Class 12: Increasing and Decreasing Functions

$\displaystyle \textbf{Question 1: }\text{Find the intervals in which the following functions are increasing or decreasing.}$
$\displaystyle \text{(i) }f(x)=10-6x-2x^2$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=10-6x-2x^2$
$\displaystyle \therefore f'(x)=-6-4x=-2(2x+3)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow -2(2x+3)>0$
$\displaystyle \Rightarrow 2x+3<0$
$\displaystyle \Rightarrow x<-\frac{3}{2}$
$\displaystyle \therefore f(x)\text{ is increasing on }\left(-\infty,-\frac{3}{2}\right).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow -2(2x+3)<0$
$\displaystyle \Rightarrow 2x+3>0$
$\displaystyle \Rightarrow x>-\frac{3}{2}$
$\displaystyle \therefore f(x)\text{ is decreasing on }\left(-\frac{3}{2},\infty\right).$
$\displaystyle \\$

$\displaystyle \text{(ii) }f(x)=x^2+2x-5$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^2+2x-5$
$\displaystyle \therefore f'(x)=2x+2=2(x+1)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow 2(x+1)>0$
$\displaystyle \Rightarrow x+1>0$
$\displaystyle \Rightarrow x>-1$
$\displaystyle \therefore f(x)\text{ is increasing on }(-1,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow 2(x+1)<0$
$\displaystyle \Rightarrow x+1<0$
$\displaystyle \Rightarrow x<-1$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-\infty,-1).$
$\displaystyle \\$

$\displaystyle \text{(iii) }f(x)=6-9x-x^2$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=6-9x-x^2$
$\displaystyle \therefore f'(x)=-2x-9=-(2x+9)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow -(2x+9)>0$
$\displaystyle \Rightarrow 2x+9<0$
$\displaystyle \Rightarrow x<-\frac{9}{2}$
$\displaystyle \therefore f(x)\text{ is increasing on }\left(-\infty,-\frac{9}{2}\right).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow -(2x+9)<0$
$\displaystyle \Rightarrow 2x+9>0$
$\displaystyle \Rightarrow x>-\frac{9}{2}$
$\displaystyle \therefore f(x)\text{ is decreasing on }\left(-\frac{9}{2},\infty\right).$
$\displaystyle \\$

$\displaystyle \text{(iv) }f(x)=2x^3-12x^2+18x+15$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=2x^3-12x^2+18x+15$
$\displaystyle \therefore f'(x)=6x^2-24x+18$
$\displaystyle =6(x^2-4x+3)$
$\displaystyle =6(x-1)(x-3)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow 6(x-1)(x-3)>0$
$\displaystyle \Rightarrow (x-1)(x-3)>0\qquad[\because\,6>0]$
$\displaystyle \Rightarrow x<1\text{ or }x>3$
$\displaystyle \therefore f(x)\text{ is increasing on }(-\infty,1)\cup(3,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow 6(x-1)(x-3)<0$
$\displaystyle \Rightarrow (x-1)(x-3)<0\qquad[\because\,6>0]$
$\displaystyle \Rightarrow 1<x<3$
$\displaystyle \therefore f(x)\text{ is decreasing on }(1,3).$
$\displaystyle \\$

$\displaystyle \text{(v) }f(x)=5+36x+3x^2-2x^3$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=5+36x+3x^2-2x^3$
$\displaystyle \therefore f'(x)=36+6x-6x^2$
$\displaystyle =-6(x^2-x-6)$
$\displaystyle =-6(x-3)(x+2)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow -6(x-3)(x+2)>0$
$\displaystyle \Rightarrow (x-3)(x+2)<0\qquad[\because\,-6<0]$
$\displaystyle \Rightarrow -2<x<3$
$\displaystyle \therefore f(x)\text{ is increasing on }(-2,3).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow -6(x-3)(x+2)<0$
$\displaystyle \Rightarrow (x-3)(x+2)>0\qquad[\because\,-6<0]$
$\displaystyle \Rightarrow x<-2\text{ or }x>3$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-\infty,-2)\cup(3,\infty).$
$\displaystyle \\$

$\displaystyle \text{(vi) }f(x)=8+36x+3x^2-2x^3$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=8+36x+3x^2-2x^3$
$\displaystyle \therefore f'(x)=36+6x-6x^2$
$\displaystyle =-6(x^2-x-6)$
$\displaystyle =-6(x-3)(x+2)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow -6(x-3)(x+2)>0$
$\displaystyle \Rightarrow (x-3)(x+2)<0\qquad[\because\,-6<0]$
$\displaystyle \Rightarrow -2<x<3$
$\displaystyle \therefore f(x)\text{ is increasing on }(-2,3).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow -6(x-3)(x+2)<0$
$\displaystyle \Rightarrow (x-3)(x+2)>0\qquad[\because\,-6<0]$
$\displaystyle \Rightarrow x<-2\text{ or }x>3$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-\infty,-2)\cup(3,\infty).$
$\displaystyle \\$

$\displaystyle \text{(vii) }f(x)=5x^3-15x^2-120x+3$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=5x^3-15x^2-120x+3$
$\displaystyle \therefore f'(x)=15x^2-30x-120$
$\displaystyle =15(x^2-2x-8)$
$\displaystyle =15(x-4)(x+2)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow 15(x-4)(x+2)>0$
$\displaystyle \Rightarrow (x-4)(x+2)>0\qquad[\because\,15>0]$
$\displaystyle \Rightarrow x<-2\text{ or }x>4$
$\displaystyle \therefore f(x)\text{ is increasing on }(-\infty,-2)\cup(4,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow 15(x-4)(x+2)<0$
$\displaystyle \Rightarrow (x-4)(x+2)<0\qquad[\because\,15>0]$
$\displaystyle \Rightarrow -2<x<4$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-2,4).$
$\displaystyle \\$

$\displaystyle \text{(viii) }f(x)=x^3-6x^2-36x+2$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^3-6x^2-36x+2$
$\displaystyle \therefore f'(x)=3x^2-12x-36$
$\displaystyle =3(x^2-4x-12)$
$\displaystyle =3(x-6)(x+2)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow 3(x-6)(x+2)>0$
$\displaystyle \Rightarrow (x-6)(x+2)>0\qquad[\because\,3>0]$
$\displaystyle \Rightarrow x<-2\text{ or }x>6$
$\displaystyle \therefore f(x)\text{ is increasing on }(-\infty,-2)\cup(6,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow 3(x-6)(x+2)<0$
$\displaystyle \Rightarrow (x-6)(x+2)<0\qquad[\because\,3>0]$
$\displaystyle \Rightarrow -2<x<6$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-2,6).$
$\displaystyle \\$

$\displaystyle \text{(ix) }f(x)=2x^3-15x^2+36x+1\hfill\text{[CBSE 2005, 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=2x^3-15x^2+36x+1$
$\displaystyle \therefore f'(x)=6x^2-30x+36$
$\displaystyle =6(x^2-5x+6)$
$\displaystyle =6(x-2)(x-3)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow 6(x-2)(x-3)>0$
$\displaystyle \Rightarrow (x-2)(x-3)>0\qquad[\because\,6>0]$
$\displaystyle \Rightarrow x<2\text{ or }x>3$
$\displaystyle \therefore f(x)\text{ is increasing on }(-\infty,2)\cup(3,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow 6(x-2)(x-3)<0$
$\displaystyle \Rightarrow (x-2)(x-3)<0\qquad[\because\,6>0]$
$\displaystyle \Rightarrow 2<x<3$
$\displaystyle \therefore f(x)\text{ is decreasing on }(2,3).$
$\displaystyle \\$

