SATHEE: Functions 3 Question 8

Functions 3 Question 8

8. Let $f : (0, 1) \to R$ be defined by $f (x) = \frac{b - x}{1 - b x}$ , where $b$ is a constant such that $0 < b < 1$ . Then,

(2011)

(a) $f$ is not invertible on $(0, 1)$

(b) $f \neq f^{- 1}$ on $(0, 1)$ and $f^{'} (b) = \frac{1}{f^{'} (0)}$

(d) $f^{- 1}$ is differentiable on $(0, 1)$

Assertion and Reason

For the following questions, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows.

(a) Statement I is true, Statement II is also true;

Statement II is the correct explanation of Statement I.

(b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I.

(d) Statement I is false; Statement II is true.

Show Answer

Answer:

Correct Answer: 8. (b)

Solution:

PLAN To check nature of function

(i) One-one To check one-one, we must check whether $f^{'} (x) > 0$ or $f^{'} (x) < 0$ in given domain.

(ii) Onto To check onto, we must check Range $=$ Codomain

Description of Situation To find range in given domain $[a, b]$ , put $f^{'} (x) = 0$ and find $x = α_{1}, α_{2}, \dots$ , $α_{n} \in [a, b]$

Now, find $f (a), f (α_{1}), f (α_{2}), \dots, f (α_{n}), f (b)$

its greatest and least values gives you range.

Now, $f : [0, 3] \to [1, 29]$

$\begin{aligned} f (x) & = 2 x^{3} - 15 x^{2} + 36 x + 1 \\ ∴ f^{'} (x) & = 6 x^{2} - 30 x + 36 = 6 (x^{2} - 5 x + 6) \\ = 6 (x - 2) (x - 3) \\ + - + + \\ 2 & 3 \end{aligned}$

For given domain $[0, 3], f (x)$ is increasing as well as decreasing $\Rightarrow$ many-one

$\begin{aligned} Now, put & f^{'} (x) & = 0 \\ \Rightarrow & x & = 2, 3 \end{aligned}$

Thus, for range $f (0) = 1, f (2) = 29, f (3) = 28$

$\Rightarrow$ Range $\in [1, 29]$

$∴$ Onto but not one-one.

Functions 3 Question 8

8. Let $f : (0, 1) \to R$ be defined by $f (x) = \frac{b - x}{1 - b x}$ , where $b$ is a constant such that $0 < b < 1$ . Then,

Assertion and Reason

Answer:

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Functions 3 Question 8

8. Let f:(0,1)→R be defined by f(x)=b−x1−bx, where b is a constant such that 0<b<1. Then,

Assertion and Reason

Answer:

Engineering Preparation

Medical Preparation

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Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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8. Let $f : (0, 1) \to R$ be defined by $f (x) = \frac{b - x}{1 - b x}$ , where $b$ is a constant such that $0 < b < 1$ . Then,