## MATHEMATICS RELATIONS QUIZ-3Dear Readers,As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background. Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.

Q1. If a function f∶[2,∞)→B defined by f(x)=x2-4 x+5 is a bijection, then B=
•  R
•  [1,∞)
•  [4,∞)
•  [5,∞)
Solution
We have,
f∶[2,∞)→B such that f(x)=x2-4 x+5
Since f is a bijection.
Therefore, B= Range of f
Now,
f(x)=x2-4x+5=5=(x-2)2+1 for all x∈[2,∞) ⇒f(x)≥1 for all x∈[2,∞)⇒ Range of f=[1,∞) Hence, B=[1,∞)

Q2.

•  (-1/2,1)
•  (-1/2,1/2)∪(1/2,1)∪(3/2,∞)
•  (-1/2-1)
•  None of these
Solution

Q3.  If f(x)=x3-x and Ï•(x)=sin⁡2x, then
•  Ï•(f(2) )=sin⁡2
•  Ï•(f(1) )=1
•  f(Ï•(Ï€/12) )=-3/8
•  f(f(1) )=2

Q4. If the function f(x) is defined by f(x)=a+bx and fr=fff.... (repeated r times), then fr (x) is equal to
•  a+br x
•  ar+br x
•  ar+bxr
•

Solution

Q5.If f is a real valued function such that f(x+y)=f(x)+f(y) and f(1)=5, then the value of f(100) is
•  200
•  300
•  350
•  500
Solution
Given, f(x+y)=f(x)+f(y)
For x=1, y=1
we get f(2)=f(1)+f(1) =2.f(1)=10
Also f(3)=f(2)+f(1)=15 ⇒ f(n)=5n
∴ f(100)=500

Q6.

•  [2, 3]
•  [2, 3)
•  [1, 2]
•  [1, 2)
Solution
The function f(x) will be defined, if -1≤(x-3)≤1⇒ 2≤x≤4 And 9-x2>0 ⇒ ∴ 2≤x

Q7.

•  R
•  [0,1]
•  (0,1]
•  [0,1)
Solution

Q8.If f:R→R, defined by f(x)=x2+1, then the values of f-1 (17) and f-1 (-3) respectively are
•  Ï•,{4,-4}
•  {3,-3},Ï•
•  {4,-4},Ï•
•  {4,-4},{2,-2}
Solution
Let f-1 (17)=x.
Then, f(x)=17⇒x2+1=17⇒x±4
Let f-1 (-3)=x
Then, f(x)=-3⇒x2+1=-3⇒x2=-4
which is not possible for any real number x

Q9.The equivalent definition of f(x)=(max.)⁡({x2,(1-x)2,2x(1-x) },) where 0≤x≤1,
•

•

•

•  None of these

Q10. If f:R→C is defined by f(x)=e2ix for x∈R, then f is (where C denotes the set of all complex numbers)
•  One-one
•  Onto
•  One-one and onto
•  Neither one-one nor onto
Solution
Given, f(x)=e2ixand f:R→C.
Function f(x) is not one-one, because after some values of x(ie,Ï€) it will give the same values.
Also, f(x) is not onto, because it has minimum and maximum values -1-i and 1+i respectively.

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