$\displaystyle \text{(x) }f(x)=2x^3+9x^2+12x+20\hfill\text{[CBSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=2x^3+9x^2+12x+20$
$\displaystyle \therefore f'(x)=6x^2+18x+12$
$\displaystyle =6(x^2+3x+2)$
$\displaystyle =6(x+1)(x+2)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow 6(x+1)(x+2)>0$
$\displaystyle \Rightarrow (x+1)(x+2)>0\qquad[\because\,6>0]$
$\displaystyle \Rightarrow x<-2\text{ or }x>-1$
$\displaystyle \therefore f(x)\text{ is increasing on }(-\infty,-2)\cup(-1,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow 6(x+1)(x+2)<0$
$\displaystyle \Rightarrow (x+1)(x+2)<0\qquad[\because\,6>0]$
$\displaystyle \Rightarrow -2<x<-1$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-2,-1).$
$\displaystyle \\$

$\displaystyle \text{(xi) }f(x)=2x^3-9x^2+12x-5$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=2x^3-9x^2+12x-5$
$\displaystyle \therefore f'(x)=6x^2-18x+12$
$\displaystyle =6(x^2-3x+2)$
$\displaystyle =6(x-1)(x-2)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow 6(x-1)(x-2)>0$
$\displaystyle \Rightarrow (x-1)(x-2)>0\qquad[\because\,6>0]$
$\displaystyle \Rightarrow x<1\text{ or }x>2$
$\displaystyle \therefore f(x)\text{ is increasing on }(-\infty,1)\cup(2,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow 6(x-1)(x-2)<0$
$\displaystyle \Rightarrow (x-1)(x-2)<0\qquad[\because\,6>0]$
$\displaystyle \Rightarrow 1<x<2$
$\displaystyle \therefore f(x)\text{ is decreasing on }(1,2).$
$\displaystyle \\$

$\displaystyle \text{(xii) }f(x)=6+12x+3x^2-2x^3$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=6+12x+3x^2-2x^3$
$\displaystyle \therefore f'(x)=12+6x-6x^2$
$\displaystyle =-6(x^2-x-2)$
$\displaystyle =-6(x-2)(x+1)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow -6(x-2)(x+1)>0$
$\displaystyle \Rightarrow (x-2)(x+1)<0\qquad[\because\,-6<0]$
$\displaystyle \Rightarrow -1<x<2$
$\displaystyle \therefore f(x)\text{ is increasing on }(-1,2).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow -6(x-2)(x+1)<0$
$\displaystyle \Rightarrow (x-2)(x+1)>0\qquad[\because\,-6<0]$
$\displaystyle \Rightarrow x<-1\text{ or }x>2$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-\infty,-1)\cup(2,\infty).$
$\displaystyle \\$

$\displaystyle \text{(xiii) }f(x)=2x^3-24x+107$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=2x^3-24x+107$
$\displaystyle \therefore f'(x)=6x^2-24$
$\displaystyle =6(x^2-4)$
$\displaystyle =6(x+2)(x-2)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow 6(x+2)(x-2)>0$
$\displaystyle \Rightarrow (x+2)(x-2)>0\qquad[\because\,6>0]$
$\displaystyle \Rightarrow x<-2\text{ or }x>2$
$\displaystyle \therefore f(x)\text{ is increasing on }(-\infty,-2)\cup(2,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow 6(x+2)(x-2)<0$
$\displaystyle \Rightarrow (x+2)(x-2)<0\qquad[\because\,6>0]$
$\displaystyle \Rightarrow -2<x<2$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-2,2).$
$\displaystyle \\$

$\displaystyle \text{(xiv) }f(x)=-2x^3-9x^2-12x+1$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=-2x^3-9x^2-12x+1$
$\displaystyle \therefore f'(x)=-6x^2-18x-12$
$\displaystyle =-6(x^2+3x+2)$
$\displaystyle =-6(x+1)(x+2)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow -6(x+1)(x+2)>0$
$\displaystyle \Rightarrow (x+1)(x+2)<0\qquad[\because\,-6<0]$
$\displaystyle \Rightarrow -2<x<-1$
$\displaystyle \therefore f(x)\text{ is increasing on }(-2,-1).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow -6(x+1)(x+2)<0$
$\displaystyle \Rightarrow (x+1)(x+2)>0\qquad[\because\,-6<0]$
$\displaystyle \Rightarrow x<-2\text{ or }x>-1$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-\infty,-2)\cup(-1,\infty).$
$\displaystyle \\$

$\displaystyle \text{(xv) }f(x)=(x-1)(x-2)^2\hfill\text{[CBSE 2001]}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=(x-1)(x-2)^2$
$\displaystyle =(x-1)(x^2-4x+4)$
$\displaystyle =x^3-5x^2+8x-4$
$\displaystyle \therefore f'(x)=3x^2-10x+8$
$\displaystyle =3x^2-6x-4x+8$
$\displaystyle =(x-2)(3x-4)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow (x-2)(3x-4)>0$
$\displaystyle \Rightarrow x<\frac{4}{3}\text{ or }x>2$
$\displaystyle \therefore f(x)\text{ is increasing on }\left(-\infty,\frac{4}{3}\right)\cup(2,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow (x-2)(3x-4)<0$
$\displaystyle \Rightarrow \frac{4}{3}<x<2$
$\displaystyle \therefore f(x)\text{ is decreasing on }\left(\frac{4}{3},2\right).$
$\displaystyle \\$

$\displaystyle \text{(xvi) }f(x)=x^3-12x^2+36x+17$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^3-12x^2+36x+17$
$\displaystyle \therefore f'(x)=3x^2-24x+36$
$\displaystyle =3(x^2-8x+12)$
$\displaystyle =3(x-2)(x-6)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow 3(x-2)(x-6)>0$
$\displaystyle \Rightarrow (x-2)(x-6)>0\qquad[\because\,3>0]$
$\displaystyle \Rightarrow x<2\text{ or }x>6$
$\displaystyle \therefore f(x)\text{ is increasing on }(-\infty,2)\cup(6,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow 3(x-2)(x-6)<0$
$\displaystyle \Rightarrow (x-2)(x-6)<0\qquad[\because\,3>0]$
$\displaystyle \Rightarrow 2<x<6$
$\displaystyle \therefore f(x)\text{ is decreasing on }(2,6).$
$\displaystyle \\$

$\displaystyle \text{(xvii) }f(x)=2x^3-24x+7$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=2x^3-24x+7$
$\displaystyle \therefore f'(x)=6x^2-24$
$\displaystyle =6(x^2-4)$
$\displaystyle =6(x+2)(x-2)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow 6(x+2)(x-2)>0$
$\displaystyle \Rightarrow (x+2)(x-2)>0\qquad[\because\,6>0]$
$\displaystyle \Rightarrow x<-2\text{ or }x>2$
$\displaystyle \therefore f(x)\text{ is increasing on }(-\infty,-2)\cup(2,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow 6(x+2)(x-2)<0$
$\displaystyle \Rightarrow (x+2)(x-2)<0\qquad[\because\,6>0]$
$\displaystyle \Rightarrow -2<x<2$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-2,2).$
$\displaystyle \\$

