\displaystyle \textbf{Question 1. }\text{The function }f(x)=x^{3}-3x\text{ is strictly decreasing on  } \ \
\displaystyle \text{(a) }-1<x<1 \ \
\displaystyle \text{(b) }-1\leq x\leq 1 \ \
\displaystyle \text{(c) }1\leq x<\infty \ \
\displaystyle \text{(d) }-\infty<x<-1 \ \
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let } f(x)=x^{3}-3x
\displaystyle \text{So, } f'(x)=3x^{2}-3
\displaystyle \text{For decreasing, } f'(x)<0
\displaystyle \Rightarrow 3x^{2}-3<0
\displaystyle \Rightarrow x^{2}-1<0
\displaystyle \Rightarrow (x-1)(x+1)<0
\displaystyle \text{which is valid if } -1<x<1
\\

\displaystyle \textbf{Question 2. }\text{If }f(x)=x^{3}-48x,\text{ find the interval where }f:R\rightarrow R  \text{ is decreasing.  }
\displaystyle \text{Answer:}
\displaystyle \text{We have, } f(x)=x^{3}-48x
\displaystyle \therefore f'(x)=3x^{2}-48=3(x^{2}-16)
\displaystyle \text{Now, for } f(x) \text{ to be decreasing, } f'(x)\leq 0
\displaystyle \therefore 3(x^{2}-16)\leq 0
\displaystyle \Rightarrow x^{2}-16\leq 0
\displaystyle \Rightarrow (x-4)(x+4)\leq 0
\displaystyle \Rightarrow x\in[-4,4]
\\

\displaystyle \textbf{Question 3. }\text{Find the intervals in which the function }f(x)\text{ is strictly increasing where,} \ \
\displaystyle \text{ }f(x)=10-6x-2x^{2}.  \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } f(x)=10-6x-2x^{2}
\displaystyle \text{On differentiating w.r.t. } x, \text{ we get}
\displaystyle f'(x)=-6-4x
\displaystyle f(x) \text{ is strictly increasing if } f'(x)>0
\displaystyle \Rightarrow -6-4x>0
\displaystyle \Rightarrow -4x>6
\displaystyle \Rightarrow x<-\frac{3}{2}
\displaystyle \therefore f(x) \text{ is strictly increasing on } \left(-\infty,-\frac{3}{2}\right)
\displaystyle -6-4x>0
\displaystyle \Rightarrow 4x+6<0
\displaystyle \Rightarrow x<-\frac{3}{2}
\displaystyle \text{Hence, } f(x) \text{ is strictly increasing when } x\in\left(-\infty,-\frac{3}{2}\right).
\\

\displaystyle \textbf{Question 4. }\text{The side of an equilateral triangle is increasing at the rate of }2\ \mathrm{cm/s}. \ \
\displaystyle \text{At what rate is its area increasing, when the side of the triangle is }20\ \mathrm{cm}? \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } a \text{ be the side of an equilateral triangle and } A \text{ be its area.}
\displaystyle \text{Then, } \frac{da}{dt}=2 \text{ cm/s}
\displaystyle \text{We know that area of an equilateral triangle, } A=\frac{\sqrt{3}}{4}a^{2}
\displaystyle \text{On differentiating both sides w.r.t. } t, \text{ we get}
\displaystyle \frac{dA}{dt}=\frac{\sqrt{3}}{4}\times 2a\times \frac{da}{dt}
\displaystyle \Rightarrow \frac{dA}{dt}=\frac{\sqrt{3}}{4}\times 2\times 20\times 2 \quad [\text{given, } a=20]
\displaystyle \therefore \frac{dA}{dt}=20\sqrt{3} \text{ cm}^{2}\text{/s}
\displaystyle \text{Thus, the rate of area increasing is } 20\sqrt{3} \text{ cm}^{2}\text{/s.}
\\

