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Functions question

2022 · 27 Jun · Shift 2 · Q37
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  5. /2022 · 27 Jun · Shift 2 · Q37

Functions question

2022 · 27 Jun · Shift 2 · Q37

JEE MainMathematicsFunctionsNumerical+4 / −1
Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Define f : S →\to→ S as f(n)={2n,if n=1,2,3,4,52n−11,if n=6,7,8,9,10f(n) = \left\{ {\begin{matrix} {2n} & , & {if\,n = 1,2,3,4,5} \\ {2n - 11} & , & {if\,n = 6,7,8,9,10} \\ \end{matrix} } \right.f(n)={2n2n−11​,,​ifn=1,2,3,4,5ifn=6,7,8,9,10​. Let g : S →\to→ S be a function such that fog(n)={n+1,if n  is oddn−1,if n  is evenfog(n) = \left\{ {\begin{matrix} {n + 1} & , & {if\,n\,\,is\,odd} \\ {n - 1} & , & {if\,n\,\,is\,even} \\ \end{matrix} } \right.fog(n)={n+1n−1​,,​ifnisoddifniseven​. Then g(10)g(1)+g(2)+g(3)+g(4)+g(5))g(10)g(1) + g(2) + g(3) + g(4) + g(5))g(10)g(1)+g(2)+g(3)+g(4)+g(5)) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 28

  1. First understand the function fff.

Given

f(n)={2n,n=1,2,3,4,52n−11,n=6,7,8,9,10 f(n)= \begin{cases} 2n, & n=1,2,3,4,5 \\ 2n-11, & n=6,7,8,9,10 \end{cases}f(n)={2n,2n−11,​n=1,2,3,4,5n=6,7,8,9,10​

So its values are:

f(1)=2,  f(2)=4,  f(3)=6,  f(4)=8,  f(5)=10f(1)=2,\; f(2)=4,\; f(3)=6,\; f(4)=8,\; f(5)=10f(1)=2,f(2)=4,f(3)=6,f(4)=8,f(5)=10 f(6)=1,  f(7)=3,  f(8)=5,  f(9)=7,  f(10)=9f(6)=1,\; f(7)=3,\; f(8)=5,\; f(9)=7,\; f(10)=9f(6)=1,f(7)=3,f(8)=5,f(9)=7,f(10)=9

Thus fff is a bijection on SSS, so ggg is determined by the condition f∘gf\circ gf∘g.

  1. Now use the given relation
(f∘g)(n)={n+1,n oddn−1,n even(f\circ g)(n)= \begin{cases} n+1,& n \text{ odd}\\ n-1,& n \text{ even} \end{cases}(f∘g)(n)={n+1,n−1,​n oddn even​

This simply swaps consecutive numbers:

(f∘g)(1)=2,  (f∘g)(2)=1,  (f∘g)(3)=4,  (f∘g)(4)=3,(f\circ g)(1)=2,\; (f\circ g)(2)=1,\; (f\circ g)(3)=4,\; (f\circ g)(4)=3,(f∘g)(1)=2,(f∘g)(2)=1,(f∘g)(3)=4,(f∘g)(4)=3, (f∘g)(5)=6,  (f∘g)(6)=5,  (f∘g)(7)=8,  (f∘g)(8)=7,(f\circ g)(5)=6,\; (f\circ g)(6)=5,\; (f\circ g)(7)=8,\; (f\circ g)(8)=7,(f∘g)(5)=6,(f∘g)(6)=5,(f∘g)(7)=8,(f∘g)(8)=7, (f∘g)(9)=10,  (f∘g)(10)=9.(f\circ g)(9)=10,\; (f\circ g)(10)=9.(f∘g)(9)=10,(f∘g)(10)=9.

So for each nnn, g(n)g(n)g(n) must be the element whose image under fff is the above value. That is,

g(n)=f−1((f∘g)(n)).g(n)=f^{-1}((f\circ g)(n)).g(n)=f−1((f∘g)(n)).
  1. Compute f−1f^{-1}f−1 from the table of fff.

From

f(6)=1,  f(1)=2,  f(7)=3,  f(2)=4,  f(8)=5,f(6)=1,\; f(1)=2,\; f(7)=3,\; f(2)=4,\; f(8)=5,f(6)=1,f(1)=2,f(7)=3,f(2)=4,f(8)=5, f(3)=6,  f(9)=7,  f(4)=8,  f(10)=9,  f(5)=10,f(3)=6,\; f(9)=7,\; f(4)=8,\; f(10)=9,\; f(5)=10,f(3)=6,f(9)=7,f(4)=8,f(10)=9,f(5)=10,

we get

f−1(1)=6,  f−1(2)=1,  f−1(3)=7,  f−1(4)=2,  f−1(5)=8,f^{-1}(1)=6,\; f^{-1}(2)=1,\; f^{-1}(3)=7,\; f^{-1}(4)=2,\; f^{-1}(5)=8,f−1(1)=6,f−1(2)=1,f−1(3)=7,f−1(4)=2,f−1(5)=8, f−1(6)=3,  f−1(7)=9,  f−1(8)=4,  f−1(9)=10,  f−1(10)=5.f^{-1}(6)=3,\; f^{-1}(7)=9,\; f^{-1}(8)=4,\; f^{-1}(9)=10,\; f^{-1}(10)=5.f−1(6)=3,f−1(7)=9,f−1(8)=4,f−1(9)=10,f−1(10)=5.
  1. Now compute the needed values of ggg.
  • g(1)=f−1(2)=1g(1)=f^{-1}(2)=1g(1)=f−1(2)=1
  • g(2)=f−1(1)=6g(2)=f^{-1}(1)=6g(2)=f−1(1)=6
  • g(3)=f−1(4)=2g(3)=f^{-1}(4)=2g(3)=f−1(4)=2
  • g(4)=f−1(3)=7g(4)=f^{-1}(3)=7g(4)=f−1(3)=7
  • g(5)=f−1(6)=3g(5)=f^{-1}(6)=3g(5)=f−1(6)=3
  • g(10)=f−1(9)=10g(10)=f^{-1}(9)=10g(10)=f−1(9)=10
  1. Evaluate the required expression.

The question asks for

g(10)g(1)+g(2)+g(3)+g(4)+g(5).g(10)g(1)+g(2)+g(3)+g(4)+g(5).g(10)g(1)+g(2)+g(3)+g(4)+g(5).

Substitute the values:

=10⋅1+6+2+7+3=10\cdot 1+6+2+7+3=10⋅1+6+2+7+3 =10+18=28.=10+18=28.=10+18=28.
  1. Final answer:
28\boxed{28}28​
  1. Comparison with stored answer.

The stored correct answer is 190190190, but the derived value is 282828. So I do not agree with the stored answer.

A likely possibility is that the intended expression may have been

g(10)(g(1)+g(2)+g(3)+g(4)+g(5)),g(10)\big(g(1)+g(2)+g(3)+g(4)+g(5)\big),g(10)(g(1)+g(2)+g(3)+g(4)+g(5)),

which would give

10(1+6+2+7+3)=10⋅19=190.10(1+6+2+7+3)=10\cdot 19=190.10(1+6+2+7+3)=10⋅19=190.

That matches the stored answer exactly. Hence the printed expression likely has a bracket/typographical ambiguity.

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