$\displaystyle \text{(xviii) }f(x)=\frac{3}{10}x^4-\frac{4}{5}x^3-3x^2+\frac{36}{5}x+11$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\frac{3}{10}x^4-\frac{4}{5}x^3-3x^2+\frac{36}{5}x+11$
$\displaystyle \therefore f'(x)=\frac{6}{5}x^3-\frac{12}{5}x^2-6x+\frac{36}{5}$
$\displaystyle =\frac{6}{5}(x^3-2x^2-5x+6)$
$\displaystyle =\frac{6}{5}(x-1)(x^2-x-6)$
$\displaystyle =\frac{6}{5}(x+2)(x-1)(x-3)$
$\displaystyle f'(x)=0\Rightarrow x=-2,\,1,\,3$
$\displaystyle \text{These points divide }\mathbb{R}\text{ into }(-\infty,-2),\,(-2,1),\,(1,3)\text{ and }(3,\infty).$
$\displaystyle f'(x)<0\text{ on }(-\infty,-2),\qquad f'(x)>0\text{ on }(-2,1)$
$\displaystyle f'(x)<0\text{ on }(1,3),\qquad f'(x)>0\text{ on }(3,\infty)$
$\displaystyle \therefore f(x)\text{ is increasing on }(-2,1)\cup(3,\infty).$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-\infty,-2)\cup(1,3).$
$\displaystyle \\$

$\displaystyle \text{(xix) }f(x)=x^4-4x$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^4-4x$
$\displaystyle \therefore f'(x)=4x^3-4$
$\displaystyle =4(x^3-1)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow 4(x^3-1)>0$
$\displaystyle \Rightarrow x^3-1>0\qquad[\because\,4>0]$
$\displaystyle \Rightarrow x^3>1$
$\displaystyle \Rightarrow x>1$
$\displaystyle \therefore f(x)\text{ is increasing on }(1,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow 4(x^3-1)<0$
$\displaystyle \Rightarrow x^3-1<0\qquad[\because\,4>0]$
$\displaystyle \Rightarrow x^3<1$
$\displaystyle \Rightarrow x<1$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-\infty,1).$
$\displaystyle \\$

$\displaystyle \text{(xx) }f(x)=\frac{x^4}{4}+\frac{2}{3}x^3-\frac{5}{2}x^2-6x+7$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\frac{x^4}{4}+\frac{2}{3}x^3-\frac{5}{2}x^2-6x+7$
$\displaystyle \therefore f'(x)=x^3+2x^2-5x-6$
$\displaystyle =(x+1)(x^2+x-6)$
$\displaystyle =(x+1)(x-2)(x+3)$
$\displaystyle f'(x)=0\Rightarrow x=-3,\,-1,\,2$
$\displaystyle \text{These points divide }\mathbb{R}\text{ into }(-\infty,-3),\,(-3,-1),\,(-1,2)\text{ and }(2,\infty).$
$\displaystyle f'(x)<0\text{ on }(-\infty,-3),\qquad f'(x)>0\text{ on }(-3,-1)$
$\displaystyle f'(x)<0\text{ on }(-1,2),\qquad f'(x)>0\text{ on }(2,\infty)$
$\displaystyle \therefore f(x)\text{ is increasing on }(-3,-1)\cup(2,\infty).$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-\infty,-3)\cup(-1,2).$
$\displaystyle \\$

$\displaystyle \text{(xxi) }f(x)=x^4-4x^3+4x^2+15$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^4-4x^3+4x^2+15$
$\displaystyle \therefore f'(x)=4x^3-12x^2+8x$
$\displaystyle =4x(x^2-3x+2)$
$\displaystyle =4x(x-1)(x-2)$
$\displaystyle f'(x)=0\Rightarrow x=0,\,1,\,2$
$\displaystyle \text{These points divide }\mathbb{R}\text{ into }(-\infty,0),\,(0,1),\,(1,2)\text{ and }(2,\infty).$
$\displaystyle f'(x)<0\text{ on }(-\infty,0),\qquad f'(x)>0\text{ on }(0,1)$
$\displaystyle f'(x)<0\text{ on }(1,2),\qquad f'(x)>0\text{ on }(2,\infty)$
$\displaystyle \therefore f(x)\text{ is increasing on }(0,1)\cup(2,\infty).$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-\infty,0)\cup(1,2).$
$\displaystyle \\$

$\displaystyle \text{(xxii) }f(x)=5x^{3/2}-3x^{5/2},\;x>0$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=5x^{3/2}-3x^{5/2},\quad x>0$
$\displaystyle \therefore f'(x)=\frac{15}{2}x^{1/2}-\frac{15}{2}x^{3/2}$
$\displaystyle =\frac{15}{2}x^{1/2}(1-x)$
$\displaystyle \text{Since }\frac{15}{2}>0\text{ and }x^{1/2}>0\text{ for }x>0,\text{ the sign of }f'(x)\text{ depends on }(1-x).$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow 1-x>0$
$\displaystyle \Rightarrow 0<x<1$
$\displaystyle \therefore f(x)\text{ is increasing on }(0,1).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow 1-x<0$
$\displaystyle \Rightarrow x>1$
$\displaystyle \therefore f(x)\text{ is decreasing on }(1,\infty).$
$\displaystyle \\$

$\displaystyle \text{(xxiii) }f(x)=x^8+6x^2$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^8+6x^2$
$\displaystyle \therefore f'(x)=8x^7+12x$
$\displaystyle =4x(2x^6+3)$
$\displaystyle \text{Since }2x^6+3>0\text{ for all }x\in\mathbb{R},\text{ the sign of }f'(x)\text{ depends only on the sign of }x.$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow x>0$
$\displaystyle \therefore f(x)\text{ is increasing on }(0,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow x<0$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-\infty,0).$
$\displaystyle \\$

$\displaystyle \text{(xxiv) }f(x)=x^3-6x^2+9x+15\hfill\text{[CBSE 2000, 2004]}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^3-6x^2+9x+15$
$\displaystyle \therefore f'(x)=3x^2-12x+9$
$\displaystyle =3(x^2-4x+3)$
$\displaystyle =3(x-1)(x-3)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow 3(x-1)(x-3)>0$
$\displaystyle \Rightarrow (x-1)(x-3)>0\qquad[\because\,3>0]$
$\displaystyle \Rightarrow x<1\text{ or }x>3$
$\displaystyle \therefore f(x)\text{ is increasing on }(-\infty,1)\cup(3,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow 3(x-1)(x-3)<0$
$\displaystyle \Rightarrow (x-1)(x-3)<0\qquad[\because\,3>0]$
$\displaystyle \Rightarrow 1<x<3$
$\displaystyle \therefore f(x)\text{ is decreasing on }(1,3).$
$\displaystyle \\$

$\displaystyle \text{(xxv) }f(x)=\{x(x-2)\}^2\hfill\text{[CBSE 2010, 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\{x(x-2)\}^2$
$\displaystyle \therefore f'(x)=2\{x(x-2)\}\frac{d}{dx}\{x(x-2)\}$
$\displaystyle =2x(x-2)(2x-2)$
$\displaystyle =4x(x-1)(x-2)$
$\displaystyle f'(x)=0\Rightarrow x=0,\,1,\,2$
$\displaystyle \text{These points divide }\mathbb{R}\text{ into }(-\infty,0),\,(0,1),\,(1,2)\text{ and }(2,\infty).$
$\displaystyle f'(x)<0\text{ on }(-\infty,0),\qquad f'(x)>0\text{ on }(0,1)$
$\displaystyle f'(x)<0\text{ on }(1,2),\qquad f'(x)>0\text{ on }(2,\infty)$
$\displaystyle \therefore f(x)\text{ is increasing on }(0,1)\cup(2,\infty).$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-\infty,0)\cup(1,2).$
$\displaystyle \\$

