Show that the function f: R→\x R:-1<x<1\ defined by f(x)=(x)/(1+ x ), x R is one-one and onto function.

MathsCBSE Class 12 · Sample paper, session 2023-24 · 5 marks · Long answer

CBSE Class 12 Mathematics sample paper 2023-24: question 33(b)

From the official CBSE sample question paper for 2023-24, section D, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Long answerRelations and Functions

Show that the function $f : R \to {x \in R : - 1 < x < 1}$ defined by $f (x) = \frac{x}{1 + ∣ x ∣}$ , $x \in R$ is one-one and onto function.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

Open one hint at a time and try again before the next.

Hint 1

Remove the modulus by taking the cases $x \geq 0$ and $x < 0$ separately.

Hint 2

In each case solve $f (x) = y$ for $x$ . Also note that the sign of $f (x)$ matches the sign of $x$ , which handles the mixed case.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

$f (x) = {\frac{x}{1 + x}, \frac{x}{1 - x}, x \geq 0 x < 0$
**Case 1 ( $x \geq 0$ ), injectivity:** If $x, y \geq 0$ and $\frac{x}{1 + x} = \frac{y}{1 + y}$ , then $x + x y = y + x y$ , so $x = y$ . Hence $f$ is one-one on $[0, \infty)$ .
1 mark
Case 1, surjectivity: For $x \geq 0$ , $f (x) \geq 0$ and $f (x) = 1 - \frac{1}{1 + x} < 1$ , so $f (x) \in [0, 1)$ . For any $y \in [0, 1)$ , take $x = \frac{y}{1 - y} \geq 0$ . Then $f (x) = \frac{y / ( 1 - y )}{1 + y / ( 1 - y )} = y$ . So $f$ maps $[0, \infty)$ onto $[0, 1)$ .
1 mark
**Case 2 ( $x < 0$ ), injectivity:** If $x, y < 0$ and $\frac{x}{1 - x} = \frac{y}{1 - y}$ , then $x - x y = y - x y$ , so $x = y$ . Hence $f$ is one-one on $(- \infty, 0)$ .
1 mark
Case 2, surjectivity: For $x < 0$ , $f (x) < 0$ and $f (x) = - 1 + \frac{1}{1 - x} > - 1$ , so $f (x) \in (- 1, 0)$ . For any $y \in (- 1, 0)$ , take $x = \frac{y}{1 + y} < 0$ . Then $f (x) = \frac{y / ( 1 + y )}{1 - y / ( 1 + y )} = y$ . So $f$ maps $(- \infty, 0)$ onto $(- 1, 0)$ .
1 mark
**Case 3 ( $x \geq 0$ , $y < 0$ ):** $f (x) \geq 0$ while $f (y) < 0$ , so $f (x) \neq = f (y)$ . (Equivalently, $\frac{x}{1 + x} = \frac{y}{1 - y}$ gives $x - y = 2 x y$ , where the LHS is $> 0$ and the RHS is $\leq 0$ , which is impossible.) So $f$ is one-one on all of $R$ . Since its range is $[0, 1) \cup (- 1, 0) = (- 1, 1)$ , the whole codomain, $f$ is also onto. Hence $f$ is one-one and onto.
1 mark

Answer: $f$ is one-one and onto from $R$ to $(- 1, 1)$ .

Where the marks are

Injectivity for $x \geq 0$ (1 mark)
Surjectivity onto $[0, 1)$ for $x \geq 0$ , with the pre-image $\frac{y}{1 - y}$ (1 mark)
Injectivity for $x < 0$ (1 mark)
Surjectivity onto $(- 1, 0)$ for $x < 0$ , with the pre-image $\frac{y}{1 + y}$ (1 mark)
Mixed-sign case and the final conclusion (1 mark)

Revise the chapter

Chapter: Relations and Functions (notes, formulas, where marks are lost and every past question)
Topic: One-One and Onto Functions

Practise this chapterBoard-style questions from Relations and Functions, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2023-24: every question, by section.

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CBSE Class 12 Mathematics sample paper 2023-24: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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