\displaystyle \textbf{Question 1: }\text{Find the intervals in which the function} \\ f(x)=2x^3-9x^2+12x+15 \text{ is (i) increasing, (ii) decreasing.} \hspace{2.0cm} [\text{CBSE 2010, 2011}]

\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle f(x)=2x^3-9x^2+12x+15
\displaystyle \Rightarrow f'(x)=6x^2-18x+12=6(x^2-3x+2)
\displaystyle \text{(i) For } f(x) \text{ to be increasing, we must have } f'(x)>0
\displaystyle \Rightarrow 6(x^2-3x+2)>0
\displaystyle \Rightarrow x^2-3x+2>0\ \ [\because\ 6>0\ \therefore\ 6(x^2-3x+2)>0\Rightarrow x^2-3x+2>0]
\displaystyle \Rightarrow (x-1)(x-2)>0
\displaystyle \Rightarrow x<1 \text{ or } x>2
\displaystyle \Rightarrow x\in(-\infty,1)\cup(2,\infty)
\displaystyle \text{So, } f(x) \text{ is increasing on } (-\infty,1)\cup(2,\infty).
\displaystyle \text{(ii) For } f(x) \text{ to be decreasing, we must have } f'(x)<0
\displaystyle \Rightarrow 6(x^2-3x+2)<0
\displaystyle \Rightarrow x^2-3x+2<0\ \ [\because\ 6>0\ \therefore\ 6(x^2-3x+2)<0\Rightarrow x^2-3x+2<0]
\displaystyle \Rightarrow (x-1)(x-2)<0
\displaystyle \Rightarrow 1<x<2\Rightarrow x\in(1,2)
\displaystyle \text{So, } f(x) \text{ is decreasing on } (1,2).

\displaystyle \textbf{Question 2: }\text{Find the intervals in which } f(x)=(x+1)^3(x-3)^3 \text{ is increasing} \\ \text{or decreasing.} \hspace{10.0cm} [\text{CBSE 2001, 2011}]

\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle f(x)=(x+1)^3(x-3)^3
\displaystyle \Rightarrow f'(x)=\left\{3(x+1)^2\frac{d}{dx}(x+1)\right\}(x-3)^3+(x+1)^3\left\{3(x-3)^2\frac{d}{dx}(x-3)\right\}
\displaystyle \Rightarrow f'(x)=3(x+1)^2(x-3)^3+3(x+1)^3(x-3)^2
\displaystyle \Rightarrow f'(x)=3(x+1)^2(x-3)^2(x+1+x-3)
\displaystyle \Rightarrow f'(x)=6(x+1)^2(x-3)^2(x-1)
\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x)>0
\displaystyle \Rightarrow 6(x+1)^2(x-3)^2(x-1)>0
\displaystyle \Rightarrow x-1>0 \text{ and } x\ne -1,3\ \ [\because\ 6(x+1)^2(x-3)^2>0 \text{ for all } x\ne -1,3]
\displaystyle \Rightarrow x>1 \text{ and } x\ne -1,3\Rightarrow x\in(1,3)\cup(3,\infty)
\displaystyle \text{So, } f(x) \text{ is increasing on } (1,3)\cup(3,\infty).
\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x)<0
\displaystyle \Rightarrow 6(x+1)^2(x-3)^2(x-1)<0
\displaystyle \Rightarrow x-1<0 \text{ and } x\ne -1,3\ \ [\because\ 6(x+1)^2(x-3)^2>0 \text{ for all } x\ne -1,3]
\displaystyle \Rightarrow x<1 \text{ and } x\ne -1,3\Rightarrow x\in(-\infty,-1)\cup(-1,1)
\displaystyle \text{So, } f(x) \text{ is decreasing on } (-\infty,-1)\cup(-1,1).

\displaystyle \textbf{Question 3: }\text{Find the intervals in which the function} \\ f(x)=\log(1+x)-\frac{2x}{2+x} \text{ is increasing or decreasing.} \hspace{4.0cm} [\text{CBSE 2012}]

\displaystyle \text{Answer:}
\displaystyle \text{We have, } f(x)=\log(1+x)-\frac{2x}{2+x}.\ \text{Clearly, } f(x) \text{ is defined for all } x \text{ satisfying } x+1>0 \text{ i.e. } x>-1.\ \text{So, domain}(f)=(-1,\infty).
\displaystyle \text{Now,}
\displaystyle f(x)=\log(1+x)-\frac{2x}{2+x}
\displaystyle \Rightarrow f'(x)=\frac{1}{1+x}\frac{d}{dx}(x+1)-\frac{(2+x)\times 2-2x(0+1)}{(2+x)^2}=\frac{1}{1+x}-\frac{4}{(2+x)^2}
\displaystyle \Rightarrow f'(x)=\frac{(2+x)^2-4(1+x)}{(2+x)^2(1+x)}=\frac{x^2}{(2+x)^2(1+x)}=\left(\frac{x}{2+x}\right)^2\frac{1}{x+1}
\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x)>0
\displaystyle \Rightarrow \left(\frac{x}{2+x}\right)^2\frac{1}{x+1}>0
\displaystyle \Rightarrow \frac{1}{x+1}>0 \text{ and } x\ne 0\ \ [\because\ \left(\frac{x}{2+x}\right)^2>0 \text{ for all } x\ne 0]
\displaystyle \Rightarrow x+1>0 \text{ and } x\ne 0
\displaystyle \Rightarrow x>-1 \text{ and } x\ne 0
\displaystyle \Rightarrow x\in(-1,0)\cup(0,\infty)
\displaystyle \text{So, } f(x) \text{ is increasing on } (-1,0)\cup(0,\infty).

\displaystyle \textbf{Question 4: }\text{Find the intervals in which } f(x)=\frac{4x^2+1}{x} \text{ is increasing} \\ \text{or decreasing.} \hspace{10.0cm} [\text{CBSE 2004}]

\displaystyle \text{Answer:}
\displaystyle \text{We have, } f(x)=\frac{4x^2+1}{x}
\displaystyle \text{Now, } f(x)=4x+\frac{1}{x}\Rightarrow f'(x)=4-\frac{1}{x^2}=\frac{4x^2-1}{x^2}
\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x)>0
\displaystyle \Rightarrow \frac{4x^2-1}{x^2}>0
\displaystyle \Rightarrow 4x^2-1>0\ \ [\because\ x^2>0]
\displaystyle \Rightarrow (2x-1)(2x+1)>0
\displaystyle \Rightarrow \left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)>0
\displaystyle \Rightarrow x<-\frac{1}{2} \text{ or } x>\frac{1}{2}
\displaystyle \Rightarrow x\in(-\infty,-\frac{1}{2})\cup(\frac{1}{2},\infty)
\displaystyle \text{So, } f(x) \text{ is increasing on } (-\infty,-\frac{1}{2})\cup(\frac{1}{2},\infty).
\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x)<0
\displaystyle \Rightarrow \frac{4x^2-1}{x^2}<0
\displaystyle \Rightarrow 4x^2-1<0\ \ [\because\ x^2>0]
\displaystyle \Rightarrow (2x-1)(2x+1)<0
\displaystyle \Rightarrow -\frac{1}{2}<x<\frac{1}{2}
\displaystyle \Rightarrow x\in(-\frac{1}{2},\frac{1}{2})
\displaystyle \text{But, domain}(f)=R-\{0\}.\ \text{So, } f(x) \text{ is decreasing on } (-\frac{1}{2},0)\cup(0,\frac{1}{2}).

\displaystyle \textbf{Question 5: }\text{Find the intervals in which the function} \\ f(x)=x^3+\frac{1}{x^3},\ x\ne 0 \text{ is (i) increasing (ii) decreasing.} \hspace{4.0cm} [\text{CBSE 2009}]

\displaystyle \text{Answer:}
\displaystyle  \text{Clearly, domain}(f)=R-\{0\}.
\displaystyle \text{Now, } f(x)=x^3+\frac{1}{x^3}
\displaystyle \Rightarrow f'(x)=3x^2-\frac{3}{x^4}=\frac{3}{x^4}(x^6-1)=\frac{3}{x^4}(x^2-1)(x^4+x^2+1)=3\left(\frac{x^4+x^2+1}{x^4}\right)(x^2-1)
\displaystyle \text{(i) For } f(x) \text{ to be increasing, we must have } f'(x)>0
\displaystyle \Rightarrow 3\left(\frac{x^4+x^2+1}{x^4}\right)(x^2-1)>0
\displaystyle \Rightarrow (x^2-1)>0\ \ [\because\ 3\left(\frac{x^4+x^2+1}{x^4}\right)>0,\ x\ne 0]
\displaystyle \Rightarrow (x-1)(x+1)>0
\displaystyle \Rightarrow x\in(-\infty,-1)\cup(1,\infty)
\displaystyle \text{So, } f(x) \text{ is increasing on } (-\infty,-1)\cup(1,\infty).
\displaystyle \text{(ii) For } f(x) \text{ to be decreasing, we must have } f'(x)<0
\displaystyle \Rightarrow 3\left(\frac{x^4+x^2+1}{x^4}\right)(x^2-1)<0
\displaystyle \Rightarrow (x^2-1)<0\ \ [\because\ 3\left(\frac{x^4+x^2+1}{x^4}\right)>0,\ x\ne 0]
\displaystyle \Rightarrow (x-1)(x+1)<0
\displaystyle \Rightarrow x\in(-1,0)\cup(0,1)
\displaystyle \text{Hence, } f(x) \text{ is decreasing on } (-1,0)\cup(0,1).

\displaystyle \textbf{Question 6: }\text{Separate the interval } [0,\frac{\pi}{2}] \text{ into sub-intervals in which} \\ f(x)=\sin^4 x+\cos^4 x \text{ is increasing or decreasing.} \hspace{5.0cm} [\text{CBSE 2000}]

\displaystyle \text{Answer:}
\displaystyle  \text{We have,}
\displaystyle f(x)=\sin^4 x+\cos^4 x
\displaystyle \Rightarrow f'(x)=4\sin^3 x\cos x-4\cos^3 x\sin x
\displaystyle \Rightarrow f'(x)=-4\sin x\cos x(\cos^2 x-\sin^2 x)
\displaystyle \Rightarrow f'(x)=-2(2\sin x\cos x)(\cos 2x)
\displaystyle \Rightarrow f'(x)=-2\sin 2x\cos 2x
\displaystyle \Rightarrow f'(x)=-\sin 4x
\displaystyle \text{We have, } 0<x<\frac{\pi}{2}\Rightarrow 0<4x<2\pi
\displaystyle \text{Since sine function is positive in the first and second quadrants and negative in the} \\ \text{third and fourth quadrants. So, we consider the following:}
\displaystyle \text{Case I}\ \ \text{When } 0<4x<\pi \text{ i.e. } 0<x<\frac{\pi}{4}
\displaystyle \text{In this case, we have } \sin 4x>0
\displaystyle \Rightarrow -\sin 4x<0\Rightarrow f'(x)<0
\displaystyle \therefore\ f(x)<0 \text{ for } 0<4x<\pi \text{ i.e. } 0<x<\frac{\pi}{4}
\displaystyle \text{So, } f(x) \text{ is decreasing on } [0,\frac{\pi}{4}].
\displaystyle \text{Case II}\ \ \text{When } \pi<4x<2\pi \text{ i.e. } \frac{\pi}{4}<x<\frac{\pi}{2}
\displaystyle \text{In this case, we have } \sin 4x<0
\displaystyle \Rightarrow -\sin 4x>0\Rightarrow f'(x)>0
\displaystyle \therefore\ f(x)>0 \text{ for } \pi<4x<2\pi \text{ i.e. } \frac{\pi}{4}<x<\frac{\pi}{2}
\displaystyle \text{So, } f(x) \text{ is increasing on } [\frac{\pi}{4},\frac{\pi}{2}].

\displaystyle \textbf{Question 7: }\text{Find the intervals in which the function } f \text{ given by} \\ f(x)=\sin x+\cos x,\ 0\le x\le 2\pi \text{ is increasing or decreasing.} \hspace{1.0cm} [\text{CBSE 2009, 2017}]

\displaystyle \text{Answer:}
\displaystyle  \text{We have,}
\displaystyle f(x)=\sin x+\cos x
\displaystyle \Rightarrow f'(x)=\cos x-\sin x=\sqrt{2}\left(\cos x\sin\frac{\pi}{4}-\sin x\cos\frac{\pi}{4}\right)=\sqrt{2}\sin\left(\frac{\pi}{4}-x\right)=-\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)
\displaystyle \text{Now, } 0\le x\le 2\pi \Rightarrow 0-\frac{\pi}{4}<x-\frac{\pi}{4}<2\pi-\frac{\pi}{4}\Rightarrow -\frac{\pi}{4}<x-\frac{\pi}{4}<\frac{7\pi}{4}
\displaystyle \text{For } f(x) \text{ to be increasing, we must have } f'(x)>0
\displaystyle \Rightarrow -\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)>0
\displaystyle \Rightarrow \sin\left(x-\frac{\pi}{4}\right)<0
\displaystyle \Rightarrow -\frac{\pi}{4}<x-\frac{\pi}{4}<0 \text{ or, } \pi<x-\frac{\pi}{4}<\frac{7\pi}{4}
\displaystyle \Rightarrow 0<x<\frac{\pi}{4} \text{ or, } \frac{5\pi}{4}<x<2\pi
\displaystyle \Rightarrow x\in\left(0,\frac{\pi}{4}\right) \text{ or, } x\in\left(\frac{5\pi}{4},2\pi\right)
\displaystyle \Rightarrow x\in(0,\frac{\pi}{4})\cup(\frac{5\pi}{4},2\pi)
\displaystyle \text{Hence, } f(x) \text{ is increasing on } (0,\frac{\pi}{4})\cup(\frac{5\pi}{4},2\pi).
\displaystyle \text{For } f(x) \text{ to be decreasing, we must have } f'(x)<0
\displaystyle \Rightarrow -\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)<0
\displaystyle \Rightarrow \sin\left(x-\frac{\pi}{4}\right)>0
\displaystyle \Rightarrow 0<x-\frac{\pi}{4}<\pi \Rightarrow \frac{\pi}{4}<x<\frac{5\pi}{4}\Rightarrow x\in\left(\frac{\pi}{4},\frac{5\pi}{4}\right)
\displaystyle \text{Hence, } f(x) \text{ is decreasing on } (\frac{\pi}{4},\frac{5\pi}{4}).

\displaystyle \textbf{Question 8: }\text{Find the intervals in which } f(x)=\sin 3x-\cos 3x,\ 0<x<\pi,\ \text{is strictly increasing or decreasing.} \hspace{6.0cm} [\text{CBSE 2016}]

\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle f(x)=\sin 3x-\cos 3x
\displaystyle \Rightarrow f'(x)=3(\cos 3x+\sin 3x)
\displaystyle \Rightarrow f'(x)=3\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos 3x+\frac{1}{\sqrt{2}}\sin 3x\right)=3\sqrt{2}\left(\sin\frac{\pi}{4}\cos 3x+\cos\frac{\pi}{4}\sin 3x\right)
\displaystyle \Rightarrow f'(x)=3\sqrt{2}\sin\left(3x+\frac{\pi}{4}\right)
\displaystyle \text{It is given that } 0<x<\pi\Rightarrow 0<3x<3\pi\Rightarrow \frac{\pi}{4}<3x+\frac{\pi}{4}<3\pi+\frac{\pi}{4}\Rightarrow \frac{\pi}{4}<3x+\frac{\pi}{4}<\frac{13\pi}{4}
\displaystyle \text{(i) For } f(x) \text{ to be strictly increasing, we must have } f'(x)>0
\displaystyle \Rightarrow 3\sqrt{2}\sin\left(3x+\frac{\pi}{4}\right)>0
\displaystyle \Rightarrow \sin\left(3x+\frac{\pi}{4}\right)>0
\displaystyle \Rightarrow \frac{\pi}{4}<3x+\frac{\pi}{4}<\pi \text{ or, } 2\pi<3x+\frac{\pi}{4}<3\pi
\displaystyle \Rightarrow 0<3x<\frac{3\pi}{4} \text{ or, } \frac{7\pi}{4}<3x<\frac{11\pi}{4}
\displaystyle \Rightarrow 0<x<\frac{\pi}{4} \text{ or, } \frac{7\pi}{12}<x<\frac{11\pi}{12}
\displaystyle \Rightarrow x\in(0,\frac{\pi}{4})\cup(\frac{7\pi}{12},\frac{11\pi}{12})
\displaystyle \text{So, } f(x) \text{ is strictly increasing on } (0,\frac{\pi}{4})\cup(\frac{7\pi}{12},\frac{11\pi}{12}).
\displaystyle \text{(ii) For } f(x) \text{ to be strictly decreasing, we must have } f'(x)<0
\displaystyle \Rightarrow 3\sqrt{2}\sin\left(3x+\frac{\pi}{4}\right)<0
\displaystyle \Rightarrow \sin\left(3x+\frac{\pi}{4}\right)<0
\displaystyle \Rightarrow \pi<3x+\frac{\pi}{4}<2\pi \text{ or, } 3\pi<3x+\frac{\pi}{4}<\frac{13\pi}{4}
\displaystyle \Rightarrow \frac{3\pi}{4}<3x<\frac{7\pi}{4} \text{ or, } \frac{11\pi}{4}<3x<3\pi
\displaystyle \Rightarrow \frac{\pi}{4}<x<\frac{7\pi}{12} \text{ or, } \frac{11\pi}{12}<x<\pi
\displaystyle \Rightarrow x\in(\frac{\pi}{4},\frac{7\pi}{12})\cup(\frac{11\pi}{12},\pi)
\displaystyle \text{So, } f(x) \text{ is strictly decreasing on } (\frac{\pi}{4},\frac{7\pi}{12})\cup(\frac{11\pi}{12},\pi).

