\displaystyle \text{LONG ANSWER QUESTIONS - 4 Marks Each}

\displaystyle \textbf{Question 93: }\text{Let }f(x)=\frac{\log 5x}{kx},\text{ where }x>0,\ k\in R^+.
\displaystyle \text{(a) Show that }f'(x)=\frac{1-\log 5x}{kx^2}.
\displaystyle \text{The graph of }f\text{ has exactly one maximum point at }P.
\displaystyle \text{(b) Find the }x\text{-coordinate of }P.
\displaystyle \text{(c) Find }f''(x).
\displaystyle \text{(d) Find the value of }x\text{ for which }f''(x)\text{ vanishes.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) }f(x)=\frac{1}{k}(\log 5x)x^{-1}.
\displaystyle f'(x)=\frac{1}{k}\left[\frac{1}{x}\cdot x^{-1}-(\log 5x)x^{-2}\right].
\displaystyle \therefore f'(x)=\frac{1-\log 5x}{kx^2}.

\displaystyle \text{(b) At the maximum point, }f'(x)=0.
\displaystyle 1-\log 5x=0.
\displaystyle \therefore \log 5x=1.
\displaystyle \therefore 5x=e.
\displaystyle \therefore x=\frac{e}{5}.
\displaystyle \text{Also, }f'(x)>0\text{ for }x<\frac{e}{5}\text{ and }f'(x)<0\text{ for }x>\frac{e}{5}.
\displaystyle \therefore \text{the }x\text{-coordinate of }P\text{ is }\frac{e}{5}.

\displaystyle \text{(c) }f'(x)=\frac{1}{k}(1-\log 5x)x^{-2}.
\displaystyle f''(x)=\frac{1}{k}\left[-x^{-3}-2(1-\log 5x)x^{-3}\right].
\displaystyle \therefore f''(x)=\frac{2\log 5x-3}{kx^3}.

\displaystyle \text{(d) For }f''(x)\text{ to vanish,}
\displaystyle 2\log 5x-3=0.
\displaystyle \therefore \log 5x=\frac32.
\displaystyle \therefore 5x=e^{3/2}.
\displaystyle \therefore x=\frac{e^{3/2}}{5}.
\displaystyle \\

\displaystyle \textbf{Question 94: }\text{A linear programming problem (LPP) is given as:}
\displaystyle \text{Maximize }Z=x+2y\text{ subject to the constraints}
\displaystyle x-y\geq0,\quad 2\geq2y-x,\quad x\geq0,\quad y\geq0.
\displaystyle \text{Based on the above information, answer the following questions.}
\displaystyle \text{(a) Find the corner points of the feasible region.}
\displaystyle \text{(b) Find the corner point where maximum occurs.}
\displaystyle \text{(c) Optimum solution does not exist. Justify your answer.}
\displaystyle \text{Answer:}
\displaystyle \text{The constraints can be written as}
\displaystyle y\leq x,\qquad y\leq\frac{x+2}{2},\qquad x\geq0,\qquad y\geq0.
\displaystyle \text{(a) The boundary lines }y=x\text{ and }y=\frac{x+2}{2}\text{ intersect when}
\displaystyle x=\frac{x+2}{2}.
\displaystyle 2x=x+2.
\displaystyle \therefore x=2,\qquad y=2.
\displaystyle \text{The origin }(0,0)\text{ also lies on the boundary of the feasible region.}
\displaystyle \therefore \text{the corner points are }(0,0)\text{ and }(2,2).

\displaystyle \text{(b) }Z(0,0)=0.
\displaystyle Z(2,2)=2+2(2)=6.
\displaystyle \text{However, the feasible region is unbounded.}
\displaystyle \therefore \text{there is no corner point at which a maximum occurs.}

\displaystyle \text{(c) Every point }(x,0),\ x\geq0,\text{ satisfies all the given constraints.}
\displaystyle \text{Along the }x\text{-axis, }Z=x+2(0)=x.
\displaystyle \text{As }x\to\infty,\quad Z\to\infty.
\displaystyle \therefore Z\text{ can be made arbitrarily large in the feasible region.}
\displaystyle \therefore \text{the LPP has no finite optimum solution.}
\displaystyle \\

\displaystyle \textbf{Question 95: }\text{There are two curves given in the first quadrant as:}
\displaystyle x^2+y^2=\pi^2,\quad y=\sin x.
\displaystyle \text{(a) What are the points of intersection of both the given curves?}
\displaystyle \text{(b) What is the value of }K,\text{ if }\int_0^\pi\sqrt{\pi^2-x^2}\,dx=\frac{\pi^3}{K}?
\displaystyle \text{(c) Sketch the region enclosed by the given curves in the first quadrant and the }y\text{-axis.}
\displaystyle \text{(d) Find the area of the region enclosed by the given curves in the first quadrant and the}
\displaystyle y\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) At the points of intersection, }y=\sin x\text{ and }x^2+y^2=\pi^2.
\displaystyle \therefore x^2+\sin^2x=\pi^2.
\displaystyle \text{For }0\leq x\leq\pi,\text{ the equation is satisfied at }x=\pi.
\displaystyle \therefore y=\sin\pi=0.
\displaystyle \therefore \text{the curves intersect at }(\pi,0).

\displaystyle \text{(b) }\int_0^\pi\sqrt{\pi^2-x^2}\,dx\text{ represents the area of a quarter-circle of radius }\pi.
\displaystyle \therefore \int_0^\pi\sqrt{\pi^2-x^2}\,dx=\frac14\pi(\pi)^2=\frac{\pi^3}{4}.
\displaystyle \text{Comparing with }\frac{\pi^3}{K},
\displaystyle \therefore K=4.

\displaystyle \text{(c) The required region is bounded by the }y\text{-axis, the quarter-circle}\displaystyle y=\sqrt{\pi^2-x^2}\text{ and the curve }y=\sin x,\quad 0\leq x\leq\pi.

\displaystyle \text{(d) Required area}=\int_0^\pi\left(\sqrt{\pi^2-x^2}-\sin x\right)dx.
\displaystyle =\int_0^\pi\sqrt{\pi^2-x^2}\,dx-\int_0^\pi\sin x\,dx.
\displaystyle =\frac{\pi^3}{4}-[-\cos x]_0^\pi.
\displaystyle =\frac{\pi^3}{4}-2.
\displaystyle \therefore \text{the required area is }\frac{\pi^3}{4}-2\text{ square units.}
\displaystyle \\

\displaystyle \textbf{Question 96: }\text{Let }\overrightarrow{\alpha}=3\widehat{i}+\widehat{j}\text{ and }\overrightarrow{\beta}=2\widehat{i}-\widehat{j}+3\widehat{k}.\text{ If }\overrightarrow{\beta}=\overrightarrow{\beta}_1-\overrightarrow{\beta}_2,\text{ where }\overrightarrow{\beta}_1\text{ is parallel to }\overrightarrow{\alpha}
\displaystyle \text{and }\overrightarrow{\beta}_2\text{ is perpendicular to }\overrightarrow{\alpha}.
\displaystyle \text{(a) Find }\overrightarrow{\beta}_1.
\displaystyle \text{(b) Find }\overrightarrow{\beta}_2.
\displaystyle \text{(c) Hence, find }\overrightarrow{\beta}_1\times\overrightarrow{\beta}_2.
\displaystyle \text{Answer:}
\displaystyle \text{(a) Since }\overrightarrow{\beta}_1\parallel\overrightarrow{\alpha},\text{ let }\overrightarrow{\beta}_1=\lambda\overrightarrow{\alpha}.
\displaystyle \therefore \overrightarrow{\beta}_1=\lambda(3\widehat{i}+\widehat{j}).
\displaystyle \text{Since }\overrightarrow{\beta}_2=\overrightarrow{\beta}_1-\overrightarrow{\beta}\text{ and }\overrightarrow{\beta}_2\perp\overrightarrow{\alpha},
\displaystyle (\overrightarrow{\beta}_1-\overrightarrow{\beta})\cdot\overrightarrow{\alpha}=0.
\displaystyle \lambda(\overrightarrow{\alpha}\cdot\overrightarrow{\alpha})=\overrightarrow{\beta}\cdot\overrightarrow{\alpha}.
\displaystyle \lambda(3^2+1^2)=(2)(3)+(-1)(1)+(3)(0).
\displaystyle 10\lambda=5\Rightarrow\lambda=\frac12.
\displaystyle \therefore \overrightarrow{\beta}_1=\frac32\widehat{i}+\frac12\widehat{j}.

