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****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.

**MATHEMATICS RELATIONS QUIZ-10**

**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

Solution

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

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(tanx ) is

Solution

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

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

Therefore, f(tanx) exists, if
0≤ tanx≤1

⇒nÏ€≤x≤nÏ€+Ï€/4,n∈Z

⇒x∈[n Ï€,nÏ€+Ï€/4],n∈Z

**Q3.**If f(x)=2x

^{6}+3x

^{4}+4x

^{2}, then f'(x) is

Solution

Since f(x) is an even function.

Since f(x) is an even function.

So f'(x) is an odd function

**Q4.**The function f(x)=x[x], is

Solution

We have, f(x)=x[x]=kx,

We have, f(x)=x[x]=kx,

when k≤x < k+1 and k∈Z

Clearly, it is not a periodic function

**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

Solution

Given, F(0)=2, F(1)=3,

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

Solution

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

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

So, f is many-one into

**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

Solution