\displaystyle \textbf{Question 1. }\text{The distance covered by a particle in time }t\text{ is given by } \\ x=3+8t-4t^{2}.\text{ After }1s,\text{ its velocity will be}
\displaystyle \text{(a) }0\text{ units/s}\qquad\text{(b) }3\text{ units/s}
\displaystyle \text{(c) }4\text{ units/s}\qquad\text{(d) }7\text{ units/s}
\displaystyle \text{Answer:}
\displaystyle x=3+8t-4t^{2}
\displaystyle \therefore v=\frac{dx}{dt}=8-8t
\displaystyle \text{At }t=1,
\displaystyle v=8-8(1)=0
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 2. }\text{If }f:R\to R\text{ be defined by }f(x)=2x+\cos x,\text{ then }f
\displaystyle \text{(a) has a minimum at }x=\pi
\displaystyle \text{(b) has a maximum at }x=0
\displaystyle \text{(c) is a decreasing function}
\displaystyle \text{(d) is an increasing function}
\displaystyle \text{Answer:}
\displaystyle f(x)=2x+\cos x
\displaystyle \therefore f'(x)=2-\sin x
\displaystyle \text{Since }-1\leq\sin x\leq1,
\displaystyle 1\leq2-\sin x\leq3
\displaystyle \therefore f'(x)>0\text{ for all }x\in R
\displaystyle \therefore f\text{ is an increasing function.}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 3. }\text{Let }g(x)=2f\left(\frac{x}{2}\right)+f(2-x)\text{ and }f''(x)<0\text{ for all }x\in(0,2).\text{ Then, }g(x)\text{ is}
\displaystyle \text{(a) increasing on }\left(\frac{4}{3},2\right)\text{ and increasing on }\left(0,\frac{4}{3}\right)
\displaystyle \text{(b) decreasing on }\left(0,\frac{4}{3}\right)\text{ and increasing on }\left(\frac{4}{3},2\right)
\displaystyle \text{(c) increasing on }\left(0,\frac{4}{3}\right)\text{ and decreasing on }\left(\frac{4}{3},2\right)
\displaystyle \text{(d) None of the above}
\displaystyle \text{Answer:}
\displaystyle g(x)=2f\left(\frac{x}{2}\right)+f(2-x)
\displaystyle \therefore g'(x)=f'\left(\frac{x}{2}\right)-f'(2-x)
\displaystyle \text{At }x=\frac{4}{3},
\displaystyle g'\left(\frac{4}{3}\right)=f'\left(\frac{2}{3}\right)-f'\left(\frac{2}{3}\right)=0
\displaystyle g''(x)=\frac{1}{2}f''\left(\frac{x}{2}\right)+f''(2-x)
\displaystyle \text{Since }f''(x)<0,\text{ we get }g''(x)<0
\displaystyle \therefore g'(x)\text{ is decreasing.}
\displaystyle \therefore g'(x)>0\text{ for }x<\frac{4}{3}\text{ and }g'(x)<0\text{ for }x>\frac{4}{3}
\displaystyle \therefore g(x)\text{ is increasing on }\left(0,\frac{4}{3}\right)\text{ and decreasing on }\left(\frac{4}{3},2\right)
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 4. }\text{If the sides of an equilateral triangle are increasing at the rate of } \\ 4\text{ cm/s, then the rate at which the area increases, when side is }5\text{ cm, is}
\displaystyle \text{(a) }10\text{ cm}^{2}/s\qquad\text{(b) }\sqrt{3}\text{ cm}^{2}/s
\displaystyle \text{(c) }10\sqrt{3}\text{ cm}^{2}/s\qquad\text{(d) }\frac{10}{3}\text{ cm}^{2}/s
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the equilateral triangle be }a
\displaystyle \text{Area }A=\frac{\sqrt{3}}{4}a^{2}
\displaystyle \therefore \frac{dA}{dt}=\frac{\sqrt{3}}{2}a\frac{da}{dt}
\displaystyle \text{Given }a=5,\ \frac{da}{dt}=4
\displaystyle \therefore \frac{dA}{dt}=\frac{\sqrt{3}}{2}\times5\times4
\displaystyle =10\sqrt{3}\text{ cm}^{2}/s
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 5. }\text{The total cost }C(x)\text{ (in Rs.) associated with the production of }