\displaystyle \textbf{Question 9: }\text{Prove that } f(\theta)=\frac{4\sin\theta}{2+\cos\theta}-\theta \text{ is an increasing function of } \theta \text{ in } \\ \left[0,\frac{\pi}{2}\right]. \hspace{12.0cm} [\text{CBSE 2011}]

\displaystyle \text{Answer:}
\displaystyle  \text{We have,}
\displaystyle f(\theta)=\frac{4\sin\theta}{2+\cos\theta}-\theta
\displaystyle \Rightarrow f'(\theta)=\frac{(2+\cos\theta)(4\cos\theta)+4\sin^2\theta}{(2+\cos\theta)^2}-1
\displaystyle \Rightarrow f'(\theta)=\frac{8\cos\theta+4}{(2+\cos\theta)^2}-1
\displaystyle \Rightarrow f'(\theta)=\frac{4\cos\theta-\cos^2\theta}{(2+\cos\theta)^2}
\displaystyle \Rightarrow f'(\theta)=\frac{\cos\theta(4-\cos\theta)}{(2+\cos\theta)^2}>0 \text{ for all } \theta\in\left(0,\frac{\pi}{2}\right)\ \ [\because\ \cos\theta>0,\ 4-\cos\theta>0 \text{ and } 2+\cos\theta>0]
\displaystyle \text{Hence, } f(\theta) \text{ is increasing on } \left[0,\frac{\pi}{2}\right].

\displaystyle \textbf{Question 10: }\text{Prove that the function } f(x)=x^2-x+1 \text{ is neither increasing} \\ \text{nor decreasing on } (-1,1). \hspace{9.0cm} [\text{CBSE 2014}]

\displaystyle \text{Answer:}
\displaystyle \text{We have, } f(x)=x^2-x+1
\displaystyle \therefore\ f'(x)=2x-1=2\left(x-\frac{1}{2}\right)
\displaystyle \text{Now, } -1<x<\frac{1}{2}\Rightarrow \left(x-\frac{1}{2}\right)<0\Rightarrow 2\left(x-\frac{1}{2}\right)<0\Rightarrow f'(x)<0
\displaystyle \text{and, } \frac{1}{2}<x<1\Rightarrow x-\frac{1}{2}>0\Rightarrow 2\left(x-\frac{1}{2}\right)>0\Rightarrow f'(x)>0
\displaystyle \text{Thus, } f'(x) \text{ does not have the same sign throughout the interval } (-1,1).
\displaystyle \text{Hence, } f(x) \text{ is neither increasing or decreasing on } (-1,1).

\displaystyle \textbf{Question 11. }\text{Find the point on the curve }y^2=8x\text{ for which the abscissa and} \\ \text{ordinate change at the same rate.} \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required point be }(x,y)
\displaystyle \text{Given, }y^2=8x\qquad\ldots\text{(i)}
\displaystyle \text{and }\frac{dy}{dt}=\frac{dx}{dt}\qquad\ldots\text{(ii)}
\displaystyle \text{From Eq. (i), we have}
\displaystyle 2y\frac{dy}{dt}=8\frac{dx}{dt}
\displaystyle \Rightarrow 2y\left(\frac{dx}{dt}\right)=8\frac{dx}{dt}\qquad\text{[using Eq. (ii)]}
\displaystyle \Rightarrow 2y=8\Rightarrow y=4
\displaystyle \therefore \text{From Eq. (i), }(4)^2=8x
\displaystyle \Rightarrow 8x=16\Rightarrow x=2
\displaystyle \therefore \text{Required point is }(2,4)
\\

\displaystyle \textbf{Question 12. }\text{The median of an equilateral triangle is increasing at } \text{the rate of }
\displaystyle 2\sqrt3\ \text{cm/s. Find the rate at which its side is }   \text{increasing.} \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let the side of an equilateral triangle }ABC\text{ be }x\text{ cm,}
\displaystyle \text{then the median of }\triangle ABC,
\displaystyle \therefore AD=\sqrt{AB^2-BD^2}
\displaystyle =\sqrt{x^2-\left(\frac{x}{2}\right)^2}
\displaystyle =\sqrt{x^2-\frac{x^2}{4}}=\sqrt{\frac{3}{4}x^2}=\frac{\sqrt3}{2}x
\displaystyle \Rightarrow \frac{d}{dt}(AD)=\frac{d}{dt}\left(\frac{\sqrt3}{2}x\right)=\frac{\sqrt3}{2}\frac{dx}{dt}
\displaystyle \text{Also, }\frac{d}{dt}(AD)=2\sqrt3\ \text{(given)}
\displaystyle \Rightarrow \frac{\sqrt3}{2}\frac{dx}{dt}=2\sqrt3
\displaystyle \Rightarrow \frac{dx}{dt}=4\text{ cm/s}
\\

\displaystyle \textbf{Question 13. }\text{Find the points on the curve }6y=x^3+2\text{ at which ordinate} \\ \text{is changing 8 times as fast as abscissa.} \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }6y=x^3+2
\displaystyle \Rightarrow 6\frac{dy}{dt}=3x^2\frac{dx}{dt}
\displaystyle \text{Also, given }\frac{dy}{dt}=8\frac{dx}{dt}
\displaystyle \therefore 6\left(8\frac{dx}{dt}\right)=3x^2\frac{dx}{dt}
\displaystyle \Rightarrow 48=3x^2\Rightarrow x^2=16\Rightarrow x=\pm4
\displaystyle \text{When }x=4\text{, then }6y=(4)^3+2
\displaystyle \Rightarrow 6y=66\Rightarrow y=11
\displaystyle \text{and when }x=-4\text{, then }6y=(-4)^3+2
\displaystyle \Rightarrow 6y=-62\Rightarrow y=-\frac{31}{3}
\displaystyle \therefore \text{Required points are }(4,11)\text{ and }\left(-4,-\frac{31}{3}\right)
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\displaystyle \textbf{Question 14. }\text{A particle moves along the curve }3y=ax^3+1\text{ such that at a point with } \\ x\text{-coordinate 1, }y\text{-coordinate is changing twice as fast at }x\text{-coordinate. Find the value of }a. \\ \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }3y=ax^3+1\qquad\ldots\text{(i)}
\displaystyle \text{and }\frac{dy}{dt}=2\frac{dx}{dt}\qquad\ldots\text{(ii)}
\displaystyle \text{Now, from Eq. (i),}
\displaystyle 3\frac{dy}{dt}=a(3x^2)\frac{dx}{dt}
\displaystyle \Rightarrow 3\left(2\frac{dx}{dt}\right)=a(3x^2)\frac{dx}{dt}\qquad\text{[using Eq. (ii)]}
\displaystyle \Rightarrow 2=ax^2
\displaystyle \text{When }x=1\text{, then}
\displaystyle a(1)^2=2\Rightarrow a=2
\\

