\displaystyle \text{DETERMINANTS}

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{Manjit wants to donate a rectangular plot of land for a school in his village. When asked to give}
\displaystyle \text{the dimensions of the plot, he said that if its length is decreased by }50\text{ m and its breadth}
\displaystyle \text{is increased by }50\text{ m, its area remains the same. If its length is decreased by }10\text{ m and}
\displaystyle \text{its breadth is decreased by }20\text{ m, its area decreases by }5300\text{ m}^2.
\displaystyle \text{Let the length and breadth of the rectangular plot be }x\text{ m and }y\text{ m respectively.} \displaystyle \text{Based on the information given above, answer the following questions.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{The equations in terms of }x\text{ and }y\text{ are}
\displaystyle \text{(a) }x-y=50,\quad 2x+y=550
\displaystyle \text{(b) }x-y=50,\quad 2x+y=550
\displaystyle \text{(c) }x+y=50,\quad 2x+y=550
\displaystyle \text{(d) }x+y=50,\quad 2x+y=550
\displaystyle \text{Answer:}
\displaystyle \text{Original area}=xy.
\displaystyle \text{From the first condition,}
\displaystyle (x-50)(y+50)=xy.
\displaystyle xy+50x-50y-2500=xy.
\displaystyle 50x-50y=2500.
\displaystyle \therefore x-y=50.\qquad\ldots(1)
\displaystyle \text{From the second condition,}
\displaystyle (x-10)(y-20)=xy-5300.
\displaystyle xy-20x-10y+200=xy-5300.
\displaystyle 20x+10y=5500.
\displaystyle \therefore 2x+y=550.\qquad\ldots(2)
\displaystyle \therefore \text{The required equations are }x-y=50\text{ and }2x+y=550.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Which of the following matrix equations represents the given information?}
\displaystyle \text{(a) }\begin{bmatrix}1&-1\\2&1\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}50\\550\end{bmatrix}
\displaystyle \text{(b) }\begin{bmatrix}1&1\\2&1\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}50\\550\end{bmatrix}
\displaystyle \text{(c) }\begin{bmatrix}1&1\\2&-1\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}50\\550\end{bmatrix}
\displaystyle \text{(d) }\begin{bmatrix}1&1\\2&1\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-50\\-550\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle x-y=50,\qquad 2x+y=550.
\displaystyle \text{Writing these equations in matrix form,}
\displaystyle \begin{bmatrix}1&-1\\2&1\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}50\\550\end{bmatrix}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The value of }x,\text{ the length of the rectangular field, is}
\displaystyle \text{(a) }150\text{ m}\qquad\text{(b) }400\text{ m}\qquad\text{(c) }200\text{ m}\qquad\text{(d) }320\text{ m}
\displaystyle \text{Answer:}
\displaystyle x-y=50,\qquad 2x+y=550.
\displaystyle D=\begin{vmatrix}1&-1\\2&1\end{vmatrix}=1(1)-(-1)(2)=3.
\displaystyle D_x=\begin{vmatrix}50&-1\\550&1\end{vmatrix}=50(1)-(-1)(550)=600.
\displaystyle \text{By Cramer's rule,}
\displaystyle x=\frac{D_x}{D}=\frac{600}{3}=200.
\displaystyle \therefore \text{The length of the rectangular field is }200\text{ m.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The value of }y,\text{ the breadth of the rectangular field, is}
\displaystyle \text{(a) }150\text{ m}\qquad\text{(b) }200\text{ m}\qquad\text{(c) }430\text{ m}\qquad\text{(d) }350\text{ m}
\displaystyle \text{Answer:}
\displaystyle D=\begin{vmatrix}1&-1\\2&1\end{vmatrix}=3.
\displaystyle D_y=\begin{vmatrix}1&50\\2&550\end{vmatrix}=1(550)-50(2)=450.
\displaystyle \text{By Cramer's rule,}
\displaystyle y=\frac{D_y}{D}=\frac{450}{3}=150.
\displaystyle \therefore \text{The breadth of the rectangular field is }150\text{ m.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{What is the area of the rectangular field?}
\displaystyle \text{(a) }60000\text{ m}^2\qquad\text{(b) }30000\text{ m}^2\qquad\text{(c) }30000\text{ m}\qquad\text{(d) }3000\text{ m}
\displaystyle \text{Answer:}
\displaystyle x=200\text{ m}\quad\text{and}\quad y=150\text{ m}.
\displaystyle \text{Area of the rectangular field}=xy.
\displaystyle =200\times150=30000\text{ m}^2.
\displaystyle \therefore \text{The area of the rectangular field is }30000\text{ m}^2.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

