## MATHEMATICS RELATIONS QUIZ-1

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. The function f(x)=log_10⁡(x+√(x2+1)) is
•  An even function
•  An odd function
•  Periodic function
•  None of these
Solution
We have, f(x)=log⁡(x+√(x2+1)) ∴f(-x)+f(x)=log⁡((x+√(x2+1)))+lg⁡((-x+√(x2+1))) ⇒f(-x)+f(x)=log⁡(-x2+x2+1)=log⁡(1=0) for all x ⇒f(-x)=-f(x) for all x ⇒f(x) is an odd function

Q2. The domain of the real valued function f(x)=√(5-4x-x2 )+x2 log⁡(x+4) is
•  -5≤x≤1
•  -5≤x and x≥1
•  -4
•  ϕ
Solution

Q3.  Let A=[-1,1] and f:A→A be defined as f(x)=x|x| for all x∈A, then f(x) is
•  Many-one into function
•  One-one into function
•  Many-one onto function
•  One-one onto function
Solution

Q4. f(x)=x+√(x2 ) is a function from R toR, then f(x) is
•  Injective
•  Surjective
•  Bijective
•  None of these
Solution
Given, f(x)=x+√(x2 ) Since, this function is not defined

Q5.If f:R→R and g:R→R are defined by f(x)=|x| and g(x)=[x-3] for x∈R, then {g(f(x) ):-8/5
•  {0, 1}
•  {1, 2}
•  {-3, -2}
•  {2, 3}
Solution

Q6. Let f(x)=|x-1|. Then,
•  f(x2 )=[f(x) ]2
•  f(|x| )=|f(x)|
•  f(x+y)=f(x)+f(y)
•  None of these
Solution
We have,
f(x2 )=|x2-1|≠|x-1|2=[f(x) ]2 f(|x| )=|(|x|-1)|≠|x-1|=|f(x) |
And,
f(x+y)=|x+y-1|≠|x-1|+|y-1|=f(x)+f(y)
Hence, none of the above given option is true

Q7.Let f:A→B and g∶B→C be bijections, then (fog)-1=
• f-1og-1
•  fog
•  g-1 of-1
•  gof

Q8.The domain of the function f(x)=log_(3+x)⁡((x2-1)) is
•  (-3,-1)∪(1,∞)
•  [-3,-1]∪[1,∞]
•  (-3,-2)∪(-2,-1)∪(1,∞)
•  [-3,-2)∪(-2,-1)∪(1,∞)

Solution

The given function is defined when x2-1;3+x>0 and 3+x≠1 ⇒ x21; 3+x>0 and x≠-2 ⇒ -1>x>1; x>-3, x≠-2 ∴ Domain of the function is D_f=(-3,-2)∪(-2,-1)∪(1,∞)

Q9.
•  x2-2 for all x≠0
•  x2-2 for all x satisfying |x|≥2
•  x2-2 for all x satisfying |x|<2 span="">
•  None of these
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