\displaystyle x\text{ units of an item is given by} C(x)=0.007x^{3}-0.003x^{2}+15x+4000.
\displaystyle \text{The marginal cost when }17\text{ units are produced, is}
\displaystyle \text{(a) Rs. }20.967\qquad\text{(b) Rs. }21.967
\displaystyle \text{(c) Rs. }81.968\qquad\text{(d) Rs. }11.967
\displaystyle \text{Answer:}
\displaystyle \text{Marginal cost }=C'(x)
\displaystyle C'(x)=0.021x^{2}-0.006x+15
\displaystyle \therefore C'(17)=0.021(17)^{2}-0.006(17)+15
\displaystyle =0.021(289)-0.102+15
\displaystyle =6.069-0.102+15
\displaystyle =20.967
\displaystyle \therefore \text{Marginal cost when }17\text{ units are produced is Rs. }20.967
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 6. }\text{The function }f(x)=4\sin^{3}x-6\sin^{2}x+12\sin x+100\text{ is strictly}
\displaystyle \text{(a) increasing in }\left[\pi,\frac{3\pi}{2}\right]\qquad\text{(b) decreasing in }\left(\frac{\pi}{2},\pi\right)
\displaystyle \text{(c) decreasing in }\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\qquad\text{(d) decreasing in }\left[0,\frac{\pi}{2}\right]
\displaystyle \text{Answer:}
\displaystyle f(x)=4\sin^{3}x-6\sin^{2}x+12\sin x+100
\displaystyle \therefore f'(x)=12\sin^{2}x\cos x-12\sin x\cos x+12\cos x
\displaystyle =12\cos x(\sin^{2}x-\sin x+1)
\displaystyle \text{Since }\sin^{2}x-\sin x+1=\left(\sin x-\frac{1}{2}\right)^{2}+\frac{3}{4}>0
\displaystyle \therefore \text{sign of }f'(x)\text{ depends on }\cos x
\displaystyle \text{In }\left[\pi,\frac{3\pi}{2}\right],\ \cos x<0
\displaystyle \therefore f(x)\text{ is decreasing in }\left[\pi,\frac{3\pi}{2}\right]
\displaystyle \text{Also, in }\left(\frac{\pi}{2},\pi\right),\ \cos x<0
\displaystyle \therefore f(x)\text{ is decreasing in }\left(\frac{\pi}{2},\pi\right)
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 7. }\text{Which of the following function is decreasing on }\left(0,\frac{\pi}{2}\right)?
\displaystyle \text{(a) }\sin2x\qquad\text{(b) }\tan x
\displaystyle \text{(c) }\cos x\qquad\text{(d) }\cos3x
\displaystyle \text{Answer:}
\displaystyle \text{For }\sin2x,\ \frac{d}{dx}(\sin2x)=2\cos2x,\text{ which changes sign in }\left(0,\frac{\pi}{2}\right)
\displaystyle \text{For }\tan x,\ \frac{d}{dx}(\tan x)=\sec^{2}x>0
\displaystyle \text{For }\cos x,\ \frac{d}{dx}(\cos x)=-\sin x<0\text{ in }\left(0,\frac{\pi}{2}\right)
\displaystyle \text{For }\cos3x,\ \frac{d}{dx}(\cos3x)=-3\sin3x,\text{ which changes sign in }\left(0,\frac{\pi}{2}\right)
\displaystyle \therefore \cos x\text{ is decreasing on }\left(0,\frac{\pi}{2}\right)
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 8. }\text{The volume of a sphere is increasing at the rate of }3\text{ cm}^{3}/s. \\ \text{Find the rate of increase of its surface area when the radius is }2\text{ cm}.
\displaystyle \text{Answer:}
\displaystyle V=\frac{4}{3}\pi r^{3}
\displaystyle \therefore \frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}
\displaystyle \text{Given, }\frac{dV}{dt}=3\text{ cm}^{3}/s,\ r=2\text{ cm}
\displaystyle 3=4\pi(2)^{2}\frac{dr}{dt}
\displaystyle \therefore \frac{dr}{dt}=\frac{3}{16\pi}
\displaystyle \text{Surface area, }S=4\pi r^{2}
\displaystyle \therefore \frac{dS}{dt}=8\pi r\frac{dr}{dt}
\displaystyle =8\pi(2)\cdot\frac{3}{16\pi}=3
\displaystyle \therefore \text{Rate of increase of surface area is }3\text{ cm}^{2}/s
\\