\displaystyle \text{(b) }\overrightarrow{\beta}_2=\overrightarrow{\beta}_1-\overrightarrow{\beta}.
\displaystyle =\left(\frac32\widehat{i}+\frac12\widehat{j}\right)-(2\widehat{i}-\widehat{j}+3\widehat{k}).
\displaystyle \therefore \overrightarrow{\beta}_2=-\frac12\widehat{i}+\frac32\widehat{j}-3\widehat{k}.

\displaystyle \text{(c) }\overrightarrow{\beta}_1\times\overrightarrow{\beta}_2
\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\\frac32&\frac12&0\\-\frac12&\frac32&-3\end{vmatrix}.
\displaystyle =-\frac32\widehat{i}+\frac92\widehat{j}+\frac52\widehat{k}.
\displaystyle \therefore \overrightarrow{\beta}_1\times\overrightarrow{\beta}_2=\frac12(-3\widehat{i}+9\widehat{j}+5\widehat{k}).
\displaystyle \\

\displaystyle \textbf{Question 97: }\text{Given }f(x)=2\log(x-2)-x^2+4x+1\text{ and }f'(x)=\frac{-k(x-p)(x-q)}{(x-k)}.
\displaystyle \text{(a) Find }k+p+q.
\displaystyle \text{(b) The Statement ``}f(x)\text{ is strictly increasing in }(-\infty,1]\cup(2,3]\text{'' is false. Justify.}
\displaystyle \text{(c) Hence, find the interval(s) in which the function is strictly increasing.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=2\log(x-2)-x^2+4x+1.
\displaystyle \text{Since }\log(x-2)\text{ is defined only when }x-2>0,
\displaystyle \therefore \text{the domain of }f\text{ is }x>2.
\displaystyle \text{(a) }f'(x)=\frac{2}{x-2}-2x+4.
\displaystyle =\frac{2-2(x-2)^2}{x-2}.
\displaystyle =\frac{-2\{(x-2)^2-1\}}{x-2}.
\displaystyle =\frac{-2(x-1)(x-3)}{x-2}.
\displaystyle \text{Comparing with }f'(x)=\frac{-k(x-p)(x-q)}{x-k},
\displaystyle k=2,\qquad p=1,\qquad q=3.
\displaystyle \therefore k+p+q=2+1+3=6.

\displaystyle \text{(b) The given statement includes }(-\infty,1].
\displaystyle \text{But }f(x)\text{ is defined only for }x>2.
\displaystyle \therefore f(x)\text{ cannot be strictly increasing on }(-\infty,1].
\displaystyle \therefore \text{the given statement is false.}

\displaystyle \text{(c) }f'(x)=\frac{-2(x-1)(x-3)}{x-2}.
\displaystyle \text{For }2<x<3,\quad x-1>0,\ x-3<0,\ x-2>0.
\displaystyle \therefore f'(x)>0.
\displaystyle \text{For }x>3,\quad x-1>0,\ x-3>0,\ x-2>0.
\displaystyle \therefore f'(x)<0.
\displaystyle \therefore f(x)\text{ is strictly increasing on }(2,3].
\displaystyle \\

\displaystyle \textbf{Question 98: }\text{Case Study}
\displaystyle \text{The length of the perimeter of a slice of a pizza in the form of a sector of circle is }20\text{ cm.}
\displaystyle r\text{ be the radius of the circle, sectorial angle be }\theta\text{ radian and }l\text{ be the length of the arc.} \displaystyle \text{Based on the above information, answer the following questions.}
\displaystyle \text{(a) Express the radius of the sector is expressed in terms of sectorial angle be }\theta\text{ radian.}
\displaystyle \text{(b) Let }A\text{ be the area of the slice. Then, express }A\text{ in terms of }r.
\displaystyle \text{(c) For the maximum value of }A,\text{ find the value of the sectorial angle.}
\displaystyle \text{(d) Maximum area of the slice of the pizza is }\underline{\hspace{3cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{The perimeter of the sector is }20\text{ cm.}
\displaystyle \therefore 2r+l=20.
\displaystyle \text{Also, }l=r\theta.
\displaystyle \therefore 2r+r\theta=20.
\displaystyle \therefore r(\theta+2)=20.

\displaystyle \text{(a) }\therefore r=\frac{20}{\theta+2}.

\displaystyle \text{(b) }A=\frac12r^2\theta.
\displaystyle \text{From }r(\theta+2)=20,\quad \theta=\frac{20}{r}-2.
\displaystyle \therefore A=\frac12r^2\left(\frac{20}{r}-2\right).
\displaystyle \therefore A=10r-r^2.

\displaystyle \text{(c) For maximum area, }\frac{dA}{dr}=0.
\displaystyle \frac{dA}{dr}=10-2r.
\displaystyle \therefore 10-2r=0\Rightarrow r=5.
\displaystyle \frac{d^2A}{dr^2}=-2<0,\text{ hence }A\text{ is maximum at }r=5.
\displaystyle \text{Now, }\theta=\frac{20}{5}-2=2.
\displaystyle \therefore \text{the sectorial angle is }2\text{ radians.}

\displaystyle \text{(d) Maximum area}=10(5)-5^2.
\displaystyle =50-25=25\text{ cm}^2.
\displaystyle \therefore \text{maximum area of the pizza slice is }25\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 99: }\text{In a classroom, a teacher explains the properties of a particular curve by saying}
\displaystyle \text{that this particular curve has beautiful ups and downs. It starts and heads down until }\pi
\displaystyle \text{radian, and then heads up again and is closely related to sine function. Both follow each}
\displaystyle \text{other at exactly }\frac{\pi}{2}\text{ radians apart as shown in the figure given below:} \displaystyle \text{Based on the above information, answer the questions that follow.}
\displaystyle \text{(a) Name the curve that the teacher explained in the classroom.}
\displaystyle \text{(b) Find the area of the curve explained in the passage from }0\text{ to }\frac{\pi}{2}.
\displaystyle \text{(c) Find the area of the shaded region.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) The curve is }y=\cos x.
\displaystyle \text{It is shifted by }\frac{\pi}{2}\text{ radians with respect to }y=\sin x.

\displaystyle \text{(b) Required area}=\int_0^{\pi/2}\cos x\,dx.
\displaystyle =[\sin x]_0^{\pi/2}.
\displaystyle =1-0=1.
\displaystyle \therefore \text{the required area is }1\text{ square unit.}

\displaystyle \text{(c) The curves }y=\cos x\text{ and }y=\sin x\text{ intersect when}
\displaystyle \sin x=\cos x.
\displaystyle \therefore \tan x=1.
\displaystyle \therefore x=\frac{\pi}{4}.
\displaystyle \text{Required shaded area}=\int_0^{\pi/4}(\cos x-\sin x)\,dx
\displaystyle +\int_{\pi/4}^{\pi/2}(\sin x-\cos x)\,dx.
\displaystyle =[\sin x+\cos x]_0^{\pi/4}+[-\cos x-\sin x]_{\pi/4}^{\pi/2}.
\displaystyle =(\sqrt2-1)+(\sqrt2-1).
\displaystyle =2(\sqrt2-1).
\displaystyle \therefore \text{the area of the shaded region is }2(\sqrt2-1)\text{ square units.}
\displaystyle \\

\displaystyle \textbf{Question 100: }\text{(a) If a real-valued function is given by:}
\displaystyle f(x)=\sqrt{25-x^2}\text{ is an onto function, then find the co-domain for }f(x).
\displaystyle \text{(b) If the domain is given to be }[-5,5],\text{ is }f(x)\text{ a one-one function?}
\displaystyle \text{(c) Find all possible values of }a\text{ for which }f(a)=4.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\sqrt{25-x^2}.

\displaystyle \text{(a) For }f(x)\text{ to be real-valued,}
\displaystyle 25-x^2\geq0.
\displaystyle \therefore -5\leq x\leq5.
\displaystyle \text{Also, }0\leq\sqrt{25-x^2}\leq5.
\displaystyle \therefore \text{the range of }f\text{ is }[0,5].
\displaystyle \text{Since }f\text{ is onto, its co-domain is }[0,5].

\displaystyle \text{(b) }f(-x)=\sqrt{25-(-x)^2}=\sqrt{25-x^2}=f(x).
\displaystyle \text{For example, }f(3)=f(-3)=4,\text{ although }3\ne-3.
\displaystyle \therefore f(x)\text{ is not one-one on }[-5,5].

\displaystyle \text{(c) }f(a)=4.
\displaystyle \therefore \sqrt{25-a^2}=4.
\displaystyle 25-a^2=16.
\displaystyle \therefore a^2=9.
\displaystyle \therefore a=\pm3.
\displaystyle \\

\displaystyle \textbf{Question 101: }\text{Find the value:}
\displaystyle \cot\left[\sum_{n=1}^{25}\cot^{-1}\left(1+\sum_{k=1}^{n}2k\right)\right].
\displaystyle \text{Answer:}
\displaystyle \cot\left[\sum_{n=1}^{25}\cot^{-1}\left(1+\sum_{k=1}^{n}2k\right)\right]
\displaystyle =\cot\left[\sum_{n=1}^{25}\cot^{-1}\left(1+2\times\frac{n(n+1)}{2}\right)\right].
\displaystyle =\cot\left[\sum_{n=1}^{25}\cot^{-1}(n^2+n+1)\right].
\displaystyle =\cot\left[\sum_{n=1}^{25}\tan^{-1}\left(\frac{1}{n^2+n+1}\right)\right].
\displaystyle =\cot\left[\sum_{n=1}^{25}\tan^{-1}\left(\frac{n+1-n}{1+n(n+1)}\right)\right].
\displaystyle =\cot\left[\sum_{n=1}^{25}\left\{\tan^{-1}(n+1)-\tan^{-1}(n)\right\}\right].
\displaystyle =\cot\left[\tan^{-1}(26)-\tan^{-1}(1)\right].
\displaystyle =\cot\left[\tan^{-1}\left(\frac{25}{27}\right)\right].
\displaystyle =\cot\left[\cot^{-1}\left(\frac{27}{25}\right)\right].
\displaystyle =\frac{27}{25}.
\displaystyle \\

\displaystyle \textbf{Question 102: }\text{If }f:[1,\infty)\to[2,\infty)\text{ is given by }f(x)=x+\frac1x,\text{ then find }\frac{d}{dx}f^{-1}(x).
\displaystyle \text{Answer:}
\displaystyle y=x+\frac1x.
\displaystyle y=\frac{x^2+1}{x}.
\displaystyle xy=x^2+1.
\displaystyle x^2-xy+1=0.
\displaystyle x=\frac{y\pm\sqrt{y^2-4}}{2}.
\displaystyle f^{-1}(y)=\frac{y\pm\sqrt{y^2-4}}{2}.
\displaystyle f^{-1}(x)=\frac{x\pm\sqrt{x^2-4}}{2}.
\displaystyle \text{Since range of inverse function is }[1,\infty),\text{ then}
\displaystyle f^{-1}(x)=\frac{x+\sqrt{x^2-4}}{2}.
\displaystyle f^{-1}(x)=\frac{x-\sqrt{x^2-4}}{2}\text{ then }f^{-1}(x)>1\text{ is possible when,}
\displaystyle \frac{x-\sqrt{x^2-4}}{2}>1.
\displaystyle x-\sqrt{x^2-4}>2.
\displaystyle (x-2)^2>x^2-4.
\displaystyle x^2+4-4x>x^2-4.
\displaystyle 8>4x.
\displaystyle x<2,
\displaystyle \text{which is not possible since }x>2\text{ (given).}
\displaystyle \therefore f^{-1}(x)=\frac{x+\sqrt{x^2-4}}{2}.
\displaystyle \frac{d}{dx}f^{-1}(x)
\displaystyle =\frac12\left[1+\frac{2x}{2\sqrt{x^2-4}}\right].
\displaystyle =\frac12\left[1+\frac{x}{\sqrt{x^2-4}}\right].
\displaystyle =\frac12\left(\frac{x+\sqrt{x^2-4}}{\sqrt{x^2-4}}\right).
\displaystyle \\

\displaystyle \textbf{Question 103: }\text{Find the acute angle between the curves }y=|x^2-1|\text{ and }
\displaystyle y=|x^2-3|\text{ at their point of} \ \text{intersection when }x>0.
\displaystyle \text{Answer:}
\displaystyle |x^2-1|=|x^2-3|.
\displaystyle (x^2-1)^2=(x^2-3)^2.
\displaystyle 2x(2x^2-4)=0.
\displaystyle 2x^2=4.
\displaystyle x\ne0\text{ as }x>0.
\displaystyle x=\pm\sqrt2.
\displaystyle \text{But,}
\displaystyle x=\sqrt2\text{ as }x>0.
\displaystyle \text{We have point of intersection as }x=\sqrt2.
\displaystyle y=|x^2-1|=x^2-1\text{ in the neighbourhood of }x=\sqrt2\text{ and }y=-(x^2-3)\text{ in the}
\displaystyle \text{neighbourhood of }x=\sqrt2.
\displaystyle \text{Now at }x=\sqrt2,
\displaystyle \frac{dy}{dx}\text{ for first curve}
\displaystyle \frac{dy}{dx}=2x=2\sqrt2\text{ and}
\displaystyle \frac{dy}{dx}\text{ for second curve}
\displaystyle \frac{dy}{dx}=-2\sqrt2.
\displaystyle \therefore \tan\theta=\left|\frac{2\sqrt2-(-2\sqrt2)}{1+(2\sqrt2)(-2\sqrt2)}\right|.
\displaystyle \tan\theta=\left|\frac{4\sqrt2}{-7}\right|=\frac{4\sqrt2}{7}.
\displaystyle \theta=\tan^{-1}\left(\frac{4\sqrt2}{7}\right).
\displaystyle \\

\displaystyle \textbf{Question 104: }\text{The curve is }y=\ln(x+1)-\ln x.\text{ The tangent to the curve at}
\displaystyle \text{the point }P(1,\ln2)\text{ meets the x-axis at }A\text{ and y-axis at }B.\text{ The normal to the curve}
\displaystyle \text{at }P\text{ meets the x-axis at }C\text{ and y-axis at }D.
\displaystyle \text{(a) Find the slope of tangent at }P\text{ and find the slope of normal at }P.
\displaystyle \text{(b) Find the equation of tangent at }P.
\displaystyle \text{(c) Find the equation of normal at }P.
\displaystyle \text{(d) Find the co-ordinates of }A\text{ and }C\text{ in terms of }\ln2.
\displaystyle \text{Answer:}

\displaystyle \text{(a) }y=\ln\left(\frac{x+1}{x}\right)
\displaystyle \frac{dy}{dx}=\frac{x}{x+1}\times\frac{x\times1-(x+1)}{x^2}
\displaystyle =\frac{-1}{x(x+1)}.
\displaystyle \text{slope of tangent at }P
\displaystyle m_1=-\frac12.
\displaystyle \text{slope of normal at }P,\quad m_2=2.

\displaystyle \text{(b) Equation of tangent at }P
\displaystyle y-\ln2=-\frac12(x-1).
\displaystyle 2y-2\ln2=-x+1.
\displaystyle x+2y=2\ln2+1.

\displaystyle \text{(c) Equation of Normal at }P
\displaystyle y-\ln2=2(x-4).
\displaystyle y-\ln2=2x-8.
\displaystyle 2x-y=8-\ln2.

\displaystyle \text{(d) Since the tangent at }P\text{ meets x axis at }A(x_1,0).
\displaystyle \text{Equation of tangent}
\displaystyle x+2y=2\ln2+1.
\displaystyle x_1=2\ln2+1.
\displaystyle \therefore A(2\ln2+1,0).

\displaystyle \text{Since, the normal meets x axis at }C(x_2,0).
\displaystyle \text{Equation of normal}
\displaystyle 2x-y=8-\ln2.
\displaystyle 2x_2-0=8-\ln2.
\displaystyle x_2=\frac{8-\ln2}{2}.
\displaystyle \therefore C\left(\frac{8-\ln2}{2},0\right).
\displaystyle \\

\displaystyle \textbf{Question 105: }\text{Solve the following differential equation:}
\displaystyle \cos^2x\frac{dy}{dx}+y=\tan x,\text{ given that }y(0)=1.\text{ Hence, find }y\left(\frac{\pi}{4}\right).
\displaystyle \text{Answer:}
\displaystyle \cos^2x\frac{dy}{dx}+y=\tan x.
\displaystyle \text{Dividing throughout by }\cos^2x,
\displaystyle \frac{dy}{dx}+y\sec^2x=\tan x\sec^2x.
\displaystyle \text{This is a linear differential equation of the form }\frac{dy}{dx}+Py=Q.
\displaystyle \therefore \text{I.F.}=e^{\int\sec^2x\,dx}=e^{\tan x}.
\displaystyle \therefore ye^{\tan x}=\int\tan x\sec^2x\,e^{\tan x}\,dx+C.
\displaystyle \text{Let }t=\tan x,\quad dt=\sec^2x\,dx.
\displaystyle \therefore ye^{\tan x}=\int te^t\,dt+C.
\displaystyle =e^t(t-1)+C.
\displaystyle \therefore ye^{\tan x}=e^{\tan x}(\tan x-1)+C.
\displaystyle \therefore y=\tan x-1+Ce^{-\tan x}.
\displaystyle \text{Given, }y(0)=1.
\displaystyle 1=0-1+C.
\displaystyle \therefore C=2.
\displaystyle \therefore y=\tan x-1+2e^{-\tan x}.

