Solve this

Question:

Let $\mathrm{f:} \mathrm{R} \rightarrow \mathrm{R}: \mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+1$ and $\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}: \mathrm{g}(\mathrm{x})=(\mathrm{x}+1)$.

Find:

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

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

(iii) (1/f) (x)

(iv) $(f / g)(x)$

 

 

Solution:

(i) Given:

$f(x)=x^{3}+1$ and $g(x)=x+1$

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

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

$=\left(x^{3}+1\right)+(x+1)$

$=x^{3}+1+x+1$

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

Therefore,

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

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

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

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

$=x^{3}+1-x-1$

$=x^{3}-x$

Therefore

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

(iii) To find $:\left(\frac{1}{f}\right)(x)$

$\left(\frac{1}{f}\right)(x)=\left(\frac{1}{f(x)}\right)$

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

Therefore,

$\left(\frac{1}{f}\right)(x)=\left(\frac{1}{x^{3}+1}\right)$

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

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

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

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

$=\left(\frac{(x+1)\left(x^{2}-x+1\right)}{x+1}\right)$ (Because $\left.\mathrm{a}^{3}+\mathrm{b}^{3}=(\mathrm{a}+\mathrm{b})\left(\mathrm{a}^{2}-\mathrm{ab}+\mathrm{b}^{2}\right)\right)$

Therefore,

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

 

Leave a comment

Close
faculty

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