Class 12: Increasing and Decreasing Functions – Exercise 17.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \text{Question 1: Find the intervals in which the following functions are increasing or decreasing.}$

$\displaystyle \text{(i) } f(x) = 10-6x-2x^2$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = 10-6x-2x^2$

$\displaystyle \Rightarrow f'(x) = -6-4x$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow -6-4x > 0$

$\displaystyle \Rightarrow -4x > 6$

$\displaystyle \Rightarrow x > \frac{-3}{2}$

$\displaystyle \Rightarrow x \in \Big( -\infty, \frac{-3}{2} \Big)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } \Big( -\infty, \frac{-3}{2} \Big)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow -6-4x < 0$

$\displaystyle \Rightarrow 4x > -6$

$\displaystyle \Rightarrow x > \frac{-3}{2}$

$\displaystyle \Rightarrow x \in \Big( \frac{-3}{2}, \infty, \Big)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } \Big( \frac{-3}{2}, \infty, \Big)$

$\\$

$\displaystyle \text{(ii) } f(x) = x^2 + 2x - 5$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = x^2 + 2x - 5$

$\displaystyle \Rightarrow f'(x) = 2x+2$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 2x+2 > 0$

$\displaystyle \Rightarrow x+1 > 0$

$\displaystyle \Rightarrow x > -1$

$\displaystyle \Rightarrow x \in ( -1, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -1, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 2x+2 < 0$

$\displaystyle \Rightarrow x+1 < 0$

$\displaystyle \Rightarrow x < -1$

$\displaystyle \Rightarrow x \in ( -\infty, -1)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } ( -\infty, -1)$

$\\$

$\displaystyle \text{(iii) } f(x) = 6 - 9x - x^2$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = 6 - 9x - x^2$

$\displaystyle \Rightarrow f'(x) = -9-2x$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow -9-2x > 0$

$\displaystyle \Rightarrow -2x > 9$

$\displaystyle \Rightarrow x < \frac{-9}{2}$

$\displaystyle \Rightarrow x \in \Big( -\infty, \frac{-9}{2} \Big)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } \Big( -\infty, \frac{-9}{2} \Big)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow -9-2x < 0$

$\displaystyle \Rightarrow 2x > -9$

$\displaystyle \Rightarrow x > \frac{-9}{2}$

$\displaystyle \Rightarrow x \in \Big(\frac{-9}{2}, \infty \Big)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } \Big(\frac{-9}{2}, \infty \Big)$

$\\$

$\displaystyle \text{(iv) } f(x) = 2x^3 - 12x^2 + 18x + 15$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = 2x^3 - 12x^2 + 18x + 15$

$\displaystyle \Rightarrow f'(x) = 6x^2-24x+18 = 6(x^2 - 4x+3) = 6(x-1)(x-3)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 6(x-1)(x-3) > 0$

$\displaystyle \Rightarrow (x-1)(x-3) > 0$

$\displaystyle \Rightarrow x > 3 \text{ or } x < 1$

$\displaystyle \Rightarrow x \in ( -\infty , 1) \cup ( 3, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -\infty , 1) \cup ( 3, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 6(x-1)(x-3) < 0$

$\displaystyle \Rightarrow (x-1)(x-3) < 0$

$\displaystyle \Rightarrow 1 < x < 3$

$\displaystyle \Rightarrow x \in (1, 3)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } (1, 3)$

$\\$

$\displaystyle \text{(v) } f(x) = 5+36x+3x^2- 2x^3$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = 5+36x+3x^2- 2x^3$

$\displaystyle \Rightarrow f'(x) = 36+ 6x-6x^2 = 6(6+x-x^2) = -6(x-3)(x+2)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow -6(x-3)(x+2) > 0$

$\displaystyle \Rightarrow (x-1)(x-3) < 0$

$\displaystyle \Rightarrow -2 < x < 3$

$\displaystyle \Rightarrow x \in (-2, 3)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } (-2, 3)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow -6(x-3)(x+2) < 0$

$\displaystyle \Rightarrow (x-1)(x-3) > 0$

$\displaystyle \Rightarrow x > 3 \text{ or } x < -2$

$\displaystyle \Rightarrow x \in ( -\infty , -2) \cup ( 3, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } ( -\infty , -2) \cup ( 3, \infty)$

$\\$

$\displaystyle \text{(vi) } f(x) = 8+36x+3x^2- 2x^3$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = 8+36x+3x^2- 2x^3$

$\displaystyle \Rightarrow f'(x) = 36+6x-6x^2 = 6(6+x-x^2) = -6(x-3)(x+2)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow -6(x-3)(x+2) > 0$

$\displaystyle \Rightarrow (x-1)(x-3) < 0$

$\displaystyle \Rightarrow -2 < x < 3$

$\displaystyle \Rightarrow x \in (-2, 3)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } (-2, 3)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow -6(x-3)(x+2) < 0$

$\displaystyle \Rightarrow (x-1)(x-3) > 0$

$\displaystyle \Rightarrow x > 3 \text{ or } x < -2$

$\displaystyle \Rightarrow x \in ( -\infty , -2) \cup ( 3, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } ( -\infty , -2) \cup ( 3, \infty)$

$\\$

$\displaystyle \text{(vii) } f(x) = 5x^3-15x^2-120x+3$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = 5x^3-15x^2-120x+3$

$\displaystyle \Rightarrow f'(x) = 15x^2-30x-120 = 15(x^2 - 2x-8) = 15(x-4)(x+2)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 15(x-4)(x+2) > 0$

$\displaystyle \Rightarrow (x-4)(x+2) > 0$

$\displaystyle \Rightarrow x > 4 \text{ or } x < -2$

$\displaystyle \Rightarrow x \in ( -\infty , -2) \cup ( 4, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -\infty , -2) \cup ( 4, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 15(x-4)(x+2) < 0$

$\displaystyle \Rightarrow (x-4)(x+2) < 0$

$\displaystyle \Rightarrow -2 < x < 4$

$\displaystyle \Rightarrow x \in (-2, 4)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } (-2, 4)$

$\\$

$\displaystyle \text{(viii) } f(x) = x^3-6x^2-36x+2$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = x^3-6x^2-36x+2$

$\displaystyle \Rightarrow f'(x) = 3x^2-12x-36 = 3(x^2 - 4x-12) = 3(x-6)(x+2)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 3(x-6)(x+2) > 0$

$\displaystyle \Rightarrow (x-6)(x+2) > 0$

$\displaystyle \Rightarrow x > 6 \text{ or } x < -2$

$\displaystyle \Rightarrow x \in ( -\infty , -2) \cup ( 6, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -\infty , -2) \cup ( 6, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 3(x-6)(x+2) < 0$

$\displaystyle \Rightarrow (x-6)(x+2) < 0$

$\displaystyle \Rightarrow -2 < x < 6$

$\displaystyle \Rightarrow x \in (-2, 6)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } (-2, 6)$

$\\$

$\displaystyle \text{(ix) } f(x) = 2x^3-15x^2+36x+1 {\hspace{7.0cm} \text{[CBSE 2005, 2010]} }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = 2x^3-15x^2+36x+1$

$\displaystyle \Rightarrow f'(x) = 6x^2-30x+36 = 6(x^2 - 5x+6) = 3(x-2)(x-3)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 3(x-2)(x-3) > 0$

$\displaystyle \Rightarrow (x-2)(x-3) > 0$

$\displaystyle \Rightarrow x > 3 \text{ or } x < 2$

$\displaystyle \Rightarrow x \in ( -\infty , 2) \cup ( 3, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -\infty , 2) \cup ( 3, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 3(x-2)(x-3) < 0$

$\displaystyle \Rightarrow (x-2)(x-3) < 0$

$\displaystyle \Rightarrow 2 < x < 3$

$\displaystyle \Rightarrow x \in (2, 3)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } (2, 3)$

$\\$

$\displaystyle \text{(x) } f(x) = 2x^3+9x^2+12x+20 {\hspace{7.0cm} \text{[CBSE 2011]} }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = 2x^3+9x^2+12x+20$

$\displaystyle \Rightarrow f'(x) = 6x^2+18x+12 = 6(x^2 +3x+2) = 6(x+1)(x+2)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 6(x+1)(x+2) > 0$

$\displaystyle \Rightarrow (x+1)(x+2) > 0$

$\displaystyle \Rightarrow x > -1 \text{ or } x < -2$

$\displaystyle \Rightarrow x \in ( -\infty , -2) \cup ( -1, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -\infty , -2) \cup ( -1, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 6(x+1)(x+2) < 0$