\displaystyle \text{At }x=\frac{\pi}{4},
\displaystyle y\left(\frac{\pi}{4}\right)=1-1+2e^{-1}.
\displaystyle \therefore y\left(\frac{\pi}{4}\right)=\frac{2}{e}.
\displaystyle \\

\displaystyle \textbf{Question 106: }\text{Evaluate }\int\frac{dx}{\sqrt[4]{(x-1)^3(x+2)^5}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=\left(\frac{x-1}{x+2}\right)^{1/4}.
\displaystyle \therefore t^4=\frac{x-1}{x+2}.
\displaystyle t^4(x+2)=x-1.
\displaystyle \therefore x=\frac{1+2t^4}{1-t^4}.
\displaystyle \therefore x-1=\frac{3t^4}{1-t^4},\qquad x+2=\frac{3}{1-t^4}.
\displaystyle \text{Also, }dx=\frac{12t^3}{(1-t^4)^2}\,dt.
\displaystyle \sqrt[4]{(x-1)^3(x+2)^5}
\displaystyle =\sqrt[4]{\frac{3^8t^{12}}{(1-t^4)^8}}.
\displaystyle =\frac{9t^3}{(1-t^4)^2}.
\displaystyle \therefore \int\frac{dx}{\sqrt[4]{(x-1)^3(x+2)^5}}
\displaystyle =\int\frac{\frac{12t^3}{(1-t^4)^2}}{\frac{9t^3}{(1-t^4)^2}}\,dt.
\displaystyle =\frac43\int dt.
\displaystyle =\frac43t+C.
\displaystyle \therefore \int\frac{dx}{\sqrt[4]{(x-1)^3(x+2)^5}}
\displaystyle =\frac43\left(\frac{x-1}{x+2}\right)^{1/4}+C.
\displaystyle \\

\displaystyle \textbf{Question 107: }\text{For any function }f(x),\text{ we have}
\displaystyle \int_a^b f(x)\,dx=\int_a^{c_1}f(x)\,dx+\int_{c_1}^{c_2}f(x)\,dx+\cdots+\int_{c_n}^b f(x)\,dx.
\displaystyle \text{Where }a<c_1<c_2<\cdots<c_n<b.
\displaystyle \text{Based on above information, evaluate: }\int_0^2|x^2+2x-3|\,dx.
\displaystyle \text{Answer:} \displaystyle x^2+2x-3=(x-1)(x+3).
\displaystyle \text{In }[0,2],\text{ the expression changes sign at }x=1.
\displaystyle \text{For }0\leq x<1,\quad x^2+2x-3<0.
\displaystyle \text{For }1<x\leq2,\quad x^2+2x-3>0.
\displaystyle \therefore \int_0^2|x^2+2x-3|\,dx
\displaystyle =-\int_0^1(x^2+2x-3)\,dx+\int_1^2(x^2+2x-3)\,dx.
\displaystyle =-\left[\frac{x^3}{3}+x^2-3x\right]_0^1+\left[\frac{x^3}{3}+x^2-3x\right]_1^2.
\displaystyle =-\left(-\frac53\right)+\left(\frac23+\frac53\right).
\displaystyle =\frac53+\frac73.
\displaystyle =4.
\displaystyle \therefore \text{the value of the integral is }4.
\displaystyle \\

\displaystyle \textbf{Question 108: }\text{A pot contains }5\text{ red and }2\text{ green balls. At random a ball is drawn from}
\displaystyle \text{this pot. If a drawn ball is green, then put a red ball in the pot. If a drawn ball is red,}
\displaystyle \text{then put a green ball in the pot. While drawn ball is not replaced in the pot. Now, we draw}
\displaystyle \text{another ball randomly. What is the probability that the second ball drawn is a red ball?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }R_1,G_1\text{ denote a red or green ball on the first draw, and }R_2\text{ a red ball on the second draw.}
\displaystyle P(R_1)=\frac57,\qquad P(G_1)=\frac27.
\displaystyle \text{If the first ball is red, it is removed and a green ball is added.}
\displaystyle \therefore P(R_2\mid R_1)=\frac47.
\displaystyle \text{If the first ball is green, it is removed and a red ball is added.}
\displaystyle \therefore P(R_2\mid G_1)=\frac67.
\displaystyle \text{By the theorem of total probability,}
\displaystyle P(R_2)=P(R_1)P(R_2\mid R_1)+P(G_1)P(R_2\mid G_1).
\displaystyle =\frac57\cdot\frac47+\frac27\cdot\frac67.
\displaystyle =\frac{20}{49}+\frac{12}{49}=\frac{32}{49}.
\displaystyle \therefore \text{the probability that the second ball is red is }\frac{32}{49}.
\displaystyle \\

\displaystyle \textbf{Question 109: }\text{Rahul and Divya were playing the snakes and ladders board game. Each}
\displaystyle \text{one had their own dice to play the game. Rahul was using a red dice, whereas Divya was}
\displaystyle \text{using a black dice. In the beginning of the game, they were using their own dice to play.}
\displaystyle \text{After some time, in order to play the game faster they both started using both the dice}
\displaystyle \text{together for playing. When Divya rolled both red and black dice together then:}
\displaystyle \text{(a) find the conditional probability of obtaining sum greater than }9,\text{ given that black}
\displaystyle \text{dice resulted in a }5.
\displaystyle \text{(b) find the conditional probability that sum of the number on the dice is not }4,\text{ given}
\displaystyle \text{that the numbers on the both the dice are different.}
\displaystyle \text{Answer:}

\displaystyle \text{(a) Let }A\text{ represent obtaining sum greater than }9\text{ and }B\text{ represents black dice resulted in a }5.
\displaystyle n(S)=36.
\displaystyle n(A)=\{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)\}=6.
\displaystyle n(B)=\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}=6.
\displaystyle n(A\cap B)=\{(5,5),(5,6)\}=2.
\displaystyle P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}.
\displaystyle =\frac{\frac{2}{36}}{\frac{6}{36}}=\frac13.

\displaystyle \text{(b) Let }A\text{ represent obtaining sum }4\text{ and }B\text{ represents that both the dice show different number.}
\displaystyle n(S)=36.
\displaystyle n(A)=\{(1,3),(2,2),(3,1)\}=3.
\displaystyle n(B)=30.
\displaystyle n(A\cap B)=\{(1,3),(3,1)\}=2.
\displaystyle P(\text{sum of the numbers showing different number is }4)=P\left(\frac{A}{B}\right).
\displaystyle =\frac{P(A\cap B)}{P(B)}=\frac{\frac{2}{36}}{\frac{30}{36}}=\frac1{15}.
\displaystyle P(\text{sum of the numbers showing different number is not }4).
\displaystyle P\left(\frac{A'}{B}\right)=1-P\left(\frac{A}{B}\right)=1-\frac1{15}=\frac{14}{15}.
\displaystyle \\

\displaystyle \textbf{Question 110: }\text{Imagine you are at a point }A,\text{ a cafe you visit often. Your friend is at}
\displaystyle \text{the point }B,\text{ a bookstore a few blocks away on a straight road. You want to meet your friend}
\displaystyle \text{at a point on the line joining the cafe and the bookstore. Another friend, who is at home on}
\displaystyle \text{the other side of the same road represented by point }P,\text{ also wants to join. You decide to}
\displaystyle \text{determine the exact meeting point by finding the foot of the perpendicular from }P\text{ on the line}
\displaystyle \text{joining the cafe and the bookstore. Given that co-ordinates of the cafe }(A)\text{ are }(1,2,4),
\displaystyle \text{of the bookstore }(B)\text{ are }(3,4,5)\text{ and the home are }(2,1,3),\text{ find the location of the}
\displaystyle \text{meeting point.}
\displaystyle \text{Answer:} \displaystyle \text{Let }Q\text{ be the foot of the perpendicular from }P(2,1,3)\text{ to the line }AB.
\displaystyle A=(1,2,4),\qquad B=(3,4,5).
\displaystyle \therefore \overrightarrow{AB}=(3-1)\widehat{i}+(4-2)\widehat{j}+(5-4)\widehat{k}.
\displaystyle =2\widehat{i}+2\widehat{j}+\widehat{k}.
\displaystyle \text{Let }Q=A+\lambda\overrightarrow{AB}.
\displaystyle \therefore Q=(1+2\lambda,2+2\lambda,4+\lambda).
\displaystyle \therefore \overrightarrow{PQ}=(-1+2\lambda)\widehat{i}+(1+2\lambda)\widehat{j}+(1+\lambda)\widehat{k}.
\displaystyle \text{Since }PQ\perp AB,\quad \overrightarrow{PQ}\cdot\overrightarrow{AB}=0.
\displaystyle 2(-1+2\lambda)+2(1+2\lambda)+(1+\lambda)=0.
\displaystyle -2+4\lambda+2+4\lambda+1+\lambda=0.
\displaystyle \therefore 1+9\lambda=0.
\displaystyle \therefore \lambda=-\frac19.
\displaystyle \therefore Q=\left(1-\frac29,2-\frac29,4-\frac19\right).
\displaystyle \therefore Q=\left(\frac79,\frac{16}{9},\frac{35}{9}\right).
\displaystyle \therefore \text{the meeting point is }\left(\frac79,\frac{16}{9},\frac{35}{9}\right).
\displaystyle \\