\displaystyle \textbf{Question 9. }\text{The sides of an equilateral triangle are increasing at the rate of }2\text{ cm/s}. \\ \text{Find the rate at which the area increases, when the side is }10\text{ cm}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the side of the equilateral triangle be }a
\displaystyle \text{Area, }A=\frac{\sqrt{3}}{4}a^{2}
\displaystyle \therefore \frac{dA}{dt}=\frac{\sqrt{3}}{2}a\frac{da}{dt}
\displaystyle \text{Given, }\frac{da}{dt}=2\text{ cm/s},\ a=10\text{ cm}
\displaystyle \therefore \frac{dA}{dt}=\frac{\sqrt{3}}{2}(10)(2)=10\sqrt{3}
\displaystyle \therefore \text{Rate of increase of area is }10\sqrt{3}\text{ cm}^{2}/s
\\

\displaystyle \textbf{Question 10. }\text{Show that the function }f(x)=4x^{3}-18x^{2}+28x-7\text{ is always increasing on }R.
\displaystyle \text{Answer:}
\displaystyle f(x)=4x^{3}-18x^{2}+28x-7
\displaystyle \therefore f'(x)=12x^{2}-36x+28
\displaystyle =4(3x^{2}-9x+7)
\displaystyle \text{Discriminant of }3x^{2}-9x+7\text{ is }(-9)^{2}-4(3)(7)
\displaystyle =81-84=-3<0
\displaystyle \text{Since the leading coefficient is positive, }3x^{2}-9x+7>0\text{ for all }x\in R
\displaystyle \therefore f'(x)>0\text{ for all }x\in R
\displaystyle \therefore f(x)\text{ is always increasing on }R
\\

\displaystyle \textbf{Question 11. }\text{Find the intervals in which the function } \\ f(x)=2x^{3}-15x^{2}+36x+17\text{ is increasing or decreasing.}
\displaystyle \text{Answer:}
\displaystyle f(x)=2x^{3}-15x^{2}+36x+17
\displaystyle \therefore f'(x)=6x^{2}-30x+36
\displaystyle =6(x^{2}-5x+6)
\displaystyle =6(x-2)(x-3)
\displaystyle f'(x)=0\implies x=2,3
\displaystyle \text{For }x<2,\ f'(x)>0
\displaystyle \text{For }2<x<3,\ f'(x)<0
\displaystyle \text{For }x>3,\ f'(x)>0
\displaystyle \therefore f(x)\text{ is increasing on }(-\infty,2)\cup(3,\infty)
\displaystyle \text{and decreasing on }(2,3)
\\

\displaystyle \textbf{Question 12. }\text{The volume of a cube increases at a constant rate. Prove that the increase} \\ \text{in its surface area varies inversely as the length of its side.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ be the length of the side of the cube.}
\displaystyle \text{Volume }V=x^{3}
\displaystyle \therefore \frac{dV}{dt}=3x^{2}\frac{dx}{dt}
\displaystyle \text{Given that }\frac{dV}{dt}=k,\text{ a constant.}
\displaystyle \therefore \frac{dx}{dt}=\frac{k}{3x^{2}}
\displaystyle \text{Surface area }S=6x^{2}
\displaystyle \therefore \frac{dS}{dt}=12x\frac{dx}{dt}
\displaystyle =12x\left(\frac{k}{3x^{2}}\right)
\displaystyle =\frac{4k}{x}
\displaystyle \therefore \frac{dS}{dt}\propto\frac{1}{x}
\displaystyle \text{Hence, the increase in surface area varies inversely as the length of the side of the cube.}
\\

\displaystyle \textbf{Question 13. }\text{Find the point on the curve }y^{2}=8x\text{ for which the abscissa and} \\ \text{ordinate change at the same rate.}
\displaystyle \text{Answer:}
\displaystyle y^{2}=8x
\displaystyle \text{Differentiating w.r.t. }t,
\displaystyle 2y\frac{dy}{dt}=8\frac{dx}{dt}
\displaystyle \therefore y\frac{dy}{dt}=4\frac{dx}{dt}
\displaystyle \text{Given that abscissa and ordinate change at the same rate,}
\displaystyle \frac{dx}{dt}=\frac{dy}{dt}
\displaystyle \therefore y=4
\displaystyle \text{Substituting in }y^{2}=8x,
\displaystyle 16=8x
\displaystyle \therefore x=2
\displaystyle \therefore \text{The required point is }(2,4)
\displaystyle \text{(Similarly, }y=-4\text{ gives the point }(2,-4)\text{).}
\\