\displaystyle \textbf{Question 15. }\text{The interval in which the function } \\ f(x)=2x^3+9x^2+12x-1\text{ is decreasing is} \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }(-1,\infty) \qquad \text{(b) }(-2,-1) \qquad \text{(c) }(-\infty,-2) \qquad \text{(d) }(-1,1)
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, }f(x)=2x^3+9x^2+12x-1
\displaystyle \Rightarrow f'(x)=6x^2+18x+12
\displaystyle \text{For }f(x)\text{ to be decreasing, }f'(x)\leq0
\displaystyle \therefore (6x^2+18x+12)\leq0
\displaystyle \Rightarrow 6(x^2+3x+2)\leq0
\displaystyle \Rightarrow (x+2)(x+1)\leq0\Rightarrow -2\leq x\leq-1
\displaystyle \therefore \text{Correct option is (b)}
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\displaystyle \textbf{Question 16. }\text{If }f(x)=a(x-\cos x)\text{ is strictly decreasing in }R\text{, then }a\text{ belongs to} \hspace{0.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }\{0\} \qquad \text{(b) }(0,\infty) \qquad \text{(c) }(-\infty,0) \qquad \text{(d) }(-\infty,\infty)
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, }f(x)=a(x-\cos x)
\displaystyle \Rightarrow f'(x)=a(1+\sin x)
\displaystyle \text{For }f(x)\text{ to be strictly decreasing,}
\displaystyle \text{we have, }f'(x)<0
\displaystyle \Rightarrow a(1+\sin x)<0
\displaystyle \Rightarrow a<0 \qquad[\because 0\leq1+\sin x\leq2]
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 17. }\text{The function }f(x)=x^3+3x\text{ is increasing in interval.} \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }(-\infty,0) \qquad \text{(b) }(0,\infty) \qquad \text{(c) }R \qquad \text{(d) }(0,1)
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, function }f(x)=x^3+3x
\displaystyle \Rightarrow f'(x)=3x^2+3=3(x^2+1)
\displaystyle \text{For }x\in R,\ 3(x^2+1)>0,\ x^2+1>0
\displaystyle \Rightarrow f'(x)>0\ \forall\ x\in R
\displaystyle \Rightarrow f(x)\text{ is increasing on }R
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 18. }\text{The real function }f(x)=2x^3-3x^2-36x+7\text{ is} \hspace{1.2cm} \text{[CBSE 2022 (Term I)]}
\displaystyle \text{(a) strictly increasing in }(-\infty,-2)\text{ and strictly decreasing in }(-2,\infty)
\displaystyle \text{(b) strictly decreasing in }(-2,3)
\displaystyle \text{(c) strictly decreasing in }(-\infty,3)\text{ and strictly increasing in }(3,\infty)
\displaystyle \text{(d) strictly decreasing in }(-\infty,-2)\cup(3,\infty)
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, }f(x)=2x^3-3x^2-36x+7
\displaystyle f'(x)=6x^2-6x-36
\displaystyle f'(x)=6(x-3)(x+2)
\displaystyle \text{On putting }f'(x)=0\text{, so }x=-2\text{ and }x=3
\displaystyle \text{Plotting points on number line}
\displaystyle \text{Here, }f\text{ is strictly increasing in }(-\infty,-2)\text{ and }(3,\infty)
\displaystyle f\text{ is strictly decreasing in }(-2,3)
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 19. }\text{The value of }b\text{ for which the function }f(x)=x+\cos x+b \\ \text{ is strictly decreasing over }R\text{ is} \hspace{1.2cm} \text{[CBSE 2022 (Term I)]}
\displaystyle \text{(a) }b<1 \qquad \text{(b) no value of }b\text{ exists} \qquad \text{(c) }b\leq1 \qquad \text{(d) }b\geq1
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, }f(x)=x+\cos x+b
\displaystyle \text{Now, }f'(x)=1-\sin x
\displaystyle \text{Since, }f(x)\text{ is strictly decreasing over }R
\displaystyle \therefore f'(x)<0
\displaystyle \Rightarrow 1-\sin x<0\Rightarrow \sin x>1
\displaystyle \text{Since, }-1\leq\sin x\leq1
\displaystyle \text{Thus, }\sin x>1\text{ is not possible.}
\displaystyle \text{Thus, for no value of }b,\ f(x)\text{ is strictly decreasing.}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 20. }\text{The function }y=x^2e^{-x}\text{ is decreasing in the interval} \hspace{0.2cm} \text{[CBSE 2022 (Term I)]}
\displaystyle \text{(a) }(0,2) \qquad \text{(b) }(2,\infty) \qquad \text{(c) }(-\infty,0) \qquad \text{(d) }(-\infty,0)\cup(2,\infty)
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, }y=x^2e^{-x}
\displaystyle \frac{dy}{dx}=2xe^{-x}-x^2e^{-x}
\displaystyle \frac{dy}{dx}=e^{-x}(2x-x^2)
\displaystyle \text{On putting }\frac{dy}{dx}=0\text{, so }x=0,\ 2
\displaystyle \text{Plotting points on number line}
\displaystyle \text{Here, }y\text{ is decreasing in the interval }(-\infty,0)\cup(2,\infty)
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 21. }\text{Show that the function }f(x)=\frac{x}{3}+\frac{3}{x}\text{ decreases in the } \text{intervals }
\displaystyle (-3,0)\cup(0,3).\hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }f(x)=\frac{x}{3}+\frac{3}{x}\Rightarrow f'(x)=\frac13-\frac{3}{x^2}
\displaystyle \Rightarrow f'(x)=\frac{x^2-9}{3x^2}
\displaystyle \text{When }x\in(-3,0)\cup(0,3)
\displaystyle f'(x)<0
\displaystyle \therefore f(x)\text{ is decreasing function in }(-3,0)\cup(0,3)
\\

\displaystyle \textbf{Question 22. }\text{Show that the function }f\text{ defined by } f(x)=(x-1)e^x+1
\displaystyle \text{ is an increasing function for all}   x>0.\hspace{2.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle  \text{We have, }f(x)=(x-1)e^x+1
\displaystyle \text{On differentiating w.r.t. }x,\text{ we get}
\displaystyle f'(x)=(x-1)e^x+e^x
\displaystyle \Rightarrow f'(x)=xe^x
\displaystyle \text{For all }x>0\Rightarrow f'(x)>0
\displaystyle \therefore f(x)\text{ is an increasing function for all }x>0.
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 23. }\text{The total cost }C(x)\text{ associated with the production of }x
\displaystyle \text{units of an item is given by}
\displaystyle C(x)=0.005x^3-0.02x^2+30x+5000.
\displaystyle \text{Find the marginal cost when }3\text{ units are produced, where by marginal cost we mean}
\displaystyle \text{the instantaneous } \text{rate of change of total cost at any level of output.} \hspace{0.2cm} \text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle  \text{We have, }C(x)=0.005x^3-0.02x^2+30x+5000
\displaystyle \text{Clearly, the marginal cost, }MC(x)=\frac{d}{dx}\,C(x)
\displaystyle =\frac{d}{dx}(0.005x^3-0.02x^2+30x+5000)
\displaystyle =0.005\times3x^2-0.02\times2x+30+0
\displaystyle =0.015x^2-0.04x+30
\displaystyle \text{Now, marginal cost when }3\text{ units are produced}
\displaystyle =MC(3)=0.015(9)-0.04(3)+30
\displaystyle =0.135-0.12+30=30.015
\\

\displaystyle \textbf{Question 24. }\text{The total revenue received from the sale of }x\text{ units of a}
\displaystyle \text{product is given by }R(x)=3x^2+36x+5\text{ in rupees.}
\displaystyle \text{Find the marginal revenue when }x=5,\text{ where by marginal revenue we mean the rate}
\displaystyle \text{of change of total }   \text{revenue with respect to the number of items sold at an instant.}
\displaystyle  \hspace{2.2cm} \text{[CBSE 2018 C]}
\displaystyle \text{Answer:}
\displaystyle  \text{Marginal revenue }(MR)=\frac{dR}{dx}=\frac{d}{dx}(3x^2+36x+5)
\displaystyle =6x+36
\displaystyle \therefore \text{When }x=5
\displaystyle \text{Marginal revenue }(MR)=6\times5+36=66
\\

\displaystyle \textbf{Question 25. }\text{The volume of a sphere is increasing at the rate of } 8\ \text{cm}^3/\text{s. Find the }
\displaystyle \text{rate at which its surface area is }   \text{increasing when the radius of the sphere is }12\ \text{cm}. \hspace{0.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }r\text{ be the radius, }V\text{ be the volume and }S\text{ be the}
\displaystyle \text{surface area of sphere.}
\displaystyle \text{Then, we have } \frac{dV}{dt}=8\text{ cm}^3/\text{s}
\displaystyle \text{To find }\frac{dS}{dt},\text{ when }r=12\text{ cm}
\displaystyle \text{Since, }V=\frac{4}{3}\pi r^3
\displaystyle \therefore \frac{dV}{dt}=\frac{4}{3}\pi\cdot3r^2\cdot\frac{dr}{dt}
\displaystyle 8=4\pi r^2\frac{dr}{dt}
\displaystyle \Rightarrow \frac{dr}{dt}=\frac{2}{\pi r^2}\text{ cm/s}\qquad \ldots(i)
\displaystyle \text{Now, }S=4\pi r^2
\displaystyle \therefore \frac{dS}{dt}=\frac{d}{dt}(4\pi r^2)=4\pi\cdot2r\cdot\frac{dr}{dt}
\displaystyle =8\pi r\times\frac{2}{\pi r^2}\qquad [\text{using Eq. (i)}]
\displaystyle =\frac{16}{r}
\displaystyle \therefore \left(\frac{dS}{dt}\right)_{r=12}=\frac{16}{12}=\frac{4}{3}\text{ cm}^2/\text{s}
\\

\displaystyle \textbf{Question 26. }\text{The volume of a sphere is increasing at the rate of } 3\ \text{cm}^3/\text{s. }
\displaystyle \text{Find the rate of increase of its surface area, }  \text{when the radius is }2\ \text{cm}. \hspace{0.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }r\text{ be the radius of the sphere and }V\text{ be its volume.}
\displaystyle \text{Then, }V=\frac{4}{3}\pi r^3
\displaystyle \text{Given, }\frac{dV}{dt}=3\text{ cm}^3/\text{s}
\displaystyle \therefore \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=3
\displaystyle \Rightarrow \frac{4}{3}\pi(3r^2)\frac{dr}{dt}=3
\displaystyle \Rightarrow (4\pi r^2)\frac{dr}{dt}=3
\displaystyle \Rightarrow \frac{dr}{dt}=\frac{3}{4\pi r^2}\qquad \ldots(i)
\displaystyle \text{Now, let }S\text{ be the surface area of sphere, then}
\displaystyle S=4\pi r^2
\displaystyle \Rightarrow \frac{dS}{dt}=4\pi(2r)\frac{dr}{dt}
\displaystyle \Rightarrow \frac{dS}{dt}=8\pi r\left(\frac{3}{4\pi r^2}\right)\qquad [\text{using Eq. (i)}]
\displaystyle \Rightarrow \left(\frac{dS}{dt}\right)=\frac{6}{r}
\displaystyle \text{When }r=2,\text{ then}
\displaystyle \frac{dS}{dt}=\frac{6}{2}=3\text{ cm}^2/\text{s}
\\

\displaystyle \textbf{Question 27. }\text{Show that the function }f(x)=4x^3-18x^2+27x-7\text{ is}
\displaystyle \text{always increasing on }R. \hspace{2.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle  \text{We have, }f(x)=4x^3-18x^2+27x-7
\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}
\displaystyle f'(x)=12x^2-36x+27
\displaystyle \Rightarrow f'(x)=3(4x^2-12x+9)
\displaystyle \Rightarrow f'(x)=3(2x-3)^2
\displaystyle \therefore f'(x)\geq0
\displaystyle \Rightarrow \text{For any }x\in R,\ (2x-3)^2\geq0
\displaystyle \text{Since, a perfect square number cannot be negative.}
\displaystyle \therefore \text{Given, function }f(x)\text{ is an increasing function}
\displaystyle \text{on }R.
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 28. }\text{The volume of a cube is increasing at the rate of }8\ \text{cm}^3/\text{s. How fast is}
\displaystyle \text{the surface area increasing when }  \text{the length of its edge is }12\ \text{cm}? \hspace{0.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }x\text{ be the length of an edge of the cube, }V\text{ be the}
\displaystyle \text{volume and }S\text{ be the surface area at any time }t.
\displaystyle \text{Then, }V=x^3\text{ and }S=6x^2.
\displaystyle \text{It is given that}
\displaystyle \frac{dV}{dt}=8\text{ cm}^3/\text{s}\Rightarrow \frac{d}{dt}(x^3)=8
\displaystyle \Rightarrow 3x^2\frac{dx}{dt}=8\Rightarrow \frac{dx}{dt}=\frac{8}{3x^2}\qquad \ldots(i)
\displaystyle \text{Now, }S=6x^2
\displaystyle \Rightarrow \frac{dS}{dt}=12x\frac{dx}{dt}\Rightarrow \frac{dS}{dt}=12x\times\frac{8}{3x^2}
\displaystyle \Rightarrow \frac{dS}{dt}=\frac{32}{x}
\displaystyle \therefore \left(\frac{dS}{dt}\right)_{x=12}=\frac{32}{12}=\frac{8}{3}\text{ cm}^2/\text{s}
\\

\displaystyle \textbf{Question 29. }\text{Find the intervals in which the function}
\displaystyle f(x)=\frac{x^4}{4}-x^3-5x^2+24x+12\text{ is} \hspace{2.2cm} \text{[CBSE 2018]}
\displaystyle \text{(i) strictly increasing \quad (ii) strictly decreasing.}
\displaystyle \text{Answer:}
\displaystyle  \text{We have, }f(x)=\frac{x^4}{4}-x^3-5x^2+24x+12
\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}
\displaystyle f'(x)=x^3-3x^2-10x+24
\displaystyle =(x-2)(x^2-x-12)
\displaystyle =(x-2)(x^2-4x+3x-12)
\displaystyle =(x-2)\{x(x-4)+3(x-4)\}
\displaystyle =(x-2)(x-4)(x+3)
\displaystyle \text{Now, put }f'(x)=0,\text{ which gives }x=2,4\text{ and }-3.
\displaystyle \text{The points }x=-3,\ x=2\text{ and }x=4\text{ divide the whole real}
\displaystyle \text{line into four disjoint intervals namely,}
\displaystyle (-\infty,-3),\ (-3,2),\ (2,4),\ (4,\infty)
\displaystyle \text{Note that}
\displaystyle \text{for }x\in(-\infty,-3),\ f'(x)<0
\displaystyle \text{for }x\in(-3,2),\ f'(x)>0
\displaystyle \text{for }x\in(2,4),\ f'(x)<0
\displaystyle \text{and for }x\in(4,\infty),\ f'(x)>0
\displaystyle \therefore f(x)\text{ is strictly increasing in the intervals }(-3,2)\text{ and}
\displaystyle (4,\infty),\text{ and strictly decreasing in the intervals }(-\infty,-3)
\displaystyle \text{and }(2,4).
\\

\displaystyle \textbf{Question 30. }\text{Find the intervals in which the function}
\displaystyle f(x)=-2x^3-9x^2-12x+1\text{ is} \hspace{2.2cm} \text{[CBSE 2018 C]}
\displaystyle \text{(i) strictly increasing \quad (ii) strictly decreasing.}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }f(x)=-2x^3-9x^2-12x+1
\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}
\displaystyle f'(x)=-6x^2-18x-12
\displaystyle \Rightarrow f'(x)=-6(x^2+3x+2)
\displaystyle \Rightarrow f'(x)=-6(x^2+2x+x+2)
\displaystyle \Rightarrow f'(x)=-6[(x+2)x+1(x+2)]
\displaystyle =-6(x+2)(x+1)
\displaystyle \text{Now, put }f'(x)=0
\displaystyle \Rightarrow -6(x+2)(x+1)=0
\displaystyle \Rightarrow x=-2,-1
\displaystyle \text{The points }x=-2\text{ and }x=-1\text{ divide the real line into}
\displaystyle \text{their disjoint intervals }(-\infty,-2),\ (-2,-1)\text{ and }(-1,\infty).
\displaystyle \text{The nature of function in these intervals are given below}
\displaystyle \text{Interval}\qquad\qquad \text{Sign of }f'(x)\qquad\qquad \text{Nature of function}
\displaystyle (-\infty,-2)\qquad (-)(-)(-)=(-)<0\qquad \text{Strictly decreasing}
\displaystyle (-2,-1)\qquad (-)(+)(-)=(+)>0\qquad \text{Strictly increasing}
\displaystyle (-1,\infty)\qquad (+)(+)(-)=(-)<0\qquad \text{Strictly decreasing}
\displaystyle \text{Hence, }f(x)\text{ is strictly increasing in the interval}
\displaystyle (-2,-1)\text{ and }f(x)\text{ is strictly decreasing in the interval}
\displaystyle (-\infty,-2)\cup(-1,\infty).
\\

\displaystyle \textbf{Question 31. }\text{The length }x\text{ of a rectangle is decreasing at the rate of } 5\ \text{cm/min }
\displaystyle \text{and the width }y\text{ is increasing at the rate of }  4\ \text{cm/min. When }x=8\ \text{cm and }
\displaystyle y=6\ \text{cm, find the rate of }   \text{change of}
\displaystyle \text{(i) the perimeter. \quad (ii) area of rectangle.} \hspace{2.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given that length }x\text{ of a rectangle is decreasing}
\displaystyle \text{at the rate of }5\text{ cm/min.}
\displaystyle \therefore \frac{dx}{dt}=-5\text{ cm/min}\qquad \ldots(i)
\displaystyle \text{Also, the breadth }y\text{ of rectangle is increasing at the}
\displaystyle \text{rate of }4\text{ cm/min}
\displaystyle \therefore \frac{dy}{dt}=4\text{ cm/min}\qquad \ldots(ii)
\displaystyle \text{(i) Here, we have to find rate of change of perimeter,}
\displaystyle \text{i.e. }\frac{dP}{dt}
\displaystyle \text{and 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)\qquad [\text{from Eqs. (i) and (ii)}]
\displaystyle =2(-1)=-2\text{ cm/min}
\displaystyle \text{Hence, the perimeter of rectangle is decreasing at}
\displaystyle \text{the rate }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 \qquad [\text{by using product rule of derivative}]
\displaystyle \text{Now, we have }x=8\text{ cm and }y=6\text{ cm}
\displaystyle \frac{dx}{dt}=-5\text{ cm/min and }\frac{dy}{dt}=4\text{ cm/min}
\displaystyle \therefore \frac{dA}{dt}=(8\times4)+[6\times(-5)]=32-30
\displaystyle \Rightarrow \frac{dA}{dt}=2\text{ cm}^2/\text{min}
\displaystyle \text{Hence, the area of rectangle is increasing at the rate}
\displaystyle 2\text{ cm}^2/\text{min}.
\\

\displaystyle \textbf{Question 32. }\text{The side of an equilateral triangle is increasing at the } \text{rate of }2\ \text{cm/s. }
\displaystyle  \text{At what rate is its area increasing, when }   \text{the side of the triangle is }20\ \text{cm?} \hspace{0.2cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }a\text{ be the side of an equilateral triangle and}
\displaystyle A\text{ be the area of an equilateral triangle.}
\displaystyle \text{Then, }\frac{da}{dt}=2\text{ cm/s}
\displaystyle \text{We know that area of an equilateral triangle,}
\displaystyle A=\frac{\sqrt3}{4}a^2
\displaystyle \text{On differentiating both sides w.r.t. }t,\text{ we get}
\displaystyle \frac{dA}{dt}=\frac{\sqrt3}{4}\times2a\times\frac{da}{dt}
\displaystyle \Rightarrow \frac{dA}{dt}=\frac{\sqrt3}{4}\times2\times20\times2\qquad [\text{given, }a=20]
\displaystyle \therefore \frac{dA}{dt}=20\sqrt3\text{ cm}^2/\text{s}
\displaystyle \text{Thus, the rate of area increasing is }20\sqrt3\text{ cm}^2/\text{s}.
\\