\displaystyle \textbf{Question 111: }\text{Two friends are planning a road trip. One friend stays in City A represented}
\displaystyle \text{by the position vector }(-2\widehat{i}+3\widehat{j}+5\widehat{k}).\text{ The trip will start from City A and proceed towards}
\displaystyle \text{City B represented by the position vector }(\widehat{i}+2\widehat{j}+3\widehat{k}).\text{ The friend living in City C is}
\displaystyle \text{represented by the position vector }7\widehat{i}-\widehat{k}\text{ and will join when the first friend passes through}
\displaystyle \text{her city.}
\displaystyle \text{(a) Find the vector equation for the straight path between the cities A and B.}
\displaystyle \text{(b) Hence, find out whether the three cities lie on the same straight path.}
\displaystyle \text{(c) If the two friends now plan to travel }\sqrt{126}\text{ units along the vector }\overrightarrow{AB}\text{ from the City C,}
\displaystyle \text{find the position vector of the destination point.}
\displaystyle \text{Answer:}

\displaystyle \text{(a) Let }\overrightarrow{a}=-2\widehat{i}+3\widehat{j}+5\widehat{k},\ \overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k},\ \overrightarrow{c}=7\widehat{i}-\widehat{k}.
\displaystyle \text{The vector equation of }AB\text{ is given by }\overrightarrow{r}=\overrightarrow{a}+\lambda(\overrightarrow{b}-\overrightarrow{a}).
\displaystyle \overrightarrow{r}=(-2\widehat{i}+3\widehat{j}+5\widehat{k})+\lambda[(\widehat{i}+2\widehat{j}+3\widehat{k})-(-2\widehat{i}+3\widehat{j}+5\widehat{k})].
\displaystyle \overrightarrow{r}=(-2\widehat{i}+3\widehat{j}+5\widehat{k})+\lambda(3\widehat{i}-\widehat{j}-2\widehat{k}).

\displaystyle \text{(b) The three cities will lie on the same straight path if they are collinear.}
\displaystyle \text{So, if }C\text{ lies on }AB,\text{ the three points are collinear.}
\displaystyle \text{When,}
\displaystyle 7\widehat{i}-\widehat{k}=(-2\widehat{i}+3\widehat{j}+5\widehat{k})+\lambda(3\widehat{i}-\widehat{j}-2\widehat{k}),
\displaystyle \text{we get,}
\displaystyle 7=-2+3\lambda,\qquad 0=3-\lambda,\qquad -1=5-2\lambda.
\displaystyle \text{The value of }\lambda=3\text{ satisfies all three equations. So, }C\text{ lies on }AB.
\displaystyle \text{Hence, we can conclude that the three cities lie on the same straight path.}

\displaystyle \text{(c) Let }P\text{ be the destination point.}
\displaystyle \text{Co-ordinates of }P\text{ are }(3r-2,-r+3,-2r+5).
\displaystyle CP=\sqrt{126}=\sqrt{(3r-2-7)^2+(3-r)^2+(5-2r+1)^2}.
\displaystyle r(r-6)=0.
\displaystyle r=0\text{ or }6.
\displaystyle \text{If }r=0,\ P\text{ is }(-2,3,5)\text{ which is City A.}
\displaystyle \text{So, for }r=6,
\displaystyle P\text{ is }(16,-3,-7).
\displaystyle \text{Position vector of }P\text{ is }16\widehat{i}-3\widehat{j}-7\widehat{k}.
\displaystyle \\

\displaystyle \textbf{Question 112: }\text{In the beautiful town of Darjeeling in the Himalayan foothills, the}
\displaystyle \text{city planning committee wants to construct two major roads to connect the various}
\displaystyle \text{neighbourhoods. The two roads are represented by the equations } \\ \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}\text{ and }\frac{x-2}{1}=\frac{y-4}{k}=\frac{z-6}{7}.
\displaystyle \text{As an in charge of the planning committee, ensure that these roads lie on the same plane to}
\displaystyle \text{facilitate efficient urban planning and infrastructure development.}
\displaystyle \text{(a) For what value of }k,\text{ will the construction meet the requirement?}
\displaystyle \text{(b) Hence, find the equation of the plane containing these lines.}
\displaystyle \text{Answer:}
\displaystyle \text{For the first line, take }A(-1,-3,-5)\text{ and direction vector }\overrightarrow{d_1}=(3,5,7).
\displaystyle \text{For the second line, take }B(2,4,6)\text{ and direction vector }\overrightarrow{d_2}=(1,k,7).
\displaystyle \overrightarrow{AB}=(2+1,4+3,6+5)=(3,7,11).

\displaystyle \text{(a) For the two lines to be coplanar,}
\displaystyle \overrightarrow{AB}\cdot(\overrightarrow{d_1}\times\overrightarrow{d_2})=0.
\displaystyle \overrightarrow{d_1}\times\overrightarrow{d_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&5&7\\1&k&7\end{vmatrix}.
\displaystyle =(35-7k)\widehat{i}-14\widehat{j}+(3k-5)\widehat{k}.
\displaystyle \therefore 3(35-7k)+7(-14)+11(3k-5)=0.
\displaystyle 105-21k-98+33k-55=0.
\displaystyle 12k-48=0.
\displaystyle \therefore k=4.

\displaystyle \text{(b) For }k=4,\quad \overrightarrow{d_2}=(1,4,7).
\displaystyle \overrightarrow{d_1}\times\overrightarrow{d_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&5&7\\1&4&7\end{vmatrix}.
\displaystyle =7\widehat{i}-14\widehat{j}+7\widehat{k}=7(\widehat{i}-2\widehat{j}+\widehat{k}).
\displaystyle \therefore \text{a normal vector to the plane is }\widehat{i}-2\widehat{j}+\widehat{k}.
\displaystyle \text{The plane passes through }A(-1,-3,-5).
\displaystyle \therefore (x+1)-2(y+3)+(z+5)=0.
\displaystyle \therefore x-2y+z=0.
\displaystyle \therefore \text{the required plane is }x-2y+z=0.
\displaystyle \\

\displaystyle \textbf{Question 113: }\text{The plane }px+y+pz=r\text{ intersects the co-ordinate axes at }A,\ B\text{ and }C.
\displaystyle \text{(a) Find the co-ordinates of the points }A,\ B\text{ and }C.
\displaystyle \text{(b) Find the co-ordinates of orthocentre of the triangle }ABC.
\displaystyle \text{Answer:}
\displaystyle \text{Given plane is }px+y+pz=r.

\displaystyle \text{(a) For the }x\text{-intercept, put }y=0,\ z=0.
\displaystyle px=r\Rightarrow x=\frac rp.
\displaystyle \therefore A=\left(\frac rp,0,0\right).
\displaystyle \text{For the }y\text{-intercept, put }x=0,\ z=0.
\displaystyle y=r.
\displaystyle \therefore B=(0,r,0).
\displaystyle \text{For the }z\text{-intercept, put }x=0,\ y=0.
\displaystyle pz=r\Rightarrow z=\frac rp.
\displaystyle \therefore C=\left(0,0,\frac rp\right).