$\displaystyle \Rightarrow (x+1)(x+2) < 0$

$\displaystyle \Rightarrow -2 < x < -1$

$\displaystyle \Rightarrow x \in (-2, -1)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } (-2, -1)$

$\\$

$\displaystyle \text{(xi) } f(x) = 2x^3-9x^2+12x-5$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = 2x^3-9x^2+12x-5$

$\displaystyle \Rightarrow f'(x) = 6x^2-18x+12 = 6(x^2 -3x+2) = 6(x-1)(x-2)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 6(x-1)(x-2) > 0$

$\displaystyle \Rightarrow (x-1)(x-2) > 0$

$\displaystyle \Rightarrow x > 2 \text{ or } x < 1$

$\displaystyle \Rightarrow x \in ( -\infty , 1) \cup ( 2, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -\infty , 1) \cup ( 2, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 6(x-1)(x-2) < 0$

$\displaystyle \Rightarrow (x-1)(x-2) < 0$

$\displaystyle \Rightarrow 1 < x < 2$

$\displaystyle \Rightarrow x \in (1, 2)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } (1, 2)$

$\\$

$\displaystyle \text{(xii) } f(x) = 6+12x+3x^2-2x^3$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = 6+12x+3x^2-2x^3$

$\displaystyle \Rightarrow f'(x) = 12+6x-6x^2 = 6(2+x-x^2) = -6(x-2)(x+1)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow -6(x-2)(x+1) > 0$

$\displaystyle \Rightarrow (x-2)(x+1) < 0$

$\displaystyle \Rightarrow -1 < x < 2$

$\displaystyle \Rightarrow x \in (-1, 2)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } (-1, 2)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow -6(x-2)(x+1) < 0$

$\displaystyle \Rightarrow (x-2)(x+1) > 0$

$\displaystyle \Rightarrow x > 2 \text{ or } x < -1$

$\displaystyle \Rightarrow x \in ( -\infty , -1) \cup ( 2, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } ( -\infty , -1) \cup ( 2, \infty)$

$\\$

$\displaystyle \text{(xiii) } f(x) = 2x^3-24x+107$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = 2x^3-24x+107$

$\displaystyle \Rightarrow f'(x) = 6x^2-24 = 6(x^2 -4) = 6(x+2)(x-2)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 6(x+2)(x-2) > 0$

$\displaystyle \Rightarrow (x+2)(x-2) > 0$

$\displaystyle \Rightarrow x > 2 \text{ or } x < -2$

$\displaystyle \Rightarrow x \in ( -\infty , -2) \cup ( 2, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -\infty , -2) \cup ( 2, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 6(x+2)(x-2) < 0$

$\displaystyle \Rightarrow (x+2)(x-2) < 0$

$\displaystyle \Rightarrow -2 < x < 2$

$\displaystyle \Rightarrow x \in (-2, 2)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } (-2, 2)$

$\\$

$\displaystyle \text{(xiv) } f(x) = -2x^3-9x^2-12x+1$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = -2x^3-9x^2-12x+1$

$\displaystyle \Rightarrow f'(x) = -6x^2-18x-12 = -6(x^2+3x+2) = -6(x+1)(x+2)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow -6(x+1)(x+2) > 0$

$\displaystyle \Rightarrow (x+1)(x+2) < 0$

$\displaystyle \Rightarrow -2 < x < -1$

$\displaystyle \Rightarrow x \in (-2, -1)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } (-2, -1)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow -6(x+1)(x+2) < 0$

$\displaystyle \Rightarrow (x+1)(x+2) > 0$

$\displaystyle \Rightarrow x > -1 \text{ or } x < -2$

$\displaystyle \Rightarrow x \in ( -\infty , -2) \cup ( -1, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } ( -\infty , -1) \cup ( 2, \infty)$

$\\$

$\displaystyle \text{(xv) } f(x) = (x-1)(x-2)^2$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = (x-1)(x-2)^2 = (x-1)(x^2-4x+4) = x^3 - 5x^2 + 8x - 4$

$\displaystyle \Rightarrow f'(x) = 3x^2 - 10x+8 = 3x^2 - 6x - 4x+8 = (x-2)(3x-4)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow (x-2)(3x-4) > 0$

$\displaystyle \Rightarrow x > 2 \text{ or } x < \frac{4}{3}$

$\displaystyle \Rightarrow x \in \Big( -\infty , \frac{4}{3}\Big) \cup ( 2, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } \Big( -\infty , \frac{4}{3}\Big) \cup ( 2, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow (x-2)(3x-4) < 0$

$\displaystyle \Rightarrow \frac{4}{3} < x < 2$

$\displaystyle \Rightarrow x \in \Big(\frac{4}{3}, 2\Big)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } \Big(\frac{4}{3}, 2\Big)$

$\\$

$\displaystyle \text{(xvi) } f(x) = x^3-12x^2+36x+17 {\hspace{7.0cm} \text{[CBSE 2001]} }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = x^3-12x^2+36x+17$

$\displaystyle \Rightarrow f'(x) = 3x^2-24x+36 = 3(x^2 -8x+12) = 3(x-2)(x-6)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 3(x-2)(x-6) > 0$

$\displaystyle \Rightarrow (x-2)(x-6) > 0$

$\displaystyle \Rightarrow x > 6 \text{ or } x < 2$

$\displaystyle \Rightarrow x \in ( -\infty , 2) \cup ( 6, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -\infty , 2) \cup ( 6, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 3(x-2)(x-6) < 0$

$\displaystyle \Rightarrow (x-2)(x-6) < 0$

$\displaystyle \Rightarrow 2 < x < 6$

$\displaystyle \Rightarrow x \in (2, 6)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } (2, 6)$

$\\$

$\displaystyle \text{(xvii) } f(x) = 2x^3-24x+7$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = 2x^3-24x+7$

$\displaystyle \Rightarrow f'(x) = 6x^2-24 = 6(x^2 -4) = 6(x+2)(x-2)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 6(x+2)(x-2) > 0$

$\displaystyle \Rightarrow (x+2)(x-2) > 0$

$\displaystyle \Rightarrow x > 2 \text{ or } x < -2$

$\displaystyle \Rightarrow x \in ( -\infty , -2) \cup ( 2, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -\infty , 2) \cup ( 6, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 6(x+2)(x-2) < 0$

$\displaystyle \Rightarrow (x+2)(x-2) < 0$

$\displaystyle \Rightarrow -2 < x < -$

$\displaystyle \Rightarrow x \in (-2, 2)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } (-2, 2)$

$\\$

$\displaystyle \text{(xviii) } f(x) = \frac{3}{10} x^4 - \frac{4}{5} x^3 - 3x^2 + \frac{36}{5} x + 11$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = \frac{3}{10} x^4 - \frac{4}{5} x^3 - 3x^2 + \frac{36}{5} x + 11 = \frac{3x^4-8x^3-30x^2+72x+110}{10}$

$\displaystyle \Rightarrow f'(x) =\frac{12x^3-24x^2-60x+72}{10} \\ \\ {\hspace{1.5cm} = \frac{12}{10}(x^3-2x^2-5x+6) = \frac{12(x-1)(x^2-x-6)}{10} = \frac{12}{10}(x-1)(x+2)(x-3) }$

$\displaystyle \text{Here } 1, 2 \text{ and } 3 \text{ are critical points. }$

$\displaystyle \text{The possible intervals are } (-\infty, -2), (-2, -) , (1, 3) \text{ and } (3, \infty).$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow \frac{12}{10}(x-1)(x+2)(x-3) > 0$

$\displaystyle \Rightarrow (x-1)(x+2)(x-3) > 0$

$\displaystyle \Rightarrow x \in ( -2, -1) \cup ( 3, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -2, -1) \cup ( 3, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow \frac{12}{10}(x-1)(x+2)(x-3) < 0$

$\displaystyle \Rightarrow (x-1)(x+2)(x-3) < 0$

$\displaystyle \Rightarrow x \in ( -\infty , -2) \cup ( 1,3)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } ( -\infty , -2) \cup ( 1,3)$

$\\$

$\displaystyle \text{(xix) } f(x) = x^4 - 4x$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = x^4 - 4x$

$\displaystyle \Rightarrow f'(x) = 4x^3-4 = 4(x^3 -1)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 4(x^3 -1) > 0$

$\displaystyle \Rightarrow (x^3 -1) > 0$

$\displaystyle \Rightarrow x^3 > 1$

$\displaystyle \Rightarrow x > 1$

$\displaystyle \Rightarrow x \in ( 1, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( 1, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 4(x^3 -1) < 0$