\displaystyle \textbf{Question 33. }\text{Find the intervals in which the function } 
\displaystyle f(x)=3x^4-4x^3-12x^2+5\text{ is}
\displaystyle \text{(i) strictly increasing.}  \ \ \ \ \ \ \ \ \ \  \text{(ii) strictly decreasing.} \hspace{2.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given function is}
\displaystyle 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, put}
\displaystyle 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\text{ or }2
\displaystyle \text{Now, we find intervals in which }f(x)\text{ is strictly}
\displaystyle \text{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 a function }f(x)\text{ is said to be strictly}
\displaystyle \text{increasing, if }f'(x)>0\text{ and it is said to be strictly}
\displaystyle \text{decreasing, if }f'(x)<0.\text{ So, the given function }f(x)\text{ is}
\displaystyle \text{(i) strictly increasing on the intervals }(-1,0)\text{ and }(2,\infty).
\displaystyle \text{(ii) strictly decreasing on the intervals }(-\infty,-1)\text{ and}
\displaystyle (0,2).
\\

\displaystyle \textbf{Question 34. }\text{Find the intervals in which }  
\displaystyle f(x)=\frac{3}{10}x^{4}-\frac{4}{5}x^{3}-3x^{2}+\frac{36x}{5}+11  \\ \text{is (i) strictly increasing and (ii) strictly decreasing.}\hspace{0.2cm}\text{[CBSE 2014]}  
\displaystyle \text{Answer:}
\displaystyle f'(x)=\frac{3}{10}\cdot4x^{3}-\frac{4}{5}\cdot3x^{2}-6x+\frac{36}{5}
\displaystyle =\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-3)(x+2)
\displaystyle \text{For critical points, }f'(x)=0
\displaystyle \Rightarrow x=-2,1,3
\displaystyle \begin{array}{|c|c|c|}\hline \text{Interval}&f'(x)=\frac{6}{5}(x+2)(x-1)(x-3)&\text{Sign of }f'(x)\\ \hline x<-2&(-)(-)(-)&-\text{ve}\\ \hline -2<x<1&(+)(-)(-)&+\text{ve}\\ \hline 1<x<3&(+)(+)(-)&-\text{ve}\\ \hline x>3&(+)(+)(+)&+\text{ve}\\ \hline \end{array}
\displaystyle \therefore f(x)\text{ is strictly increasing on }(-2,1)\cup(3,\infty)
\displaystyle \text{and strictly decreasing on }(-\infty,-2)\cup(1,3).
\\

\displaystyle \textbf{Question 35. }\text{The sides of an equilateral triangle are increasing at the rate of }2\text{ cm/s.}  
\displaystyle \text{Find the rate at which the area increases, when the side is }10\text{ cm.}\hspace{0.2cm}\text{[CBSE 2014]}  
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the equilateral triangle be }a\text{ cm.}
\displaystyle \text{Given, }\frac{da}{dt}=2\text{ cm/s}
\displaystyle \text{Area of an equilateral triangle, }A=\frac{\sqrt{3}}{4}a^{2}
\displaystyle \text{Differentiating both sides w.r.t. }t,\text{ we get}
\displaystyle \frac{dA}{dt}=\frac{\sqrt{3}}{4}\cdot2a\frac{da}{dt}
\displaystyle \frac{dA}{dt}=\frac{\sqrt{3}}{2}a\frac{da}{dt}
\displaystyle \text{When }a=10\text{ cm and }\frac{da}{dt}=2\text{ cm/s,}
\displaystyle \frac{dA}{dt}=\frac{\sqrt{3}}{2}\times10\times2
\displaystyle =10\sqrt{3}\text{ cm}^{2}\text{/s}
\displaystyle \therefore \text{The area increases at the rate of }10\sqrt{3}\text{ cm}^{2}\text{/s.}
\\

\displaystyle \textbf{Question 36. }\text{Find the value(s) of }x\text{ for which } y=[x(x-2)]^2
\displaystyle \text{is an increasing function.} \hspace{2.2cm} \text{[CBSE 2014; CBSE 2010]}
\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)
\displaystyle =4x(x-2)(x-1)
\displaystyle \text{On putting }\frac{dy}{dx}=0,\text{ we get}
\displaystyle 4x(x-2)(x-1)=0
\displaystyle \Rightarrow x=0,1\text{ and }2
\displaystyle \text{Now, we find interval in which }f(x)\text{ is strictly}
\displaystyle \text{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}
\displaystyle x=0,1\text{ and }2.
\displaystyle \text{Hence, }y\text{ is increasing in }[0,1]\cup[2,\infty).
\\

\displaystyle \textbf{Question 37. }\text{Find the intervals in which }f(x)=\frac{3}{2}x^{4}-4x^{3}-45x^{2}+51
\displaystyle \text{is (i) strictly increasing and (ii) strictly decreasing.}\hspace{2.2cm}\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle f'(x)=\frac{3}{2}\cdot4x^{3}-12x^{2}-90x
\displaystyle =6x^{3}-12x^{2}-90x
\displaystyle =6x(x^{2}-2x-15)
\displaystyle =6x(x-5)(x+3)
\displaystyle \text{For critical points, }f'(x)=0
\displaystyle \Rightarrow x=-3,0,5
\displaystyle \begin{array}{|c|c|c|}\hline \text{Interval}&f'(x)=6x(x+3)(x-5)&\text{Sign of }f'(x)\\ \hline x<-3&(-)(-)(-)&-\text{ve}\\ \hline -3<x<0&(-)(+)(-)&+\text{ve}\\ \hline 0<x<5&(+)(+)(-)&-\text{ve}\\ \hline x>5&(+)(+)(+)&+\text{ve}\\ \hline \end{array}
\displaystyle \therefore f(x)\text{ is strictly increasing on }(-3,0)\cup(5,\infty)
\displaystyle \text{and strictly decreasing on }(-\infty,-3)\cup(0,5).
\\

\displaystyle \textbf{Question 38. }\text{A ladder }5\ \text{m long is leaning against a wall. Bottom of ladder is }
\displaystyle \text{pulled along the ground away from wall at } \text{the rate of }2\ \text{m/s. How fast is the }
\displaystyle \text{height on the wall } \text{decreasing, when the foot of ladder is }4\ \text{m away from the wall.}
\displaystyle   \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }AC\text{ be the ladder, }BC=x\text{ and height of the wall,}
\displaystyle AB=y.
\displaystyle \text{As the ladder is pulled along} \\ \text{the ground away from}
\displaystyle \text{the wall at the rate of }2\text{ m/s.}
\displaystyle \text{So,}
\displaystyle \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=x^2+y^2=25\qquad \ldots(i)
\displaystyle \Rightarrow (4)^2+y^2=25\Rightarrow 16+y^2=25
\displaystyle \Rightarrow y^2=9\Rightarrow y=\sqrt9
\displaystyle \Rightarrow y=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 \ldots(ii)
\displaystyle \text{[dividing both sides by 2]}
\displaystyle \text{On substituting the values of }x,\ y\text{ and }\frac{dx}{dt}\text{ in Eq. (ii),}
\displaystyle \text{we get}
\displaystyle (4\times2)+3\times\frac{dy}{dt}=0
\displaystyle \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}
\displaystyle \text{of }\frac{8}{3}\text{ m/s.}
\\

\displaystyle \textbf{Question 39. }\text{Show that }y=\log(1+x)-\frac{2x}{2+x},\ x>-1\text{ is an}
\displaystyle \text{increasing function of }x,\text{ throughout its domain.} \hspace{2.2cm} \text{[CBSE 2012]}
\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}(1)-\frac{(2+x)\cdot2-2x\cdot1}{(2+x)^2}
\displaystyle \qquad [\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 =\frac{x^2}{(1+x)(2+x)^2}\qquad \ldots(i)
\displaystyle \text{Now, }x^2\text{ and }(2+x)^2\text{ are always positive, and also }1+x>0
\displaystyle \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 40. }\text{Find the intervals in which the function given by}
\displaystyle f(x)=\sin x+\cos x,\ 0\leq x\leq2\pi\text{ is}
\displaystyle \text{(i) increasing.}
\displaystyle \text{(ii) decreasing.} \hspace{2.2cm} \text{[CBSE 2012 C; CBSE 2011]}
\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
\displaystyle \Rightarrow x=\frac{\pi}{4},\frac{5\pi}{4},\text{ as }0\leq x\leq2\pi
\displaystyle \text{Now, we find the intervals in which }f(x)\text{ is strictly}
\displaystyle \text{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{\sqrt3}{2}-\frac12=\frac{\sqrt3-1}{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-\left(-1\right)=1 & +\text{ve} \\ \hline \end{array}
\displaystyle \text{Note that }f'(x)>0\text{ in }\left(0,\frac{\pi}{4}\right),\ f'(x)<0\text{ in }\left(\frac{\pi}{4},\frac{5\pi}{4}\right)
\displaystyle \text{and }f'(x)>0\text{ in }\left(\frac{5\pi}{4},2\pi\right).
\displaystyle \text{Since, }f(x)\text{ is a trigonometric function, so it is}
\displaystyle \text{continuous at }x=0,\frac{\pi}{4},\frac{5\pi}{4}\text{ and }2\pi.
\displaystyle \text{Hence, the function is}
\displaystyle \text{(i) increasing in }\left[0,\frac{\pi}{4}\right]\text{ and }\left[\frac{5\pi}{4},2\pi\right]
\displaystyle \text{(ii) decreasing in }\left[\frac{\pi}{4},\frac{5\pi}{4}\right]
\\

\displaystyle \textbf{Question 41. }\text{Find the intervals in which the function given by}
\displaystyle f(x)=x^4-8x^3+22x^2-24x+21\text{ is}
\displaystyle \text{(i) increasing.} \ \ \ \ \ \ \ \ \ \  \text{(ii) decreasing.} \hspace{2.2cm} \text{[CBSE 2012 C]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, function is}
\displaystyle f(x)=x^4-8x^3+22x^2-24x+21
\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}
\displaystyle f'(x)=4x^3-24x^2+44x-24
\displaystyle =4(x^3-6x^2+11x-6)
\displaystyle =4(x-1)(x-2)(x-3)
\displaystyle \text{On putting }f'(x)=0
\displaystyle \Rightarrow 4(x-1)(x-2)(x-3)=0\Rightarrow x=1,2,3
\displaystyle \text{So, the possible intervals are }(-\infty,1),\ (1,2),\ (2,3)
\displaystyle \text{and }(3,\infty).
\displaystyle \text{For interval }(-\infty,1),\ f'(x)<0
\displaystyle \text{For interval }(1,2),\ f'(x)>0
\displaystyle \text{For interval }(2,3),\ f'(x)<0
\displaystyle \text{For interval }(3,\infty),\ f'(x)>0
\displaystyle \text{Also, as }f(x)\text{ is a polynomial function, so it is}
\displaystyle \text{continuous at }x=1,2,3.
\displaystyle \text{Hence,}
\displaystyle \text{(i) function increases in }[1,2]\text{ and }[3,\infty).
\displaystyle \text{(ii) function decreases in }(-\infty,1]\text{ and }[2,3].
\\

\displaystyle \textbf{Question 42. }\text{Sand is pouring from the pipe at the rate of }12\ \text{cm}^3/\text{s.}
\displaystyle \text{The falling sand forms a cone on a ground in such a way that the height of cone is }
\displaystyle \text{always one-sixth of the radius }  \text{of the base. How fast is the height of sand cone }
\displaystyle \text{increasing when the height is }4\ \text{cm?} \hspace{2.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }V\text{ be the volume of cone, }h\text{ be the height and}
\displaystyle r\text{ be the radius of base of the cone.}
\displaystyle \text{Given, }\frac{dV}{dt}=12\text{ cm}^3/\text{s}\qquad \ldots(i)
\displaystyle \text{Also, height of cone }=\frac{1}{6}\times(\text{radius of base of cone})
\displaystyle \therefore h=\frac{1}{6}r\text{ or }r=6h\qquad \ldots(ii)
\displaystyle \text{We know that volume of cone is given by}
\displaystyle V=\frac{1}{3}\pi r^2h\qquad \ldots(iii)
\displaystyle \text{On putting }r=6h\text{ from Eq. (ii) in Eq. (iii), we get}
\displaystyle V=\frac{1}{3}\pi(6h)^2h
\displaystyle \Rightarrow V=\frac{\pi}{3}\cdot36h^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}=12\pi\times3h^2\cdot\frac{dh}{dt}
\displaystyle \Rightarrow \frac{dV}{dt}=36\pi h^2\cdot\frac{dh}{dt}
\displaystyle \text{On putting }\frac{dV}{dt}=12\text{ cm}^3/\text{s and }h=4\text{ cm, we get}
\displaystyle 12=36\pi\times16\times\frac{dh}{dt}
\displaystyle \Rightarrow \frac{dh}{dt}=\frac{12}{36\pi\times16}
\displaystyle \therefore \frac{dh}{dt}=\frac{1}{48\pi}\text{ cm/s}
\displaystyle \text{Hence, the height of sand cone is increasing at the rate}
\displaystyle \text{of }\frac{1}{48\pi}\text{ cm/s.}
\\

\displaystyle \textbf{Question 43. }\text{Show that the function }f(x)=x^3-3x^2+3x,\ x\in R\text{ is}
\displaystyle \text{increasing on }R. \hspace{2.2cm} \text{[CBSE 2011 C]}
\displaystyle \text{Answer:}
\displaystyle  \text{We know that a continuous function }y=f(x)\text{ is said}
\displaystyle \text{to be increasing on }R,\text{ if }\frac{dy}{dx}\geq0,\ \forall x\in R.
\displaystyle \text{Given, }y=x^3-3x^2+3x
\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}
\displaystyle \frac{dy}{dx}=3x^2-6x+3
\displaystyle \Rightarrow \frac{dy}{dx}=3(x^2-2x+1)
\displaystyle \Rightarrow \frac{dy}{dx}=3(x-1)^2
\displaystyle \text{Now, }3(x-1)^2\geq0\text{ for all real values of }x,
\displaystyle \text{i.e. }\forall x\in R,
\displaystyle \frac{dy}{dx}\geq0,\ \forall x\in R
\displaystyle \text{Hence, the given function is increasing on }R.
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 44. }\text{Find the intervals in which the function } \\  f(x)=(x-1)^3(x-2)^2\text{ is}
\displaystyle \text{(i) increasing.}  \hspace{1.0cm} \text{(ii) decreasing.} \hspace{2.2cm} \text{[CBSE 2011 C]}
\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 \qquad \left[\because \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\right]
\displaystyle \Rightarrow f'(x)=(x-1)^3\cdot2(x-2)+(x-2)^2\cdot3(x-1)^2
\displaystyle \Rightarrow f'(x)=(x-1)^2(x-2)\left[2(x-1)+3(x-2)\right]
\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 \text{Either }(x-1)^2=0\text{ or }x-2=0\text{ or }5x-8=0
\displaystyle \therefore x=1,\frac{8}{5},2
\displaystyle \text{Now, we find intervals and check in which interval}
\displaystyle f(x)\text{ is strictly increasing and strictly decreasing.}
\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{We know that a function }f(x)\text{ is said to be a strictly}
\displaystyle \text{increasing function, if }f'(x)>0\text{ and strictly}
\displaystyle \text{decreasing, if }f'(x)<0.
\displaystyle \text{So, the given function }f(x)\text{ is increasing on the}
\displaystyle \text{intervals }(-\infty,1),\left(1,\frac{8}{5}\right)\text{ and }(2,\infty)\text{ and decreasing on}
\displaystyle \left(\frac{8}{5},2\right).
\displaystyle \text{Since, }f(x)\text{ is a polynomial function, so it is}
\displaystyle \text{continuous at }x=1,\frac{8}{5},2.\text{ Hence, }f(x)\text{ is}
\displaystyle \text{(i) increasing on intervals }\left(-\infty,\frac{8}{5}\right]\text{ and }[2,\infty).
\displaystyle \text{(ii) decreasing on interval }\left[\frac{8}{5},2\right].
\\

\displaystyle \textbf{Question 45. }\text{Find the intervals in which the function}
\displaystyle f(x)=2x^3+9x^2+12x+20\text{ is}
\displaystyle \text{(i) increasing.}  \hspace{1.0cm} \text{(ii) decreasing.} \hspace{2.2cm} \text{[CBSE 2011 C]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, function is}
\displaystyle 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{Put }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+1)(x+2)=0
\displaystyle \Rightarrow x+1=0\text{ or }x+2=0
\displaystyle \therefore x=-2,-1
\displaystyle \text{Now, we find intervals and check in which interval}
\displaystyle 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 a function }f(x)\text{ is said to be a strictly}
\displaystyle \text{increasing function, if }f'(x)>0\text{ and strictly}
\displaystyle \text{decreasing, if }f'(x)<0.\text{ So, given function is}
\displaystyle \text{increasing on intervals }(-\infty,-2)\text{ and }(-1,\infty)\text{ and}
\displaystyle \text{decreasing on interval }(-2,-1).
\displaystyle \text{Since, }f(x)\text{ is a polynomial function, so it is}
\displaystyle \text{continuous at }x=-1,-2.
\displaystyle \text{Hence, given function is}
\displaystyle \text{(i) increasing on intervals }(-\infty,-2]\text{ and }[-1,\infty).
\displaystyle \text{(ii) decreasing on interval }[-2,-1].
\\

\displaystyle \textbf{Question 46. }\text{Find the intervals in which }f(x)=2x^{3}-9x^{2}+12x-15\text{ is}
\displaystyle \text{(i) increasing and (ii) decreasing.}\hspace{2.2cm}\text{[CBSE 2011 C]}
\displaystyle \text{Answer:}
\displaystyle f'(x)=6x^{2}-18x+12
\displaystyle =6(x^{2}-3x+2)
\displaystyle =6(x-1)(x-2)
\displaystyle \text{For critical points, }f'(x)=0
\displaystyle \Rightarrow x=1,2
\displaystyle \begin{array}{|c|c|c|}\hline \text{Interval}&f'(x)=6(x-1)(x-2)&\text{Sign of }f'(x)\\ \hline x<1&(+)(-)(-)&+\text{ve}\\ \hline 1<x<2&(+)(+)(-)&-\text{ve}\\ \hline x>2&(+)(+)(+)&+\text{ve}\\ \hline \end{array}
\displaystyle \therefore f(x)\text{ is increasing on }(-\infty,1]\cup[2,\infty)
\displaystyle \text{and decreasing on }[1,2].
\\

