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Functions question

2017 · 9 Apr · Shift 1 · Q30
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Functions question

2017 · 9 Apr · Shift 1 · Q30

JEE MainMathematicsFunctionsMCQ+4 / −1
The function f : N →\to→ N defined by f (x) = x −-− 5 [x5],\left[ {{x \over 5}} \right],[5x​], Where N is the set of natural numbers and [x] denotes the greatest integer less than or equal to x, is :
  1. A
    one-one and onto
  2. B
    one-one but not onto.
  3. C
    onto but not one-one.
  4. D
    neither one-one nor onto.
View written solutionFree

Correct answer: D

  1. Understand the function

Given

f(x)=x−5⌊x5⌋,f(x)=x-5\left\lfloor \frac{x}{5}\right\rfloor,f(x)=x−5⌊5x​⌋,

with domain N\mathbb NN and codomain N\mathbb NN.

This expression gives the remainder when xxx is divided by 555.

Indeed, if

x=5q+r,0≤r<5,x=5q+r, \qquad 0\le r<5,x=5q+r,0≤r<5,

then

⌊x5⌋=q,\left\lfloor \frac{x}{5}\right\rfloor=q,⌊5x​⌋=q,

and so

f(x)=x−5q=(5q+r)−5q=r.f(x)=x-5q=(5q+r)-5q=r.f(x)=x−5q=(5q+r)−5q=r.

Thus,

f(x)=x(mod5),f(x)=x\pmod 5,f(x)=x(mod5),

with possible values 0,1,2,3,40,1,2,3,40,1,2,3,4.

  1. Check whether the function is one-one

A function is one-one if different inputs give different outputs.

Take two different natural numbers:

1 and 6.1 \text{ and } 6.1 and 6.

Then

f(1)=1−5⌊15⌋=1−5(0)=1,f(1)=1-5\left\lfloor \frac{1}{5}\right\rfloor=1-5(0)=1,f(1)=1−5⌊51​⌋=1−5(0)=1,

and

f(6)=6−5⌊65⌋=6−5(1)=1.f(6)=6-5\left\lfloor \frac{6}{5}\right\rfloor=6-5(1)=1.f(6)=6−5⌊56​⌋=6−5(1)=1.

So,

f(1)=f(6)f(1)=f(6)f(1)=f(6)

but 1≠61\ne 61=6.

Hence, fff is not one-one.

  1. Check whether the function is onto

For onto, every element of the codomain N\mathbb NN must be attained as an output.

But the outputs of fff are only among

{0,1,2,3,4}.\{0,1,2,3,4\}.{0,1,2,3,4}.

So values like 5,6,7,…5,6,7,\dots5,6,7,… are never obtained.

Therefore, if codomain is N\mathbb NN, the function is not onto.

Note: Even if one takes N={1,2,3,… }\mathbb N=\{1,2,3,\dots\}N={1,2,3,…}, then 000 is not in the codomain, but still the image is only {1,2,3,4}\{1,2,3,4\}{1,2,3,4} for multiples excluded as output issue; in any case, all natural numbers are not covered. So it is not onto.

  1. Conclusion

The function is neither one-one nor onto.

So the correct option is:

D\boxed{\text{D}}D​
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