Consider f : R - -4 3 R - 4 3 given by f(x) = 4x + 3 3x + 4 . Sho… - Vidyadip

Question

Consider $\text{f : R} - \left\{-\frac{4}{3}\right\} \rightarrow \text{R} - \left\{\frac{4}{3}\right\} \text{given by f(x)} = \frac{\text{4x + 3}}{\text{3x + 4}}.$ Show that f is bijective. Find the inverse of f and hence find $f ^{–1}(0)$ and x such that $f ^{–1}(x) = 2$.

✓

Answer

$\text{Let } \text{x}_{1}, \text{x}_{2} \in \text{R} - \left\{-\frac{4}{3}\right\} \text{and } \text{f(x}_{1}) = \text{f(x}_{2})$
$\Rightarrow \frac{\text{4x}_{1} + 3}{\text{3x}_{1} + 4} = \frac{\text{4x}_{2} + 3}{\text{3x}_{2} + 4} \Rightarrow \text{(4x}_{1} + 3) \text{(3x}_{2} + 4) = \text{(3x}_{1} + 4) \text{(4x}_{2} + 3)$
$\Rightarrow \text{12x}_{1} \text{x}_{2} + \text{16x}_{1} + \text{9x}_{2} + 12 = 12_{1} \text{x}_{2} + 16\text{x}_{2} + 9\text{x}_{1} + 12$
$\Rightarrow 16(\text{x}_{1} - \text{x}_{2}) - 9 \text{(x}_{1} - \text{x}_{2}) = 0 \Rightarrow \text{x}_{1} - \text{x}_{2} = 0 \Rightarrow \text{x}_{1} = \text{x}_{2}$
Hence f is a 1–1 function
$\text{Let} \text{ y} = \frac{\text{4x + 3}}{\text{3x + 4}}, \text{for y} \in \text{R} - \left\{\frac{4}{3}\right\}$
$\text{3xy + 4y = 4x + 3} \Rightarrow \text{4x – 3xy = 4y – 3}$
$\Rightarrow \text{x} = \frac{\text{4y - 3}}{\text{4 - 3y}} \therefore \forall \text{ y} \in \text{R} - \left\{\frac{4}{3}\right\}, \text{x} \in \text{R} - \left\{-\frac{4}{3}\right\}$
Hence f is ONTO and so bijective
$\text{and } \text{f}^{-1} \text{(y)} = \frac{\text{4y - 3}}{\text{4 - 3y}} ; \text{y} \in \text{R} - \left\{\frac{4}{3}\right\}$
$\text{f}^{-1} (0) = - \frac{3}{4}$
$\text{and } \text{f}^{-1} \text{(x)} = 2 \Rightarrow \frac{\text{4x - 3}}{\text{4 - 3x}} = 2$
$\Rightarrow \text{4x - 3 = 8 - 6x}$
$\Rightarrow \text{10x} = 11 \Rightarrow \text{x} = \frac{11}{10}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from RELATIONS AND FUNCTIONS→RELATIONS AND FUNCTIONS — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If $\sin(\text{xy})+\frac{\text{y}}{\text{x}}=\text{x}^2-\text{y}^2,$ find $\frac{\text{dy}}{\text{dx}}$

Differentiate the following functions with respect to x:
$\log\sqrt{\frac{\text{x}-1}{\text{x}+1}}$

Evaluate $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin ^2 x d x$.

Write the minimum value of $\text{f}(\text{x})=\text{x}^\frac{1}{\text{x}}.$

An item is manufactured by three machines A, B and C. Out of the total number of items manufactured during a specified period, 50% are manufactured on machine A, 30% on B and 20% on C, 2% of the items produced on A and 2% of items produced on B are defective and 3% of these produced on C are defective. All the items stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?

A manufacturer has three machine operators A, B and C. The first operator A produces $1 \%$ of defective items, whereas the other two operators B and C produces $5 \%$ and $7 \%$ defective items respectively. A is on the job for $50 \%$ of the time, B on the job $30 \%$ of the time and C on the job for $20 \%$ of the time. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by A ?

Find the equation of the plane passing through the line of intersection of the planes $2x - y = 0$ and $3z - y= 0$ and perpendicular to the plane $4x + 5y - 3z = 8$.

Show that $\text{f}(\text{x})=\sin\text{x}-\cos\text{x}$ is an increasing function on $\Big(-\frac{\pi}{4},\frac{\pi}{4}\Big).$

Solve the following differential equation:
$\left(1+x^2\right) d y+2 x y\ d x=\cot x\ d x ; x \neq 0$

Differentiate $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$ with respect to $\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big),$ if $-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$