\displaystyle \text{(b) Let the orthocentre of }\triangle ABC\text{ be }H(x,y,z).
\displaystyle \overrightarrow{BC}=\left(0,-r,\frac rp\right).
\displaystyle \text{Since }AH\perp BC,
\displaystyle \left(x-\frac rp,y,z\right)\cdot\left(0,-r,\frac rp\right)=0.
\displaystyle -ry+\frac rpz=0.
\displaystyle \therefore z=py.\qquad ...(1)
\displaystyle \overrightarrow{AC}=\left(-\frac rp,0,\frac rp\right).
\displaystyle \text{Since }BH\perp AC,
\displaystyle (x,y-r,z)\cdot\left(-\frac rp,0,\frac rp\right)=0.
\displaystyle -x+z=0.
\displaystyle \therefore x=z.\qquad ...(2)
\displaystyle \text{From (1) and (2), }x=z=py.
\displaystyle \text{Since }H\text{ lies on the plane, }px+y+pz=r.
\displaystyle p(py)+y+p(py)=r.
\displaystyle (1+2p^2)y=r.
\displaystyle \therefore y=\frac{r}{1+2p^2}.
\displaystyle \therefore x=z=\frac{pr}{1+2p^2}.
\displaystyle \therefore H=\left(\frac{pr}{1+2p^2},\frac{r}{1+2p^2},\frac{pr}{1+2p^2}\right).
\displaystyle \\

\displaystyle \textbf{Question 114: }\text{From any point }P(2,1,2)\text{ perpendiculars }PM\text{ and } \\ PN\text{ are drawn to }ZX\text{ and }XY\text{ planes.}
\displaystyle \text{(a) If }O\text{ is the origin, find the equation of the plane }OMN.
\displaystyle \text{(b) Find }\theta,\text{ if }\theta\text{ is the angle made by }OP\text{ with the plane }OMN.
\displaystyle \text{(c) If }\alpha,\beta\text{ and }\gamma\text{ are the angles made by }OP\text{ with the co-ordinate planes, prove that}
\displaystyle \mathrm{cosec}^2\theta=\mathrm{cosec}^2\alpha+\mathrm{cosec}^2\beta+\mathrm{cosec}^2\gamma.
\displaystyle \text{Answer:}
\displaystyle P=(2,1,2).
\displaystyle \text{Since }PM\perp ZX\text{ plane, }M=(2,0,2).
\displaystyle \text{Since }PN\perp XY\text{ plane, }N=(2,1,0).

\displaystyle \text{(a) }\overrightarrow{OM}=2\widehat{i}+2\widehat{k},\qquad \overrightarrow{ON}=2\widehat{i}+\widehat{j}.
\displaystyle \text{A normal vector to plane }OMN\text{ is }\overrightarrow{OM}\times\overrightarrow{ON}.
\displaystyle \overrightarrow{OM}\times\overrightarrow{ON}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&0&2\\2&1&0\end{vmatrix}.
\displaystyle =-2\widehat{i}+4\widehat{j}+2\widehat{k}.
\displaystyle \therefore \text{a normal vector is }\widehat{i}-2\widehat{j}-\widehat{k}.
\displaystyle \text{Since the plane passes through the origin,}
\displaystyle \therefore x-2y-z=0.

\displaystyle \text{(b) }\overrightarrow{OP}=2\widehat{i}+\widehat{j}+2\widehat{k}.
\displaystyle \text{A normal vector to the plane is }\overrightarrow{n}=\widehat{i}-2\widehat{j}-\widehat{k}.
\displaystyle \sin\theta=\frac{|\overrightarrow{OP}\cdot\overrightarrow{n}|}{|\overrightarrow{OP}||\overrightarrow{n}|}.
\displaystyle =\frac{|2-2-2|}{\sqrt{4+1+4}\sqrt{1+4+1}}.
\displaystyle =\frac{2}{3\sqrt6}.
\displaystyle \therefore \theta=\sin^{-1}\left(\frac{2}{3\sqrt6}\right).

\displaystyle \text{(c) }|\overrightarrow{OP}|=\sqrt{2^2+1^2+2^2}=3.
\displaystyle \text{Let }\alpha,\beta,\gamma\text{ be the angles made by }OP\text{ with the }YZ,\ ZX,\ XY\text{ planes respectively.}
\displaystyle \sin\alpha=\frac23,\qquad \sin\beta=\frac13,\qquad \sin\gamma=\frac23.
\displaystyle \therefore \mathrm{cosec}^2\alpha+\mathrm{cosec}^2\beta+\mathrm{cosec}^2\gamma
\displaystyle =\frac94+9+\frac94=\frac{27}{2}.
\displaystyle \text{Also, }\sin\theta=\frac{2}{3\sqrt6}.
\displaystyle \therefore \mathrm{cosec}^2\theta=\frac{54}{4}=\frac{27}{2}.
\displaystyle \therefore \mathrm{cosec}^2\theta=\mathrm{cosec}^2\alpha+\mathrm{cosec}^2\beta+\mathrm{cosec}^2\gamma.
\displaystyle \\

\displaystyle \textbf{Question 115: }\text{Draw a rough sketch and find the area enclosed by the curve }
\displaystyle y=-x^2\text{ and the line} \ x+y+2=0.
\displaystyle \text{Answer:}
\displaystyle \text{The given curves are }y=-x^2\text{ and }y=-x-2.\displaystyle \text{At the points of intersection,}
\displaystyle -x^2=-x-2.
\displaystyle x^2-x-2=0.
\displaystyle (x-2)(x+1)=0.
\displaystyle \therefore x=-1\text{ or }x=2.
\displaystyle \text{For }x=-1,\quad y=-1.
\displaystyle \text{For }x=2,\quad y=-4.
\displaystyle \therefore \text{the points of intersection are }(-1,-1)\text{ and }(2,-4).
\displaystyle \text{For }-1\leq x\leq2,\text{ the curve }y=-x^2\text{ lies above the line }y=-x-2.
\displaystyle \therefore \text{Required area}=\int_{-1}^{2}\{(-x^2)-(-x-2)\}\,dx.
\displaystyle =\int_{-1}^{2}(-x^2+x+2)\,dx.
\displaystyle =\left[-\frac{x^3}{3}+\frac{x^2}{2}+2x\right]_{-1}^{2}.
\displaystyle =\left(-\frac83+2+4\right)-\left(\frac13+\frac12-2\right).
\displaystyle =\frac{10}{3}+\frac76.
\displaystyle =\frac{27}{6}=\frac92.
\displaystyle \therefore \text{the area enclosed by the curve and the line is }\frac92\text{ square units.}
\displaystyle \\

\displaystyle \textbf{Question 116: }\text{Find the area of the region bounded by the x-axis, part of the curve }
\displaystyle y=1+\frac{8}{x^2},\text{ and} \ \text{the ordinates }x=2\text{ and }x=4.\text{ If the ordinate at }x=a
\displaystyle \text{ divides the area into two equal parts,} \ \text{find the value of }a.
\displaystyle \text{Answer:} \displaystyle \text{Total area}=\int_2^4\left(1+\frac{8}{x^2}\right)dx.
\displaystyle =\left[x-\frac{8}{x}\right]_2^4.
\displaystyle =\left(4-\frac84\right)-\left(2-\frac82\right).
\displaystyle =2-(-2)=4\text{ square units.}
\displaystyle \therefore \text{each equal part has area }2\text{ square units.}
\displaystyle \therefore \int_2^a\left(1+\frac{8}{x^2}\right)dx=2.
\displaystyle \left[x-\frac{8}{x}\right]_2^a=2.
\displaystyle a-\frac8a-\left(2-\frac82\right)=2.
\displaystyle a-\frac8a+2=2.
\displaystyle a-\frac8a=0.
\displaystyle a^2=8.
\displaystyle \text{Since }2<a<4,\quad a=2\sqrt2.
\displaystyle \therefore \text{the required value of }a\text{ is }2\sqrt2.
\displaystyle \\

\displaystyle \textbf{Question 117: }\text{Find the area enclosed between the co-ordinate axis and the curve }
\displaystyle y^2=4a(x+\lambda)\text{ in the}  \ \text{second quadrant.}
\displaystyle \text{Answer:} \displaystyle \text{Given, }y^2=4a(x+\lambda).
\displaystyle \text{In the second quadrant, }y=2\sqrt{a(x+\lambda)}.
\displaystyle \text{The curve meets the x-axis when }y=0.
\displaystyle \therefore x=-\lambda.
\displaystyle \text{The curve meets the y-axis when }x=0.
\displaystyle \therefore y=2\sqrt{a\lambda}.
\displaystyle \text{Required area}=\int_{-\lambda}^{0}y\,dx.
\displaystyle =2\sqrt{a}\int_{-\lambda}^{0}\sqrt{x+\lambda}\,dx.
\displaystyle =2\sqrt{a}\left[\frac{2}{3}(x+\lambda)^{3/2}\right]_{-\lambda}^{0}.
\displaystyle =\frac{4\sqrt{a}}{3}\left[\lambda^{3/2}-0\right].
\displaystyle =\frac{4}{3}\sqrt{a}\lambda^{3/2}.
\displaystyle \therefore \text{the required area is }\frac{4}{3}\sqrt{a}\lambda^{3/2}\text{ square units.}
\displaystyle \\