$\displaystyle \Rightarrow (x^3 -1) < 0$

$\displaystyle \Rightarrow x^3 < 1$

$\displaystyle \Rightarrow x < 1$

$\displaystyle \Rightarrow x \in ( -\infty , 1)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } ( -\infty , 1)$

$\\$

$\displaystyle \text{(xx) } f(x) = \frac{x^4}{4} + \frac{2}{3} x^3 - \frac{5}{2} x^2 - 6x + 7$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = \frac{x^4}{4} + \frac{2}{3} x^3 - \frac{5}{2} x^2 - 6x + 7 = \frac{3x^4+8x^3-30x^2-72x+84}{12}$

$\displaystyle \Rightarrow f'(x) =\frac{12x^3+24x^2-60x-72}{12} \\ \\ {\hspace{1.5cm} = (x^3+2x^2-5x-6) = (x+1)(x^2+x-6) = (x+1)(x-2)(x+3) }$

$\displaystyle \text{Here } -1, 2 \text{ and } -3 \text{ are critical points. }$

$\displaystyle \text{The possible intervals are } (-\infty, -3), (-3, -1) , (-1, 2) \text{ and } (2, \infty)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow (x+1)(x-2)(x+3) > 0$

$\displaystyle \Rightarrow x \in ( -3, -2) \cup ( 2, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -3, -2) \cup ( 2, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow (x+1)(x-2)(x+3) < 0$

$\displaystyle \Rightarrow x \in ( -\infty, -3) \cup ( -1, -2)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } ( -\infty, -3) \cup ( -1, -2)$

$\\$

$\displaystyle \text{(xxi) } f(x) = x^4 - 4x^3 + 4x^2 + 14$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = x^4 - 4x^3 + 4x^2 + 14$

$\displaystyle \Rightarrow f'(x) = 4x^3 - 12x^3+8x = 4x(x-1)(x-2)$

$\displaystyle \text{Here } 0, 1 \text{ and } 2 \text{ are critical points. }$

$\displaystyle \text{The possible intervals are } (-\infty, 0), (0, 1) , (1, 2) \text{ and } (2, \infty)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 4x(x-1)(x-2) > 0$

$\displaystyle \Rightarrow x(x-1)(x-2) > 0$

$\displaystyle \Rightarrow x \in ( 0, 1) \cup ( 2, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( 0, 1) \cup ( 2, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 4x(x-1)(x-2) < 0$

$\displaystyle \Rightarrow x(x-1)(x-2) < 0$

$\displaystyle \Rightarrow x \in ( -\infty, 0) \cup ( 1, 2)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } ( -\infty, 0) \cup ( 1, 2)$

$\\$

$\displaystyle \text{(xxii) } f(x) = 5x^{\frac{3}{2}} -3x^{\frac{5}{2}}, x> 0$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = 5x^{\frac{3}{2}} -3x^{\frac{5}{2}}$

$\displaystyle \Rightarrow f'(x) = \frac{15}{2} x^{\frac{1}{2}}- \frac{15}{2} x^{\frac{3}{2}} = \frac{15}{2} x^{\frac{1}{2}} (1-x)$

$\displaystyle \text{Here } 0 \text{ and } 1 \text{ are the roots. }$

$\displaystyle \text{The possible intervals are } (-\infty, 0), (0, 1) \text{ and } (1, \infty)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow \frac{15}{2} x^{\frac{1}{2}} (1-x) > 0$

$\displaystyle \Rightarrow x^{\frac{1}{2}} (1-x) > 0$

$\displaystyle \Rightarrow x \in ( 0, 1)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( 0, 1)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow \frac{15}{2} x^{\frac{1}{2}} (1-x) < 0$

$\displaystyle \Rightarrow x^{\frac{1}{2}} (1-x) < 0$

$\displaystyle \Rightarrow x \in (1, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } (1, \infty)$

$\\$

$\displaystyle \text{(xxiii) } f(x) = x^8+6x^2$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = x^8+6x^2$

$\displaystyle \Rightarrow f'(x) = 8x^7+12x = 4x(2x^6+3)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 4x(2x^6+3) > 0$

$\displaystyle \Rightarrow x(2x^6+3) > 0$

$\displaystyle \Rightarrow x \in ( 0, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( 0, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 4x(2x^6+3) < 0$

$\displaystyle \Rightarrow x(2x^6+3) < 0$

$\displaystyle \Rightarrow x \in (-\infty , 0)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } (-\infty , 0)$

$\\$

$\displaystyle \text{(xxiv) } f(x) = x^3-6x^2+ 9x+ 15 {\hspace{6.0cm} \text{[CBSE 2000, 2004]} }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = x^3-6x^2+ 9x+ 15$

$\displaystyle \Rightarrow f'(x) = 3x^2-12x+9 = 3(x-1)(x-3)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 3(x-1)(x-3) > 0$

$\displaystyle \Rightarrow (x-1)(x-3) > 0$

$\displaystyle \Rightarrow x > 3 \text{ or } x < 1$

$\displaystyle \Rightarrow x \in ( -\infty, 1) \cup ( 3, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -\infty, 1) \cup ( 3, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 3(x-1)(x-3) < 0$

$\displaystyle \Rightarrow (x-1)(x-3) < 0$

$\displaystyle \Rightarrow 1 < x < 3$

$\displaystyle \Rightarrow x \in ( 1, 3)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } ( 1, 3)$

$\\$

$\displaystyle \text{(xxv) } f(x) = \{ x(x-2)\}^2 {\hspace{8.0cm} \text{[CBSE 2010, 2014]} }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = \{ x(x-2)\}^2 = (x^2 - 2x)^2 = x^4 + 4x^2 - 4x^3$

$\displaystyle \Rightarrow f'(x) = 4x^3 + 8x-12x^2 = 4x(x^2-3x+2) = 4x(x-1)(x-2)$

$\displaystyle \text{Here } 0, 1 \text{ and } 2 \text{ are the critical points. }$

$\displaystyle \text{The possible intervals are } (-\infty, 0), (0, 1), (1, 2) \text{ and } (2, \infty)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 4x(x-1)(x-2) > 0$

$\displaystyle \Rightarrow x(x-1)(x-2) > 0$

$\displaystyle \Rightarrow x \in ( 0, 1) \cup (2, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( 0, 1) \cup (2, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 4x(x-1)(x-2) < 0$

$\displaystyle \Rightarrow x(x-1)(x-2) < 0$

$\displaystyle \Rightarrow x \in ( -\infty, 0) \cup (1, 2)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } ( -\infty, 0) \cup (1, 2)$

$\\$

$\displaystyle \text{(xxvi) } f(x) =3x^4 - 4x^3 - 12x^2 + 5 {\hspace{7.0cm} \text{[CBSE 2014]} }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) =3x^4 - 4x^3 - 12x^2 + 5$

$\displaystyle \Rightarrow f'(x) =12x^3-12x^2 - 24x=12x(x+1)(x-2)$

$\displaystyle \text{Here } 0, -1 \text{ and } 2 \text{ are the critical points. }$

$\displaystyle \text{The possible intervals are } (-\infty, -1), (-1, 0), (0, 2) \text{ and } (2, \infty)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 12x(x+1)(x-2) > 0$

$\displaystyle \Rightarrow x(x+1)(x-2) > 0$

$\displaystyle \Rightarrow x \in ( -1, 0) \cup (2, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -1, 0) \cup (2, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 12x(x+1)(x-2) < 0$

$\displaystyle \Rightarrow x(x+1)(x-2) < 0$

$\displaystyle \Rightarrow x \in (-\infty, -1) \cup (0, 2)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } (-\infty, -1) \cup (0, 2)$

$\\$

$\displaystyle \text{(xxvii) } f(x) = \frac{3}{2} x^4 - 4x^3 - 45x^2 + 51 {\hspace{7.0cm} \text{[CBSE 2014]} }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = \frac{3}{2} x^4 - 4x^3 - 45x^2 + 51$

$\displaystyle \Rightarrow f'(x) =6x^2-12x^2-90x = 6x(x^2 - 2x - 15)= 6x(x-5)(x+3)$

$\displaystyle \text{Here } -3, 0 \text{ and } 5 \text{ are the critical points. }$

$\displaystyle \text{The possible intervals are } (-\infty, -3), (-3, 0), (0, 5) \text{ and } (5, \infty)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 6x(x-5)(x+3) > 0$

$\displaystyle \Rightarrow x(x-5)(x+3) > 0$

$\displaystyle \Rightarrow x \in ( -3, 0) \cup (5, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -3, 0) \cup (5, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 6x(x-5)(x+3) < 0$

$\displaystyle \Rightarrow x(x-5)(x+3) < 0$

$\displaystyle \Rightarrow x \in ( -\infty, -3) \cup (0, 5)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } ( -\infty, -3) \cup (0, 5)$

$\\$

$\displaystyle \text{(xxviii) } f(x) = \log(2+x) - \frac{2x}{2+x}, x \in R {\hspace{7.0cm} \text{[CBSE 2014]} }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = \log(2+x) - \frac{2x}{2+x}$

$\displaystyle \Rightarrow f'(x) =\frac{1}{(2+x)}- \frac{[(2+x)2 - 2x]}{(2+x)^2} = \frac{(2+x) - (4+2x-2x)}{(2+x)^2} = \frac{x-2}{(2+x)^2}$

$\displaystyle \text{Here } 2 \text{ is the critical point. }$

$\displaystyle \text{The possible intervals are } (-\infty, 2) \text{ and } (2, \infty)$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow \frac{x-2}{(2+x)^2} > 0$

$\displaystyle \Rightarrow (x-2) > 0$

$\displaystyle \Rightarrow x> 2$

$\displaystyle \Rightarrow x \in ( 2, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( 2, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow \frac{x-2}{(2+x)^2} < 0$

$\displaystyle \Rightarrow (x-2) < 0$

$\displaystyle \Rightarrow x < 2$

$\displaystyle \Rightarrow x \in ( -\infty , 2)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } ( -\infty , 2)$

$\\$

$\displaystyle \text{Question 2: Determine the value of } x \text{ for which the function } f(x) = x^2 - 6x + 9 \text{ is increasing or decreasing. Also, find the coordinates of the point on the curve } y = x^2 - 6x + 9 \text{ where the normal is parallel to the line } y = x + 5$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = x^2 - 6x + 9$

$\displaystyle \Rightarrow f'(x) =2x-6$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 2x-6 > 0$

$\displaystyle \Rightarrow x > 3$

$\displaystyle \Rightarrow x \in ( 3, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( 3, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 2x-6 < 0$

$\displaystyle \Rightarrow x < 3$

$\displaystyle \Rightarrow x \in ( -\infty , 3)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -\infty , 3)$

$\displaystyle \text{Let } (x, y) \text{ be the coordinates on the given curve where the normal to the curve} \\ \text{is parallel to the given line. }$

$\displaystyle \text{Slope of the given line }= 1.$

$\displaystyle \text{Slope of the tangent }= \Big( \frac{dy}{dx} \Big)_{(x, y)} = 2x - 6$

$\displaystyle \text{Slope of the normal }= \frac{-1}{\text{Slope of the tangent }} = \frac{-1}{2x-6}$

$\displaystyle \text{Now, Slope of the normal = Slope of the given line }$

$\displaystyle \frac{-1}{2x-6} = 1$

$\displaystyle \Rightarrow -1 = 2x-6$

$\displaystyle \Rightarrow 2x = 5$

$\displaystyle \Rightarrow x = \frac{5}{2}$

$\displaystyle \text{Given curve is } y = x^2 - 6x + 9 = \frac{25}{4}- 15 + 6 = \frac{1}{4}$

$\displaystyle (x, y) = \Big( \frac{5}{2} , \frac{1}{4} \Big)$

$\displaystyle \text{Hence the coordinates are } \Big( \frac{5}{2} , \frac{1}{4} \Big)$

$\\$

$\displaystyle \text{Question 3: Find the intervals in which } f(x) = \sin x - \cos x , \text{ where } \\ 0 < x < 2 \pi \text{ is increasing or decreasing. }$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) =\sin x - \cos x , \text{ where } \\ 0 < x < 2 \pi$

$\displaystyle \Rightarrow f'(x) =\cos x + \sin x$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow \cos x + \sin x > 0$

$\displaystyle \Rightarrow \sin x > - \cos x$

$\displaystyle \Rightarrow \tan x > - 1$

$\displaystyle \Rightarrow x \in \Big( 0, \frac{3\pi}{4} \Big) \cup \Big( \frac{7\pi}{4} , 2\pi \Big)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } \Big( 0, \frac{3\pi}{4} \Big) \cup \Big( \frac{7\pi}{4} , 2\pi \Big)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow \cos x + \sin x < 0$

$\displaystyle \Rightarrow \sin x < - \cos x$

$\displaystyle \Rightarrow \tan x < - 1$

$\displaystyle \Rightarrow x \in \Big( \frac{3\pi}{4} , \frac{7\pi}{4} \Big)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on }\Big( \frac{3\pi}{4} , \frac{7\pi}{4} \Big)$

$\\$

$\displaystyle \text{Question 4: Show that } f(x) = e^{2x} \text{ is increasing on } R. {\hspace{1.0cm} \text{[CBSE 2000, 2010]} }$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = e^{2x}$

$\displaystyle \Rightarrow f'(x) = 2e^{2x}$

$\displaystyle \text{Now } x \in R$

$\displaystyle \text{Since the value of } e^{2x} \text{ is always positive for any real value of } x , e^{2x}> 0$

$\displaystyle \Rightarrow 2e^{2x} > 0$

$\displaystyle \Rightarrow f'(x) > 0$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } R.$

$\\$

$\displaystyle \text{Question 5: Show that } f(x) = e^{\frac{1}{x}}, x \neq 0 \text{ is a decreasing function for all } \\ x \neq 0.$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = f(x) = e^{\frac{1}{x}}, x \neq 0$

$\displaystyle \Rightarrow f'(x) = e^{\frac{1}{x}} \frac{d}{dx} \Big( \frac{1}{x} \Big) = e^{\frac{1}{x}} \Big( \frac{-1}{x^2} \Big) = - \frac{e^{\frac{1}{x}}}{x^2}$

$\displaystyle \text{Now } x \in R$

$\displaystyle \text{Here, } e^{\frac{1}{x}} > 0 \text{ and } x^2 > 0 \text{ for any real value of } x \neq 0.$

$\displaystyle \therefore f(x) = - \frac{e^{\frac{1}{x}}}{x^2}< 0 \ \ \forall \ x \in R \ x \neq 0.$

$\displaystyle \text{Hence, } f(x) \text{ is a decreasing function.}$

$\\$

$\displaystyle \text{Question 6: Show that } f(x) = \log_a x , 0 < a < 1 \text{ is a decreasing function for all } \\ x > 0.$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \log_a x = \frac{\log x}{\log a}$

$\displaystyle \Rightarrow f'(x) = \frac{1}{x \log a}$

$\displaystyle \text{Since } 0 < a <1 \text{ and } x > 0, \ f'(x) = \frac{1}{x \log a} < 0$

$\displaystyle \text{Hence, } f(x) \text{ is a decreasing for all } x>0.$

$\\$

$\displaystyle \text{Question 7: Show that } f(x) = \sin x \text{ is increasing on } \Big(0, \frac{\pi}{2} \Big) \text{ and decreasing on } \Big(\frac{\pi}{2} , \pi \Big) \\ \text{and neither increasing or decreasing in } (0, \pi).$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \sin x$

$\displaystyle \Rightarrow f'(x) = \cos x$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow \cos x > 0$

$\displaystyle \Rightarrow x \in \Big( 0, \frac{\pi}{2} \Big)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } \Big( 0, \frac{\pi}{2} \Big).$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow \cos x < 0$

$\displaystyle \Rightarrow x \in \Big( \frac{\pi}{2}, \pi \Big)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } \Big( \frac{\pi}{2}, \pi \Big).$

$\displaystyle \text{Since } f(x) \text{ is increasing on } \Big( 0, \frac{\pi}{2} \Big) \text{ and is decreasing on } \Big( \frac{\pi}{2}, \pi \Big) \\ \text{it is neither increasing or decreasing in } (0, \pi).$

$\\$

$\displaystyle \text{Question 8: Show that } f(x) = \log \sin x \text{ is increasing on } \Big(0, \frac{\pi}{2} \Big) \text{ and decreasing on } \Big(\frac{\pi}{2} , \pi \Big).$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \log \sin x$

$\displaystyle \text{Domain of }\log \sin x \text{ is } (0, \pi).$

$\displaystyle \Rightarrow f'(x) = \frac{1}{\sin x} \cos x = \cot x$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow \cot x > 0$

$\displaystyle \Rightarrow x \in \Big( 0, \frac{\pi}{2} \Big) \ [\because \text{ Cot function is positive in first quadrant} ]$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } \Big( 0, \frac{\pi}{2} \Big).$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow \cot x < 0$

