RELATIONS-QUIZ-10

MATHEMATICS RELATIONS QUIZ-10

Dear 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 X={1,2,3,4}, then one-one onto mappings f:X→X such that f(1)=1,f(2)≠2,f(4)≠4 are given by

•  f={(1,1),(2,3),(3,4),(4,2) }
•  f={(1,2),(2,4),(3,3),(4,2)}
•  f={(1,2),(2,4),(3,2),(4,3)}
•  None of these

Solution
Clearly, mapping f given in option (a) satisfies the given conditions

Q2. If f(x) is defined on [0,1], then the domain of definition of f(tan⁡x ) is

•  [n π,n π+π/4],n∈Z
•  [2 n π,2nπ+π/4],n∈Z
•  [nπ-π/4,nπ+π/4],n∈Z]
•  None of these

Solution
It is given that f(x) is defined on [0,1]. 

Therefore, f(tan⁡x) exists, if 0≤ tan⁡x≤1 

 ⇒nπ≤x≤nπ+π/4,n∈Z

⇒x∈[n π,nπ+π/4],n∈Z

Q3.  If f(x)=2x⁶+3x⁴+4x², then f'(x) is

•  An even function
•  An odd function
•  Neither even nor odd
•  None of the above

Solution
Since f(x) is an even function. 

So f'(x) is an odd function

Q4. The function f(x)=x[x], is

•  Periodic with period 1
•  Periodic with period 2
•  Periodic with indeterminate period
•  Not-periodic

Solution
We have, f(x)=x[x]=kx, 

when k≤x < k+1 and k∈Z 

 Clearly, it is not a periodic function

Q5.

•  f is bijective
•  f is one-one but not onto
•  f is onto but not one-one
•  None of the above

Solution
 

Q6. The period of the function f(θ)=4+4sin³θ-3 sin⁡θ is

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Solution

Q7.If a function F is such that F(0)=2,F(1)=3,F(n+2)=2F(n)-F(n+1) for n≠0, then F(5) is equal to

• -7
•  -3
•  7
•  13

Solution
Given, F(0)=2, F(1)=3, 

 Since, F(n+2)=2F(n)-F(n+1) 

 At n=0, F(0+2)=2F(0)-F(1) 

 ⇒ F(2)=2(2)-3=1 

 At n=1,F(1+2)=2F(1)-F(2) 

 ⇒ F(3)=2(3)-1=5 

 At n=2,F(2+2)=2F(2)-F(3)

⇒F(4)=2(1)-5=-3 

 At n=3, F(3+2)=2F(3)-F(4)=2(5)-(-3) 

 ⇒ F(5)=13

Q8. The function f:R→R, defined by f(x)=[x], where [x] denotes the greatest integer less than or equal to x, is

•  One-one
•  Onto
•  One-one and onto
•  Neither one-one nor onto

Solution
We have, f(x)=[x]=k for k≤x< k+1, where k∈Z 

 So, f is many-one into

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Solution

Q10. Let f:N→Y be a function defined as f(x)=4x+3 where Y={y∈N:y=4x+3 for somex∈N}. Show that f is invertible and its inverse is

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Solution