$\displaystyle \text{(xxvi) }f(x)=3x^4-4x^3-12x^2+5\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=3x^4-4x^3-12x^2+5$
$\displaystyle \therefore f'(x)=12x^3-12x^2-24x$
$\displaystyle =12x(x^2-x-2)$
$\displaystyle =12x(x+1)(x-2)$
$\displaystyle f'(x)=0\Rightarrow x=-1,\,0,\,2$
$\displaystyle \text{These points divide }\mathbb{R}\text{ into }(-\infty,-1),\,(-1,0),\,(0,2)\text{ and }(2,\infty).$
$\displaystyle f'(x)<0\text{ on }(-\infty,-1),\qquad f'(x)>0\text{ on }(-1,0)$
$\displaystyle f'(x)<0\text{ on }(0,2),\qquad f'(x)>0\text{ on }(2,\infty)$
$\displaystyle \therefore f(x)\text{ is increasing on }(-1,0)\cup(2,\infty).$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-\infty,-1)\cup(0,2).$
$\displaystyle \\$

$\displaystyle \text{(xxvii) }f(x)=\frac{3}{2}x^4-4x^3-45x^2+51\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\frac{3}{2}x^4-4x^3-45x^2+51$
$\displaystyle \therefore f'(x)=6x^3-12x^2-90x$
$\displaystyle =6x(x^2-2x-15)$
$\displaystyle =6x(x-5)(x+3)$
$\displaystyle f'(x)=0\Rightarrow x=-3,\,0,\,5$
$\displaystyle \text{These points divide }\mathbb{R}\text{ into }(-\infty,-3),\,(-3,0),\,(0,5)\text{ and }(5,\infty).$
$\displaystyle f'(x)<0\text{ on }(-\infty,-3),\qquad f'(x)>0\text{ on }(-3,0)$
$\displaystyle f'(x)<0\text{ on }(0,5),\qquad f'(x)>0\text{ on }(5,\infty)$
$\displaystyle \therefore f(x)\text{ is increasing on }(-3,0)\cup(5,\infty).$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-\infty,-3)\cup(0,5).$
$\displaystyle \\$

