APPLICATION OF DERIVATIVES

(SUM 34 TO 37)

ALL EDUCATION BOARD

 

Question 34.
Show that the function $$f(x)=\sin x$$ is
  (i) strictly increasing on $$\left(0, \frac{\pi}{2}\right)$$
  
  (ii) strictly decreasing on $$\left(\frac{\pi}{2}, \pi\right)$$
  
  
 Solution.    
  Here $$f(x)=\sin x$$
  
 $$\therefore \quad f^{\prime}(x)=\cos x$$
  
 (i)
 For$$0<x<\frac{\pi}{2}, \cos x>0$$ i.e., $$f^{\prime}(x)>0$$
  
 $$\therefore f(x)$$ is strictly increasing on$$\left(0, \frac{\pi}{2}\right)$$.
  
 $$\therefore f(x)$$ is strictly increasing on$$\left(0, \frac{\pi}{2}\right)$$.
  
  (ii)
 $$For \frac{\pi}{2}<x<\pi, \cos x<0` i.e., f^{\prime}(x)<0$$
  
 `\therefore f(x)` is strictly decreasing on `\left(\frac{\pi}{2}, \pi\right)`.
  
 Hence, `f(x)` is strictly increasing on `\left(0, \frac{\pi}{2}\right)`
and strictly decreasing on `\left(\frac{\pi}{2}, \pi\right)`.
For `\frac{\pi}{2}<x<\pi, \cos x<0` i.e., `f^{\prime}(x)<0`
 `\therefore f(x)` is strictly decreasing on `\left(\frac{\pi}{2}, \pi\right)`.
  
 Hence, `f(x)` is strictly increasing on `\left(0, \frac{\pi}{2}\right)`
and strictly decreasing on `\left(\frac{\pi}{2}, \pi\right)`.$$
  
  
 Question 35
 (i) Prove that the function `\log \sin x` is strictly increasing on
`\left(0, \frac{\pi}{2}\right)` and strictly decreasing
on `\left(\frac{\pi}{2}, \pi\right)`.
  
 (ii Prove that the function `\log \cos x` is strictly decreasing
on `0, \frac{\pi}{2}` and strictly increasing on `\left(\frac{\pi}{2}, \pi\right)`.
  
 Solution.
 (i)       Let `f(x)=\log \sin x`
  
 `\therefore \quad f^{\prime}(x)=\frac{1}{\sin x} \cdot \cos x=\cot x`
  
 For `0<x<\frac{\pi}{2}, \quad \cot x>0` i.e., `f^{\prime}(x)>0`
  
 `\therefore f(x)` is strictly increasing for `0<x<\frac{\pi}{2}`
  
 For `\frac{\pi}{2}<x<\pi, \quad \cot x<0` i.e., `\quad f^{\prime}(x)<0`
  
 `\therefore f(x)` is strictly decreasing for `\frac{\pi}{2}<x<\pi`
  
 Hence, `f(x)` is strictly increasing on `\left(0, \frac{\pi}{2}\right)`
and strictly decreasing on `\left(\frac{\pi}{2}, \pi\right)`
 
 (ii)           Let `\quad f(x)=\log \cos x`
  
 `\therefore \quad f^{\prime}(x)`
`=\frac{1}{\cos x} \cdot(-\sin x)=-\tan x`
  
 For `0<x<\frac{\pi}{2}, \quad \tan x>0`
  
 For `\frac{\pi}{2}<x<\pi, \quad \tan x<0`
 ``
 `f^{\prime}(x)=-\tan x>0`
 ``
 `\therefore(x)` is strictly increasing on
`\left(\frac{\pi}{2}, \pi\right)`.
  
 Hence, `f(x)` is strictly decreasing on `\left(0, \frac{\pi}{2}\right)`
and strictly increasing on `\left(\frac{\pi}{2}, \pi\right)`
  
 `\therefore \quad f^{\prime}(x)=-\tan x<0`
  
 `\therefore f(x)` is strictly decreasing on
 `\left(0, \frac{\pi}{2}\right)`
  
 Question 36. Prove that
`y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta` is an   
  increasing function of `\theta` on `\left[0, \frac{\pi}{2}\right]`.
  
 Solution.      Proceeding as in above example, we have
 ``
 `\frac{d y}{d \theta}`=`\frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^2}`
 ``
  
 Now, for `0 \leq \theta \leq \frac{\pi}{2}, \cos \theta`
lies between 0 and 1 (both inclusive)
 ``
  `\therefore \quad 4-\cos \theta  >0 \text { and }(2+\cos \theta)>0 `
 `\frac{d y}{d \theta}  =\frac{(+)(+)}{+}=+v e`
  
 ``
  Hence `y`  is  an  increasing function of 0 on 
`[0, \frac{\pi}{2}\right]`
  
 Question 37. Show that `y=\log (1+x)-\frac{2 x}{x+2}` is
an increasing function of `x` for all values of `x>-1`.
  
 Solution.
 Let `f(x)=\log (1+x)-\frac{2 x}{x+2}`


`f^{\prime}(x)`=`\frac{1}{1+x}-2\left[\frac{(x+2) \cdot 1-x \cdot1}{(x+2)^2}\right]`


`=\frac{1}{1+x}-\frac{4}{(x+2)^2} `
  =`\frac{(x+2)^2-4(x+1)}{(x+1)(x+2)^2}`
 
=`\frac{x^2}{(x+1)(x+2)^2}`
  
 ``For the function to be increasing. `f^{\prime}(x)>0`
 Now, `x^2` and `(x+2)^2` being square quantities are
always positive. Hence `f^{\prime}(x)` will be +ve
if `\frac{1}{x+1}>0` i.e., if `x+1>0 \Rightarrow x>-1`.
  
 Hence, `y=\log (1+x)-\frac{2 x}{x+2}` is an increasing
function of `x` for all value of `x>-1`.