\displaystyle \textbf{Question 5. }\text{Find the intervals in which the function } \\ f(x)=3x^4-4x^3-12x^2+5 \text{ is}
\displaystyle \text{(i) strictly increasing,} \ \            \displaystyle \text{(ii) strictly decreasing.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, function is } f(x)=3x^{4}-4x^{3}-12x^{2}+5
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle f'(x)=12x^{3}-12x^{2}-24x
\displaystyle \text{For strictly increasing or strictly decreasing, putting } f'(x)=0, \text{ we get}
\displaystyle 12x^{3}-12x^{2}-24x=0
\displaystyle \Rightarrow 12x(x^{2}-x-2)=0
\displaystyle \Rightarrow 12x(x^{2}-2x+x-2)=0
\displaystyle \Rightarrow 12x(x+1)(x-2)=0
\displaystyle \therefore x=0,-1,2
\displaystyle \text{Now, we find intervals in which } f(x) \text{ is strictly increasing or strictly decreasing.}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Interval}&f'(x)=12x(x+1)(x-2)&\text{Sign of } f'(x)\\ \hline x<-1&(-)(-)(-)&-\text{ve}\\ \hline -1<x<0&(-)(+)(-)&+\text{ve}\\ \hline 0<x<2&(+)(+)(-)&-\text{ve}\\ \hline x>2&(+)(+)(+)&+\text{ve}\\ \hline \end{array}
\displaystyle \text{We know that } f(x) \text{ is strictly increasing if } f'(x)>0 \text{ and strictly decreasing if } f'(x)<0.
\displaystyle \text{So, } f(x) \text{ is strictly increasing on } (-1,0) \text{ and } (2,\infty).
\displaystyle \text{Also, } f(x) \text{ is strictly decreasing on } (-\infty,-1) \text{ and } (0,2).
\\

\displaystyle \textbf{Question 6. }\text{Find the value(s) of }x \text{ for which } y=[x(x-2)]^2 \text{ is an increasing function.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, function is } y=[x(x-2)]^{2}=[x^{2}-2x]^{2}
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=2(x^{2}-2x)\frac{d}{dx}(x^{2}-2x)
\displaystyle =2(x^{2}-2x)(2x-2)=4x(x-2)(x-1)
\displaystyle \text{On putting } \frac{dy}{dx}=0, \text{ we get}
\displaystyle 4x(x-2)(x-1)=0 \Rightarrow x=0,1,2
\displaystyle \text{Now, we find interval in which } f(x) \text{ is strictly increasing or strictly decreasing.}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Interval}&\frac{dy}{dx}=4x(x-2)(x-1)&\text{Sign of } f'(x)\\ \hline (-\infty,0)&(-)(-)(-)&-\text{ve}\\ \hline (0,1)&(+)(-)(-)&+\text{ve}\\ \hline (1,2)&(+)(-)(+)&-\text{ve}\\ \hline (2,\infty)&(+)(+)(+)&+\text{ve}\\ \hline \end{array}
\displaystyle \text{Hence, } y \text{ is strictly increasing in } (0,1) \text{ and } (2,\infty).
\displaystyle \text{Also, } y \text{ is a polynomial function, so it is continuous at } x=0,1,2.
\displaystyle \text{Hence, } y \text{ is increasing in } [0,1]\cup[2,\infty).
\\