$\displaystyle \text{(xxviii) }f(x)=\log(2+x)-\frac{2x}{2+x}\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\log(2+x)-\frac{2x}{2+x}$
$\displaystyle \text{Since }2+x>0,\text{ the domain of }f\text{ is }(-2,\infty).$
$\displaystyle \therefore f'(x)=\frac{1}{2+x}-\frac{2(2+x)-2x}{(2+x)^2}$
$\displaystyle =\frac{2+x-4}{(2+x)^2}$
$\displaystyle =\frac{x-2}{(x+2)^2}$
$\displaystyle \text{Since }(x+2)^2>0\text{ for }x>-2,\text{ the sign of }f'(x)\text{ depends on }x-2.$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow x-2>0$
$\displaystyle \Rightarrow x>2$
$\displaystyle \therefore f(x)\text{ is increasing on }(2,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow x-2<0$
$\displaystyle \Rightarrow -2<x<2$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-2,2).$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Determine the values of }x\text{ for which the function }f(x)=x^2-6x+9\text{ is increasing or decreasing. Also, find the coordinates of the point on the curve}$
$\displaystyle y=x^2-6x+9\text{ where the normal is parallel to the line }y=x+5.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^2-6x+9$
$\displaystyle \therefore f'(x)=2x-6$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow 2x-6>0$
$\displaystyle \Rightarrow x>3$
$\displaystyle \therefore f(x)\text{ is increasing on }(3,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow 2x-6<0$
$\displaystyle \Rightarrow x<3$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-\infty,3).$
$\displaystyle \text{The slope of the line }y=x+5\text{ is }1.$
$\displaystyle \text{Since the normal is parallel to this line, its slope is }1.$
$\displaystyle \text{Therefore, the slope of the tangent is }-1.$
$\displaystyle \therefore \frac{dy}{dx}=2x-6=-1$
$\displaystyle \Rightarrow 2x=5$
$\displaystyle \Rightarrow x=\frac{5}{2}$
$\displaystyle \text{On the curve }y=x^2-6x+9,$
$\displaystyle y=\left(\frac{5}{2}\right)^2-6\left(\frac{5}{2}\right)+9$
$\displaystyle =\frac{25}{4}-15+9$
$\displaystyle =\frac{1}{4}$
$\displaystyle \therefore \text{The required point is }\left(\frac{5}{2},\frac{1}{4}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the intervals in which }f(x)=\sin x-\cos x,\text{ where }0<x<2\pi,\text{ is increasing or decreasing.}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\sin x-\cos x,\qquad x\in(0,2\pi)$
$\displaystyle \therefore f'(x)=\cos x+\sin x$
$\displaystyle =\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow \sin\left(x+\frac{\pi}{4}\right)>0$
$\displaystyle \Rightarrow x+\frac{\pi}{4}\in(0,\pi)\cup(2\pi,3\pi)$
$\displaystyle \Rightarrow x\in\left(-\frac{\pi}{4},\frac{3\pi}{4}\right)\cup\left(\frac{7\pi}{4},\frac{11\pi}{4}\right)$
$\displaystyle \text{Since }x\in(0,2\pi),$
$\displaystyle \therefore f(x)\text{ is increasing on }\left(0,\frac{3\pi}{4}\right)\cup\left(\frac{7\pi}{4},2\pi\right).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow \sin\left(x+\frac{\pi}{4}\right)<0$
$\displaystyle \Rightarrow x+\frac{\pi}{4}\in(\pi,2\pi)$
$\displaystyle \Rightarrow x\in\left(\frac{3\pi}{4},\frac{7\pi}{4}\right)$
$\displaystyle \therefore f(x)\text{ is decreasing on }\left(\frac{3\pi}{4},\frac{7\pi}{4}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Show that }f(x)=e^{2x}\text{ is increasing on }\mathbb{R}.\hfill\text{[CBSE 2000, 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=e^{2x}$
$\displaystyle \therefore f'(x)=2e^{2x}$
$\displaystyle \text{Since }e^{2x}>0\text{ for all }x\in\mathbb{R},$
$\displaystyle \therefore 2e^{2x}>0$
$\displaystyle \Rightarrow f'(x)>0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \therefore f(x)\text{ is increasing on }\mathbb{R}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Show that }f(x)=e^{1/x},\;x\neq0,\text{ is a decreasing function for all }x\neq0.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=e^{\frac{1}{x}},\qquad x\neq0$
$\displaystyle \therefore f'(x)=e^{\frac{1}{x}}\cdot\frac{d}{dx}\left(\frac{1}{x}\right)$
$\displaystyle =e^{\frac{1}{x}}\left(-\frac{1}{x^2}\right)$
$\displaystyle =-\frac{e^{\frac{1}{x}}}{x^2}$
$\displaystyle \text{Since }e^{\frac{1}{x}}>0\text{ and }x^2>0\text{ for all }x\in\mathbb{R},\,x\neq0,$
$\displaystyle \therefore f'(x)<0\text{ for all }x\neq0.$
$\displaystyle \therefore f(x)\text{ is a decreasing function for all }x\neq0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Show that }f(x)=\log_a x,\;0<a<1,\text{ is a decreasing function for all }x>0.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\log_a x$
$\displaystyle =\frac{\log x}{\log a}$
$\displaystyle \therefore f'(x)=\frac{1}{x\log a}$
$\displaystyle \text{Since }0<a<1,\text{ we have }\log a<0.$
$\displaystyle \text{Also, }x>0.$
$\displaystyle \therefore x\log a<0$
$\displaystyle \Rightarrow \frac{1}{x\log a}<0$
$\displaystyle \Rightarrow f'(x)<0\text{ for all }x>0.$
$\displaystyle \therefore f(x)\text{ is decreasing for all }x>0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Show that }f(x)=\sin x\text{ is increasing on }\left(0,\frac{\pi}{2}\right)\text{ and decreasing on }\left(\frac{\pi}{2},\pi\right),$
$\displaystyle \text{and neither increasing nor decreasing on }(0,\pi).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\sin x,\qquad x\in(0,\pi)$
$\displaystyle \therefore f'(x)=\cos x$
$\displaystyle \text{For }x\in\left(0,\frac{\pi}{2}\right),\;\cos x>0.$
$\displaystyle \therefore f'(x)>0$
$\displaystyle \Rightarrow f(x)\text{ is increasing on }\left(0,\frac{\pi}{2}\right).$
$\displaystyle \text{For }x\in\left(\frac{\pi}{2},\pi\right),\;\cos x<0.$
$\displaystyle \therefore f'(x)<0$
$\displaystyle \Rightarrow f(x)\text{ is decreasing on }\left(\frac{\pi}{2},\pi\right).$
$\displaystyle \text{Since }f(x)\text{ is increasing on }\left(0,\frac{\pi}{2}\right)\text{ and decreasing on }\left(\frac{\pi}{2},\pi\right),$
$\displaystyle \therefore f(x)\text{ is neither increasing nor decreasing on }(0,\pi).$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Show that }f(x)=\log(\sin x)\text{ is increasing on }\left(0,\frac{\pi}{2}\right)\text{ and decreasing on }\left(\frac{\pi}{2},\pi\right).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\log(\sin x),\qquad x\in(0,\pi)$
$\displaystyle \therefore f'(x)=\frac{1}{\sin x}\cdot\cos x$
$\displaystyle =\cot x$
$\displaystyle \text{For }x\in\left(0,\frac{\pi}{2}\right),\;\cot x>0.$
$\displaystyle \therefore f'(x)>0$
$\displaystyle \Rightarrow f(x)\text{ is increasing on }\left(0,\frac{\pi}{2}\right).$
$\displaystyle \text{For }x\in\left(\frac{\pi}{2},\pi\right),\;\cot x<0.$
$\displaystyle \therefore f'(x)<0$
$\displaystyle \Rightarrow f(x)\text{ is decreasing on }\left(\frac{\pi}{2},\pi\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Show that }f(x)=x-\sin x\text{ is increasing for all }x\in\mathbb{R}.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x-\sin x$
$\displaystyle \therefore f'(x)=1-\cos x$
$\displaystyle \text{Since }\cos x\le1\text{ for all }x\in\mathbb{R},$
$\displaystyle \therefore 1-\cos x\ge0$
$\displaystyle \Rightarrow f'(x)\ge0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \therefore f(x)\text{ is increasing for all }x\in\mathbb{R}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Show that }f(x)=x^3-15x^2+75x-50\text{ is an increasing function for all }x\in\mathbb{R}.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^3-15x^2+75x-50$
$\displaystyle \therefore f'(x)=3x^2-30x+75$
$\displaystyle =3(x^2-10x+25)$
$\displaystyle =3(x-5)^2$
$\displaystyle \text{Since }(x-5)^2\ge0\text{ for all }x\in\mathbb{R},$
$\displaystyle \therefore f'(x)=3(x-5)^2\ge0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \therefore f(x)\text{ is an increasing function for all }x\in\mathbb{R}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Show that }f(x)=\cos^2x\text{ is a decreasing function on }\left(0,\frac{\pi}{2}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\cos^2x$
$\displaystyle \therefore f'(x)=2\cos x(-\sin x)$
$\displaystyle =-\sin2x$
$\displaystyle \text{Now, }0<x<\frac{\pi}{2}$
$\displaystyle \Rightarrow 0<2x<\pi$
$\displaystyle \Rightarrow \sin2x>0$
$\displaystyle \Rightarrow -\sin2x<0$
$\displaystyle \therefore f'(x)<0$
$\displaystyle \therefore f(x)\text{ is decreasing on }\left(0,\frac{\pi}{2}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Show that }f(x)=\sin x\text{ is an increasing function on }\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\sin x$
$\displaystyle \therefore f'(x)=\cos x$
$\displaystyle \text{For }x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),\;\cos x>0.$
$\displaystyle \therefore f'(x)>0$
$\displaystyle \therefore f(x)\text{ is increasing on }\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Show that }f(x)=\cos x\text{ is a decreasing function on }(0,\pi), \\ \text{ increasing on }(-\pi,0)\text{ and neither increasing nor decreasing on }(-\pi,\pi).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\cos x,\qquad x\in(-\pi,\pi)$
$\displaystyle \therefore f'(x)=-\sin x$
$\displaystyle \text{For }x\in(-\pi,0),\;\sin x<0.$
$\displaystyle \therefore -\sin x>0$
$\displaystyle \Rightarrow f(x)\text{ is increasing on }(-\pi,0).$
$\displaystyle \text{For }x\in(0,\pi),\;\sin x>0.$
$\displaystyle \therefore -\sin x<0$
$\displaystyle \Rightarrow f(x)\text{ is decreasing on }(0,\pi).$
$\displaystyle \text{Since }f(x)\text{ is increasing on }(-\pi,0)\text{ and decreasing on }(0,\pi),$
$\displaystyle \therefore f(x)\text{ is neither increasing nor decreasing on }(-\pi,\pi).$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Show that }f(x)=\tan x\text{ is an increasing function on }\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\tan x$
$\displaystyle \therefore f'(x)=\sec^2x$
$\displaystyle \text{Since }\sec^2x>0\text{ for all }x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$
$\displaystyle \therefore f'(x)>0\text{ for all }x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$
$\displaystyle \therefore f(x)\text{ is increasing on }\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Show that }f(x)=\tan^{-1}(\sin x+\cos x)\text{ is decreasing on }\left(\frac{\pi}{4},\frac{\pi}{2}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\tan^{-1}(\sin x+\cos x)$
$\displaystyle \therefore f'(x)=\frac{\cos x-\sin x}{1+(\sin x+\cos x)^2}$
$\displaystyle =\frac{\cos x-\sin x}{1+1+2\sin x\cos x}$
$\displaystyle =\frac{\cos x-\sin x}{2+\sin2x}$
$\displaystyle \text{For }\frac{\pi}{4}<x<\frac{\pi}{2},\text{ we have }\frac{\pi}{2}<2x<\pi.$
$\displaystyle \therefore \sin2x>0$
$\displaystyle \Rightarrow 2+\sin2x>0$
$\displaystyle \text{Also, for }\frac{\pi}{4}<x<\frac{\pi}{2},\;\cos x<\sin x.$
$\displaystyle \therefore \cos x-\sin x<0$
$\displaystyle \Rightarrow f'(x)=\frac{\cos x-\sin x}{2+\sin2x}<0$
$\displaystyle \therefore f(x)\text{ is decreasing on }\left(\frac{\pi}{4},\frac{\pi}{2}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{Show that the function }f(x)=\sin\left(2x+\frac{\pi}{4}\right)\text{ is decreasing on }\left(\frac{3\pi}{8},\frac{5\pi}{8}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\sin\left(2x+\frac{\pi}{4}\right)$
$\displaystyle \therefore f'(x)=2\cos\left(2x+\frac{\pi}{4}\right)$
$\displaystyle \text{For }\frac{3\pi}{8}<x<\frac{5\pi}{8},$
$\displaystyle \Rightarrow \frac{3\pi}{4}<2x<\frac{5\pi}{4}$
$\displaystyle \Rightarrow \pi<2x+\frac{\pi}{4}<\frac{3\pi}{2}$
$\displaystyle \Rightarrow \cos\left(2x+\frac{\pi}{4}\right)<0$
$\displaystyle \Rightarrow 2\cos\left(2x+\frac{\pi}{4}\right)<0$
$\displaystyle \therefore f'(x)<0\text{ for all }x\in\left(\frac{3\pi}{8},\frac{5\pi}{8}\right).$
$\displaystyle \therefore f(x)\text{ is decreasing on }\left(\frac{3\pi}{8},\frac{5\pi}{8}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{Show that the function }f(x)=\cot^{-1}(\sin x+\cos x)\text{ is decreasing on }\left(0,\frac{\pi}{4}\right)$
$\displaystyle \text{and increasing on }\left(\frac{\pi}{4},\frac{\pi}{2}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\cot^{-1}(\sin x+\cos x)$
