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Functions question

2018 · 15 Apr · Shift 2 · Q28
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Functions question

2018 · 15 Apr · Shift 2 · Q28

JEE MainMathematicsFunctionsMCQ+4 / −1
Let f : A →\to→ B be a function defined as f(x) = x−1x−2,{{x - 1} \over {x - 2}},x−2x−1​, Where A = R −-−{2} and B = R −-− {1}. Then f is :
  1. A
    invertible and f−1(y)=3y−1y−1{f^{ - 1}}(y) = {{3y - 1} \over {y - 1}}f−1(y)=y−13y−1​
  2. B
    invertible and f−1(y)=2y−1y−1{f^{ - 1}}\left( y \right) = {{2y - 1} \over {y - 1}}f−1(y)=y−12y−1​
  3. C
    invertible and f−1(y)=2y+1y−1{f^{ - 1}}\left( y \right) = {{2y + 1} \over {y - 1}}f−1(y)=y−12y+1​
  4. D
    not invertible
View written solutionFree

Correct answer: B

  1. Given function and sets

We have

f:A→B,f(x)=x−1x−2f:A\to B,\qquad f(x)=\frac{x-1}{x-2}f:A→B,f(x)=x−2x−1​

with

A=R∖{2},B=R∖{1}.A=\mathbb R\setminus\{2\},\qquad B=\mathbb R\setminus\{1\}.A=R∖{2},B=R∖{1}.

We must determine whether fff is invertible, and if yes, find f−1f^{-1}f−1.

  1. Check whether the range lies in BBB

Let

y=x−1x−2.y=\frac{x-1}{x-2}.y=x−2x−1​.

Can y=1y=1y=1 occur?

If

x−1x−2=1,\frac{x-1}{x-2}=1,x−2x−1​=1,

then

x−1=x−2,x-1=x-2,x−1=x−2,

which gives

−1=−2,-1=-2,−1=−2,

a contradiction.

So f(x)≠1f(x)\neq 1f(x)=1 for every x∈Ax\in Ax∈A. Hence indeed

f(A)⊆R∖{1}=B.f(A)\subseteq \mathbb R\setminus\{1\}=B.f(A)⊆R∖{1}=B.
  1. Check injectivity

Assume

f(x1)=f(x2).f(x_1)=f(x_2).f(x1​)=f(x2​).

Then

x1−1x1−2=x2−1x2−2.\frac{x_1-1}{x_1-2}=\frac{x_2-1}{x_2-2}.x1​−2x1​−1​=x2​−2x2​−1​.

Cross-multiplying,

(x1−1)(x2−2)=(x2−1)(x1−2).(x_1-1)(x_2-2)=(x_2-1)(x_1-2).(x1​−1)(x2​−2)=(x2​−1)(x1​−2).

Expanding both sides:

x1x2−2x1−x2+2=x1x2−2x2−x1+2.x_1x_2-2x_1-x_2+2=x_1x_2-2x_2-x_1+2.x1​x2​−2x1​−x2​+2=x1​x2​−2x2​−x1​+2.

Cancel x1x2x_1x_2x1​x2​ and 222:

−2x1−x2=−2x2−x1.-2x_1-x_2=-2x_2-x_1.−2x1​−x2​=−2x2​−x1​.

So

−x1+x2=0-x_1+x_2=0−x1​+x2​=0

which gives

x1=x2.x_1=x_2.x1​=x2​.

Thus fff is injective.

  1. Check surjectivity onto BBB

Let any y∈B=R∖{1}y\in B=\mathbb R\setminus\{1\}y∈B=R∖{1}. We solve

y=x−1x−2y=\frac{x-1}{x-2}y=x−2x−1​

for xxx.

Multiply both sides:

y(x−2)=x−1.y(x-2)=x-1.y(x−2)=x−1.

So

yx−2y=x−1.yx-2y=x-1.yx−2y=x−1.

Bring xxx-terms together:

yx−x=2y−1.yx-x=2y-1.yx−x=2y−1.

Factor:

x(y−1)=2y−1.x(y-1)=2y-1.x(y−1)=2y−1.

Hence

x=2y−1y−1.x=\frac{2y-1}{y-1}.x=y−12y−1​.

This is well-defined because y≠1y\neq 1y=1.

Now check whether this xxx can be 222:

2y−1y−1=2\frac{2y-1}{y-1}=2y−12y−1​=2

would imply

2y−1=2y−2,2y-1=2y-2,2y−1=2y−2,

so

−1=−2,-1=-2,−1=−2,

impossible. Thus x≠2x\neq 2x=2, so x∈Ax\in Ax∈A.

Therefore, for every y∈By\in By∈B, there exists x∈Ax\in Ax∈A such that f(x)=yf(x)=yf(x)=y. Hence fff is surjective onto BBB.

So fff is invertible.

  1. Find the inverse

From the solving step above,

x=2y−1y−1.x=\frac{2y-1}{y-1}.x=y−12y−1​.

Therefore,

f−1(y)=2y−1y−1.f^{-1}(y)=\frac{2y-1}{y-1}.f−1(y)=y−12y−1​.
  1. Match with the options

The correct statement is:

f is invertible and f−1(y)=2y−1y−1.f \text{ is invertible and } f^{-1}(y)=\frac{2y-1}{y-1}.f is invertible and f−1(y)=y−12y−1​.

So the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.

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