\displaystyle \textbf{Question 7. }\text{A ladder }5\ \mathrm{m} \text{ long is leaning against a wall. Bottom of ladder is} \\ \text{pulled along the ground away from wall} \text{at the rate of }2\ \mathrm{m/s}. \text{ How fast is the} \\ \text{height on the wall decreasing, when the foot of ladder is }4\ \mathrm{m} \text{ away from the wall?} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } AC \text{ be the ladder, } BC=x \text{ and height of the wall } AB=y.
\displaystyle \text{As the ladder is pulled along the ground away from the wall at the rate of } 2 \text{ m/s,}
\displaystyle \text{so, } \frac{dx}{dt}=2 \text{ m/s}
\displaystyle \text{To find } \frac{dy}{dt}, \text{ when } x=4.
\displaystyle \text{In right angled } \triangle ABC, \\ \text{by Pythagoras theorem, we get}
\displaystyle (AB)^{2}+(BC)^{2}=(AC)^{2}
\displaystyle \Rightarrow x^{2}+y^{2}=25 \qquad \cdots(i)
\displaystyle \Rightarrow (4)^{2}+y^{2}=25 \Rightarrow 16+y^{2}=25
\displaystyle \Rightarrow y^{2}=9 \Rightarrow y=\sqrt{9}=3
\displaystyle \text{On differentiating both sides of Eq. (i) w.r.t. } t, \text{ we get}
\displaystyle 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0
\displaystyle \Rightarrow x\frac{dx}{dt}+y\frac{dy}{dt}=0 \qquad \cdots(ii)
\displaystyle \text{On substituting the values of } x,\ y \text{ and } \frac{dx}{dt} \text{ in Eq. (ii), we get}
\displaystyle (4\times 2)+3\times \frac{dy}{dt}=0 \Rightarrow 8+3\frac{dy}{dt}=0
\displaystyle \therefore \frac{dy}{dt}=-\frac{8}{3} \text{ m/s}
\displaystyle \text{Hence, the height of the wall is decreasing at the rate of } \frac{8}{3} \text{ m/s.}
\\

\displaystyle \textbf{Question 8. }\text{Show that } y=\log(1+x)-\frac{2x}{2+x},\ x>-1 \text{ is an increasing function of }x, \\ \text{throughout its domain.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, function is } y=\log(1+x)-\frac{2x}{2+x}
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=\frac{1}{1+x}-\frac{(2+x)\cdot 2-2x\cdot 1}{(2+x)^{2}}
\displaystyle \text{[by using quotient rule of derivative]}
\displaystyle =\frac{1}{1+x}-\frac{4+2x-2x}{(2+x)^{2}}
\displaystyle =\frac{(2+x)^{2}-4(1+x)}{(1+x)(2+x)^{2}}
\displaystyle =\frac{4+x^{2}+4x-4-4x}{(1+x)(2+x)^{2}}
\displaystyle \therefore \frac{dy}{dx}=\frac{x^{2}}{(1+x)(2+x)^{2}} \qquad \cdots(i)
\displaystyle \text{Now, } x^{2},\ (2+x)^{2} \text{ are always positive and } (1+x)>0 \text{ for } x>-1
\displaystyle \text{From Eq. (i), } \frac{dy}{dx}>0 \text{ for } x>-1
\displaystyle \text{Hence, the function increases for } x>-1
\\

\displaystyle \textbf{Question 9. }\text{Find the intervals in which the function given by }  \\ f(x)=\sin x+\cos x,\ 0\leq x\leq 2\pi \text{ is} \ \
\displaystyle \text{(i) increasing,} \ \           \displaystyle \text{(ii) decreasing.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, function is } f(x)=\sin x+\cos x
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle f'(x)=\cos x-\sin x
\displaystyle \text{Now, put } f'(x)=0 \Rightarrow \cos x-\sin x=0
\displaystyle \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4},\frac{5\pi}{4}, \text{ as } 0\leq x\leq 2\pi
\displaystyle \text{Now, we find the intervals in which } f(x) \text{ is strictly increasing or strictly decreasing.}
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Interval}&\text{Test value}&f'(x)=\cos x-\sin x&\text{Sign of } f'(x)\\ \hline 0<x<\frac{\pi}{4}&x=\frac{\pi}{6}&\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}&+\text{ve}\\ \hline \frac{\pi}{4}<x<\frac{5\pi}{4}&x=\frac{\pi}{2}&0-1=-1&-\text{ve}\\ \hline \frac{5\pi}{4}<x<2\pi&x=\frac{3\pi}{2}&0-(-1)=1&+\text{ve}\\ \hline \end{array}
\displaystyle \text{Here, } f'(x)>0 \text{ in } \left(0,\frac{\pi}{4}\right) \text{ and } \left(\frac{5\pi}{4},2\pi\right), \text{ and } f'(x)<0 \text{ in } \left(\frac{\pi}{4},\frac{5\pi}{4}\right)
\displaystyle \text{Since } f(x) \text{ is continuous,}
\displaystyle \text{Hence, } f(x) \text{ is increasing in } \left[0,\frac{\pi}{4}\right] \cup \left[\frac{5\pi}{4},2\pi\right]
\displaystyle \text{and decreasing in } \left[\frac{\pi}{4},\frac{5\pi}{4}\right]
\\

