x) = 2x + x
f (2x) + f (−x) − f (x) = 4x + 2 x + x −2x − 2x − x
= 2x
Q4. If f (x) = cos [π^{2}] x + cos[−π^{2}] x, then find the function of the angle.
a) −1
b) 0
c) 1
d) 2
a) −1
Solution:
f (x) = cos [π^{2}] x + cos[−π^{2}] x
f (x) = cos (9x) + cos (−10x)
= cos (9x) + cos (10x)
= 2 cos (19x / 2) cos (x / 2)
f (π / 2) = 2 cos (19π / 4) cos (π / 4);
f (π / 2) = 2 * −1 / √2 * 1/ √2
= −1
Q5. In a college of 300 students, every student reads 5 newspapers and every newspaper is read by 60 students. The number of newspapers is ________.
a) 25
b) 35
c) 30
d) 20
a) 25
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
Syllabus and Previous Year Papers 

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