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Functions question

2019 · 9 Jan · Shift 1 · Q32
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Functions question

2019 · 9 Jan · Shift 1 · Q32

JEE MainMathematicsFunctionsMCQ+4 / −1
For x∈R−{0,1}x \in R - \left\{ {0,1} \right\}x∈R−{0,1}, Let f1(x) = 1x1\over xx1​, f2 (x) = 1 – x and f3 (x) = 11−x1 \over {1 - x}1−x1​ be three given functions. If a function, J(x) satisfies (f2 o J o f1) (x) = f3 (x) then J(x) is equal to :
  1. A
    f1 (x)
  2. B
    1x1 \over xx1​ f3 (x)
  3. C
    f2 (x)
  4. D
    f3 (x)
View written solutionFree

Correct answer: D

  1. We are given

f1(x)=1x,f2(x)=1−x,f3(x)=11−xf_1(x)=\frac{1}{x},\qquad f_2(x)=1-x,\qquad f_3(x)=\frac{1}{1-x}f1​(x)=x1​,f2​(x)=1−x,f3​(x)=1−x1​

and

(f2∘J∘f1)(x)=f3(x).(f_2\circ J\circ f_1)(x)=f_3(x).(f2​∘J∘f1​)(x)=f3​(x).

This means

f2(J(f1(x)))=f3(x).f_2\big(J(f_1(x))\big)=f_3(x).f2​(J(f1​(x)))=f3​(x).

So,

f2(J(1x))=11−x.f_2\left(J\left(\frac{1}{x}\right)\right)=\frac{1}{1-x}.f2​(J(x1​))=1−x1​.

  1. Since f2(y)=1−yf_2(y)=1-yf2​(y)=1−y, we get

1−J(1x)=11−x.1-J\left(\frac{1}{x}\right)=\frac{1}{1-x}.1−J(x1​)=1−x1​.

Hence,

J(1x)=1−11−x.J\left(\frac{1}{x}\right)=1-\frac{1}{1-x}.J(x1​)=1−1−x1​.

Now simplify:

1−11−x=1−x−11−x=−x1−x=xx−1.1-\frac{1}{1-x}=\frac{1-x-1}{1-x}=\frac{-x}{1-x}=\frac{x}{x-1}.1−1−x1​=1−x1−x−1​=1−x−x​=x−1x​.

Thus,

J(1x)=xx−1.J\left(\frac{1}{x}\right)=\frac{x}{x-1}.J(x1​)=x−1x​.

  1. Put

t=1x  ⟹  x=1t.t=\frac{1}{x} \implies x=\frac{1}{t}.t=x1​⟹x=t1​.

Then

J(t)=1t1t−1.J(t)=\frac{\frac{1}{t}}{\frac{1}{t}-1}.J(t)=t1​−1t1​​.

Simplify:

J(t)=1t1−tt=11−t.J(t)=\frac{\frac{1}{t}}{\frac{1-t}{t}}=\frac{1}{1-t}.J(t)=t1−t​t1​​=1−t1​.

Therefore,

J(x)=11−x=f3(x).J(x)=\frac{1}{1-x}=f_3(x).J(x)=1−x1​=f3​(x).

  1. Check with options:
  • A: f1(x)=1xf_1(x)=\frac{1}{x}f1​(x)=x1​
  • B: 1xf3(x)=1x(1−x)\frac{1}{x}f_3(x)=\frac{1}{x(1-x)}x1​f3​(x)=x(1−x)1​
  • C: f2(x)=1−xf_2(x)=1-xf2​(x)=1−x
  • D: f3(x)=11−xf_3(x)=\frac{1}{1-x}f3​(x)=1−x1​

So the correct option is D.

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