\displaystyle \textbf{Question 10. }\text{Sand is pouring from the pipe at the rate of }12\ \mathrm{cm^3/s}. \text{ The falling} \\ \text{sand forms a cone on a ground }  \text{in such a way that the height of cone is always } \\ \text{one-sixth of radius of the base. How fast is the height of sand cone }  \text{increasing when} \\ \text{the height is }4\ \mathrm{cm}? \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } V \text{ be the volume of cone, } h \text{ be height and } r \text{ be radius of base.}
\displaystyle \text{Given, } \frac{dV}{dt}=12 \text{ cm}^{3}\text{/s} \qquad \cdots(i)
\displaystyle \text{Also, height of cone } =\frac{1}{6} \text{ (radius of base of cone)}
\displaystyle \therefore h=\frac{r}{6} \text{ or } r=6h \qquad \cdots(ii)
\displaystyle \text{We know that volume of cone is } V=\frac{1}{3}\pi r^{2}h \qquad \cdots(iii)
\displaystyle \text{On putting } r=6h \text{ in Eq. (iii), we get}
\displaystyle V=\frac{1}{3}\pi (6h)^{2}h=\frac{1}{3}\pi \cdot 36h^{3}
\displaystyle \Rightarrow V=12\pi h^{3}
\displaystyle \text{On differentiating both sides w.r.t. } t, \text{ we get}
\displaystyle \frac{dV}{dt}=36\pi h^{2}\frac{dh}{dt}
\displaystyle \text{On putting } \frac{dV}{dt}=12 \text{ and } h=4, \text{ we get}
\displaystyle 12=36\pi (16)\frac{dh}{dt}
\displaystyle \Rightarrow \frac{dh}{dt}=\frac{12}{36\pi \cdot 16}=\frac{1}{48\pi} \text{ cm/s}
\displaystyle \text{Hence, the height of sand cone is increasing at the rate of } \frac{1}{48\pi} \text{ cm/s.}
\\

\displaystyle \textbf{Question 11. }\text{Find the intervals in which the function } \\ f(x)=(x-1)^3(x-2)^2 \text{ is} \ \
\displaystyle \text{(i) increasing,} \ \           \displaystyle \text{(ii) decreasing.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } f(x)=(x-1)^{3}(x-2)^{2}
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle f'(x)=(x-1)^{3}\frac{d}{dx}(x-2)^{2}+(x-2)^{2}\frac{d}{dx}(x-1)^{3}
\displaystyle \text{[by product rule]}
\displaystyle \Rightarrow f'(x)=(x-1)^{3}\cdot 2(x-2)+(x-2)^{2}\cdot 3(x-1)^{2}
\displaystyle =(x-1)^{2}(x-2)\left[2(x-1)+3(x-2)\right]
\displaystyle =(x-1)^{2}(x-2)(2x-2+3x-6)
\displaystyle \Rightarrow f'(x)=(x-1)^{2}(x-2)(5x-8)
\displaystyle \text{Now, put } f'(x)=0 \Rightarrow (x-1)^{2}(x-2)(5x-8)=0
\displaystyle \Rightarrow x=1,\ \frac{8}{5},\ 2
\displaystyle \text{Now, we find intervals and check sign of } f'(x)
\displaystyle \begin{array}{|c|c|c|}\hline \text{Interval}&f'(x)=(x-1)^{2}(x-2)(5x-8)&\text{Sign of } f'(x)\\ \hline x<1&(+)(-)(-)&+\text{ve}\\ \hline 1<x<\frac{8}{5}&(+)(-)(-)&+\text{ve}\\ \hline \frac{8}{5}<x<2&(+)(-)(+)&-\text{ve}\\ \hline x>2&(+)(+)(+)&+\text{ve}\\ \hline \end{array}
\displaystyle \text{Hence, } f(x) \text{ is increasing on } (-\infty,\tfrac{8}{5}) \cup (2,\infty)
\displaystyle \text{and decreasing on } (\tfrac{8}{5},2)
\displaystyle \text{We know that } f(x) \text{ is strictly increasing if } f'(x)>0 \text{ and strictly decreasing if } f'(x)<0.
\displaystyle \text{So, the given function } f(x) \text{ is increasing on } (-\infty,1),\ (1,\tfrac{8}{5}) \text{ and } (2,\infty)
\displaystyle \text{and decreasing on } (\tfrac{8}{5},2).
\displaystyle \text{Since } f(x) \text{ is a polynomial function, it is continuous at } x=1,\tfrac{8}{5},2.
\displaystyle \text{Hence, } f(x) \text{ is increasing on } (-\infty,\tfrac{8}{5}] \cup [2,\infty)
\displaystyle \text{and decreasing on } [\tfrac{8}{5},2].
\\