\displaystyle \textbf{Question 47. }\text{Find the intervals in which }f(x)=2x^{3}-15x^{2}+36x+17
\displaystyle \text{is increasing or decreasing.}\hspace{2.2cm}\text{[CBSE 2010 C]}
\displaystyle \text{Answer:}
\displaystyle f'(x)=6x^{2}-30x+36
\displaystyle =6(x^{2}-5x+6)
\displaystyle =6(x-2)(x-3)
\displaystyle \text{For critical points, }f'(x)=0
\displaystyle \Rightarrow x=2,3
\displaystyle \begin{array}{|c|c|c|}\hline \text{Interval}&f'(x)=6(x-2)(x-3)&\text{Sign of }f'(x)\\ \hline x<2&(+)(-)(-)&+\text{ve}\\ \hline 2<x<3&(+)(+)(-)&-\text{ve}\\ \hline x>3&(+)(+)(+)&+\text{ve}\\ \hline \end{array}
\displaystyle \therefore f(x)\text{ is increasing on }(-\infty,2]\cup[3,\infty)
\displaystyle \text{and decreasing on }[2,3].
\\

\displaystyle \textbf{Question 48. }\text{Find the intervals in which }f(x)=2x^{3}-9x^{2}+12x+15\text{ is}
\displaystyle \text{(i) increasing and (ii) decreasing.}\hspace{2.2cm}\text{[CBSE 2010 C]}
\displaystyle \text{Answer:}
\displaystyle f'(x)=6x^{2}-18x+12
\displaystyle =6(x^{2}-3x+2)
\displaystyle =6(x-1)(x-2)
\displaystyle \text{For critical points, }f'(x)=0
\displaystyle \Rightarrow x=1,2
\displaystyle \begin{array}{|c|c|c|}\hline \text{Interval}&f'(x)=6(x-1)(x-2)&\text{Sign of }f'(x)\\ \hline x<1&(+)(-)(-)&+\text{ve}\\ \hline 1<x<2&(+)(+)(-)&-\text{ve}\\ \hline x>2&(+)(+)(+)&+\text{ve}\\ \hline \end{array}
\displaystyle \therefore f(x)\text{ is increasing on }(-\infty,1]\cup[2,\infty)
\displaystyle \text{and decreasing on }[1,2].
\\

\displaystyle \textbf{Question 49. }\text{Find the intervals in which the function}
\displaystyle f(x)=(x-1)^3(x-2)^2\text{ is}
\displaystyle \text{(i) strictly increasing.} \hspace{1.0cm} \text{(ii) strictly decreasing.} \hspace{2.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, function is }f(x)=(x-1)^3(x-2)^2.
\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}
\displaystyle f'(x)=3(x-1)^2(x-2)^2+2(x-2)(x-1)^3
\displaystyle =(x-1)^2(x-2)[3(x-2)+2(x-1)]
\displaystyle =(x-1)^2(x-2)(5x-8)
\displaystyle \text{(i) For strictly increasing }f'(x)>0
\displaystyle \text{We get positive }f'(x)\text{ in the interval}
\displaystyle (-\infty,\tfrac{8}{5})\cup(2,\infty).
\displaystyle \text{(ii) For strictly decreasing }f'(x)<0,\text{ we get}
\displaystyle \text{negative }f'(x)\text{ in the interval }\left(\tfrac{8}{5},2\right).
\\

\displaystyle \textbf{Question 50. }\text{Prove that }y=\frac{4\sin\theta}{2+\cos\theta}-\theta\text{ is an increasing } \text{function in }\left(0,\frac{\pi}{2}\right). 
\displaystyle \hspace{2.2cm} \text{[CBSE 2016; CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, function is}
\displaystyle y=\frac{4\sin\theta}{2+\cos\theta}-\theta\qquad \ldots(i)
\displaystyle \text{We know that a function }y=f(x)\text{ is said to be an}
\displaystyle \text{increasing function, if }\frac{dy}{dx}\geq0,\text{ for all values of }x.
\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 \qquad [\text{by using quotient rule of derivative}]
\displaystyle =\frac{(2+\cos\theta)(4\cos\theta)-4\sin\theta(0-\sin\theta)}{(2+\cos\theta)^2}-1
\displaystyle =\left[\frac{8\cos\theta+4\cos^2\theta+4\sin^2\theta}{(2+\cos\theta)^2}\right]-1
\displaystyle =\left[\frac{8\cos\theta+4(\cos^2\theta+\sin^2\theta)}{(2+\cos\theta)^2}\right]-\frac{(2+\cos\theta)^2}{(2+\cos\theta)^2}
\displaystyle \left[\because (a+b)^2=a^2+b^2+2ab\right]
\displaystyle =\frac{8\cos\theta+4-4-\cos^2\theta-4\cos\theta}{(2+\cos\theta)^2}
\displaystyle \left[\because \sin^2\theta+\cos^2\theta=1\right]
\displaystyle =\frac{4\cos\theta-\cos^2\theta}{(2+\cos\theta)^2}
\displaystyle =\frac{\cos\theta(4-\cos\theta)}{(2+\cos\theta)^2}\qquad \ldots(ii)
\displaystyle \text{Now, as }\cos\theta>0,\ \forall\theta\in\left(0,\frac{\pi}{2}\right)\text{ and }(2+\cos\theta)^2\text{ being a}
\displaystyle \text{perfect square is always positive for all }\theta\in\left(0,\frac{\pi}{2}\right).
\displaystyle \text{Also, for }\theta\in\left(0,\frac{\pi}{2}\right),\text{ we know that }0<\cos\theta<1.
\displaystyle \therefore 4-\cos\theta>0,\ \forall\theta\in\left(0,\frac{\pi}{2}\right)
\displaystyle \text{Thus, we conclude that}
\displaystyle \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 \therefore y\text{ is an increasing function in }\left(0,\frac{\pi}{2}\right)
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 51. }\text{Find the intervals in which the function } f(x)=\sin3x-\cos3x,\ 
\displaystyle 0<x<\pi,\text{ is strictly increasing or }   \text{strictly decreasing.} \hspace{0.2cm} \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }f(x)=\sin3x-\cos3x,\ 0<x<\pi
\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}
\displaystyle f'(x)=3\cos3x+3\sin3x
\displaystyle \text{On putting }f'(x)=0,\text{ we get}
\displaystyle 3\cos3x=-3\sin3x\Rightarrow \tan3x=-1
\displaystyle \Rightarrow 3x=\frac{3\pi}{4},\frac{7\pi}{4},\frac{11\pi}{4}
\displaystyle \left[\because \tan\theta\text{ is negative in II and IV quadrants}\right]
\displaystyle \Rightarrow x=\frac{\pi}{4},\frac{7\pi}{12},\frac{11\pi}{12}
\displaystyle \text{Now, we find intervals and check in which intervals }f(x)\text{ is strictly increasing or strictly decreasing.}
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Interval} & \text{Test Value} & f'(x)=3(\cos3x+\sin3x) & \text{Sign of }f'(x) \\ \hline 0<x<\frac{\pi}{4} & x=\frac{\pi}{6} & 3\left(\cos\frac{\pi}{2}+\sin\frac{\pi}{2}\right)=3(0+1)=3 & +\text{ve} \\ \hline \frac{\pi}{4}<x<\frac{7\pi}{12} & x=\frac{\pi}{3} & 3(\cos\pi+\sin\pi)=3(-1+0)=-3 & -\text{ve} \\ \hline \frac{7\pi}{12}<x<\frac{11\pi}{12} & x=\frac{3\pi}{4} & 3\left(\cos\frac{9\pi}{4}+\sin\frac{9\pi}{4}\right)=3\left(\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\right)=3\sqrt2 & +\text{ve} \\ \hline \frac{11\pi}{12}<x<\pi & x=\frac{23\pi}{24} & 3\left(\cos\frac{23\pi}{8}+\sin\frac{23\pi}{8}\right) & \\ & & =3\left[\cos\left(3\pi-\frac{\pi}{8}\right)+\sin\left(3\pi-\frac{\pi}{8}\right)\right] & \\ & & =3\left(-\cos\frac{\pi}{8}+\sin\frac{\pi}{8}\right) & -\text{ve} \\ \hline \end{array}
\displaystyle \text{Here, we see that }f'(x)>0,\text{ for }0<x<\frac{\pi}{4}\text{ and }\frac{7\pi}{12}<x<\frac{11\pi}{12},
\displaystyle \text{so, }f(x)\text{ is strictly increasing in the intervals }\left(0,\frac{\pi}{4}\right)\text{ and }
\displaystyle \left(\frac{7\pi}{12},\frac{11\pi}{12}\right).
\displaystyle \text{While, }f'(x)<0\text{ in }\frac{\pi}{4}<x<\frac{7\pi}{12}\text{ and }\frac{11\pi}{12}<x<\pi.
\displaystyle \text{So, }f(x)\text{ is strictly decreasing in the intervals }\left(\frac{\pi}{4},\frac{7\pi}{12}\right)\text{ and }\left(\frac{11\pi}{12},\pi\right).
\\

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