$\displaystyle \Rightarrow x \in \Big( \frac{\pi}{2}, \pi \Big) \ [\because \text{ Cot function is negative in second quadrant} ]$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } \Big( \frac{\pi}{2}, \pi \Big).$

$\\$

$\displaystyle \text{Question 9: Show that } f(x) = x - \sin x \text{ is increasing on for all } x \in R.$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = x - \sin x$

$\displaystyle \Rightarrow f'(x) = 1 - \cos x$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 1 - \cos x > 0$

$\displaystyle \Rightarrow \cos x < 1$

$\displaystyle \Rightarrow f'(x) \geq 0 \text{ for all } x \in R \ \ [ \because \cos x \leq 1]$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } x \in R.$

$\\$

$\displaystyle \text{Question 10: Show that } f(x) = x^3 - 15x^2+75x-50 \text{ is an increasing function for all } \\ x \in R.$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = x^3 - 15x^2+75x-50$

$\displaystyle \Rightarrow f'(x) = 3x^2 - 30x + 75 = 3(x^2-10x+25) = 3 (x-5)^2 > 0 \ \ \forall x \in R$

$\displaystyle [ \because \text{ square of any function is always greater than 0} ]$

$\displaystyle \text{So } f(x) \text{ is increasing function for all } x \in R$

$\\$

$\displaystyle \text{Question 11: Show that } f(x) = \cos^2 x \text{ is decreasing function on } (0, \frac{\pi}{2}).$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \cos^2 x$

$\displaystyle \Rightarrow f'(x) = 2 \cos x (- \sin x) = - \sin 2x$

$\displaystyle \text{Now, } 0 < x < \frac{\pi}{2}$

$\displaystyle \Rightarrow 0 < 2x < \pi$

$\displaystyle \Rightarrow \sin 2x > 0 \ [\because \text{ Sine function is positive in first and second quadrant} ]$

$\displaystyle \Rightarrow -\sin 2x < 0$

$\displaystyle \Rightarrow f'(x) < 0$

$\displaystyle \text{So, } f(x) \text{ is decreasing function on } \Big(0, \frac{\pi}{2} \Big)$

$\\$

$\displaystyle \text{Question 12: Show that } f(x) = \sin x \text{ is an increasing function on } (-\frac{\pi}{2}, \frac{\pi}{2}).$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \sin x$

$\displaystyle \Rightarrow f'(x) = \cos x > 0 \ \forall \ x \in \Big( - \frac{\pi}{2}, \frac{\pi}{2} \Big) \\ \ [\because \text{Cos function is positive in first and fourth quadrant} ]$

$\displaystyle \text{So } f(x) \text{ is increasing function for all } \Big( - \frac{\pi}{2}, \frac{\pi}{2} \Big)$

$\\$

$\displaystyle \text{Question 13: Show that } f(x) = \cos x \text{ is decreasing function on } (0, \pi), \text{ increasing in } \\ (-\pi, 0) \text{and neither increasing or decreasing in } (-\pi, \pi).$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \cos x$

$\displaystyle \text{Domain of } \cos x \text{ is } (-\pi, \pi).$

$\displaystyle \Rightarrow f'(x) = - \sin x$

$\displaystyle \text{For } x \in ( -\pi, 0 ), \sin x < 0 \ [\because \text{Sine function is negative in third and fourth quadrant} ]$

$\displaystyle \Rightarrow - \sin x > 0$

$\displaystyle \Rightarrow f'(x) > 0$

$\displaystyle \text{So, } f(x) \text{ is increasing function in } ( -\pi, 0 )$

$\displaystyle \text{For } x \in ( 0, \pi ), \sin x > 0 \ [\because \text{Sine function is positive in first and second quadrant} ]$

$\displaystyle \Rightarrow - \sin x <0$

$\displaystyle \Rightarrow f'(x) < 0$

$\displaystyle \text{So, } f(x) \text{ is decreasing function in } ( 0, \pi )$

$\displaystyle \text{Thus } f(x) \text{is neither increasing or decreasing in } (-\pi, \pi).$

$\\$

$\displaystyle \text{Question 14: Show that } f(x) = \tan x \text{ is an increasing function on } (-\frac{\pi}{2}, \frac{\pi}{2}).$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \tan x$

$\displaystyle \Rightarrow f'(x) = \sec^2 x$

$\displaystyle \text{Here, } -\frac{\pi}{2} < x < \frac{\pi}{2}$

$\displaystyle \Rightarrow \sec x > 0 \ [\because \text{Sec function is positive in first and four quadrant} ]$

$\displaystyle \Rightarrow \sec^2 x > 0$

$\displaystyle \Rightarrow f'(x) > 0, \ \forall \ x \in \Big( -\frac{\pi}{2} < x < \frac{\pi}{2} \Big)$

$\displaystyle \text{Thus } f(x) \text{is increasing on } \Big( -\frac{\pi}{2} < x < \frac{\pi}{2} \Big).$

$\\$

$\displaystyle \text{Question 15: Show that } f(x) = \tan^{-1} (\sin x + \cos x) \text{ is a decreasing function on interval } (\frac{\pi}{4}, \frac{\pi}{2}).$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \tan^{-1} (\sin x + \cos x)$

$\displaystyle \Rightarrow f'(x) = \frac{1}{1+(\sin x + \cos x)^2}(\cos x - \sin x) \\ \\ {\hspace{1.5cm}= \frac{1}{1+1 + 2 \sin x \cos x }(\cos x - \sin x)} \\ \\ {\hspace{1.5cm}= \frac{(\cos x - \sin x)}{2 + \sin 2x}}$

$\displaystyle \text{Here, } \frac{\pi}{4} < x < \frac{\pi}{2}$

$\displaystyle \Rightarrow \frac{\pi}{2} < 2x < \pi$

$\displaystyle \Rightarrow \sin 2x > 0$

$\displaystyle \Rightarrow 2+ \sin 2x > 0$

$\displaystyle \text{Here, } \frac{\pi}{4} < x < \frac{\pi}{2}$

$\displaystyle \Rightarrow \cos x < \sin x$

$\displaystyle \Rightarrow \cos x - \sin x < 0$

$\displaystyle f'(x) = \frac{(\cos x - \sin x)}{2 + \sin 2x} < 0 \ \forall x \in \Big(\frac{\pi}{4}, \frac{\pi}{2}\Big)$

$\displaystyle \text{Thus } f(x) \text{is decreasing on } \Big(\frac{\pi}{4}, \frac{\pi}{2}\Big).$

$\\$

$\displaystyle \text{Question 16: Show that the function } f(x) = \sin \Big( 2x+ \frac{\pi}{4} \Big) \text{ is decreasing on } \Big(\frac{3\pi}{8}, \frac{5\pi}{8}\Big).$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \sin ( 2x+ \frac{\pi}{4})$

$\displaystyle \Rightarrow f'(x) = 2 \cos ( 2x+ \frac{\pi}{4})$

$\displaystyle \text{Here, } \frac{3\pi}{8} < x < \frac{5\pi}{8}$

$\displaystyle \Rightarrow \frac{3\pi}{4} < 2x < \frac{5\pi}{4}$

$\displaystyle \Rightarrow \pi < 2x + \frac{\pi}{4}< \frac{3\pi}{2}$

$\displaystyle \Rightarrow \cos \Big( 2x + \frac{\pi}{4} \Big) < 0$

$\displaystyle \Rightarrow 2\cos \Big( 2x + \frac{\pi}{4} \Big) < 0$

$\displaystyle \Rightarrow f'(x) < 0 \ \forall \ x \in (\frac{3\pi}{8}, \frac{5\pi}{8})$

$\displaystyle \text{Thus } f(x) \text{is decreasing on } (\frac{3\pi}{8}, \frac{5\pi}{8}).$

$\\$

$\displaystyle \text{Question 17: Show that the function } f(x) = \cot^{-1} (\sin x + \cos x) \text{ is decreasing on } (0, \frac{\pi}{4}) \\ \text{and increasing on } (\frac{\pi}{4}, \frac{\pi}{2}).$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \cot^{-1} (\sin x + \cos x)$

$\displaystyle \Rightarrow f'(x) = \frac{-1}{1+(\sin x + \cos x)^2} \times (\cos x - \sin x) \\ \\ {\hspace{1.5cm} = \frac{(\sin x - \cos x)}{1 + 1 + 2 \sin x \cos x}} \\ \\ {\hspace{1.5cm} = \frac{(\sin x - \cos x)}{2 + 2 \sin x \cos x }} \\ \\ {\hspace{1.5cm} = \frac{1}{2} \times \frac{(\sin x - \cos x)}{1 + \sin x \cos x }}$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow \frac{1}{2} \times \frac{(\sin x - \cos x)}{1 + \sin x \cos x } < 0$

$\displaystyle \Rightarrow \frac{(\sin x - \cos x)}{1 + \sin x \cos x } < 0$

$\displaystyle \Rightarrow \sin x - \cos x < 0 \ \ [\text{In first quadrant}]$

$\displaystyle \Rightarrow \sin x < \cos x$

$\displaystyle \Rightarrow \tan x < 1$

$\displaystyle \Rightarrow 0 < x < \frac{\pi}{4}$

$\displaystyle \text{Thus } f(x) \text{is decreasing on } \Big(0, \frac{\pi}{4} \Big).$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow \frac{1}{2} \times \frac{(\sin x - \cos x)}{1 + \sin x \cos x } > 0$

$\displaystyle \Rightarrow \frac{(\sin x - \cos x)}{1 + \sin x \cos x } > 0$

$\displaystyle \Rightarrow \sin x - \cos x > 0 \ \ [\text{In first quadrant}]$

$\displaystyle \Rightarrow \sin x > \cos x$

$\displaystyle \Rightarrow \tan x > 1$

$\displaystyle \Rightarrow \frac{\pi}{4} < x < \frac{\pi}{2}$

$\displaystyle \text{Thus } f(x) \text{is increasing on } \Big(\frac{\pi}{4}, \frac{\pi}{2} \Big).$

$\\$

$\displaystyle \text{Question 18: Show that } f(x) = (x-1) e^x + 1 \text{ is an increasing function for all } \\ x > 0.$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = (x-1) e^x + 1$

$\displaystyle \Rightarrow f'(x) =(x-1) e^x + e^x = x e^x$

$\displaystyle \text{Given } x > 0$

$\displaystyle \text{We know } e^x > 0$

$\displaystyle \Rightarrow x e^x > 0$

$\displaystyle \Rightarrow f'(x) > 0 , \ \forall \ x > 0$

$\displaystyle \text{So, } f(x) \text{ is an increasing function for all } \ \ x > 0.$

$\\$

$\displaystyle \text{Question 19: Show that } f(x) = x^2 - x + 1 \text{ is neither increasing or decreasing on } (0, 1).$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = x^2 - x + 1$

$\displaystyle \Rightarrow f'(x) = 2x-1$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow 2x-1 > 0$

$\displaystyle \Rightarrow 2x > 1$

$\displaystyle \Rightarrow x > \frac{1}{2}$

$\displaystyle \Rightarrow x \in \Big(\frac{1}{2}, 1 \Big)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } \Big(\frac{1}{2}, 1 \Big)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow 2x-1 < 0$

$\displaystyle \Rightarrow 2x < 1$

$\displaystyle \Rightarrow x < \frac{1}{2}$

$\displaystyle \Rightarrow x \in \Big(0, \frac{1}{2} \Big)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } \Big(0, \frac{1}{2} \Big)$

$\displaystyle \text{Since } f(x) \text{ is increasing on } \Big(\frac{1}{2}, 1 \Big) \text{ and is decreasing on } \Big(0, \frac{1}{2} \Big), f(x) \\ \text{is neither increasing or decreasing on } (0, 1).$

$\\$

$\displaystyle \text{Question 20: Show that the function} f(x) = x^9+4x^7 \text{ is an increasing function for all } \\ x \in R.$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = x^9+4x^7$

$\displaystyle \Rightarrow f'(x) =9x^8+28x^6 \geq 0 \ \forall \ x \in R \ [\because x^7, x^6 \geq 0 \ \forall \ x \in R ]$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on R. }$

$\\$

$\displaystyle \text{Question 21: Prove that the function } f(x) = x^3 - 6x^2 + 12x - 18 \\ \text{is increasing on } R. {\hspace{9.0cm} \text{[CBSE 2002C]} }$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = x^3 - 6x^2 + 12x - 18$

$\displaystyle \Rightarrow f'(x) =3x^2- 12x + 12 = 3 (x-2)^2 \geq 0 \ \forall \ x \in R \ [\because 3 > 0 (x-2)^2 \geq 0 ]$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on R. }$

$\\$

$\displaystyle \text{Question 22: State when a function } f(x) \text{ is said to be increasing on an interval } [a, b]. \\ \text{Test whether the function } f(x) = x^2 - 6x + 3 \text{ is increasing on the interval } [4, 6].$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = x^2 - 6x + 3$

$\displaystyle \Rightarrow f'(x) = 2x - 6 = 2 ( x - 3)$

$\displaystyle \text{Now,} f'(4) = 2 ( 4 - 3 ) = 2$

$\displaystyle \therefore f'(4) > 0$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } x=4$

$\displaystyle \text{Now,} f'(6) = 2 ( 6 - 3 ) = 6$

$\displaystyle \therefore f'(6) > 0$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } x=6$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } [4, 6]$

$\\$

$\displaystyle \text{Question 23: Show that } f(x) = \sin x - \cos x \text{ is an increasing function on interval } (-\frac{\pi}{4}, \frac{\pi}{4}).$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \sin x - \cos x$

$\displaystyle \Rightarrow f'(x) = \cos x + \sin x \\ \\ {\hspace{1.5cm} = \sqrt{2} \Big( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x \Big) } \\ \\ {\hspace{1.5cm} = \sqrt{2} \Big( \sin \frac{\pi}{4} \cos x + \cos \frac{\pi}{4} \sin x \Big) } \\ \\ {\hspace{1.5cm} = \sqrt{2} \sin \Big( \frac{\pi}{4}+ x \Big) }$

$\displaystyle \text{Now, } x \in \Big( -\frac{\pi}{4}, \frac{5\pi}{4} \Big)$

$\displaystyle \text{Here, } -\frac{\pi}{4} < x < \frac{\pi}{4}$

$\displaystyle \Rightarrow 0 < x + \frac{\pi}{4}< \frac{\pi}{2}$

$\displaystyle \Rightarrow \sin 0 < \sin \Big(x + \frac{\pi}{4} \Big) < \sin \Big( \frac{\pi}{2} \Big)$

$\displaystyle \Rightarrow 0 < \sin \Big(x + \frac{\pi}{4} \Big) < 1$

$\displaystyle \Rightarrow \sqrt{2} \sin \Big(x + \frac{\pi}{4} \Big) > 0$

$\displaystyle f'(x) > 0$

$\displaystyle \text{Thus } f(x) \text{is increasing on } \Big(-\frac{\pi}{4}, \frac{\pi}{4} \Big).$

$\\$

$\displaystyle \text{Question 24: Show that } f(x) = \tan^{-1} x - x \text{ is a decreasing function on } R.$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \tan^{-1} x - x$

$\displaystyle \Rightarrow f'(x) = \frac{1}{1+x^2}-1 = \frac{1-1-x^2}{1+x^2} = \frac{-x^2}{1+x^2}$

$\displaystyle \text{We know, } x^2 \geq 0 \text{ and } (1+x^2) \geq 0 , \ \forall \ x\in R$

$\displaystyle \Rightarrow f'(x) < 0 \ \forall \ x\in R$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } R.$

$\\$

$\displaystyle \text{Question 25: Determine whether} f(x) = -\frac{x}{2} + \sin x \text{ is a increasing or decreasing on } \\ \\ \Big(-\frac{\pi}{3}, \frac{\pi}{3} \Big).$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = -\frac{x}{2} + \sin x$

$\displaystyle \Rightarrow f'(x) = \frac{-1}{2}+ \cos x$

$\displaystyle \text{Here, } -\frac{\pi}{3} < x < \frac{\pi}{3}$

$\displaystyle \Rightarrow \cos x > \frac{1}{2}$

$\displaystyle \Rightarrow -\frac{1}{2}+ \cos x > 0$

$\displaystyle \Rightarrow f'(x) < 0 \ \forall \ x\in \Big(-\frac{\pi}{3}, \frac{\pi}{3} \Big)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } \Big(-\frac{\pi}{3}, \frac{\pi}{3} \Big).$

$\\$

$\displaystyle \text{Question 26: Find the intervals in which } f(x) = \log (1+x) - \frac{x}{1+x} \text{ is increasing or decreasing.} \\ {\hspace{1.0cm} \text{[CBSE 2002C]} }$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \log (1+x) - \frac{x}{1+x}$

$\displaystyle \text{Domain for } f(x) \text{ is } (-1, \infty)$

$\displaystyle \Rightarrow f'(x) = \frac{1}{1+x} - \Bigg\{ \frac{1+x-x}{(1+x)^2} \Bigg\} = \frac{1}{1+x} - \frac{1}{(1+x)^2} = \frac{x}{(1+x)^2}$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow \frac{x}{(1+x)^2} > 0$

$\displaystyle \Rightarrow x > 0 \ [ \because (1+x)^2 > 0 \ \ \text{Domain: } (-1, \infty) ]$

$\displaystyle \Rightarrow x \in (0, \infty)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } (0, \infty)$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow \frac{x}{(1+x)^2} < 0$

$\displaystyle \Rightarrow x < 0 \ [ \because (1+x)^2 > 0 \ \ \text{Domain: } (-1, \infty) ]$

$\displaystyle \Rightarrow x \in (-1, 0)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } (-1, 0)$

$\\$

$\displaystyle \text{Question 27: Show that the function } f \text{ given by } f(x) = (x+2)e^{-x} \\ \text{ is increasing or decreasing.} {\hspace{8.0cm} \text{[CBSE 2002C]} }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When } (x-a)(x-b)> 0 \text{ with } a < b, \ x < a \text{ or } x> b.$

$\displaystyle \text{When } (x-a)(x-b)< 0 \text{ with } a < b, \ a < x < b.$

$\displaystyle f(x) = (x+2)e^{-x}$

$\displaystyle \Rightarrow f'(x) = - e^{-x} (x+2) + e^{-x} = -x e^{-x}-2e^{-x}+ e^{-x} \\ {\hspace{5.5cm} = -x e^{-x}- e^{-x}= e^{-x} (-x-1) }$

$\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x) > 0$

$\displaystyle \Rightarrow e^{-x} (-x-1) > 0$

$\displaystyle \Rightarrow -x-1 > 0 \ [e^{-x} > 0 \ \forall \ x \in R ]$

$\displaystyle \Rightarrow -x > 1$

$\displaystyle \Rightarrow x < -1$

$\displaystyle \Rightarrow x \in ( -\infty, 1 )$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } ( -\infty, 1 )$

$\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x) < 0$

$\displaystyle \Rightarrow e^{-x} (-x-1) < 0$

$\displaystyle \Rightarrow -x-1 < 0 \ [e^{-x} > 0 \ \forall \ x \in R ]$

$\displaystyle \Rightarrow -x < 1$

$\displaystyle \Rightarrow x > -1$

$\displaystyle \Rightarrow x \in ( -1, \infty )$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } ( -1, \infty )$

$\\$

$\displaystyle \text{Question 28: Show that the function } f \text{ given by } f(x) = 10^{x} \text{ is increasing for all } x.$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = 10^{x}$

$\displaystyle f'(x) = 10^{x} \log 10 > 0 \ \forall \ x \in R$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } R$

$\\$

$\displaystyle \text{Question 29: Prove that the function } f \text{ given by } f(x) = x - [x] \text{ is increasing in } (0, 1).$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = x - [x]$

$\displaystyle \text{Let } x_-1, x_2 \in (0, 1) \text{ such that } x_1 < x_2. \text{ Then, }$

$\displaystyle [x_1] = [x_2] = 0$

$\displaystyle \text{Now, } x_1 < x_2$

$\displaystyle \Rightarrow x_1 - [x_1] < x_2 - [x_2]$

$\displaystyle \Rightarrow f(x_1) < f(x_2)$

$\displaystyle \Rightarrow \therefore x_1 < x_2$

$\displaystyle f(x_1) < f(x_2) \ \forall \ x_1, x_2 \in (0, 1)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } (0, 1)$

$\\$

$\displaystyle \text{Question 30: Prove that the following functions are increasing on } R: {\hspace{0.0cm} \text{[CBSE 2017]} }$

$\displaystyle \text{(i) } f(x) = 3x^5+ 40x^3 + 240x$

$\displaystyle \text{(ii) } f(x) = 4x^3-18x^2+27x-27$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle f(x) = 3x^5+ 40x^3 + 240x$

$\displaystyle f'(x) = 15x^4+ 120x^2 + 240 = 15(x^2+4)^2 > 0 \ \forall \ x \in R \ [ \because 15 > 0, (x^2+4)^2> 0 ]$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } R$

$\displaystyle \text{(ii) }$

$\displaystyle f(x) = 4x^3-18x^2+27x-27$

$\displaystyle f'(x) = 12x^2-36x+27 = 3(4x^2-12x+9) \\ {\hspace{4.5cm} = 3(2x-3)^2 > 0 \ \forall \ x \in R \ [ \because 3 > 0, (2x-3)^2> 0 ] }$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } R$

$\\$

$\displaystyle \text{Question 31: Prove that the function } f \text{ given by } f(x) = \log \cos x \text{ is strictly increasing on } (-\frac{\pi}{2}, 0) \\ \text{and is strictly decreasing on } (0, \frac{\pi}{2}).$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \log \cos x$

$\displaystyle f'(x) = \frac{1}{\cos x} (-\sin x) = - \tan x$

$\displaystyle \text{Now, } x \in \Big( -\frac{\pi}{2}, 0 \Big)$

$\displaystyle \Rightarrow \tan x < 0$

$\displaystyle \Rightarrow - \tan x > 0$

$\displaystyle \Rightarrow f'(x) > 0$

$\displaystyle \text{Hence, } f(x) \text{ is strictly increasing on } \Big (-\frac{\pi}{2}, 0 \Big)$

$\displaystyle \text{Now, } x \in \Big( 0, \frac{\pi}{2} \Big)$

$\displaystyle \Rightarrow \tan x > 0$

$\displaystyle \Rightarrow - \tan x < 0$

$\displaystyle \Rightarrow f'(x) < 0$

$\displaystyle \text{Hence, } f(x) \text{ is strictly decreasing on } \Big(0, \frac{\pi}{2} \Big)$

$\\$

$\displaystyle \text{Question 32: Prove that the function } f \text{ given by } f(x) = x^3 - 3x^2 + 4x \text{and is strictly decreasing on } R.$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = x^3 - 3x^2 + 4x$

$\displaystyle f'(x) = 3x^2- 6x + 4 = 3(x^2 - 2x) + 4 = 3(x-1)^2 + 1 > 0 \ \forall x \in R$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } R$

$\\$

$\displaystyle \text{Question 33: Prove that the function } f(x) = \cos x \text{ is: }$

$\displaystyle \text{(i) strictly decreasing in } (0, \pi)$

$\displaystyle \text{(ii) strictly increasing in } (\pi,. 2\pi)$

$\displaystyle \text{(iii) neither increasing nor decreasing in } (0, 2\pi)$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \cos x$

$\displaystyle f'(x) = - \sin x$

$\displaystyle \text{(i) }$

$\displaystyle 0 < x < \pi$

$\displaystyle \Rightarrow \sin x > 0$

$\displaystyle \Rightarrow -\sin x < 0$

$\displaystyle \Rightarrow f'(x) < 0 \ \forall \ x \in (0, \pi)$

$\displaystyle \text{Hence, } f(x) \text{ is strictly decreasing on } (0, \pi).$

$\displaystyle \text{(ii) }$

$\displaystyle \pi < x < 2\pi$

$\displaystyle \Rightarrow \sin x < 0$

$\displaystyle \Rightarrow -\sin x > 0$

$\displaystyle \Rightarrow f'(x) > 0 \ \forall \ x \in (\pi, 2\pi)$

$\displaystyle \text{Hence, } f(x) \text{ is strictly increasing on } (\pi, 2\pi).$

$\displaystyle \text{(iii) }$

$\displaystyle f(x) \text{ is strictly decreasing on } (0, \pi) \text{ and is strictly increasing on } (\pi, 2\pi).$

$\displaystyle \text{Therefore } f(x) \text{ is neither increasing nor decreasing in } (0, 2\pi)$

$\\$

$\displaystyle \text{Question 34: Show that} f(x) = x^2 - x \sin x \text{ is an increasing function on } (0, \frac{\pi}{2})$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = x^2 - x \sin x$

$\displaystyle f'(x) = 2x - x \cos x- \sin x$

$\displaystyle \text{Here, } 0 < x < \frac{\pi}{2}$

$\displaystyle \Rightarrow 0 < \sin x < 1 \text{ and } 0 < \cos x < 1$

$\displaystyle \Rightarrow 2x - x \cos x- \sin x > 0$

$\displaystyle \Rightarrow f'(x) > 0 \ \forall \ x \in (0, \frac{\pi}{2})$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } (0, \frac{\pi}{2}).$

$\\$

$\displaystyle \text{Question 35: Find the values(s) of } a \text{ for which the function } f(x) = x^3 - ax \\ \text{is a decreasing function on } R.$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = x^3 - ax$

$\displaystyle f'(x) = 3x^2 - a$

$\displaystyle \text{Given } f(x) \text{ is increasing on } R.$

$\displaystyle \Rightarrow f'(x) \geq 0 \ \forall \ x \in R$

$\displaystyle \Rightarrow 3x^2 - a \geq 0 \ \forall \ x \in R$

$\displaystyle \Rightarrow 3x^2 \geq a \ \forall \ x \in R$

$\displaystyle \text{Hence, the lease value of } 3x^2 \text{ is } 0.$

$\displaystyle \therefore a \leq 0$

$\\$

$\displaystyle \text{Question 36: Find the values(s) of } b \text{ for which the function } \\ f(x) = \sin x - bx + c \text{ is a decreasing function on } R.$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = \sin x - bx + c$

$\displaystyle f'(x) = \cos x - b$

$\displaystyle \text{Given } f(x) \text{ is decreasing on } R.$

$\displaystyle \Rightarrow f'(x) < 0 \ \forall \ x \in R$

$\displaystyle \Rightarrow \cos x - b < 0 \ \forall \ x \in R$

$\displaystyle \Rightarrow \cos x < b \ \forall \ x \in R$

$\displaystyle \therefore b \geq 1 \ [\because -1 \leq \cos x \leq 1 ]$

$\\$

$\displaystyle \text{Question 37: Show that} f(x) = x + \cos x - a \text{ is a increasing function on } R \text{ for all values of } \\ a.$

$\displaystyle \text{Answer:}$

$\displaystyle f(x) = x + \cos x - a$

$\displaystyle f'(x) = 1 - \sin x$

$\displaystyle \text{We know, } \sin x \leq 1 \ \forall \ x \in R$

$\displaystyle \Rightarrow -\sin x \geq -1 \ \forall \ x \in R$

$\displaystyle \Rightarrow 1-\sin x \geq 0 \ \forall \ x \in R$

$\displaystyle \Rightarrow f'(x) \geq 0 \ \forall \ x \in R$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } R \text{ for all values of } a.$

$\\$

$\displaystyle \text{Question 38: Let } f \text{ defined on } [0, 1] \text{ be twice differentiable such that } \\ |f''(x)| \leq 1 \text{ for all } x \in [0, 1]. \text{ If } f(0) = f(1), \text{ then show that } |f''(x)| < 1 \text{ for all } \\ x \in [0, 1].$

$\displaystyle \text{Answer:}$

$\displaystyle \text{If a function is continuous and differentiable and } f(0) = f(1) \text{ in given domain } x \in [0, 1], \text{ then by Rolle's Theorem: }$

$\displaystyle f(x) = 0 \text{ for some } x \in [0, 1]$

$\displaystyle \text{Given: } |f''(x)| \leq 1$

$\displaystyle \text{On integrating both sides we get, }$

$\displaystyle |f'(x)| \leq x$

$\displaystyle \text{Now within the interval } x \in [0, 1], \text{ we get } |f'(x)| < 1.$

$\\$

$\displaystyle \text{Question 39: Find the intervals in which } f(x) \text{ is increasing or decreasing: } {\hspace{0.0cm} \text{[CBSE 2014]} }$

$\displaystyle \text{(i) } f(x) = x |x|, \ x \in R$

$\displaystyle \text{(ii) } f(x) = \sin x + | \sin x |, 0 \leq x \leq 2\pi$

$\displaystyle \text{(iii) } f(x) = \sin x (1 + \cos x ), 0 < x < \frac{\pi}{2}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle f(x) = x |x|, \ x \in R$

$\displaystyle \text{Case 1: When } x \geq 0$

$\displaystyle f(x) = x |x| = x(x) = x^2$

$\displaystyle \Rightarrow f'(x) = 2x \geq 0 \ \forall x \geq 0$

$\displaystyle \text{Hence, } f(x) \text{ is increasing for } x \geq 0.$

$\displaystyle \text{Case 1: When } x < 0$

$\displaystyle f(x) = x |x| = x(-x) = -x^2$

$\displaystyle \Rightarrow f'(x) = -2x \geq 0 \ \forall x < 0$

$\displaystyle \text{Hence, } f(x) \text{ is increasing for } x < 0$

$\displaystyle \text{Hence, } f(x) \text{ is increasing for } x \in R.$

$\displaystyle \text{(ii) }$

$\displaystyle f(x) = \sin x + | \sin x |, 0 \leq x \leq 2\pi$

$\displaystyle \text{Case 1: When } x \in ( 0, \pi)$

$\displaystyle f(x) = \sin x + \sin x = 2 \sin x$

$\displaystyle \Rightarrow f'(x) = 2 \cos x$

$\displaystyle \text{As, } \cos x > 0 \text{ for } x \in \Big( 0, \frac{\pi}{2} \Big) \text{ and } \cos x < 0 \text{ for } x \in \Big(\frac{\pi}{2}, \pi \Big)$

$\displaystyle \text{So, } f'(x) > 0 \text{ for } x \in \Big( 0, \frac{\pi}{2} \Big) \text{ and } f'(x) < 0 \text{ for } x \in \Big(\frac{\pi}{2}, \pi \Big)$

$\displaystyle \text{Therefore, } f(x) \text{ is increasing on } \Big( 0, \frac{\pi}{2} \Big) \text{ and } f(x) \text{ is decreasing on } \Big(\frac{\pi}{2}, \pi \Big)$

$\displaystyle \text{Case 2: When } x \in ( \pi, 2\pi)$

$\displaystyle f(x) = \sin x - \sin x = 0$

$\displaystyle \Rightarrow f'(x) = 0$

$\displaystyle \text{Hence } f(x) \text{ is neither increasing or decreasing on } ( \pi, 2\pi).$

$\displaystyle \text{(iii) }$

$\displaystyle f(x) = \sin x (1 + \cos x ) = \sin x + \sin x \cos x$

$\displaystyle f'(x) = \cos x + \sin x ( - \sin x ) + \cos^2 x$

$\displaystyle = \cos x - \sin^2 x + \cos^2 x$

$\displaystyle = \cos x + \cos^2 x - ( 1- \cos^2 x)$

$\displaystyle = 2 \cos^2 x + \cos x - 1$

$\displaystyle = 2 \cos^2 x + 2\cos x - \cos x - 1$

$\displaystyle = 2 \cos x ( \cos x + 1 ) - ( \cos x + 1)$

$\displaystyle = ( \cos x + 1 )(2 \cos x - 1)$

$\displaystyle \text{For } f(x) \text{ to be increasing we must have, }$

$\displaystyle f'(x) > 0$

$\displaystyle \Rightarrow ( \cos x + 1 )(2 \cos x - 1) > 0$

$\displaystyle \text{This is only possible if }$

$\displaystyle (2 \cos x - 1) > 0 \text{ and } ( \cos x + 1 ) > 0$

$\displaystyle \Rightarrow (2 \cos x - 1) > 0 \text{ and } ( \cos x + 1 ) > 0$

$\displaystyle \Rightarrow \cos x > \frac{1}{2} \text{ and } \cos x > - 1$

$\displaystyle \Rightarrow x \in \Big( 0, \frac{\pi}{3} \Big) \text{ and } x \in \Big( 0, \frac{\pi}{2} \Big)$

$\displaystyle \text{Hence, } x \in \Big( 0, \frac{\pi}{3} \Big)$

$\displaystyle \text{Hence, } f(x) \text{ is increasing on } \Big( 0, \frac{\pi}{3} \Big).$

$\displaystyle \text{For } f(x) \text{ to be decreasing we must have, }$

$\displaystyle f'(x) < 0$

$\displaystyle \Rightarrow ( \cos x + 1 )(2 \cos x - 1) < 0$

$\displaystyle \text{This is only possible if }$

$\displaystyle (2 \cos x - 1) < 0 \text{ and } ( \cos x + 1 ) > 0$

$\displaystyle \Rightarrow (2 \cos x - 1) < 0 \text{ and } ( \cos x + 1 ) > 0$

$\displaystyle \Rightarrow \cos x < \frac{1}{2} \text{ and } \cos x > - 1$

$\displaystyle \Rightarrow x \in \Big( \frac{\pi}{3}, \frac{\pi}{2} \Big) \text{ and } x \in \Big( 0, \frac{\pi}{2} \Big)$

$\displaystyle \text{Hence, } x \in \Big( \frac{\pi}{3}, \frac{\pi}{2} \Big)$

$\displaystyle \text{Hence, } f(x) \text{ is decreasing on } \Big( \frac{\pi}{3}, \frac{\pi}{2} \Big).$

$\\$

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Class 12: Increasing and Decreasing Functions – Exercise 17.2

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