\displaystyle \textbf{Question 12. }\text{Find the intervals in which the function }\\  f(x)=2x^3+9x^2+12x+20 \text{ is} \ \
\displaystyle \text{(i) increasing,} \ \           \displaystyle \text{(ii) decreasing.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, function is } f(x)=2x^{3}+9x^{2}+12x+20
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle f'(x)=6x^{2}+18x+12
\displaystyle \text{On putting } f'(x)=0, \text{ we get}
\displaystyle 6x^{2}+18x+12=0
\displaystyle \Rightarrow 6(x^{2}+3x+2)=0
\displaystyle \Rightarrow 6(x+1)(x+2)=0
\displaystyle \Rightarrow x=-2,-1
\displaystyle \text{Now, we find intervals and check in which interval } f(x) \text{ is strictly increasing and strictly decreasing.}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Interval}&f'(x)=6(x+1)(x+2)&\text{Sign of } f'(x)\\ \hline x<-2&(+)(-)(-)&+\text{ve}\\ \hline -2<x<-1&(+)(-)(+)&-\text{ve}\\ \hline x>-1&(+)(+)(+)&+\text{ve}\\ \hline \end{array}
\displaystyle \text{We know that } f(x) \text{ is increasing if } f'(x)>0 \text{ and decreasing if } f'(x)<0.
\displaystyle \text{So, } f(x) \text{ is increasing on } (-\infty,-2) \text{ and } (-1,\infty)
\displaystyle \text{and decreasing on } (-2,-1).
\displaystyle \text{Since } f(x) \text{ is a polynomial function, it is continuous at } x=-2,-1.
\displaystyle \text{Hence, } f(x) \text{ is increasing on } (-\infty,-2] \cup [-1,\infty)
\displaystyle \text{and decreasing on } [-2,-1].
\\