\displaystyle \textbf{Question 118: }\text{A study was conducted to investigate the relationship between the number of}
\displaystyle \text{hours a student studies per week }(X)\text{ and their scores on a standardized test }(Y).\text{ The}
\displaystyle \text{following statistical data was collected from a sample of }50\text{ students.}
\displaystyle \begin{array}{c|cc}&X&Y\\\hline\text{Mean}&15&75\\\text{Standard Deviation (SD)}&4&10\end{array}
\displaystyle \text{The correlation coefficient between }X\text{ and }Y\text{ is }0.65.
\displaystyle \text{(a) Estimate the test score for a student who studies }20\text{ hours per week.}
\displaystyle \text{(b) If the pass mark is }40,\text{ then how many hours does a student need to study to pass?}
\displaystyle \text{Answer:}
\displaystyle \bar X=15,\quad \bar Y=75,\quad \sigma_X=4,\quad \sigma_Y=10,\quad r=0.65.

\displaystyle \text{(a) The regression equation of }Y\text{ on }X\text{ is}
\displaystyle Y-\bar Y=r\frac{\sigma_Y}{\sigma_X}(X-\bar X).
\displaystyle Y-75=0.65\left(\frac{10}{4}\right)(X-15).
\displaystyle Y-75=1.625(X-15).
\displaystyle \text{For }X=20,
\displaystyle Y-75=1.625(20-15)=8.125.
\displaystyle \therefore Y=83.125.
\displaystyle \therefore \text{the estimated test score is }83.125\text{ or approximately }83.13.

\displaystyle \text{(b) The regression equation of }X\text{ on }Y\text{ is}
\displaystyle X-\bar X=r\frac{\sigma_X}{\sigma_Y}(Y-\bar Y).
\displaystyle X-15=0.65\left(\frac{4}{10}\right)(Y-75).
\displaystyle X-15=0.26(Y-75).
\displaystyle \text{For the pass mark }Y=40,
\displaystyle X-15=0.26(40-75).
\displaystyle X-15=-9.1.
\displaystyle \therefore X=5.9.
\displaystyle \therefore \text{a student needs to study approximately }5.9\text{ hours per week to score }40.
\displaystyle \\

\displaystyle \textbf{Question 119: }\text{The corner points of the feasible region determined by the system}
\displaystyle \text{of linear constraints are as shown below:} \displaystyle A(0,8),\quad B(4,10),\quad C(6,8),\quad D(6,5),\quad E(4,0).
\displaystyle \text{Answer the following questions:}
\displaystyle \text{(a) Let }Z=3x-4y\text{ be the objective function. Find the maximum and minimum value}
\displaystyle \text{of }Z\text{ and also the corresponding points at which the maximum and minimum value occurs.}
\displaystyle \text{(b) Let }Z=px+qy\text{ where }p,q>0\text{ be the objective function. Find the condition on }p
\displaystyle \text{and }q\text{ so that the maximum value of }Z\text{ occurs at }B(4,10)\text{ and }C(6,8).
\displaystyle \text{(c) State the number of optimal solutions in this case.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) For }Z=3x-4y,
\displaystyle \begin{array}{c|c}\text{Corner Point}&Z=3x-4y\\\hline A(0,8)&-32\\B(4,10)&-28\\C(6,8)&-14\\D(6,5)&-2\\E(4,0)&12\end{array}
\displaystyle \therefore \text{the maximum value of }Z\text{ is }12\text{ at }E(4,0).
\displaystyle \therefore \text{the minimum value of }Z\text{ is }-32\text{ at }A(0,8).

\displaystyle \text{(b) Since the maximum value occurs at both }B(4,10)\text{ and }C(6,8),
\displaystyle Z_B=Z_C.
\displaystyle 4p+10q=6p+8q.
\displaystyle 2q=2p.
\displaystyle \therefore p=q.
\displaystyle \text{Since }p,q>0,\text{ the required condition is }p=q>0.

\displaystyle \text{(c) When }p=q,\text{ the objective function has the same maximum value at }B\text{ and }C.
\displaystyle \text{Hence, every point on the line segment }BC\text{ gives the same maximum value of }Z.
\displaystyle \therefore \text{there are infinitely many optimal solutions.}
\displaystyle \\

\displaystyle \textbf{Question 120: }\text{A movie cinema is considering significantly reducing the price of their}
\displaystyle \text{popcorn as they believe their customers spend more on drinks when they buy popcorn. They}
\displaystyle \text{recorded the following data of the daily revenue from popcorn, }x,\text{ and the daily revenue}
\displaystyle \text{from drinks, }y,\text{ over }8\text{ randomly selected days.}
\displaystyle \begin{array}{c|c}\text{Popcorn revenue }(x)&\text{Drinks revenue }(y)\\\hline14&22\\12&23\\12&17\\14&24\\16&18\\10&25\\13&23\\12&24\end{array}
\displaystyle \text{(a) Find }\bar{x},\bar{y}.
\displaystyle \text{(b) Using }\bar{x},\bar{y},\text{ find regression coefficient of }y\text{ on }x.
\displaystyle \text{(c) The equation of the regression line }y\text{ on }x\text{ is in the form }y=a+bx.\text{ Calculate the}
\displaystyle \text{values of }a\text{ and }b.
\displaystyle \text{Answer:}

\displaystyle \text{(a)}
\displaystyle \begin{array}{c|c|c|c}x&y&(x-\bar{x})^2&(x-\bar{x})(y-\bar{y})\\\hline14&22&1&0\\12&23&1&-1\\13&17&0&0\\14&24&1&2\\16&18&9&-12\\10&25&9&-9\\13&23&0&0\\12&24&1&-2\end{array}
\displaystyle \sum x=104.
\displaystyle \sum y=176.
\displaystyle \sum(x-\bar{x})^2=22.
\displaystyle \sum(x-\bar{x})(y-\bar{y})=-22.

\displaystyle \text{(b)}
\displaystyle \bar{x}=\frac{104}{8}=13.
\displaystyle \bar{y}=\frac{176}{8}=22.
\displaystyle \therefore b_{yx}=\frac{\sum(x-\bar{x})(y-\bar{y})}{\sum(x-\bar{x})^2}=\frac{-22}{22}=-1.
\displaystyle \therefore \text{The line of regression of }y\text{ on }x\text{ is given by}

\displaystyle \text{(c) }y-22=-1(x-13).
\displaystyle y=-x+35.
\displaystyle \text{Comparing }y=-x+35\text{ with }y=a+bx\text{ we get,}
\displaystyle a=35\text{ and }b=-1.
\displaystyle \\

\displaystyle \textbf{Question 121: }\text{A part of the graph of the function }
\displaystyle f(x)=2x^3-3x^2-12x+8,\ x\in R\text{ is shown below:} \ \text{Answer the following questions.} \displaystyle \text{(a) Explain why }f\text{ does not have an inverse.}
\displaystyle \text{(b) The domain of }f\text{ is now restricted to }a\leq x\leq b\text{ where }a<0\text{ and }
\displaystyle b>0,\text{ and }b\text{ are} \ \text{chosen so that }f\text{ has an inverse and the interval }[a,b]
\displaystyle \text{ is as large as possible. Find the} \ \text{domain and range of }f^{-1}.
\displaystyle \text{Answer:}
\displaystyle \text{(a) From the graph, some horizontal lines intersect the curve at more than one point.}
\displaystyle \therefore f\text{ fails the horizontal line test and is not one-one on }R.
\displaystyle \therefore f\text{ does not have an inverse on }R.

\displaystyle \text{(b) }f(x)=2x^3-3x^2-12x+8.
\displaystyle f'(x)=6x^2-6x-12.
\displaystyle =6(x+1)(x-2).
\displaystyle f'(x)=0\Rightarrow x=-1\text{ or }x=2.
\displaystyle \text{For }-1<x<2,\quad f'(x)<0.
\displaystyle \therefore f\text{ is strictly decreasing on }[-1,2].
\displaystyle \text{Hence, the largest required interval containing negative and positive values is }[-1,2].
\displaystyle f(-1)=2(-1)^3-3(-1)^2-12(-1)+8=15.
\displaystyle f(2)=2(2)^3-3(2)^2-12(2)+8=-12.
\displaystyle \therefore \text{the range of }f\text{ on }[-1,2]\text{ is }[-12,15].
\displaystyle \therefore \text{Domain of }f^{-1}=[-12,15].
\displaystyle \therefore \text{Range of }f^{-1}=[-1,2].
\displaystyle \\

\displaystyle \textbf{Question 122: }\text{In a Kabaddi league, two matches are being played between Jaipur and Delhi.}
\displaystyle \text{It is assumed that the outcomes of two games are independent. The probability of Jaipur}
\displaystyle \text{winning, drawing and losing the game against Delhi are }\frac12,\ \frac3{10},\text{ and }\frac15\text{ respectively. Each}
\displaystyle \text{team gets }5\text{ points for win, }3\text{ points for draw and }0\text{ point for loss in a game. After}
\displaystyle \text{two games, find the probability that:}
\displaystyle \text{(a) Jaipur has more points than Delhi.}
\displaystyle \text{(b) Jaipur and Delhi have equal points.}
\displaystyle \text{Answer:}

\displaystyle P(\text{Winning of Delhi})=P(\text{Losing of Jaipur})=\frac15.
\displaystyle P(\text{Losing of Delhi})=P(\text{Winning of Jaipur})=\frac12.
\displaystyle P(\text{Drawing of Delhi})=P(\text{drawing of Jaipur})=\frac3{10}.
\displaystyle \text{Let }X\text{ be the point of Jaipur after two games and }Y\text{ be the points of Delhi after two games.}
\displaystyle \text{According to the given information,}
\displaystyle \begin{array}{c|ccccc}X&10&8&6&3&0\\\hline Y&0&3&6&8&10\end{array}

\displaystyle \text{(a) Now, }P(X>Y)
\displaystyle =P(\text{1st match win by Jaipur})P(\text{2nd match win by Jaipur})
\displaystyle +P(\text{1st match win by Jaipur})P(\text{2nd match draw by Jaipur})
\displaystyle +P(\text{1st match draw by Jaipur})P(\text{2nd match win by Jaipur}).
\displaystyle =\frac12\cdot\frac12+\frac12\cdot\frac3{10}+\frac3{10}\cdot\frac12=\frac{11}{20}.

\displaystyle \text{(b) Now, }P(X=Y)
\displaystyle =P(\text{1st match win by Jaipur})P(\text{2nd match win by Delhi})
\displaystyle +P(\text{1st match win by Delhi})P(\text{2nd match win by Jaipur})
\displaystyle +P(\text{1st match draw by Jaipur})P(\text{2nd match draw by Delhi}).
\displaystyle =\frac12\cdot\frac15+\frac15\cdot\frac12+\frac3{10}\cdot\frac3{10}=\frac{29}{100}.
\displaystyle \\

\displaystyle \textbf{Question 123: }\text{Two drones are being used for soil analysis over an area of farmland.}
\displaystyle \text{Drone A has been programmed to fly on the path given by } \\ \overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda(\widehat{i}-2\widehat{j}+2\widehat{k})\text{ and drone B}
\displaystyle \text{has been programmed to fly on the path }\overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu(3\widehat{i}-2\widehat{j}-2\widehat{k}).
\displaystyle \text{At what points on their respective paths should they reach, so that they will be closest to}
\displaystyle \text{each other?}
\displaystyle \text{Answer:} \displaystyle \text{Let }P\text{ be a point on the path of Drone A and }Q\text{ be a point on the path of Drone B.}
\displaystyle P=(6+\lambda,\ 2-2\lambda,\ 2+2\lambda).
\displaystyle Q=(-4+3\mu,\ -2\mu,\ -1-2\mu).
\displaystyle \therefore \overrightarrow{PQ}=(-10+3\mu-\lambda)\widehat{i}+(-2-2\mu+2\lambda)\widehat{j}
\displaystyle +(-3-2\mu-2\lambda)\widehat{k}.
\displaystyle \text{The direction vectors of the two paths are}
\displaystyle \overrightarrow{d_1}=\widehat{i}-2\widehat{j}+2\widehat{k},\qquad \overrightarrow{d_2}=3\widehat{i}-2\widehat{j}-2\widehat{k}.
\displaystyle \text{For the closest points, }\overrightarrow{PQ}\perp\overrightarrow{d_1}\text{ and }\overrightarrow{PQ}\perp\overrightarrow{d_2}.
\displaystyle \therefore \overrightarrow{PQ}\cdot\overrightarrow{d_1}=0.
\displaystyle (-10+3\mu-\lambda)-2(-2-2\mu+2\lambda)+2(-3-2\mu-2\lambda)=0.
\displaystyle \therefore \mu-3\lambda=4.\qquad ...(1)
\displaystyle \text{Also, }\overrightarrow{PQ}\cdot\overrightarrow{d_2}=0.
\displaystyle 3(-10+3\mu-\lambda)-2(-2-2\mu+2\lambda)-2(-3-2\mu-2\lambda)=0.
\displaystyle \therefore 17\mu-3\lambda=20.\qquad ...(2)
\displaystyle \text{From (1), }\mu=4+3\lambda.
\displaystyle \text{Substituting in (2),}
\displaystyle 17(4+3\lambda)-3\lambda=20.
\displaystyle 48\lambda=-48.
\displaystyle \therefore \lambda=-1,\qquad \mu=1.
\displaystyle \therefore P=(6-1,\ 2+2,\ 2-2)=(5,4,0).
\displaystyle \therefore Q=(-4+3,\ -2,\ -1-2)=(-1,-2,-3).
\displaystyle \therefore \text{the closest points are }(5,4,0)\text{ and }(-1,-2,-3).
\displaystyle \\

\displaystyle \textbf{Question 124: }\text{A manufacturing company produces two types of cell phones, Android and iOS.}
\displaystyle \text{The company has resources to make at the most }300\text{ sets a week. It takes Rs. }1800\text{ to make}
\displaystyle \text{an Android set and Rs. }2700\text{ to make an iOS set. The company cannot spend more than}
\displaystyle \text{Rs. }648000\text{ a week to make cell phones. The company makes a profit of Rs. }510\text{ per Android}
\displaystyle \text{and Rs. }675\text{ per iOS set. If }x\text{ and }y\text{ denote, respectively, the number of Android sets and}
\displaystyle \text{iOS sets made each week, then formulate this problem as a Linear Programming Problem}
\displaystyle \text{(LPP) given that the objective is to maximize the profit. Based on it, answer the questions}
\displaystyle \text{that follow.}
\displaystyle \text{(a) What will be the maximum profit function on }x\text{ and }y\text{ sets?}
\displaystyle \text{(b) What will be the values of your objective function in the feasible region?}
\displaystyle \text{(c) At what point the maximum profit will occur?}
\displaystyle \text{(d) What is the weekly cost (in Rs.) of manufacturing the sets?}
\displaystyle \text{Answer:} \displaystyle \text{Let }x\text{ and }y\text{ be the number of Android and iOS sets produced per week.}
\displaystyle \text{Since at most }300\text{ sets can be produced,}
\displaystyle x+y\leq300.
\displaystyle \text{Also, }1800x+2700y\leq648000.
\displaystyle \therefore 2x+3y\leq720.
\displaystyle \text{Further, }x\geq0,\qquad y\geq0.

\displaystyle \text{(a) Profit on }x\text{ Android sets}=510x.
\displaystyle \text{Profit on }y\text{ iOS sets}=675y.
\displaystyle \therefore Z=510x+675y.
\displaystyle \text{The objective is to maximize }Z=510x+675y.

\displaystyle \text{(b) The boundary lines are }x+y=300\text{ and }2x+3y=720.
\displaystyle \text{Their point of intersection is obtained from}
\displaystyle x+y=300,\qquad 2x+3y=720.
\displaystyle \therefore y=120,\qquad x=180.
\displaystyle \text{Hence, the corner points are }(0,0),(300,0),(180,120)\text{ and }(0,240).
\displaystyle \begin{array}{c|c}\text{Corner Point}&Z=510x+675y\\\hline(0,0)&0\\(300,0)&153000\\(180,120)&172800\\(0,240)&162000\end{array}

\displaystyle \text{(c) The maximum value of }Z\text{ is Rs. }172800\text{ at }(180,120).
\displaystyle \therefore \text{the company should produce }180\text{ Android sets and }120\text{ iOS sets.}

\displaystyle \text{(d) Weekly manufacturing cost}=1800(180)+2700(120).
\displaystyle =324000+324000.
\displaystyle =\text{Rs. }648000.
\displaystyle \therefore \text{the weekly cost of manufacturing the sets is Rs. }648000.
\displaystyle \\

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