\displaystyle \textbf{Question 14. }\text{Let }I\text{ be an interval disjoint from }(-1,1).\text{ Prove that function } \\ f(x)=\left(x+\frac{1}{x}\right)\text{ is strictly increasing on }I.
\displaystyle \text{Answer:}
\displaystyle f(x)=x+\frac{1}{x}
\displaystyle \therefore f'(x)=1-\frac{1}{x^{2}}
\displaystyle =\frac{x^{2}-1}{x^{2}}
\displaystyle \text{Since }I\text{ is disjoint from }(-1,1),\text{ we have }x\leq-1\text{ or }x\geq1
\displaystyle \therefore x^{2}\geq1
\displaystyle \therefore x^{2}-1\geq0
\displaystyle \therefore f'(x)\geq0
\displaystyle \therefore f(x)\text{ is strictly increasing on }I
\\

\displaystyle \textbf{Question 15. }\text{Find the values of }k\text{ for which }f(x)=kx^{3}-9kx^{2}+9x+3\text{ is increasing on }R.
\displaystyle \text{Answer:}
\displaystyle f(x)=kx^{3}-9kx^{2}+9x+3
\displaystyle \therefore f'(x)=3kx^{2}-18kx+9
\displaystyle =3(kx^{2}-6kx+3)
\displaystyle \text{For }f(x)\text{ to be increasing on }R,\ f'(x)\geq0\text{ for all }x\in R
\displaystyle \therefore kx^{2}-6kx+3\geq0\text{ for all }x\in R
\displaystyle \text{This is possible when }k>0\text{ and discriminant }\leq0
\displaystyle (-6k)^{2}-4(k)(3)\leq0
\displaystyle 36k^{2}-12k\leq0
\displaystyle 12k(3k-1)\leq0
\displaystyle \therefore 0<k\leq\frac{1}{3}
\displaystyle \text{Also, for }k=0,\ f'(x)=9>0
\displaystyle \therefore 0\leq k\leq\frac{1}{3}
\\

\displaystyle \textbf{Question 16. }\text{Find the maximum value of the function } \\ f(x)=x^{3}+2x^{2}-4x+6.
\displaystyle \text{Answer:}
\displaystyle f(x)=x^{3}+2x^{2}-4x+6
\displaystyle \therefore f'(x)=3x^{2}+4x-4
\displaystyle =3x^{2}+6x-2x-4
\displaystyle =(3x-2)(x+2)
\displaystyle f'(x)=0\implies x=\frac{2}{3},-2
\displaystyle f''(x)=6x+4
\displaystyle f''(-2)=-12+4=-8<0
\displaystyle \therefore f(x)\text{ has a maximum at }x=-2
\displaystyle \text{Maximum value }=f(-2)
\displaystyle =(-2)^{3}+2(-2)^{2}-4(-2)+6
\displaystyle =-8+8+8+6=14
\displaystyle \therefore \text{Maximum value is }14
\\

\displaystyle \textbf{Question 17. }\text{Find the equation of tangent and normal to the curve } \\ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\text{ at the point }(\sqrt{2}a,b).
\displaystyle \text{Answer:}
\displaystyle \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
\displaystyle \text{Differentiating w.r.t. }x,
\displaystyle \frac{2x}{a^{2}}-\frac{2y}{b^{2}}\frac{dy}{dx}=0
\displaystyle \therefore \frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}
\displaystyle \text{At }(\sqrt{2}a,b),
\displaystyle \frac{dy}{dx}=\frac{b^{2}(\sqrt{2}a)}{a^{2}b}=\frac{\sqrt{2}b}{a}
\displaystyle \therefore \text{Slope of tangent }m_{t}=\frac{\sqrt{2}b}{a}
\displaystyle \therefore \text{Equation of tangent is}
\displaystyle y-b=\frac{\sqrt{2}b}{a}(x-\sqrt{2}a)
\displaystyle \therefore y-b=\frac{\sqrt{2}b}{a}x-2b
\displaystyle \therefore y=\frac{\sqrt{2}b}{a}x-b
\displaystyle \therefore \sqrt{2}bx-ay-ab=0
\displaystyle \text{Slope of normal }m_{n}=-\frac{1}{m_{t}}=-\frac{a}{\sqrt{2}b}
\displaystyle \therefore \text{Equation of normal is}
\displaystyle y-b=-\frac{a}{\sqrt{2}b}(x-\sqrt{2}a)
\displaystyle \therefore \sqrt{2}b(y-b)=-a(x-\sqrt{2}a)
\displaystyle \therefore ax+\sqrt{2}by-\sqrt{2}a^{2}-\sqrt{2}b^{2}=0
\displaystyle \therefore ax+\sqrt{2}by=\sqrt{2}(a^{2}+b^{2})
\\

