Consider the functions

I. e-x

II. x2-sinx

III. x3+1

Which of the above functions is/are increasing everywhere in [0,1]?

A.

III only

B.

II only

C.

II and III only

D.

I and III only

Solution:

To determine the nature of a function whether it is increasing or decreasing, we observe the first derivative of the function, i.e. \frac{d}{d x} f(x) 

{If:} \frac{\mathrm{d}}{\mathrm{dx}} f(\mathrm{x})<_{-\mathrm{ve}\ \rightarrow\ { decreasing\ nature }}^{+\mathrm{ve}\ \rightarrow { \ increasing \ nature }}

 

For: (I) e^{-x}: f(x)=e^{-x}

 

         \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{f}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{-\mathrm{x}}\right)=-\mathrm{e}^{-\mathrm{x}}

 

 (II) &x^{2}-\sin x: f(x)=x^{2}-\sin x \\ &\frac{d}{d x} f(x)=\frac{d}{d x}\left(x^{2}-\sin x\right)=\frac{d}{d x} x^{2}-\frac{d}{d x} \sin x=2 x-\cos x

 

{ (III) } \sqrt{x^{3}+1}: f(x)=\sqrt{x^{3}+1}=\left(x^{3}+1\right)^{1 / 2}

 

           \frac{d}{d x} f(x)=\frac{d}{d x}\left(x^{3}+1\right)^{1/2}  (By applying substitution and further solving)

 

            =\frac{3 x^{2}}{2 \sqrt{x^{3}+1}}

 

Comparing the functions I, II, III between the range [0, 1]: