\displaystyle \text{VERY LONG ANSWER QUESTIONS - 6 Marks Each}

\displaystyle \textbf{Question 125: }\text{Rajesh wants to purchase some fruits from fruit market. }4\text{ kilograms (kgs) apples,}
\displaystyle 3\text{ kgs grapes and }2\text{ kgs oranges cost him Rs. }600,\ 2\text{ kgs apples, }4\text{ kgs grapes and }6\text{ kgs}
\displaystyle \text{oranges cost him Rs. }900,\text{ and }6\text{ kgs apples, }2\text{ kgs grapes and }3\text{ kgs oranges cost him}
\displaystyle \text{Rs. }700.\text{ Using the given information, answer the following questions.}
\displaystyle \text{(a) Express the given data in the form of a set of simultaneous equations.}
\displaystyle \text{(b) Solve the set of simultaneous equations formed in sub part (a) by matrix method.}
\displaystyle \text{(c) Hence, find how much Rajesh has to pay per kilogram (Kg) for each fruit.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the cost per kg of apples, grapes and oranges be Rs. }x,\text{ Rs. }y\text{ and Rs. }z\text{ respectively.}
\displaystyle \text{(a) According to the given information,}
\displaystyle 4x+3y+2z=600
\displaystyle 2x+4y+6z=900
\displaystyle 6x+2y+3z=700

\displaystyle \text{(b) The above equations can be written in matrix form as }AX=B,
\displaystyle \begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}600\\900\\700\end{bmatrix}
\displaystyle \text{where }A=\begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}600\\900\\700\end{bmatrix}.
\displaystyle |A|=\begin{vmatrix}4&3&2\\2&4&6\\6&2&3\end{vmatrix}
\displaystyle =4(12-12)-3(6-36)+2(4-24)
\displaystyle =0+90-40=50\ne0.
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{adj}A= \begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}
\displaystyle \therefore A^{-1}=\frac{1}{|A|}\text{adj}A
\displaystyle =\frac1{50} \begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}.
\displaystyle \text{Now, }AX=B
\displaystyle \therefore X=A^{-1}B
\displaystyle =\frac1{50} \begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix} \begin{bmatrix}600\\900\\700\end{bmatrix}
\displaystyle =\frac1{50} \begin{bmatrix}-4500+7000\\18000-14000\\-12000+9000+7000\end{bmatrix}
\displaystyle =\frac1{50}\begin{bmatrix}2500\\4000\\4000\end{bmatrix} =\begin{bmatrix}50\\80\\80\end{bmatrix}.
\displaystyle \therefore x=50,\qquad y=80,\qquad z=80.

\displaystyle \text{(c) Hence, Rajesh has to pay Rs. }50\text{ per kg for apples, Rs. }80\text{ per kg for grapes}
\displaystyle \text{and Rs. }80\text{ per kg for oranges.}
\displaystyle \\

\displaystyle \textbf{Question 126: }\text{In a raffle draw, }1000\text{ raffle tickets are sold for Rs. }1\text{ each. Each has an equal}
\displaystyle \text{chance of winning. First prize is Rs. }300,\text{ second prize is Rs. }200,\text{ and third prize is Rs. }100.
\displaystyle \text{Let }X\text{ denote the net gain from the purchase of one ticket.}
\displaystyle \text{(a) Construct the probability distribution of }X.
\displaystyle \text{(b) Find the probability of winning any money in the purchase of one ticket.}
\displaystyle \text{(c) Find the expected value of }X\text{ and interpret its meaning.}
\displaystyle \text{Answer:}
\displaystyle \text{Since each ticket costs Rs. }1,\text{ the net gain is equal to the prize money received minus Rs. }1.

\displaystyle \text{If the ticket wins the first prize, the net gain is}
\displaystyle X=300-1=299.
\displaystyle \text{If the ticket wins the second prize, the net gain is}
\displaystyle X=200-1=199.
\displaystyle \text{If the ticket wins the third prize, the net gain is}
\displaystyle X=100-1=99.
\displaystyle \text{If the ticket does not win any prize, the net gain is}
\displaystyle X=0-1=-1.

\displaystyle \text{There is one ticket for each of the three prizes and }997\text{ tickets which do not win a prize.}

\displaystyle \text{(a) Therefore, the probability distribution of }X\text{ is}
\displaystyle \begin{array}{c|cccc} X&299&199&99&-1\\ \hline P(X)&\frac{1}{1000}&\frac{1}{1000}&\frac{1}{1000}&\frac{997}{1000} \end{array}

\displaystyle \text{Also, }\frac{1}{1000}+\frac{1}{1000}+\frac{1}{1000}+\frac{997}{1000}=1,
\displaystyle \text{which verifies that the above is a probability distribution.}

\displaystyle \text{(b) Winning any money means winning any one of the three prizes.}
\displaystyle \therefore P(\text{winning any money})
\displaystyle =P(X=299)+P(X=199)+P(X=99)
\displaystyle =\frac{1}{1000}+\frac{1}{1000}+\frac{1}{1000}
\displaystyle =\frac{3}{1000}.

\displaystyle \text{Hence, the probability of winning any money is }\frac{3}{1000}.

\displaystyle \text{(c) The expected value of }X\text{ is given by}
\displaystyle E(X)=\sum xP(X=x).
\displaystyle \therefore E(X)=299\left(\frac{1}{1000}\right)+199\left(\frac{1}{1000}\right)
\displaystyle \qquad\qquad+99\left(\frac{1}{1000}\right)+(-1)\left(\frac{997}{1000}\right)
\displaystyle =\frac{299+199+99-997}{1000}
\displaystyle =\frac{-400}{1000}=-0.4.

\displaystyle \therefore E(X)=-\text{Rs. }0.40.

\displaystyle \text{Thus, in the long run, a person buying such raffle tickets can expect an average loss of}
\displaystyle \text{Rs. }0.40\text{ or }40\text{ paise per ticket.}
\displaystyle \\

\displaystyle \textbf{Question 127: }\text{Ram and Shyam play a game with a coin. Ram stakes Rs. }1.00\text{ and throws the}
\displaystyle \text{coin }4\text{ times. If he throws }4\text{ heads, he gets his stake and Rs. }3.00\text{ from Shyam. If he throws}
\displaystyle \text{only three heads and they are consecutive, he gets his stake and Rs. }2.00\text{ from Shyam. If he}
\displaystyle \text{throws only }2\text{ heads and they are consecutive, he gets his stake and Rs. }1.00\text{ from Shyam.}
\displaystyle \text{In all other cases, Shyam takes the stake money. Is this game fair? Provide reasons for your answer.}
\displaystyle \text{Answer:}
\displaystyle \text{Since the coin is tossed }4\text{ times, the total number of equally likely outcomes is}
\displaystyle 2^4=16.

\displaystyle \text{Ram stakes Rs. }1\text{ at the beginning of the game. If he wins, his stake is returned to him.}
\displaystyle \text{Therefore, his net gains in the three winning cases are Rs. }3,\text{ Rs. }2\text{ and Rs. }1\text{ respectively.}
\displaystyle \text{If he loses, Shyam takes his stake and hence Ram's net gain is Rs. }-1.

\displaystyle \text{Case 1: Ram throws }4\text{ heads.}
\displaystyle \text{The only favourable outcome is }HHHH.
\displaystyle \therefore P(\text{4 heads})=\frac{1}{16}.
\displaystyle \text{In this case, Ram's net gain is Rs. }3.

\displaystyle \text{Case 2: Ram throws exactly }3\text{ consecutive heads.}
\displaystyle \text{The favourable outcomes are }HHHT\text{ and }THHH.
\displaystyle \therefore P(\text{3 consecutive heads})=\frac{2}{16}.
\displaystyle \text{In this case, Ram's net gain is Rs. }2.

\displaystyle \text{Case 3: Ram throws exactly }2\text{ heads and they are consecutive.}
\displaystyle \text{The favourable outcomes are }HHTT,\ THHT\text{ and }TTHH.
\displaystyle \therefore P(\text{2 consecutive heads})=\frac{3}{16}.
\displaystyle \text{In this case, Ram's net gain is Rs. }1.

\displaystyle \text{The number of outcomes favourable to Ram is }1+2+3=6.
\displaystyle \text{Therefore, the remaining number of outcomes in which Ram loses is}
\displaystyle 16-6=10.
\displaystyle \therefore P(\text{Ram loses})=\frac{10}{16}.
\displaystyle \text{In this case, Ram's net gain is Rs. }-1.

\displaystyle \text{Let }X\text{ denote Ram's net gain. The probability distribution of }X\text{ is}
\displaystyle \begin{array}{c|cccc} X&3&2&1&-1\\ \hline P(X)&\frac{1}{16}&\frac{2}{16}&\frac{3}{16}&\frac{10}{16} \end{array}

\displaystyle \text{The expected net gain of Ram is}
\displaystyle E(X)=\sum xP(X=x)
\displaystyle =3\left(\frac{1}{16}\right)+2\left(\frac{2}{16}\right)+1\left(\frac{3}{16}\right)
\displaystyle \qquad\qquad+(-1)\left(\frac{10}{16}\right)
\displaystyle =\frac{3+4+3-10}{16}
\displaystyle =0.

\displaystyle \text{Thus, Ram's expected net gain is zero. Hence, Shyam's expected net gain is also zero.}
\displaystyle \text{Therefore, neither player has an advantage over the other in the long run.}
\displaystyle \therefore \text{The game is fair.}
\displaystyle \\

\displaystyle \textbf{Question 128: }\text{The fuel cost per hour for running a ship is proportional to the square of the}
\displaystyle \text{speed generated in knots. The fuel cost is Rs. }75/\text{h at }10\text{ knots and the fixed charges amount}
\displaystyle \text{to Rs. }1000/\text{h.}
\displaystyle \text{(a) Given that the fuel cost per hour is }k\text{ times the square of the speed, the ship generates}
\displaystyle \text{in km per hour, then what is the value of }k\text{?}
\displaystyle \text{(b) What will be the cost per unit distance?}
\displaystyle \text{(c) Determine the most economical speed to run the ship.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }v\text{ be the speed of the ship in knots.}

\displaystyle \text{(a) Since the fuel cost per hour is proportional to the square of the speed,}
\displaystyle \text{Fuel cost per hour}=kv^2,
\displaystyle \text{where }k\text{ is the constant of proportionality.}
\displaystyle \text{At a speed of }10\text{ knots, the fuel cost is Rs. }75\text{ per hour.}
\displaystyle \therefore 75=k(10)^2
\displaystyle \therefore 75=100k
\displaystyle \therefore k=\frac{75}{100}=\frac34.
\displaystyle \text{Hence, the value of }k\text{ is }\frac34.

\displaystyle \text{(b) The fixed charges for running the ship are Rs. }1000\text{ per hour.}
\displaystyle \text{Therefore, the total cost of running the ship per hour is}
\displaystyle C=\frac34v^2+1000.

\displaystyle \text{Since the ship travels }v\text{ units of distance in one hour, the cost per unit distance is}
\displaystyle \frac{C}{v}=\frac{\frac34v^2+1000}{v}
\displaystyle =\frac34v+\frac{1000}{v}.
\displaystyle \text{Hence, the cost per unit distance is Rs. }\left(\frac34v+\frac{1000}{v}\right).

\displaystyle \text{(c) Let }f(v)=\frac34v+\frac{1000}{v}\text{ denote the cost per unit distance.}
\displaystyle \text{For the most economical speed, the cost per unit distance must be minimum.}
\displaystyle \therefore f'(v)=\frac34-\frac{1000}{v^2}.

\displaystyle \text{For a stationary value, }f'(v)=0.
\displaystyle \therefore \frac34-\frac{1000}{v^2}=0
\displaystyle \therefore \frac34=\frac{1000}{v^2}
\displaystyle \therefore 3v^2=4000
\displaystyle \therefore v^2=\frac{4000}{3}
\displaystyle \therefore v=\sqrt{\frac{4000}{3}}=\frac{20\sqrt{30}}{3}.
\displaystyle \text{Since speed is positive, we take }v=\frac{20\sqrt{30}}{3}\text{ knots.}

\displaystyle \text{Now, }f''(v)=\frac{2000}{v^3}.
\displaystyle \text{Since }v>0,\quad f''(v)>0.
\displaystyle \therefore f(v)\text{ is minimum at }v=\frac{20\sqrt{30}}{3}.

\displaystyle \text{Hence, the most economical speed of the ship is }\frac{20\sqrt{30}}{3}\text{ knots,}
\displaystyle \text{i.e., approximately }36.51\text{ knots.}
\displaystyle \\

\displaystyle \textbf{Question 129: }\text{Evaluate: }\int\frac{\tan\frac{x}{4}}{1-\sin\frac{x}{4}}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=\frac{x}{4}.
\displaystyle \therefore x=4u\quad\text{and}\quad dx=4\,du.

\displaystyle \text{Therefore, the given integral becomes}
\displaystyle I=4\int\frac{\tan u}{1-\sin u}\,du.

\displaystyle \text{Using }\tan u=\frac{\sin u}{\cos u},\text{ we get}
\displaystyle I=4\int\frac{\sin u}{\cos u(1-\sin u)}\,du.

\displaystyle \text{Multiplying the numerator and denominator by }(1+\sin u),
\displaystyle I=4\int\frac{\sin u(1+\sin u)} {\cos u(1-\sin u)(1+\sin u)}\,du.

\displaystyle \text{Since }(1-\sin u)(1+\sin u)=1-\sin^2u=\cos^2u,
\displaystyle I=4\int\frac{\sin u(1+\sin u)}{\cos^3u}\,du.

\displaystyle =4\int\left(\frac{\sin u}{\cos^3u} +\frac{\sin^2u}{\cos^3u}\right)du.

\displaystyle \text{Now, }\frac{\sin u}{\cos^3u}=\tan u\sec^2u,
\displaystyle \text{and }\frac{\sin^2u}{\cos^3u} =\frac{1-\cos^2u}{\cos^3u}=\sec^3u-\sec u.

\displaystyle \therefore I=4\int\left(\tan u\sec^2u+\sec^3u-\sec u\right)du.

\displaystyle \text{Using the standard results}
\displaystyle \int\tan u\sec^2u\,du=\frac12\sec^2u,
\displaystyle \int\sec^3u\,du=\frac12\left[\sec u\tan u +\ln|\sec u+\tan u|\right],
\displaystyle \text{and }\int\sec u\,du=\ln|\sec u+\tan u|,
\displaystyle \text{we get}
\displaystyle I=4\left[\frac12\sec^2u+\frac12\sec u\tan u -\frac12\ln|\sec u+\tan u|\right]+C.

\displaystyle \therefore I=2\sec^2u+2\sec u\tan u -2\ln|\sec u+\tan u|+C.

\displaystyle \text{Substituting }u=\frac{x}{4},\text{ we obtain}
\displaystyle \int\frac{\tan\frac{x}{4}}{1-\sin\frac{x}{4}}\,dx
\displaystyle =2\sec^2\frac{x}{4} +2\sec\frac{x}{4}\tan\frac{x}{4} -2\ln\left|\sec\frac{x}{4}+\tan\frac{x}{4}\right|+C.
\displaystyle \\

\displaystyle \textbf{Question 130: }\text{Evaluate the following integral: } \int\left(\frac{\cot x+\cot^3x}{1+\cot^3x}\right)dx
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{\cot x+\cot^3x}{1+\cot^3x}\,dx.

\displaystyle \text{Taking }\cot x\text{ common in the numerator,}
\displaystyle I=\int\frac{\cot x(1+\cot^2x)}{1+\cot^3x}\,dx.

\displaystyle \text{Using }1+\cot^2x=\mathrm{cosec}^2x,
\displaystyle I=\int\frac{\cot x\,\mathrm{cosec}^2x}{1+\cot^3x}\,dx.

\displaystyle \text{Let }t=\cot x.
\displaystyle \therefore dt=-\mathrm{cosec}^2x\,dx.

\displaystyle \text{Hence,}
\displaystyle I=-\int\frac{t}{1+t^3}\,dt.

\displaystyle \text{Now, }1+t^3=(t+1)(t^2-t+1).
\displaystyle \therefore \frac{t}{1+t^3} =\frac{t}{(t+1)(t^2-t+1)}.

\displaystyle \text{Resolving into partial fractions, let}
\displaystyle \frac{t}{(t+1)(t^2-t+1)} =\frac{A}{t+1}+\frac{Bt+C}{t^2-t+1}.

\displaystyle \therefore t=A(t^2-t+1)+(Bt+C)(t+1).

\displaystyle \text{Comparing coefficients of like powers of }t,\text{ we get}
\displaystyle A+B=0,\qquad -A+B+C=1,\qquad A+C=0.

\displaystyle \text{Solving these equations,}
\displaystyle A=-\frac13,\qquad B=\frac13,\qquad C=\frac13.

\displaystyle \therefore \frac{t}{1+t^3} =-\frac{1}{3(t+1)}+\frac{t+1}{3(t^2-t+1)}.

\displaystyle \text{Thus,}
\displaystyle I=\frac13\int\frac{dt}{t+1} -\frac13\int\frac{t+1}{t^2-t+1}\,dt.

\displaystyle \text{To evaluate the second integral, write}
\displaystyle t+1=\frac12(2t-1)+\frac32.

\displaystyle \therefore I=\frac13\int\frac{dt}{t+1} -\frac16\int\frac{2t-1}{t^2-t+1}\,dt -\frac12\int\frac{dt}{t^2-t+1}.

\displaystyle \text{Now, }t^2-t+1=\left(t-\frac12\right)^2+\frac34.

\displaystyle \therefore \int\frac{dt}{t^2-t+1} =\frac{2}{\sqrt3}\tan^{-1}\left(\frac{2t-1}{\sqrt3}\right).

\displaystyle \text{Therefore,}
\displaystyle I=\frac13\ln|t+1| -\frac16\ln(t^2-t+1) -\frac1{\sqrt3}\tan^{-1}\left(\frac{2t-1}{\sqrt3}\right)+C.

\displaystyle \text{Substituting }t=\cot x,\text{ we obtain}
\displaystyle I=\frac13\ln|\cot x+1| -\frac16\ln(\cot^2x-\cot x+1)
\displaystyle \qquad-\frac1{\sqrt3} \tan^{-1}\left(\frac{2\cot x-1}{\sqrt3}\right)+C.

\displaystyle \therefore \int\left(\frac{\cot x+\cot^3x}{1+\cot^3x}\right)dx
\displaystyle =\frac13\ln|\cot x+1| -\frac16\ln(\cot^2x-\cot x+1) -\frac1{\sqrt3}\tan^{-1}\left(\frac{2\cot x-1}{\sqrt3}\right)+C.
\displaystyle \\

\displaystyle \textbf{Question 131: }\text{Let }f(x)= \begin{cases} a+\sin^{-1}(x+b),&x\geq1\\ x,&x<1 \end{cases} ,\ f'(1)\text{ exists.}
\displaystyle \text{(a) The statement ``}f(x)\text{ is continuous at }x=1\text{'' is true. Justify.}
\displaystyle \text{(b) Hence, find a relation between }a\text{ and }b.
\displaystyle \text{(c) Find }f'(x).
\displaystyle \text{(d) Hence, find the values of }a\text{ and }b.
\displaystyle \text{Answer:}

\displaystyle \text{(a) It is given that }f'(1)\text{ exists.}
\displaystyle \text{We know that if a function is differentiable at a point, then it is continuous at that point.}
\displaystyle \therefore f(x)\text{ is continuous at }x=1.

\displaystyle \text{Hence, }\lim_{x\to1^-}f(x)=f(1)=\lim_{x\to1^+}f(x).

\displaystyle \text{(b) For }x<1,\quad f(x)=x.
\displaystyle \therefore \lim_{x\to1^-}f(x)=\lim_{x\to1^-}x=1.

\displaystyle \text{Also, since }x\geq1,\quad f(1)=a+\sin^{-1}(1+b).

\displaystyle \text{Since }f(x)\text{ is continuous at }x=1,
\displaystyle f(1)=\lim_{x\to1^-}f(x).
\displaystyle \therefore a+\sin^{-1}(1+b)=1.

\displaystyle \text{Thus, the required relation between }a\text{ and }b\text{ is}
\displaystyle a+\sin^{-1}(1+b)=1.

\displaystyle \text{(c) For }x<1,\quad f(x)=x.
\displaystyle \therefore f'(x)=1.

\displaystyle \text{For }x>1,\quad f(x)=a+\sin^{-1}(x+b).
\displaystyle \therefore f'(x)=\frac{1}{\sqrt{1-(x+b)^2}}.

\displaystyle \text{Since }f'(1)\text{ exists, the left-hand and right-hand derivatives at }x=1\text{ are equal.}

\displaystyle \text{The left-hand derivative at }x=1\text{ is}
\displaystyle f'_-(1)=\lim_{h\to0^-}\frac{f(1+h)-f(1)}{h}.

\displaystyle \text{Since }f(x)=x\text{ for }x<1\text{ and }f(1)=1,
\displaystyle f'_-(1)=\lim_{h\to0^-}\frac{(1+h)-1}{h}=1.

\displaystyle \therefore f'(1)=1.

\displaystyle \text{Thus,}
\displaystyle f'(x)= \begin{cases} \dfrac{1}{\sqrt{1-(x+b)^2}},&x>1\\ 1,&x\leq1 \end{cases}

\displaystyle \text{(d) Since }f'(1)\text{ exists, the right-hand derivative must also be equal to }1.
\displaystyle \therefore \frac{1}{\sqrt{1-(1+b)^2}}=1.

\displaystyle \therefore \sqrt{1-(1+b)^2}=1.
\displaystyle \therefore 1-(1+b)^2=1.
\displaystyle \therefore (1+b)^2=0.
\displaystyle \therefore b=-1.

\displaystyle \text{Using the relation obtained in part (b),}
\displaystyle a+\sin^{-1}(1+b)=1.

\displaystyle \text{Substituting }b=-1,\text{ we get}
\displaystyle a+\sin^{-1}(0)=1.
\displaystyle \therefore a=1.

\displaystyle \text{Hence, }a=1\text{ and }b=-1.
\displaystyle \\

\displaystyle \textbf{Question 132: }\text{Let }f(x)=\cos x+\sqrt{3}\sin x,\quad 0\leq x\leq2\pi.
\displaystyle \text{The following diagram shows the graph of }f. \displaystyle \text{The }y\text{-intercept is at }(0,1)\text{ and intersects }x\text{-axis at }C\text{ and }D.\text{ There is a minimum}
\displaystyle \text{point at }A(p,q)\text{ and a maximum point at }B.
\displaystyle \text{Based on the above information, answer the questions that follow.}
\displaystyle \text{(a) Write }f'(x)\text{ in the form of }\lambda\cos(x+\mu).
\displaystyle \text{(b) Find the value of }q.
\displaystyle \text{(c) Find the coordinate of the point }B.
\displaystyle \text{(d) Find the interval }f(x)\text{ is decreasing.}
\displaystyle \text{(e) Find the slope of the tangent to the curve at }D.
\displaystyle \text{(f) Find the slope of the normal to the curve at }C.
\displaystyle \text{Answer:}

\displaystyle \text{Given, }f(x)=\cos x+\sqrt{3}\sin x.

\displaystyle \text{(a) Differentiating }f(x)\text{ with respect to }x,
\displaystyle f'(x)=-\sin x+\sqrt{3}\cos x.

\displaystyle \text{We express }f'(x)\text{ in the form }\lambda\cos(x+\mu).
\displaystyle \lambda\cos(x+\mu)=\lambda(\cos x\cos\mu-\sin x\sin\mu).

\displaystyle \text{Comparing this with }\sqrt{3}\cos x-\sin x,\text{ we get}
\displaystyle \lambda\cos\mu=\sqrt{3},\qquad \lambda\sin\mu=1.

\displaystyle \text{Squaring and adding,}
\displaystyle \lambda^2(\cos^2\mu+\sin^2\mu)=3+1=4.
\displaystyle \therefore \lambda=2.

\displaystyle \text{Also, }\tan\mu=\frac{1}{\sqrt{3}}.
\displaystyle \therefore \mu=\frac{\pi}{6}.

\displaystyle \therefore f'(x)=2\cos\left(x+\frac{\pi}{6}\right).

\displaystyle \text{Hence, }\lambda=2\text{ and }\mu=\frac{\pi}{6}.

\displaystyle \text{(b) To find the minimum value, first write }f(x)\text{ in a suitable form.}
\displaystyle f(x)=\cos x+\sqrt{3}\sin x
\displaystyle =2\left(\frac12\cos x+\frac{\sqrt{3}}2\sin x\right)
\displaystyle =2\cos\left(x-\frac{\pi}{3}\right).

\displaystyle \text{Since }-1\leq\cos\left(x-\frac{\pi}{3}\right)\leq1,
\displaystyle -2\leq f(x)\leq2.

\displaystyle \therefore \text{the minimum value of }f(x)\text{ is }-2.
\displaystyle \text{Since }A(p,q)\text{ is the minimum point,}
\displaystyle \therefore q=-2.

\displaystyle \text{(c) At the maximum point }B,\text{ the maximum value of }f(x)\text{ is }2.
\displaystyle 2\cos\left(x-\frac{\pi}{3}\right)=2.
\displaystyle \therefore \cos\left(x-\frac{\pi}{3}\right)=1.

\displaystyle \text{For }0\leq x\leq2\pi,\text{ this gives}
\displaystyle x-\frac{\pi}{3}=0.
\displaystyle \therefore x=\frac{\pi}{3}.

\displaystyle \text{Thus, the coordinate of the maximum point is}
\displaystyle B\left(\frac{\pi}{3},2\right).

\displaystyle \text{(d) The function }f(x)\text{ is decreasing when }f'(x)<0.
\displaystyle \therefore 2\cos\left(x+\frac{\pi}{6}\right)<0.
\displaystyle \therefore \cos\left(x+\frac{\pi}{6}\right)<0.

\displaystyle \text{Now, }\cos\theta<0\text{ for }\frac{\pi}{2}<\theta<\frac{3\pi}{2}.
\displaystyle \therefore \frac{\pi}{2}<x+\frac{\pi}{6}<\frac{3\pi}{2}.

\displaystyle \text{Subtracting }\frac{\pi}{6}\text{ throughout,}
\displaystyle \frac{\pi}{3}<x<\frac{4\pi}{3}.

\displaystyle \therefore f(x)\text{ is decreasing in the interval }\left(\frac{\pi}{3},\frac{4\pi}{3}\right).

\displaystyle \text{(e) The points }C\text{ and }D\text{ lie on the }x\text{-axis. Therefore, }f(x)=0.
\displaystyle 2\cos\left(x-\frac{\pi}{3}\right)=0.
\displaystyle \therefore \cos\left(x-\frac{\pi}{3}\right)=0.

\displaystyle \therefore x-\frac{\pi}{3}=\frac{\pi}{2},\ \frac{3\pi}{2}.
\displaystyle \therefore x=\frac{5\pi}{6},\ \frac{11\pi}{6}.

\displaystyle \text{From the graph, }C\text{ is the first intercept and }D\text{ is the second intercept.}
\displaystyle \therefore C\left(\frac{5\pi}{6},0\right),\qquad D\left(\frac{11\pi}{6},0\right).

\displaystyle \text{The slope of the tangent at }D\text{ is }f'\left(\frac{11\pi}{6}\right).
\displaystyle f'\left(\frac{11\pi}{6}\right) =2\cos\left(\frac{11\pi}{6}+\frac{\pi}{6}\right)
\displaystyle =2\cos(2\pi)=2.

\displaystyle \therefore \text{the slope of the tangent to the curve at }D\text{ is }2.

\displaystyle \text{(f) The slope of the tangent at }C\text{ is}
\displaystyle f'\left(\frac{5\pi}{6}\right) =2\cos\left(\frac{5\pi}{6}+\frac{\pi}{6}\right)
\displaystyle =2\cos\pi=-2.

\displaystyle \text{The slope of the normal is the negative reciprocal of the slope of the tangent.}
\displaystyle \therefore m_{\text{normal}}=-\frac{1}{-2}=\frac12.

\displaystyle \therefore \text{the slope of the normal to the curve at }C\text{ is }\frac12.
\displaystyle \\

\displaystyle \textbf{Question 133: }\text{Consider the following differential equation and answer the questions.}
\displaystyle \left[x\cos\left(\frac{y}{x}\right)+y\sin\left(\frac{y}{x}\right)\right]y\,dx
\displaystyle -\left[y\sin\left(\frac{y}{x}\right)-x\cos\left(\frac{y}{x}\right)\right]x\,dy=0.
\displaystyle \text{(a) Transform the above equation in the form }\frac{dy}{dx}=f\left(\frac{y}{x}\right).
\displaystyle \text{(b) Use appropriate substitution to transform it into variable separable form.}
\displaystyle \text{(c) Write the differential equation in variable separable form.}
\displaystyle \text{(d) Prove that the solution of the differential equation is }\sec\left(\frac{y}{x}\right)=xyc.
\displaystyle \text{(e) Find the solution if }x=1,\ y=1.
\displaystyle \text{Answer:}

\displaystyle \text{(a) Given differential equation is}
\displaystyle \left[x\cos\left(\frac{y}{x}\right)+y\sin\left(\frac{y}{x}\right)\right]y\,dx
\displaystyle -\left[y\sin\left(\frac{y}{x}\right)-x\cos\left(\frac{y}{x}\right)\right]x\,dy=0.

\displaystyle \text{Transposing the second term to the right-hand side, we get}
\displaystyle \left[y\sin\left(\frac{y}{x}\right)-x\cos\left(\frac{y}{x}\right)\right]x\,dy
\displaystyle =\left[x\cos\left(\frac{y}{x}\right)+y\sin\left(\frac{y}{x}\right)\right]y\,dx.

\displaystyle \text{Dividing both sides by }x\left[y\sin\left(\frac{y}{x}\right)-x\cos\left(\frac{y}{x}\right)\right]dx,
\displaystyle \frac{dy}{dx}= \frac{y\left[x\cos\left(\frac{y}{x}\right)+y\sin\left(\frac{y}{x}\right)\right]} {x\left[y\sin\left(\frac{y}{x}\right)-x\cos\left(\frac{y}{x}\right)\right]}.

\displaystyle \text{Dividing the numerator and denominator appropriately by }x,\text{ we obtain}
\displaystyle \frac{dy}{dx}= \frac{\frac{y}{x}\left[\cos\left(\frac{y}{x}\right)+\frac{y}{x}\sin\left(\frac{y}{x}\right)\right]} {\frac{y}{x}\sin\left(\frac{y}{x}\right)-\cos\left(\frac{y}{x}\right)}.

\displaystyle \text{Thus, }\frac{dy}{dx}\text{ is a function of }\frac{y}{x}\text{ only, and hence the equation is homogeneous.}

\displaystyle \text{(b) Put }\frac{y}{x}=v,\text{ so that }y=vx.
\displaystyle \text{Differentiating with respect to }x,
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}.

\displaystyle \text{Substituting }\frac{y}{x}=v\text{ and }\frac{dy}{dx}=v+x\frac{dv}{dx}\text{ in the equation,}
\displaystyle v+x\frac{dv}{dx}=\frac{v(\cos v+v\sin v)}{v\sin v-\cos v}.

\displaystyle \therefore x\frac{dv}{dx} =\frac{v(\cos v+v\sin v)-v(v\sin v-\cos v)} {v\sin v-\cos v}.

\displaystyle \text{Simplifying the numerator,}
\displaystyle x\frac{dv}{dx} =\frac{v\cos v+v^2\sin v-v^2\sin v+v\cos v} {v\sin v-\cos v}.

\displaystyle \therefore x\frac{dv}{dx} =\frac{2v\cos v}{v\sin v-\cos v}.

\displaystyle \text{(c) Separating the variables, we get}
\displaystyle \frac{v\sin v-\cos v}{2v\cos v}\,dv=\frac{dx}{x}.

\displaystyle \text{Equivalently,}
\displaystyle \frac12\left(\tan v-\frac1v\right)dv=\frac{dx}{x}.

\displaystyle \text{This is the required variable separable form.}

\displaystyle \text{(d) Integrating both sides,}
\displaystyle \frac12\int\tan v\,dv-\frac12\int\frac{dv}{v} =\int\frac{dx}{x}.

\displaystyle \text{Using }\int\tan v\,dv=-\ln|\cos v|,
\displaystyle -\frac12\ln|\cos v|-\frac12\ln|v|=\ln|x|+C.

\displaystyle \text{Multiplying throughout by }-2,
\displaystyle \ln|\cos v|+\ln|v|=-2\ln|x|+C_1.

\displaystyle \therefore \ln|v\cos v|=\ln\left|\frac{C_2}{x^2}\right|.

\displaystyle \therefore v\cos v=\frac{C_2}{x^2}.

\displaystyle \text{Since }v=\frac{y}{x},
\displaystyle \frac{y}{x}\cos\left(\frac{y}{x}\right)=\frac{C_2}{x^2}.

\displaystyle \therefore xy\cos\left(\frac{y}{x}\right)=C_2.

\displaystyle \text{Taking the reciprocal and absorbing the reciprocal constant into }c,
\displaystyle \sec\left(\frac{y}{x}\right)=xyc.

\displaystyle \text{Hence, the required solution is }\sec\left(\frac{y}{x}\right)=xyc.

\displaystyle \text{(e) It is given that }x=1\text{ and }y=1.
\displaystyle \text{Substituting these values in }\sec\left(\frac{y}{x}\right)=xyc,
\displaystyle \sec(1)=(1)(1)c.

\displaystyle \therefore c=\sec(1).

\displaystyle \text{Substituting the value of }c\text{ in the general solution, we get}
\displaystyle \sec\left(\frac{y}{x}\right)=xy\sec(1).

\displaystyle \text{Hence, the required particular solution is}
\displaystyle \sec\left(\frac{y}{x}\right)=xy\sec(1).
\displaystyle \\

\displaystyle \textbf{Question 134: }\text{Sonia watches a painting which has its bottom edge }2\text{ metres (m) above eye level and}
\displaystyle \text{its top edge is }3\text{ m above eye level as shown in the diagram.} \displaystyle \text{Based on the above information answer the questions that follow.}
\displaystyle \text{(a) Given }\alpha\text{ and }\theta\text{ as shown in the diagram, find }\tan\alpha\text{ and }\tan(\alpha+\theta).
\displaystyle \text{(b) Find }\theta\text{ in terms of }x\text{ only.}
\displaystyle \text{(c) Find }\frac{d\theta}{dx}.
\displaystyle \text{(d) Find }x\text{ so that }\frac{d\theta}{dx}=0.
\displaystyle \text{(e) Use }1^{\text{st}}\text{ derivative test, find the distance Sonia should stand from the wall to}
\displaystyle \text{maximize her viewing angle of the painting.}
\displaystyle \text{Answer:}

\displaystyle \text{Let Sonia be at a horizontal distance }x\text{ m from the wall.}
\displaystyle \text{The bottom and top edges of the painting are }2\text{ m and }3\text{ m above her eye level respectively.}

\displaystyle \text{(a) From the right-angled triangle formed by the line of sight to the bottom edge,}
\displaystyle \tan\alpha=\frac{\text{perpendicular}}{\text{base}}=\frac{2}{x}.

\displaystyle \text{Similarly, the angle of elevation of the top edge is }(\alpha+\theta).
\displaystyle \therefore \tan(\alpha+\theta)=\frac{3}{x}.

\displaystyle \text{Hence, }\tan\alpha=\frac{2}{x}\qquad\text{and}\qquad\tan(\alpha+\theta)=\frac{3}{x}.

\displaystyle \text{(b) From }\tan\alpha=\frac{2}{x},\text{ we get}
\displaystyle \alpha=\tan^{-1}\left(\frac{2}{x}\right).

\displaystyle \text{Also, from }\tan(\alpha+\theta)=\frac{3}{x},\text{ we get}
\displaystyle \alpha+\theta=\tan^{-1}\left(\frac{3}{x}\right).

\displaystyle \text{Subtracting the first equation from the second,}
\displaystyle \theta=\tan^{-1}\left(\frac{3}{x}\right)-\tan^{-1}\left(\frac{2}{x}\right).

\displaystyle \text{Thus, }\theta=\tan^{-1}\left(\frac{3}{x}\right)-\tan^{-1}\left(\frac{2}{x}\right).

\displaystyle \text{(c) Differentiating }\theta\text{ with respect to }x,
\displaystyle \frac{d\theta}{dx}=\frac{d}{dx}\left[\tan^{-1}\left(\frac{3}{x}\right)\right]
\displaystyle \qquad-\frac{d}{dx}\left[\tan^{-1}\left(\frac{2}{x}\right)\right].

\displaystyle \text{Using }\frac{d}{dx}(\tan^{-1}u)=\frac{1}{1+u^2}\frac{du}{dx},\text{ we get}
\displaystyle \frac{d\theta}{dx} =\frac{1}{1+\frac{9}{x^2}}\left(-\frac{3}{x^2}\right) -\frac{1}{1+\frac{4}{x^2}}\left(-\frac{2}{x^2}\right).

\displaystyle \therefore \frac{d\theta}{dx} =-\frac{3}{x^2+9}+\frac{2}{x^2+4}.

\displaystyle \text{Taking the LCM and simplifying,}
\displaystyle \frac{d\theta}{dx} =\frac{2(x^2+9)-3(x^2+4)}{(x^2+4)(x^2+9)}.

\displaystyle \therefore \frac{d\theta}{dx} =\frac{2x^2+18-3x^2-12}{(x^2+4)(x^2+9)}.

\displaystyle \therefore \frac{d\theta}{dx} =\frac{6-x^2}{(x^2+4)(x^2+9)}.

\displaystyle \text{(d) For a stationary value of }\theta,
\displaystyle \frac{d\theta}{dx}=0.

\displaystyle \therefore \frac{6-x^2}{(x^2+4)(x^2+9)}=0.

\displaystyle \text{Since }(x^2+4)(x^2+9)\ne0,\text{ we must have}
\displaystyle 6-x^2=0.

\displaystyle \therefore x^2=6.
\displaystyle \therefore x=\sqrt{6},\text{ since }x>0.

\displaystyle \text{(e) We now apply the first derivative test to determine whether }\theta\text{ is maximum.}

\displaystyle \frac{d\theta}{dx}=\frac{6-x^2}{(x^2+4)(x^2+9)}.

\displaystyle \text{For }0<x<\sqrt{6},\quad 6-x^2>0.
\displaystyle \text{Also, }(x^2+4)(x^2+9)>0.
\displaystyle \therefore \frac{d\theta}{dx}>0.

\displaystyle \text{Hence, }\theta\text{ is increasing for }0<x<\sqrt{6}.

\displaystyle \text{For }x>\sqrt{6},\quad 6-x^2<0,
\displaystyle \text{while }(x^2+4)(x^2+9)>0.
\displaystyle \therefore \frac{d\theta}{dx}<0.

\displaystyle \text{Hence, }\theta\text{ is decreasing for }x>\sqrt{6}.

\displaystyle \text{Therefore, }\frac{d\theta}{dx}\text{ changes from positive to negative at }x=\sqrt{6}.
\displaystyle \therefore \theta\text{ is maximum when }x=\sqrt{6}.

\displaystyle \text{Hence, Sonia should stand }\sqrt{6}\text{ m from the wall to maximize the viewing angle.}
\displaystyle \text{That is, the required distance is approximately }2.45\text{ m.}
\displaystyle \\

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

Subscribe to get the latest posts sent to your email.