\displaystyle \textbf{Question 13. }\text{The length }x \text{ of a rectangle is decreasing at the rate of }5\ \mathrm{cm/min} \\ \text{and the width }y \text{ is increasing } \text{at the rate of }4\ \mathrm{cm/min}. \text{ When }x=8\ \mathrm{cm} \text{ and } \\ y=6 \mathrm{cm}, \text{ find the rate of change of} \ \
\displaystyle \text{(i) the perimeter,} \ \
\displaystyle \text{(ii) area of rectangle.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given that length } x \text{ of a rectangle is decreasing at the rate of } 5 \text{ cm/min.}
\displaystyle \therefore \frac{dx}{dt}=-5 \text{ cm/min} \qquad \cdots(i)
\displaystyle \text{Also, breadth } y \text{ of rectangle is increasing at the rate of } 4 \text{ cm/min.}
\displaystyle \therefore \frac{dy}{dt}=4 \text{ cm/min} \qquad \cdots(ii)
\displaystyle \text{(i) Here, we have to find rate of change of perimeter, i.e. } \frac{dP}{dt}
\displaystyle \text{We know that perimeter } P=2(x+y)
\displaystyle \text{On differentiating both sides w.r.t. } t, \text{ we get}
\displaystyle \frac{dP}{dt}=2\left(\frac{dx}{dt}+\frac{dy}{dt}\right)
\displaystyle \Rightarrow \frac{dP}{dt}=2(-5+4)=2(-1)=-2 \text{ cm/min}
\displaystyle \text{Hence, the perimeter of rectangle is decreasing at the rate of } 2 \text{ cm/min.}
\displaystyle \text{(ii) Here, we have to find rate of change of area } \frac{dA}{dt}
\displaystyle \text{We know that area of rectangle } A=xy
\displaystyle \text{On differentiating both sides w.r.t. } t, \text{ we get}
\displaystyle \frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}
\displaystyle \text{Now, } x=8 \text{ cm and } y=6 \text{ cm}
\displaystyle \therefore \frac{dA}{dt}=8(4)+6(-5)=32-30=2 \text{ cm}^{2}\text{/min}
\displaystyle \text{Hence, the area of rectangle is increasing at the rate of } 2 \text{ cm}^{2}\text{/min.}
\\

\displaystyle \textbf{Question 14. }\text{Prove that } y=\frac{4\sin\theta}{2+\cos\theta}-\theta \text{ is a strictly increasing function in } \left(0,\frac{\pi}{2}\right). \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given function is } y=\frac{4\sin\theta}{2+\cos\theta}-\theta \qquad \cdots(i)
\displaystyle \text{We know that a function is increasing if } \frac{dy}{d\theta}>0
\displaystyle \text{On differentiating both sides of Eq. (i) w.r.t. } \theta, \text{ we get}
\displaystyle \frac{dy}{d\theta}=\frac{(2+\cos\theta)\frac{d}{d\theta}(4\sin\theta)-4\sin\theta\frac{d}{d\theta}(2+\cos\theta)}{(2+\cos\theta)^{2}}-1
\displaystyle \text{[by using quotient rule]}
\displaystyle =\frac{(2+\cos\theta)(4\cos\theta)-4\sin\theta(0-\sin\theta)}{(2+\cos\theta)^{2}}-1
\displaystyle =\frac{8\cos\theta+4\cos^{2}\theta+4\sin^{2}\theta}{(2+\cos\theta)^{2}}-1
\displaystyle =\frac{8\cos\theta+4(\cos^{2}\theta+\sin^{2}\theta)}{(2+\cos\theta)^{2}}-1
\displaystyle =\frac{8\cos\theta+4}{(2+\cos\theta)^{2}}-1
\displaystyle =\frac{8\cos\theta+4-(2+\cos\theta)^{2}}{(2+\cos\theta)^{2}}
\displaystyle =\frac{8\cos\theta+4-(4+4\cos\theta+\cos^{2}\theta)}{(2+\cos\theta)^{2}}
\displaystyle =\frac{4\cos\theta-\cos^{2}\theta}{(2+\cos\theta)^{2}}
\displaystyle =\frac{\cos\theta(4-\cos\theta)}{(2+\cos\theta)^{2}}
\displaystyle \text{Now, for } \theta\in\left(0,\frac{\pi}{2}\right),\ \cos\theta>0 \text{ and } (2+\cos\theta)^{2}>0
\displaystyle \text{Also, } 0<\cos\theta<1 \Rightarrow 4-\cos\theta>0
\displaystyle \therefore \frac{\cos\theta(4-\cos\theta)}{(2+\cos\theta)^{2}}>0,\ \forall \theta\in\left(0,\frac{\pi}{2}\right)
\displaystyle \Rightarrow \frac{dy}{d\theta}>0,\ \forall \theta\in\left(0,\frac{\pi}{2}\right)
\displaystyle \text{Since } y \text{ is continuous, } y \text{ is increasing in } \left[0,\frac{\pi}{2}\right]
\\

\displaystyle \textbf{Question 15. }\text{Find the intervals in which the function } f(x)=\sin 3x-\cos 3x, \\ 0<x<\pi, \text{ is strictly increasing or }  \text{strictly decreasing.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } f(x)=\sin 3x-\cos 3x,\ 0<x<\pi
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle f'(x)=3\cos 3x+3\sin 3x=3(\cos 3x+\sin 3x)
\displaystyle \text{On putting } f'(x)=0, \text{ we get}
\displaystyle \cos 3x+\sin 3x=0 \Rightarrow \tan 3x=-1
\displaystyle \Rightarrow 3x=\frac{3\pi}{4},\frac{7\pi}{4},\frac{11\pi}{4}
\displaystyle \Rightarrow x=\frac{\pi}{4},\frac{7\pi}{12},\frac{11\pi}{12}
\displaystyle \text{Now, we find intervals and check sign of } f'(x)
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Interval}&\text{Test value}&f'(x)=3(\cos 3x+\sin 3x)&\text{Sign}\\ \hline 0<x<\frac{\pi}{4}&x=\frac{\pi}{6}&3(0+1)=3&+\text{ve}\\ \hline \frac{\pi}{4}<x<\frac{7\pi}{12}&x=\frac{\pi}{3}&3(-1+0)=-3&-\text{ve}\\ \hline \frac{7\pi}{12}<x<\frac{11\pi}{12}&x=\frac{3\pi}{4}&3\left(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\right)=0 & \approx \text{+ve (take nearby)}\\ \hline \frac{11\pi}{12}<x<\pi&x=\frac{23\pi}{24}&3(\cos\frac{23\pi}{8}+\sin\frac{23\pi}{8})<0&-\text{ve}\\ \hline \end{array}
\displaystyle \text{Correcting sign using nearby points:}
\displaystyle f'(x)>0 \text{ in } \left(0,\frac{\pi}{4}\right) \text{ and } \left(\frac{7\pi}{12},\frac{11\pi}{12}\right)
\displaystyle f'(x)<0 \text{ in } \left(\frac{\pi}{4},\frac{7\pi}{12}\right) \text{ and } \left(\frac{11\pi}{12},\pi\right)
\displaystyle \text{Hence, } f(x) \text{ is increasing in } \left(0,\frac{\pi}{4}\right)\cup\left(\frac{7\pi}{12},\frac{11\pi}{12}\right)
\displaystyle \text{and decreasing in } \left(\frac{\pi}{4},\frac{7\pi}{12}\right)\cup\left(\frac{11\pi}{12},\pi\right)
\\

\displaystyle \textbf{Question 16. }\text{Prove that the function } f \text{ defined by } f(x)=x^2-x+1 \text{ is neither} \\ \text{increasing nor decreasing in }(-1,1).   \text{Hence, find the intervals in which } f(x) \text{ is} \ \
\displaystyle \text{(i) strictly increasing,} \ \
\displaystyle \text{(ii) strictly decreasing.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, function is } f(x)=x^{2}-x+1
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle f'(x)=2x-1
\displaystyle \text{On putting } f'(x)=0
\displaystyle \Rightarrow 2x-1=0 \Rightarrow x=\frac{1}{2}
\displaystyle \text{Now, we find intervals in which } f(x) \text{ is strictly increasing or strictly decreasing.}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Interval}&f'(x)=2x-1&\text{Sign of } f'(x)\\ \hline x<\frac{1}{2}&(-)&-\text{ve}\\ \hline x>\frac{1}{2}&(+)&+\text{ve}\\ \hline \end{array}
\displaystyle \text{Here, } f(x) \text{ is strictly increasing on } \left(\frac{1}{2},\infty\right)
\displaystyle \text{and } f(x) \text{ is strictly decreasing on } \left(-\infty,\frac{1}{2}\right).
\displaystyle \text{Hence, } f(x) \text{ is neither increasing nor decreasing in } (-1,1).

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