$\displaystyle \therefore f'(x)=-\frac{\cos x-\sin x}{1+(\sin x+\cos x)^2}$
$\displaystyle =\frac{\sin x-\cos x}{1+\sin^2x+\cos^2x+2\sin x\cos x}$
$\displaystyle =\frac{\sin x-\cos x}{2+2\sin x\cos x}$
$\displaystyle \text{For }0<x<\frac{\pi}{2},\;2+2\sin x\cos x>0.$
$\displaystyle \text{Therefore, the sign of }f'(x)\text{ depends only on }\sin x-\cos x.$
$\displaystyle \text{For }0<x<\frac{\pi}{4},\;\sin x<\cos x.$
$\displaystyle \therefore f'(x)<0$
$\displaystyle \Rightarrow f(x)\text{ is decreasing on }\left(0,\frac{\pi}{4}\right).$
$\displaystyle \text{For }\frac{\pi}{4}<x<\frac{\pi}{2},\;\sin x>\cos x.$
$\displaystyle \therefore f'(x)>0$
$\displaystyle \Rightarrow f(x)\text{ is increasing on }\left(\frac{\pi}{4},\frac{\pi}{2}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{Show that }f(x)=(x-1)e^x+1\text{ is an increasing function for all }x>0.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=(x-1)e^x+1$
$\displaystyle \therefore f'(x)=(x-1)e^x+e^x$
$\displaystyle =xe^x$
$\displaystyle \text{Since }x>0\text{ and }e^x>0\text{ for all real }x,$
$\displaystyle \therefore xe^x>0$
$\displaystyle \Rightarrow f'(x)>0\text{ for all }x>0.$
$\displaystyle \therefore f(x)\text{ is increasing for all }x>0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{Show that the function }x^2-x+1\text{ is neither increasing nor decreasing on }(0,1).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^2-x+1$
$\displaystyle \therefore f'(x)=2x-1$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow 2x-1>0$
$\displaystyle \Rightarrow x>\frac12$
$\displaystyle \therefore f(x)\text{ is increasing on }\left(\frac12,\infty\right).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow 2x-1<0$
$\displaystyle \Rightarrow x<\frac12$
$\displaystyle \therefore f(x)\text{ is decreasing on }\left(-\infty,\frac12\right).$
$\displaystyle \text{Hence, }f(x)\text{ is decreasing on }\left(0,\frac12\right)\text{ and increasing on }\left(\frac12,1\right).$
$\displaystyle \therefore f(x)\text{ is neither increasing nor decreasing on }(0,1).$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{Show that }f(x)=x^9+4x^7+11\text{ is an increasing function for all } \\ x\in\mathbb{R}.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^9+4x^7+11$
$\displaystyle \therefore f'(x)=9x^8+28x^6$
$\displaystyle =x^6(9x^2+28)$
$\displaystyle \text{Since }x^6\ge0\text{ and }9x^2+28>0\text{ for all }x\in\mathbb{R},$
$\displaystyle \therefore f'(x)\ge0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \therefore f(x)\text{ is increasing on }\mathbb{R}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{Prove that the function }f(x)=x^3-6x^2+12x-18\text{ is increasing on }\mathbb{R}. \\ \hfill\text{[CBSE 2002C]}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^3-6x^2+12x-18$
$\displaystyle \therefore f'(x)=3x^2-12x+12$
$\displaystyle =3(x^2-4x+4)$
$\displaystyle =3(x-2)^2$
$\displaystyle \text{Since }3>0\text{ and }(x-2)^2\ge0\text{ for all }x\in\mathbb{R},$
$\displaystyle \therefore f'(x)\ge0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \therefore f(x)\text{ is increasing on }\mathbb{R}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{State when a function }f(x)\text{ is said to be increasing on an interval }[a,b]. \\ \text{ Test whether the function }f(x)=x^2-6x+3\text{ is increasing on }[4,6].$
$\displaystyle \text{Answer:}$
$\displaystyle \text{A function }f(x)\text{ is said to be increasing on }[a,b]\text{ if }f'(x)\ge0\text{ for all }x\in(a,b).$
$\displaystyle f(x)=x^2-6x+3$
$\displaystyle \therefore f'(x)=2x-6$
$\displaystyle =2(x-3)$
$\displaystyle \text{For }x\in[4,6],\;x-3>0.$
$\displaystyle \therefore f'(x)=2(x-3)>0\text{ for all }x\in(4,6).$
$\displaystyle \therefore f(x)\text{ is increasing on }[4,6].$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{Show that }f(x)=\sin x-\cos x\text{ is an increasing function on }\left(-\frac{\pi}{4},\frac{\pi}{4}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\sin x-\cos x$
$\displaystyle \therefore f'(x)=\cos x+\sin x$
$\displaystyle =\cos x(1+\tan x)$
$\displaystyle \text{For }-\frac{\pi}{4}<x<\frac{\pi}{4},\;\cos x>0.$
$\displaystyle \text{Also, }-\frac{\pi}{4}<x<\frac{\pi}{4}\Rightarrow-1<\tan x<1.$
$\displaystyle \therefore 1+\tan x>0$
$\displaystyle \Rightarrow \cos x(1+\tan x)>0$
$\displaystyle \therefore f'(x)>0\text{ for all }x\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right).$
$\displaystyle \therefore f(x)\text{ is increasing on }\left(-\frac{\pi}{4},\frac{\pi}{4}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{Show that }f(x)=\tan^{-1}x-x\text{ is a decreasing function on }\mathbb{R}.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\tan^{-1}x-x$
$\displaystyle \therefore f'(x)=\frac{1}{1+x^2}-1$
$\displaystyle =\frac{1-(1+x^2)}{1+x^2}$
$\displaystyle =-\frac{x^2}{1+x^2}$
$\displaystyle \text{Since }x^2\ge0\text{ and }1+x^2>0\text{ for all }x\in\mathbb{R},$
$\displaystyle \therefore -\frac{x^2}{1+x^2}\le0$
$\displaystyle \Rightarrow f'(x)\le0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \therefore f(x)\text{ is decreasing on }\mathbb{R}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{Determine whether }f(x)=-\frac{x}{2}+\sin x\text{ is increasing or decreasing on }\left(-\frac{\pi}{3},\frac{\pi}{3}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=-\frac{x}{2}+\sin x$
$\displaystyle \therefore f'(x)=-\frac12+\cos x$
$\displaystyle \text{For }-\frac{\pi}{3}<x<\frac{\pi}{3},\;\cos x>\frac12.$
$\displaystyle \therefore -\frac12+\cos x>0$
$\displaystyle \Rightarrow f'(x)>0\text{ for all }x\in\left(-\frac{\pi}{3},\frac{\pi}{3}\right).$
$\displaystyle \therefore f(x)\text{ is increasing on }\left(-\frac{\pi}{3},\frac{\pi}{3}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{Find the intervals in which }f(x)=\log(1+x)-\frac{x}{1+x}\text{ is increasing or decreasing.}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\log(1+x)-\frac{x}{1+x}$
$\displaystyle \text{Since }1+x>0,\text{ the domain of }f\text{ is }(-1,\infty).$
$\displaystyle \therefore f'(x)=\frac{1}{1+x}-\frac{(1+x)-x}{(1+x)^2}$
$\displaystyle =\frac{1}{1+x}-\frac{1}{(1+x)^2}$
$\displaystyle =\frac{(1+x)-1}{(1+x)^2}$
$\displaystyle =\frac{x}{(1+x)^2}$
$\displaystyle \text{Since }(1+x)^2>0\text{ for all }x\in(-1,\infty),\text{ the sign of }f'(x)\text{ depends only on }x.$
$\displaystyle \text{For }f(x)\text{ to be increasing, we must have }f'(x)>0.$
$\displaystyle \Rightarrow x>0$
$\displaystyle \therefore f(x)\text{ is increasing on }(0,\infty).$
$\displaystyle \text{For }f(x)\text{ to be decreasing, we must have }f'(x)<0.$
$\displaystyle \Rightarrow -1<x<0$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-1,0).$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{Find the intervals in which }f(x)=(x+2)e^{-x}\text{ is increasing or decreasing. } \\ \;[\text{CBSE 2000C}]$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=(x+2)e^{-x}$
$\displaystyle \therefore f'(x)=e^{-x}+(x+2)(-e^{-x})$
$\displaystyle =e^{-x}-xe^{-x}-2e^{-x}$
$\displaystyle =e^{-x}(-x-1)$
$\displaystyle \text{Since }e^{-x}>0\text{ for all }x\in\mathbb{R},\text{ the sign of }f'(x)\text{ depends only on }(-x-1).$
$\displaystyle \text{Critical point: }-x-1=0\Rightarrow x=-1.$
$\displaystyle \text{For }x<-1,\;-x-1>0\Rightarrow f'(x)>0.$
$\displaystyle \text{For }x>-1,\;-x-1<0\Rightarrow f'(x)<0.$
$\displaystyle \therefore f(x)\text{ is increasing on }(-\infty,-1).$
$\displaystyle \therefore f(x)\text{ is decreasing on }(-1,\infty).$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{Show that the function }f(x)=10^x\text{ is increasing for all }x.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=10^x$
$\displaystyle \therefore f'(x)=10^x\log10$
$\displaystyle \text{Since }10^x>0\text{ and }\log10>0\text{ for all }x\in\mathbb{R},$
$\displaystyle \therefore f'(x)>0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \therefore f(x)\text{ is increasing on }\mathbb{R}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{Prove that the function }f(x)=x-[x]\text{ is increasing in }(0,1).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x-[x]$
$\displaystyle \text{Let }x_1,x_2\in(0,1)\text{ such that }x_1<x_2.$
$\displaystyle \text{Since }0<x_1,x_2<1,\;[x_1]=[x_2]=0.$
$\displaystyle \therefore x_1-[x_1]<x_2-[x_2]$
$\displaystyle \Rightarrow f(x_1)<f(x_2).$
$\displaystyle \therefore f(x)\text{ is increasing on }(0,1).$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{Prove that the following functions are increasing on }\mathbb{R}: \\ \text{(i) }f(x)=3x^5+40x^3+240x,\qquad \text{(ii) }f(x)=4x^3-18x^2+27x-27.\;[\text{CBSE 2017}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle f(x)=3x^5+40x^3+240x$
$\displaystyle \therefore f'(x)=15x^4+120x^2+240$
$\displaystyle =15(x^4+8x^2+16)$
$\displaystyle =15(x^2+4)^2$
$\displaystyle \text{Since }15>0\text{ and }(x^2+4)^2>0\text{ for all }x\in\mathbb{R},$
$\displaystyle \therefore f'(x)>0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \therefore f(x)\text{ is increasing on }\mathbb{R}.$
$\displaystyle \text{(ii)}$
$\displaystyle f(x)=4x^3-18x^2+27x-27$
$\displaystyle \therefore f'(x)=12x^2-36x+27$
$\displaystyle =3(4x^2-12x+9)$
$\displaystyle =3(2x-3)^2$
$\displaystyle \text{Since }3>0\text{ and }(2x-3)^2\ge0\text{ for all }x\in\mathbb{R},$
$\displaystyle \therefore f'(x)\ge0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \therefore f(x)\text{ is increasing on }\mathbb{R}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: }\text{Prove that the function }f(x)=\log(\cos x)\text{ is strictly increasing on } \\ \left(-\frac{\pi}{2},0\right)\text{ and strictly decreasing on }\left(0,\frac{\pi}{2}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\log(\cos x)$
$\displaystyle \therefore f'(x)=\frac{1}{\cos x}(-\sin x)$
$\displaystyle =-\tan x$
$\displaystyle \text{For }x\in\left(-\frac{\pi}{2},0\right),\;\tan x<0.$
$\displaystyle \therefore -\tan x>0$
$\displaystyle \Rightarrow f'(x)>0.$
$\displaystyle \therefore f(x)\text{ is strictly increasing on }\left(-\frac{\pi}{2},0\right).$
$\displaystyle \text{For }x\in\left(0,\frac{\pi}{2}\right),\;\tan x>0.$
$\displaystyle \therefore -\tan x<0$
$\displaystyle \Rightarrow f'(x)<0.$
$\displaystyle \therefore f(x)\text{ is strictly decreasing on }\left(0,\frac{\pi}{2}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 32: }\text{Prove that the function }f(x)=x^3-3x^2+4x\text{ is strictly increasing on }\mathbb{R}.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^3-3x^2+4x$
$\displaystyle \therefore f'(x)=3x^2-6x+4$
$\displaystyle =3(x^2-2x)+4$
$\displaystyle =3(x^2-2x+1)-3+4$
$\displaystyle =3(x-1)^2+1$
$\displaystyle \text{Since }3(x-1)^2+1>0\text{ for all }x\in\mathbb{R},$
$\displaystyle \therefore f'(x)>0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \therefore f(x)\text{ is strictly increasing on }\mathbb{R}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 33: }\text{Prove that the function }f(x)=\cos x \\ \text{ is (i) strictly decreasing in }(0,\pi),\;(ii)\text{ strictly increasing in }(\pi,2\pi), \; \\ (iii)\text{ neither increasing nor decreasing in }(0,2\pi).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\cos x$
$\displaystyle \therefore f'(x)=-\sin x$
$\displaystyle \text{(i) For }0<x<\pi,$
$\displaystyle \sin x>0$
$\displaystyle \Rightarrow -\sin x<0$
$\displaystyle \Rightarrow f'(x)<0\text{ for all }x\in(0,\pi).$
$\displaystyle \therefore f(x)\text{ is strictly decreasing on }(0,\pi).$
$\displaystyle \text{(ii) For }\pi<x<2\pi,$
$\displaystyle \sin x<0$
$\displaystyle \Rightarrow -\sin x>0$
$\displaystyle \Rightarrow f'(x)>0\text{ for all }x\in(\pi,2\pi).$
$\displaystyle \therefore f(x)\text{ is strictly increasing on }(\pi,2\pi).$
$\displaystyle \text{(iii) Since }f(x)\text{ is strictly decreasing on }(0,\pi)\text{ and strictly increasing on }(\pi,2\pi),$
$\displaystyle \therefore f(x)\text{ is neither increasing nor decreasing on }(0,2\pi).$
$\displaystyle \\$