 

\displaystyle \text{CONTINUITY AND DIFFERENTIABILITY}

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{The relation between the height of a plant }y\text{ (in cm) and its exposure to sunlight is governed}
\displaystyle \text{by the equation }y=4x-\frac{1}{2}x^2,\text{ where }x\text{ is the number of days the plant is exposed to sunlight.} \displaystyle \text{Based on the above information, answer the following questions.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{The rate of growth of the plant with respect to sunlight is}
\displaystyle \text{(a) }4x-\frac{1}{2}x^2\qquad\text{(b) }4-x\qquad\text{(c) }x-4\qquad\text{(d) }x-\frac{1}{2}x^2
\displaystyle \text{Answer:}
\displaystyle y=4x-\frac{1}{2}x^2.
\displaystyle \text{Differentiating with respect to }x,
\displaystyle \frac{dy}{dx}=4-\frac{1}{2}(2x).
\displaystyle \therefore \frac{dy}{dx}=4-x.
\displaystyle \therefore \text{The rate of growth of the plant is }4-x.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{What is the number of days it will take for the plant to grow to the maximum height?}
\displaystyle \text{(a) }4\qquad\text{(b) }6\qquad\text{(c) }7\qquad\text{(d) }10
\displaystyle \text{Answer:}
\displaystyle \text{For maximum height, }\frac{dy}{dx}=0.
\displaystyle 4-x=0.
\displaystyle \therefore x=4.
\displaystyle \frac{d^2y}{dx^2}=-1<0.
\displaystyle \therefore y\text{ is maximum at }x=4.
\displaystyle \therefore \text{The plant reaches its maximum height after }4\text{ days.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{What is the maximum height of the plant?}
\displaystyle \text{(a) }12\text{ cm}\qquad\text{(b) }10\text{ cm}\qquad\text{(c) }8\text{ cm}\qquad\text{(d) }6\text{ cm}
\displaystyle \text{Answer:}
\displaystyle \text{From Question 2, the height is maximum when }x=4.
\displaystyle y=4x-\frac{1}{2}x^2.
\displaystyle \text{At }x=4,
\displaystyle y=4(4)-\frac{1}{2}(4)^2.
\displaystyle =16-8=8\text{ cm}.
\displaystyle \therefore \text{The maximum height of the plant is }8\text{ cm.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{What will be the height of the plant after }2\text{ days?}
\displaystyle \text{(a) }4\text{ cm}\qquad\text{(b) }6\text{ cm}\qquad\text{(c) }8\text{ cm}\qquad\text{(d) }10\text{ cm}
\displaystyle \text{Answer:}
\displaystyle y=4x-\frac{1}{2}x^2.
\displaystyle \text{At }x=2,
\displaystyle y=4(2)-\frac{1}{2}(2)^2.
\displaystyle =8-2=6\text{ cm}.
\displaystyle \therefore \text{The height of the plant after }2\text{ days is }6\text{ cm.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If the height of the plant is }\frac{7}{2}\text{ cm, the number of days it has been exposed}
\displaystyle \text{to sunlight is}
\displaystyle \text{(a) }2\qquad\text{(b) }3\qquad\text{(c) }4\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle y=4x-\frac{1}{2}x^2.
\displaystyle \text{Given }y=\frac{7}{2},
\displaystyle \frac{7}{2}=4x-\frac{1}{2}x^2.
\displaystyle 7=8x-x^2.
\displaystyle x^2-8x+7=0.
\displaystyle (x-1)(x-7)=0.
\displaystyle \therefore x=1\text{ or }x=7.
\displaystyle \text{Among the given options, }x=1.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 2}

\displaystyle P(x)=-5x^2+125x+37500\text{ is the total profit function of a company, where }x\text{ is the}
\displaystyle \text{production of the company. Based on the above information, answer the following questions.} \displaystyle \\

\displaystyle \textbf{Question 1: }\text{What will be the production when the profit is maximum?}
\displaystyle \text{(a) }37500\qquad\text{(b) }12.5\qquad\text{(c) }-12.5\qquad\text{(d) }-37500
\displaystyle \text{Answer:}
\displaystyle P(x)=-5x^2+125x+37500.
\displaystyle P'(x)=-10x+125.
\displaystyle \text{For maximum profit, }P'(x)=0.
\displaystyle -10x+125=0.
\displaystyle \therefore x=12.5.
\displaystyle P''(x)=-10<0.
\displaystyle \therefore P(x)\text{ is maximum at }x=12.5.
\displaystyle \therefore \text{The production is }12.5\text{ units.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{What will be the maximum profit?}
\displaystyle \text{(a) Rs. }38,28,125\qquad\text{(b) Rs. }38,281.25\qquad\text{(c) Rs. }39,000\qquad\text{(d) None}
\displaystyle \text{Answer:}
\displaystyle \text{From Question 1, the profit is maximum at }x=12.5.
\displaystyle P(12.5)=-5(12.5)^2+125(12.5)+37500.
\displaystyle =-781.25+1562.5+37500.
\displaystyle =38281.25.
\displaystyle \therefore \text{The maximum profit is Rs. }38,281.25.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In which interval is the profit strictly increasing?}
\displaystyle \text{(a) }(12.5,\infty)\qquad\text{(b) For all real numbers}
\displaystyle \text{(c) For all positive real numbers}\qquad\text{(d) }(0,12.5)
\displaystyle \text{Answer:}
\displaystyle P'(x)=-10x+125.
\displaystyle \text{For }P(x)\text{ to be strictly increasing, }P'(x)>0.
\displaystyle -10x+125>0.
\displaystyle x<12.5.
\displaystyle \text{Since production cannot be negative, }x>0.
\displaystyle \therefore 0<x<12.5.
\displaystyle \therefore \text{The profit is strictly increasing on }(0,12.5).
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{When the production is }2\text{ units, what will be the profit of the company?}
\displaystyle \text{(a) Rs. }37,500\qquad\text{(b) Rs. }37,730\qquad\text{(c) Rs. }37,770\qquad\text{(d) None}
\displaystyle \text{Answer:}
\displaystyle P(x)=-5x^2+125x+37500.
\displaystyle \text{At }x=2,
\displaystyle P(2)=-5(2)^2+125(2)+37500.
\displaystyle =-20+250+37500.
\displaystyle =37730.
\displaystyle \therefore \text{The profit of the company is Rs. }37,730.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{What will be the production of the company when the profit is Rs. }38,250\text{?}
\displaystyle \text{(a) }15\qquad\text{(b) }30\qquad\text{(c) }2\qquad\text{(d) Data is not sufficient to find}
\displaystyle \text{Answer:}
\displaystyle P(x)=-5x^2+125x+37500.
\displaystyle \text{Given }P(x)=38250,
\displaystyle -5x^2+125x+37500=38250.
\displaystyle -5x^2+125x-750=0.
\displaystyle x^2-25x+150=0.
\displaystyle (x-10)(x-15)=0.
\displaystyle \therefore x=10\text{ or }x=15.
\displaystyle \therefore \text{The production can be }10\text{ units or }15\text{ units.}
\displaystyle \\

\displaystyle \textbf{Case Study - 3}

\displaystyle \text{A potter made a mud vessel whose shape is based on the function }f(x)=|x-3|+|x-2|,
\displaystyle \text{where }f(x)\text{ represents the height of the pot. Based on the above information, answer the}
\displaystyle \text{following questions.} \displaystyle \\

\displaystyle \textbf{Question 1: }\text{When }x>4,\text{ what will be the height in terms of }x\text{?}
\displaystyle \text{(a) }x-2\qquad\text{(b) }x-3\qquad\text{(c) }2x-5\qquad\text{(d) }5-2x
\displaystyle \text{Answer:}
\displaystyle f(x)=|x-3|+|x-2|.
\displaystyle \text{For }x>4,\text{ both }x-3\text{ and }x-2\text{ are positive.}
\displaystyle \therefore f(x)=(x-3)+(x-2).
\displaystyle =2x-5.
\displaystyle \therefore \text{The height is }2x-5.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Will the slope vary with the value of }x\text{?}
\displaystyle \text{(a) Yes}\qquad\text{(b) No}
\displaystyle \text{Answer:}
\displaystyle f(x)=\begin{cases}5-2x,&x<2,\\1,&2\leq x<3,\\2x-5,&x\geq3.\end{cases}
\displaystyle \therefore f'(x)=\begin{cases}-2,&x<2,\\0,&2<x<3,\\2,&x>3.\end{cases}
\displaystyle \text{Thus, the slope has different values in different intervals of }x.
\displaystyle \therefore \text{The slope varies with the value of }x.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{What is }\frac{dy}{dx}\text{ at }x=3\text{?}
\displaystyle \text{(a) }2\qquad\text{(b) }-2\qquad\text{(c) Function is not differentiable}\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \text{For }2<x<3,\quad f(x)=1.
\displaystyle \therefore f'_-(3)=0.
\displaystyle \text{For }x>3,\quad f(x)=2x-5.
\displaystyle \therefore f'_+(3)=2.
\displaystyle \text{Since }f'_-(3)\neq f'_+(3),\ f'(3)\text{ does not exist.}
\displaystyle \therefore f(x)\text{ is not differentiable at }x=3.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{When the value of }x\text{ lies between }2\text{ and }3,\text{ the function is}
\displaystyle \text{(a) }2x-5\qquad\text{(b) }5-2x\qquad\text{(c) }x\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \text{For }2<x<3,\quad x-3<0\text{ and }x-2>0.
\displaystyle \therefore |x-3|=3-x\quad\text{and}\quad|x-2|=x-2.
\displaystyle f(x)=(3-x)+(x-2).
\displaystyle =1.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If the potter tries to make a pot using the function }f(x)=[x],\text{ will he get a}
\displaystyle \text{pot? Why?}
\displaystyle \text{(a) Yes, because it is a continuous function}
\displaystyle \text{(b) Yes, because it is not continuous}
\displaystyle \text{(c) No, because it is a continuous function}
\displaystyle \text{(d) No, because it is not continuous}
\displaystyle \text{Answer:}
\displaystyle f(x)=[x]\text{ is the greatest integer function.}
\displaystyle \text{The greatest integer function is discontinuous at every integer.}
\displaystyle \therefore f(x)=[x]\text{ does not give a continuous profile for the pot.}
\displaystyle \therefore \text{The potter will not get a continuous pot shape.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 4}

\displaystyle \text{The shape of a toy is given by }f(x)=6(2x^4-x^2).\text{ To make the toy beautiful, two sticks}
\displaystyle \text{perpendicular to each other are placed at the point }(2,3)\text{ above the toy. Based on the}
\displaystyle \text{above information, answer the following questions.} \displaystyle \\

\displaystyle \textbf{Question 1: }\text{Which value from the following may be the abscissa of a critical point?}
\displaystyle \text{(a) }\pm\frac14\qquad\text{(b) }\pm\frac12\qquad\text{(c) }\pm1\qquad\text{(d) None}
\displaystyle \text{Answer:}
\displaystyle f(x)=6(2x^4-x^2)=12x^4-6x^2.
\displaystyle f'(x)=48x^3-12x.
\displaystyle =12x(4x^2-1).
\displaystyle =12x(2x-1)(2x+1).
\displaystyle \text{At a critical point, }f'(x)=0.
\displaystyle 12x(2x-1)(2x+1)=0.
\displaystyle \therefore x=0,\quad x=\frac12,\quad x=-\frac12.
\displaystyle \therefore \text{Among the given options, }\pm\frac12\text{ may be the abscissa of a critical point.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the slope of the normal based on the position of the stick.}
\displaystyle \text{(a) }360\qquad\text{(b) }-360\qquad\text{(c) }\frac{1}{360}\qquad\text{(d) }-\frac{1}{360}
\displaystyle \text{Answer:}
\displaystyle f'(x)=48x^3-12x.
\displaystyle \text{At }x=2,\text{ slope of the tangent}=f'(2).
\displaystyle =48(2)^3-12(2).
\displaystyle =384-24=360.
\displaystyle \text{Slope of normal}=-\frac{1}{\text{slope of tangent}}.
\displaystyle =-\frac{1}{360}.
\displaystyle \therefore \text{The slope of the normal is }-\frac{1}{360}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{What will be the equation of the tangent if it passes through the point }(2,3)\text{?}
\displaystyle \text{(a) }x+360y=1082\qquad\text{(b) }y=360x-717
\displaystyle \text{(c) }x=717y+360\qquad\text{(d) None}
\displaystyle \text{Answer:}
\displaystyle f(x)=6(2x^4-x^2)=12x^4-6x^2.
\displaystyle f'(x)=48x^3-12x.
\displaystyle \text{At }x=2,
\displaystyle f'(2)=48(2)^3-12(2)=384-24=360.
\displaystyle \therefore \text{Slope of the tangent}=360.
\displaystyle \text{Since the tangent passes through }(2,3),
\displaystyle y-3=360(x-2).
\displaystyle y-3=360x-720.
\displaystyle \therefore y=360x-717.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the second order derivative of the function at }x=5\text{.}
\displaystyle \text{(a) }598\qquad\text{(b) }1176\qquad\text{(c) }3588\qquad\text{(d) }3312
\displaystyle \text{Answer:}
\displaystyle f'(x)=48x^3-12x.
\displaystyle f''(x)=144x^2-12.
\displaystyle f''(5)=144(5)^2-12.
\displaystyle =144(25)-12.
\displaystyle =3600-12=3588.
\displaystyle \therefore \text{The second order derivative at }x=5\text{ is }3588.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In which of the following intervals will }f(x)\text{ be increasing?}
\displaystyle \text{(a) }\left(-\infty,-\frac12\right)\cup\left(\frac12,\infty\right)
\displaystyle \text{(b) }\left(-\frac12,0\right)\cup\left(\frac12,\infty\right)
\displaystyle \text{(c) }\left(0,\frac12\right)\cup\left(\frac12,\infty\right)
\displaystyle \text{(d) }\left(-\infty,-\frac12\right)\cup\left(0,\frac12\right)
\displaystyle \text{Answer:}
\displaystyle f'(x)=12x(2x-1)(2x+1).
\displaystyle f'(x)=0\text{ at }x=-\frac12,\ 0,\ \frac12.
\displaystyle f'(x)>0\text{ for }-\frac12<x<0\text{ and }x>\frac12.
\displaystyle \therefore f(x)\text{ is increasing on }\left(-\frac12,0\right)\cup\left(\frac12,\infty\right).
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 5}

\displaystyle \text{A bridge connects two hills }100\text{ feet apart. The arch of the bridge is parabolic, and its}
\displaystyle \text{highest point is }10\text{ feet above the road at the middle of the bridge. Based on the above}
\displaystyle \text{information, answer the following questions.} \displaystyle \\

\displaystyle \textbf{Question 1: }\text{The equation of the parabola designed on the bridge is}
\displaystyle \text{(a) }x^2=250y\qquad\text{(b) }x^2=-250y\qquad\text{(c) }y^2=250x\qquad\text{(d) }y^2=250y
\displaystyle \text{Answer:}
\displaystyle \text{The parabola has its vertex at the highest point and opens downwards.}
\displaystyle \therefore \text{Its equation is of the form }x^2=-4ay.
\displaystyle \text{The bridge is }100\text{ ft wide and its highest point is }10\text{ ft above the road.}
\displaystyle \therefore \text{Relative to the vertex, a point on the parabola is }(50,-10).
\displaystyle (50)^2=-4a(-10).
\displaystyle 2500=40a.
\displaystyle \therefore 4a=250.
\displaystyle \therefore x^2=-250y.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The value of the integral }\int_{-50}^{50}\frac{x^2}{250}\,dx\text{ is}
\displaystyle \text{(a) }\frac{1000}{3}\qquad\text{(b) }\frac{250}{3}\qquad\text{(c) }1200\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \int_{-50}^{50}\frac{x^2}{250}\,dx =\frac{1}{250}\int_{-50}^{50}x^2\,dx.
\displaystyle \text{Since }x^2\text{ is an even function,}
\displaystyle =\frac{2}{250}\int_0^{50}x^2\,dx.
\displaystyle =\frac{2}{250}\left[\frac{x^3}{3}\right]_0^{50}.
\displaystyle =\frac{2}{250}\times\frac{125000}{3}.
\displaystyle =\frac{1000}{3}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The integrand in }\int_{-50}^{50}x^2\,dx\text{ is a/an}
\displaystyle \text{(a) even function}\qquad\text{(b) odd function}\qquad\text{(c) neither odd nor even}\qquad\text{(d) none}
\displaystyle \text{Answer:}
\displaystyle f(x)=x^2.
\displaystyle f(-x)=(-x)^2=x^2=f(x).
\displaystyle \therefore x^2\text{ is an even function.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The area formed by the curve }x^2=250y,\text{ the }x\text{-axis and the level }y=10\text{ is}
\displaystyle \text{(a) }\frac{1000\sqrt2}{3}\qquad\text{(b) }\frac43\qquad\text{(c) }\frac{1000}{3}\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle x^2=250y\implies y=\frac{x^2}{250}.
\displaystyle \text{At }y=10,
\displaystyle x^2=250(10)=2500.
\displaystyle \therefore x=\pm50.
\displaystyle \text{Required area}=\int_{-50}^{50}\frac{x^2}{250}\,dx.
\displaystyle =\frac{2}{250}\int_0^{50}x^2\,dx.
\displaystyle =\frac{2}{250}\left[\frac{x^3}{3}\right]_0^{50}.
\displaystyle =\frac{2}{250}\times\frac{125000}{3}=\frac{1000}{3}.
\displaystyle \therefore \text{The required area is }\frac{1000}{3}\text{ square feet.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The area formed between }x^2=250y,\text{ the }y\text{-axis, }y=2\text{ and }y=4\text{ is}
\displaystyle \text{(a) }\frac{1000}{3}\qquad\text{(b) }0\qquad\text{(c) }\frac{1000\sqrt2}{3}\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle x^2=250y\implies x=\sqrt{250y}\text{ for the region to the right of the }y\text{-axis.}
\displaystyle \therefore \text{Area}=\int_2^4\sqrt{250y}\,dy.
\displaystyle =\sqrt{250}\left[\frac{2}{3}y^{3/2}\right]_2^4.
\displaystyle =\frac{2\sqrt{250}}{3}\left(8-2\sqrt2\right).
\displaystyle =\frac{80\sqrt{10}-40\sqrt5}{3}.
\displaystyle \text{This value is not among the given options.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \text{DIFFERENTIAL EQUATION}

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{A veterinary doctor was examining a sick cat brought to the hospital by a pet lover. When the cat}
\displaystyle \text{was brought to the hospital, it was already dead. The pet lover wanted to find its time of death.}
\displaystyle \text{The temperature of the cat at }11.30\text{ p.m. was }94.6^\circ\text{F. After one hour, its temperature}
\displaystyle \text{was }93.4^\circ\text{F. The room temperature was maintained at }70^\circ\text{F. The normal temperature}
\displaystyle \text{of the cat when alive was taken as }98.6^\circ\text{F.}
\displaystyle \text{The doctor estimated the time of death using Newton's law of cooling, governed by}
\displaystyle \frac{dT}{dt}=k(T-70),\text{ where }k\text{ is a constant of proportionality and }T\text{ is the temperature at time }t.
\displaystyle \text{Based on the above information, answer the following questions.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{State the degree of the above differential equation.}
\displaystyle \text{Answer:}
\displaystyle \frac{dT}{dt}=k(T-70).
\displaystyle \text{The highest order derivative is }\frac{dT}{dt},\text{ and its power is }1.
\displaystyle \therefore \text{The degree of the differential equation is }1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Which method of solving a differential equation helps in calculating the time of death?}
\displaystyle \text{(a) Variable separable method}\qquad\text{(b) Solving homogeneous differential equation}
\displaystyle \text{(c) Solving linear differential equation}\qquad\text{(d) All of the above}
\displaystyle \text{Answer:}
\displaystyle \frac{dT}{dt}=k(T-70).
\displaystyle \frac{dT}{T-70}=k\,dt.
\displaystyle \text{The variables }T\text{ and }t\text{ can be separated directly.}
\displaystyle \therefore \text{The variable separable method can be used.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If the temperature were measured }2\text{ hours after }11.30\text{ p.m., would the estimated}
\displaystyle \text{time of death change? (Yes/No)}
\displaystyle \text{Answer:}
\displaystyle \text{No.}
\displaystyle \text{According to Newton's law of cooling, all temperature observations follow the same cooling law.}
\displaystyle \text{Hence, under the same ideal conditions, another accurate observation should give the same}
\displaystyle \text{estimated time of death.}
\displaystyle \therefore \text{The answer is No.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The solution of the differential equation }\frac{dT}{dt}=k(T-70)\text{ is given by}
\displaystyle \text{(a) }\log|T-70|=kt+C\qquad\text{(b) }\log|T-70|=\log|kt|+C
\displaystyle \text{(c) }T-70=kt+C\qquad\text{(d) }T-70=ktC
\displaystyle \text{Answer:}
\displaystyle \frac{dT}{dt}=k(T-70).
\displaystyle \frac{dT}{T-70}=k\,dt.
\displaystyle \text{Integrating both sides,}
\displaystyle \int\frac{dT}{T-70}=\int k\,dt.
\displaystyle \log|T-70|=kt+C.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }t=0\text{ when }T=72,\text{ then the value of }C\text{ is}
\displaystyle \text{(a) }-2\qquad\text{(b) }0\qquad\text{(c) }2\qquad\text{(d) }\log2
\displaystyle \text{Answer:}
\displaystyle \log|T-70|=kt+C.
\displaystyle \text{Putting }t=0\text{ and }T=72,
\displaystyle \log|72-70|=k(0)+C.
\displaystyle \log2=C.
\displaystyle \therefore C=\log2.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 2}

\displaystyle \text{Polio drops are to be given to }50\text{ thousand children in a district. The rate at which the drops}
\displaystyle \text{are administered is directly proportional to the number of children who have not yet received them.}
\displaystyle \text{By the end of the second week, half the children have been given the polio drops. The number who}
\displaystyle \text{will have received the drops by the end of the third week can be estimated using the differential}
\displaystyle \text{equation }\frac{dy}{dx}=k(50-y),\text{ where }x\text{ denotes the number of weeks and }y\text{ denotes the number}
\displaystyle \text{of children, in thousands, who have been given the drops.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{State the order of the above given differential equation.}
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}=k(50-y).
\displaystyle \text{The highest order derivative present in the differential equation is }\frac{dy}{dx}.
\displaystyle \therefore \text{The order of the differential equation is }1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Which method of solving a differential equation can be used to solve}
\displaystyle \frac{dy}{dx}=k(50-y)\text{?}
\displaystyle \text{(a) Variable separable method}\qquad\text{(b) Solving homogeneous differential equation}
\displaystyle \text{(c) Solving linear differential equation}\qquad\text{(d) All of the above}
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}=k(50-y).
\displaystyle \frac{dy}{50-y}=k\,dx.
\displaystyle \text{Since the variables }x\text{ and }y\text{ can be separated, the variable separable method can be used.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The solution of the differential equation }\frac{dy}{dx}=k(50-y)\text{ is given by}
\displaystyle \text{(a) }\log|50-y|=kx+C\qquad\text{(b) }-\log|50-y|=kx+C
\displaystyle \text{(c) }\log|50-y|=\log|kx|+C\qquad\text{(d) }50-y=kx+C
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}=k(50-y).
\displaystyle \frac{dy}{50-y}=k\,dx.
\displaystyle \text{Integrating both sides,}
\displaystyle \int\frac{dy}{50-y}=\int k\,dx.
\displaystyle -\log|50-y|=kx+C.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The value of }C\text{ in the particular solution, given that }y(0)=0\text{ and }k=0.049,\text{ is}
\displaystyle \text{(a) }\log50\qquad\text{(b) }\log\frac{1}{50}\qquad\text{(c) }50\qquad\text{(d) }-50
\displaystyle \text{Answer:}
\displaystyle -\log|50-y|=kx+C.
\displaystyle \text{Putting }x=0\text{ and }y=0,
\displaystyle -\log|50|=k(0)+C.
\displaystyle C=-\log50.
\displaystyle =\log\left(\frac{1}{50}\right).
\displaystyle \therefore C=\log\frac{1}{50}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Which of the following solutions may be used to find the number of children who}
\displaystyle \text{have been given the polio drops?}
\displaystyle \text{(a) }y=50-e^{kx}\qquad\text{(b) }y=50-e^{-kx}
\displaystyle \text{(c) }y=50(1-e^{-kx})\qquad\text{(d) }y=50(e^{-kx}-1)
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}=k(50-y).
\displaystyle -\log|50-y|=kx+C.
\displaystyle \log|50-y|=-kx+C_1.
\displaystyle 50-y=Ae^{-kx}.
\displaystyle \text{Since }y(0)=0,
\displaystyle 50=A.
\displaystyle \therefore 50-y=50e^{-kx}.
\displaystyle \therefore y=50(1-e^{-kx}).
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

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