\displaystyle \textbf{Question 18. }\text{Find the equation of tangent to the curve }y=\sqrt{3x-2}, \\ \text{which is parallel to the line }4x-2y+5=0.
\displaystyle \text{Answer:}
\displaystyle y=\sqrt{3x-2}
\displaystyle \therefore \frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}}
\displaystyle \text{The given line is }4x-2y+5=0
\displaystyle \therefore y=2x+\frac{5}{2}
\displaystyle \therefore \text{Slope of given line }=2
\displaystyle \text{For tangent parallel to given line,}
\displaystyle \frac{3}{2\sqrt{3x-2}}=2
\displaystyle \therefore 3=4\sqrt{3x-2}
\displaystyle \therefore \sqrt{3x-2}=\frac{3}{4}
\displaystyle \therefore y=\frac{3}{4}
\displaystyle \therefore 3x-2=\frac{9}{16}
\displaystyle \therefore x=\frac{41}{48}
\displaystyle \therefore \text{Point of contact is }\left(\frac{41}{48},\frac{3}{4}\right)
\displaystyle \therefore \text{Equation of tangent is}
\displaystyle y-\frac{3}{4}=2\left(x-\frac{41}{48}\right)
\displaystyle \therefore y=2x-\frac{41}{24}+\frac{3}{4}
\displaystyle \therefore y=2x-\frac{23}{24}
\displaystyle \therefore 48x-24y-23=0
\\

\displaystyle \textbf{Question 19. }\text{Find the equation of tangent and normal to the curve } \\ x=1-\cos\theta,\ y=\theta-\sin\theta\text{ at }\theta=\frac{\pi}{4}.
\displaystyle \text{Answer:}
\displaystyle x=1-\cos\theta,\quad y=\theta-\sin\theta
\displaystyle \frac{dx}{d\theta}=\sin\theta,\quad \frac{dy}{d\theta}=1-\cos\theta
\displaystyle \therefore \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{1-\cos\theta}{\sin\theta}
\displaystyle \text{At }\theta=\frac{\pi}{4},
\displaystyle x=1-\frac{1}{\sqrt{2}},\quad y=\frac{\pi}{4}-\frac{1}{\sqrt{2}}
\displaystyle m_{t}=\frac{1-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=\sqrt{2}-1
\displaystyle \therefore \text{Equation of tangent is}
\displaystyle y-\left(\frac{\pi}{4}-\frac{1}{\sqrt{2}}\right)=(\sqrt{2}-1)\left(x-\left(1-\frac{1}{\sqrt{2}}\right)\right)
\displaystyle \text{Slope of normal }m_{n}=-\frac{1}{\sqrt{2}-1}=-(\sqrt{2}+1)
\displaystyle \therefore \text{Equation of normal is}
\displaystyle y-\left(\frac{\pi}{4}-\frac{1}{\sqrt{2}}\right)=-(\sqrt{2}+1)\left(x-\left(1-\frac{1}{\sqrt{2}}\right)\right)
\\