$\displaystyle \textbf{Question 34: }\text{Show that }f(x)=x^2-x\sin x\text{ is an increasing function on }\left(0,\frac{\pi}{2}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^2-x\sin x$
$\displaystyle \therefore f'(x)=2x-x\cos x-\sin x$
$\displaystyle =x(1-\cos x)+(x-\sin x)$
$\displaystyle \text{For }0<x<\frac{\pi}{2},\;1-\cos x>0\text{ and }x-\sin x>0.$
$\displaystyle \therefore x(1-\cos x)>0\text{ and }x-\sin x>0.$
$\displaystyle \Rightarrow f'(x)>0\text{ for all }x\in\left(0,\frac{\pi}{2}\right).$
$\displaystyle \therefore f(x)\text{ is increasing on }\left(0,\frac{\pi}{2}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 35: }\text{Find the value(s) of }a\text{ for which }f(x)=x^3-ax \\ \text{ is an increasing function on }\mathbb{R}.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x^3-ax$
$\displaystyle \therefore f'(x)=3x^2-a$
$\displaystyle \text{For }f(x)\text{ to be increasing on }\mathbb{R},\text{ we must have}$
$\displaystyle f'(x)\ge0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \Rightarrow 3x^2-a\ge0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \Rightarrow a\le3x^2\text{ for all }x\in\mathbb{R}.$
$\displaystyle \text{The least value of }3x^2\text{ is }0.$
$\displaystyle \therefore a\le0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 36: }\text{Find the values of }b\text{ for which the function } \\ f(x)=\sin x-bx+c\text{ is a decreasing function on }\mathbb{R}.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\sin x-bx+c$
$\displaystyle \therefore f'(x)=\cos x-b$
$\displaystyle \text{For }f(x)\text{ to be decreasing on }\mathbb{R},\text{ we must have}$
$\displaystyle f'(x)\le0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \Rightarrow \cos x-b\le0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \Rightarrow \cos x\le b\text{ for all }x\in\mathbb{R}.$
$\displaystyle \text{Since the greatest value of }\cos x\text{ is }1,$
$\displaystyle \therefore b\ge1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 37: }\text{Show that }f(x)=x+\cos x-a\text{ is an increasing function on }\mathbb{R} \\ \text{ for all values of }a.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=x+\cos x-a$
$\displaystyle \therefore f'(x)=1-\sin x$
$\displaystyle \text{Since }\sin x\le1\text{ for all }x\in\mathbb{R},$
$\displaystyle \therefore 1-\sin x\ge0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \Rightarrow f'(x)\ge0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \text{Since }f'(x)\text{ is independent of }a,$
$\displaystyle \therefore f(x)\text{ is increasing on }\mathbb{R}\text{ for all }a\in\mathbb{R}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 38: }\text{Let }f\text{ be defined on }[0,1]\text{ and be twice differentiable such that } \\ |f''(x)|\le1\text{ for all }x\in[0,1].\text{ If }f(0)=f(1),\text{ then show that }|f'(x)|<1 \\ \text{ for all }x\in[0,1].$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }f\text{ is continuous on }[0,1],\text{ differentiable on }(0,1)\text{ and }f(0)=f(1),$
$\displaystyle \text{by Rolle's Theorem, there exists }c\in(0,1)\text{ such that }f'(c)=0.$
$\displaystyle \text{Let }x\in[0,1].$
$\displaystyle \text{Applying the Mean Value Theorem to }f'\text{ on the interval with endpoints }c\text{ and }x,$
$\displaystyle f'(x)-f'(c)=f''(\xi)(x-c)$
$\displaystyle \text{for some }\xi\text{ lying between }c\text{ and }x.$
$\displaystyle \therefore |f'(x)|=|f'(x)-f'(c)|$
$\displaystyle =|f''(\xi)|\,|x-c|$
$\displaystyle \le |x-c|$
$\displaystyle \text{Since }c\in(0,1)\text{ and }x\in[0,1],\;|x-c|<1.$
$\displaystyle \therefore |f'(x)|<1\text{ for all }x\in[0,1].$
$\displaystyle \\$

