Solve this

Question:

Let $f: R \rightarrow R: f(x)=x+1$ and $g: R \rightarrow R: g(x)=2 x-3 .$

Find

(i) $(f+g)(x)$

(ii) $(f-g)(x)$

(iii) (fg) (x)

(iv)(f/g) (x)

 

 

Solution:

(i) Given:

$f(x)=x+1$ and $g(x)=2 x-3$

(i) To find: $(f+g)(x)$

$(f+g)(x)=f(x)+g(x)$

$=(x+1)+(2 x-3)$

$=x+1+2 x-3$

$=3 x-2$

Therefore,

$(f+g)(x)=3 x-2$

(ii) To find: $(f-g)(x)$

$(f-g)(x)=f(x)-g(x)$

$=(x+1)-(2 x-3)$

$=x+1-2 x+3$

$=4-x$

Therefore

$(f-g)(x)=4-x$

(iii) To find: (fg)(x)

$(f g)(x)=f(x) \cdot g(x)$

$=(x+1)(2 x-3)$

$=x(2 x)-3(x)+1(2 x)-1(3)$

$=2 x^{2}-3 x+2 x-3$

$=2 x^{2}-x-3$

Therefore,

$(f g)(x)=2 x^{2}-x-3$

(iv) To find $:\left(\frac{f}{g}\right)(x)$

Sol. $\left(\frac{f}{g}\right)(x)=\left(\frac{f(x)}{g(x)}\right)$

$=\left(\frac{x+1}{2 x-3}\right)$

Therefore,

$\left(\frac{f}{g}\right)(x)=\left(\frac{x+1}{2 x-3}\right)$

 

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now