\displaystyle \textbf{Question 20. }\text{Show that a cylinder of a given volume which is open at the top has} \\ \text{minimum total surface area, when its height is equal to the radius of its base.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }r\text{ be the radius and }h\text{ be the height of the cylinder.}
\displaystyle \text{Volume }V=\pi r^{2}h
\displaystyle \therefore h=\frac{V}{\pi r^{2}}
\displaystyle \text{Since cylinder is open at the top, total surface area }S=\pi r^{2}+2\pi rh
\displaystyle \therefore S=\pi r^{2}+2\pi r\cdot\frac{V}{\pi r^{2}}
\displaystyle =\pi r^{2}+\frac{2V}{r}
\displaystyle \frac{dS}{dr}=2\pi r-\frac{2V}{r^{2}}
\displaystyle \text{For minimum surface area, }\frac{dS}{dr}=0
\displaystyle \therefore 2\pi r-\frac{2V}{r^{2}}=0
\displaystyle \therefore \pi r^{3}=V
\displaystyle \text{But }V=\pi r^{2}h
\displaystyle \therefore \pi r^{3}=\pi r^{2}h
\displaystyle \therefore h=r
\displaystyle \text{Also, }\frac{d^{2}S}{dr^{2}}=2\pi+\frac{4V}{r^{3}}>0
\displaystyle \therefore S\text{ is minimum when }h=r
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 21. }\text{A man }2\text{ m tall, walks at a uniform speed of }6\text{ km/h away from a} \\ \text{lamp post }6\text{ m high. Find the rate at which the length of his shadow increases.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ be the distance of the man from the lamp post and }y\text{ be the length of his shadow.}
\displaystyle \text{By similarity of triangles,}
\displaystyle \frac{6}{x+y}=\frac{2}{y}
\displaystyle \therefore 6y=2x+2y
\displaystyle \therefore 4y=2x
\displaystyle \therefore y=\frac{x}{2}
\displaystyle \text{Differentiating w.r.t. }t,
\displaystyle \frac{dy}{dt}=\frac{1}{2}\frac{dx}{dt}
\displaystyle \text{Given }\frac{dx}{dt}=6\text{ km/h}
\displaystyle \therefore \frac{dy}{dt}=3\text{ km/h}
\displaystyle \therefore \text{Length of shadow increases at the rate of }3\text{ km/h}
\\

\displaystyle \textbf{Question 22. }\text{Find the intervals in which the function }f(x)=\sin^{4}x+\cos^{4}x \\ \text{is strictly increasing or strictly decreasing.}
\displaystyle \text{Answer:}
\displaystyle f(x)=\sin^{4}x+\cos^{4}x
\displaystyle =(\sin^{2}x+\cos^{2}x)^{2}-2\sin^{2}x\cos^{2}x
\displaystyle =1-\frac{1}{2}\sin^{2}2x
\displaystyle \therefore f'(x)=-\frac{1}{2}\cdot2\sin2x\cdot2\cos2x
\displaystyle =-\sin4x
\displaystyle \text{For strictly increasing, }f'(x)>0
\displaystyle \therefore -\sin4x>0\implies \sin4x<0
\displaystyle \therefore x\in\left(\frac{(2n-1)\pi}{4},\frac{n\pi}{2}\right),\ n\in Z
\displaystyle \text{For strictly decreasing, }f'(x)<0
\displaystyle \therefore -\sin4x<0\implies \sin4x>0
\displaystyle \therefore x\in\left(\frac{n\pi}{2},\frac{(2n+1)\pi}{4}\right),\ n\in Z
\\

\displaystyle \textbf{Question 23. }\text{Find two positive numbers }x\text{ and }y\text{ such that }x+y=60 \\ \text{and }xy^{3}\text{ is maximum.}
\displaystyle \text{Answer:}
\displaystyle x+y=60
\displaystyle \therefore x=60-y
\displaystyle \text{Let }P=xy^{3}
\displaystyle \therefore P=(60-y)y^{3}
\displaystyle =60y^{3}-y^{4}
\displaystyle \frac{dP}{dy}=180y^{2}-4y^{3}
\displaystyle =4y^{2}(45-y)
\displaystyle \frac{dP}{dy}=0\implies y=45
\displaystyle \therefore x=60-45=15
\displaystyle \frac{d^{2}P}{dy^{2}}=360y-12y^{2}
\displaystyle \left.\frac{d^{2}P}{dy^{2}}\right|_{y=45}=360(45)-12(45)^{2}<0
\displaystyle \therefore xy^{3}\text{ is maximum when }x=15,\ y=45
\\

\displaystyle \textbf{Question 24. }\text{Find the area of the greatest rectangle that can be inscribed in an ellipse } \\ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
\displaystyle \text{Answer:}
\displaystyle \text{Let the vertex of the rectangle in the first quadrant be }(x,y).
\displaystyle \text{Then, area of rectangle }A=4xy
\displaystyle \text{Since }(x,y)\text{ lies on the ellipse,}
\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
\displaystyle \therefore y=b\sqrt{1-\frac{x^{2}}{a^{2}}}
\displaystyle \therefore A=4xb\sqrt{1-\frac{x^{2}}{a^{2}}}
\displaystyle \text{For maximum area, maximise }A^{2}
\displaystyle A^{2}=16b^{2}x^{2}\left(1-\frac{x^{2}}{a^{2}}\right)
\displaystyle \frac{d}{dx}(A^{2})=16b^{2}\left(2x-\frac{4x^{3}}{a^{2}}\right)
\displaystyle =32b^{2}x\left(1-\frac{2x^{2}}{a^{2}}\right)
\displaystyle \frac{d}{dx}(A^{2})=0\implies x=\frac{a}{\sqrt{2}}
\displaystyle \therefore y=b\sqrt{1-\frac{1}{2}}=\frac{b}{\sqrt{2}}
\displaystyle \therefore \text{Greatest area }=4\cdot\frac{a}{\sqrt{2}}\cdot\frac{b}{\sqrt{2}}=2ab
\\

\displaystyle \textbf{Question 25. }\text{Show that the height of a closed right circular cylinder of given} \\ \text{surface and maximum volume is equal to diameter of base.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }r\text{ be the radius and }h\text{ be the height of the cylinder.}
\displaystyle \text{Total surface area }S=2\pi r^{2}+2\pi rh
\displaystyle \therefore h=\frac{S-2\pi r^{2}}{2\pi r}
\displaystyle \text{Volume }V=\pi r^{2}h
\displaystyle \therefore V=\pi r^{2}\left(\frac{S-2\pi r^{2}}{2\pi r}\right)
\displaystyle =\frac{Sr}{2}-\pi r^{3}
\displaystyle \frac{dV}{dr}=\frac{S}{2}-3\pi r^{2}
\displaystyle \text{For maximum volume, }\frac{dV}{dr}=0
\displaystyle \therefore \frac{S}{2}=3\pi r^{2}
\displaystyle \therefore S=6\pi r^{2}
\displaystyle \text{But }S=2\pi r^{2}+2\pi rh
\displaystyle \therefore 6\pi r^{2}=2\pi r^{2}+2\pi rh
\displaystyle \therefore 4\pi r^{2}=2\pi rh
\displaystyle \therefore h=2r
\displaystyle \therefore \text{Height is equal to diameter of base.}
\displaystyle \text{Also, }\frac{d^{2}V}{dr^{2}}=-6\pi r<0
\displaystyle \therefore V\text{ is maximum. Hence proved.}
\\

\displaystyle \textbf{Question 26. }\text{Prove that radius of right circular cylinder of greatest curved surface} \\ \text{area which can be inscribed in a given cone is half of that of the cone.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }R\text{ and }H\text{ be the radius and height of the given cone.}
\displaystyle \text{Let }r\text{ and }h\text{ be the radius and height of the inscribed cylinder.}
\displaystyle \text{By similarity of triangles,}
\displaystyle \frac{R-r}{h}=\frac{R}{H}
\displaystyle \therefore h=\frac{H(R-r)}{R}
\displaystyle \text{Curved surface area of cylinder }S=2\pi rh
\displaystyle \therefore S=2\pi r\cdot\frac{H(R-r)}{R}
\displaystyle =\frac{2\pi H}{R}(Rr-r^{2})
\displaystyle \frac{dS}{dr}=\frac{2\pi H}{R}(R-2r)
\displaystyle \text{For greatest curved surface area, }\frac{dS}{dr}=0
\displaystyle \therefore R-2r=0
\displaystyle \therefore r=\frac{R}{2}
\displaystyle \text{Also, }\frac{d^{2}S}{dr^{2}}=-\frac{4\pi H}{R}<0
\displaystyle \therefore S\text{ is greatest when }r=\frac{R}{2}
\displaystyle \therefore \text{Radius of cylinder is half of radius of cone. Hence proved.}
\\

\displaystyle \textbf{Question 27. }\text{Find the intervals in which the function }f\text{ given by } \\ f(x)=4x^{3}-6x^{2}-72x+30\text{ is strictly increasing and strictly decreasing.}
\displaystyle \text{Answer:}
\displaystyle f(x)=4x^{3}-6x^{2}-72x+30
\displaystyle \therefore f'(x)=12x^{2}-12x-72
\displaystyle =12(x^{2}-x-6)
\displaystyle =12(x-3)(x+2)
\displaystyle f'(x)=0\implies x=3,-2
\displaystyle \text{For }x<-2,\ f'(x)>0
\displaystyle \text{For }-2<x<3,\ f'(x)<0
\displaystyle \text{For }x>3,\ f'(x)>0
\displaystyle \therefore f(x)\text{ is strictly increasing on }(-\infty,-2)\cup(3,\infty)
\displaystyle \text{and strictly decreasing on }(-2,3)
\\

\displaystyle \textbf{Question 28. }\text{A wire of length }36\text{ cm is cut into two pieces. One of the pieces } \\ \text{is turned in the form of a square and other in the form of an equilateral triangle.} \\ \text{Find the length of each pieces, so that the sum of the areas of the two be minimum.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ cm be used for the square and }(36-x)\text{ cm be used for the equilateral triangle.}
\displaystyle \text{Side of square }=\frac{x}{4}
\displaystyle \text{Area of square }=\left(\frac{x}{4}\right)^{2}=\frac{x^{2}}{16}
\displaystyle \text{Side of equilateral triangle }=\frac{36-x}{3}
\displaystyle \text{Area of equilateral triangle }=\frac{\sqrt{3}}{4}\left(\frac{36-x}{3}\right)^{2}
\displaystyle =\frac{\sqrt{3}}{36}(36-x)^{2}
\displaystyle \text{Let total area be }A
\displaystyle A=\frac{x^{2}}{16}+\frac{\sqrt{3}}{36}(36-x)^{2}
\displaystyle \frac{dA}{dx}=\frac{x}{8}-\frac{\sqrt{3}}{18}(36-x)
\displaystyle \text{For minimum area, }\frac{dA}{dx}=0
\displaystyle \therefore \frac{x}{8}=\frac{\sqrt{3}}{18}(36-x)
\displaystyle \therefore 18x=8\sqrt{3}(36-x)
\displaystyle \therefore x(18+8\sqrt{3})=288\sqrt{3}
\displaystyle \therefore x=\frac{288\sqrt{3}}{18+8\sqrt{3}}
\displaystyle =\frac{144\sqrt{3}}{9+4\sqrt{3}}
\displaystyle \therefore 36-x=\frac{324}{9+4\sqrt{3}}
\displaystyle \therefore \text{Length for square }=\frac{144\sqrt{3}}{9+4\sqrt{3}}\text{ cm}
\displaystyle \text{and length for equilateral triangle }=\frac{324}{9+4\sqrt{3}}\text{ cm}
\\

\displaystyle \textbf{Question 29. }\text{Prove that the area of a right angled triangle of given hypotenuse is} \\ \text{maximum, when the triangle is isosceles.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the sides containing the right angle be }x\text{ and }y,\text{ and hypotenuse be }c.
\displaystyle \therefore x^{2}+y^{2}=c^{2}
\displaystyle \text{Area }A=\frac{1}{2}xy
\displaystyle \therefore y=\sqrt{c^{2}-x^{2}}
\displaystyle A=\frac{1}{2}x\sqrt{c^{2}-x^{2}}
\displaystyle \text{For maximum area, maximise }A^{2}
\displaystyle A^{2}=\frac{1}{4}x^{2}(c^{2}-x^{2})
\displaystyle \frac{d}{dx}(A^{2})=\frac{1}{4}(2c^{2}x-4x^{3})
\displaystyle =\frac{x}{2}(c^{2}-2x^{2})
\displaystyle \frac{d}{dx}(A^{2})=0\implies c^{2}=2x^{2}
\displaystyle \therefore x^{2}=\frac{c^{2}}{2}
\displaystyle \text{Now, }y^{2}=c^{2}-x^{2}=c^{2}-\frac{c^{2}}{2}=\frac{c^{2}}{2}
\displaystyle \therefore x^{2}=y^{2}
\displaystyle \therefore x=y
\displaystyle \therefore \text{The area is maximum when the right angled triangle is isosceles.}
\displaystyle \text{Hence proved.}
\\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.