$\displaystyle \textbf{Question 39: }\text{Find the intervals in which }f(x)\text{ is increasing or decreasing: } \\ \text{(i) }f(x)=x|x|,\;x\in\mathbb{R},\qquad \text{(ii) }f(x)=\sin x+|\sin x|,\;0<x\le2\pi, \\ \text{(iii) }f(x)=\sin x(1+\cos x),\;0<x<\frac{\pi}{2}.\;[\text{CBSE 2014}]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle f(x)=x|x|$
$\displaystyle f(x)=\begin{cases}-x^2,&x<0,\\x^2,&x\ge0.\end{cases}$
$\displaystyle \therefore f'(x)=\begin{cases}-2x,&x<0,\\0,&x=0,\\2x,&x>0.\end{cases}$
$\displaystyle \text{For }x<0,\;-2x>0,\text{ and for }x>0,\;2x>0.$
$\displaystyle \text{Also, }f'(0)=0.$
$\displaystyle \therefore f'(x)\ge0\text{ for all }x\in\mathbb{R}.$
$\displaystyle \therefore f(x)\text{ is increasing on }\mathbb{R}.$
$\displaystyle \text{(ii)}$
$\displaystyle f(x)=\sin x+|\sin x|,\qquad 0<x\le2\pi$
$\displaystyle f(x)=\begin{cases}2\sin x,&0<x<\pi,\\0,&\pi\le x\le2\pi.\end{cases}$
$\displaystyle \text{For }0<x<\pi,\;f'(x)=2\cos x.$
$\displaystyle \text{For }0<x<\frac{\pi}{2},\;\cos x>0\Rightarrow f'(x)>0.$
$\displaystyle \text{For }\frac{\pi}{2}<x<\pi,\;\cos x<0\Rightarrow f'(x)<0.$
$\displaystyle \text{Also, }f(x)=0\text{ for }\pi\le x\le2\pi.$
$\displaystyle \therefore f(x)\text{ is increasing on }\left(0,\frac{\pi}{2}\right).$
$\displaystyle \therefore f(x)\text{ is decreasing on }\left(\frac{\pi}{2},2\pi\right].$
$\displaystyle \text{On }[\pi,2\pi],\;f(x)\text{ is constant and hence both increasing and decreasing.}$
$\displaystyle \text{(iii)}$
$\displaystyle f(x)=\sin x(1+\cos x),\qquad 0<x<\frac{\pi}{2}$
$\displaystyle =\sin x+\sin x\cos x$
$\displaystyle \therefore f'(x)=\cos x+\cos^2x-\sin^2x$
$\displaystyle =\cos x+\cos2x$
$\displaystyle =2\cos^2x+\cos x-1$
$\displaystyle =(2\cos x-1)(\cos x+1)$
$\displaystyle \text{Since }\cos x+1>0\text{ for }0<x<\frac{\pi}{2},\text{ the sign of }f'(x)\text{ depends on }2\cos x-1.$
$\displaystyle f'(x)>0\Rightarrow 2\cos x-1>0$
$\displaystyle \Rightarrow \cos x>\frac12$
$\displaystyle \Rightarrow 0<x<\frac{\pi}{3}$
$\displaystyle \therefore f(x)\text{ is increasing on }\left(0,\frac{\pi}{3}\right).$
$\displaystyle f'(x)<0\Rightarrow 2\cos x-1<0$
$\displaystyle \Rightarrow \cos x<\frac12$
$\displaystyle \Rightarrow \frac{\pi}{3}<x<\frac{\pi}{2}$
$\displaystyle \therefore f(x)\text{ is decreasing on }\left(\frac{\pi}{3},\frac{\pi}{2}\right).$
$\displaystyle \\$