\displaystyle \text{VERY SHORT ANSWER QUESTIONS - 1 Mark Each}

\displaystyle \textbf{Question 46: }\text{If the domain of the function }f(x)=\sqrt{\cos^{-1}(3x)+\frac{\pi}{4}}\text{ is }[a,b],
\displaystyle \text{then find the value of }a+b.
\displaystyle \text{Answer:}
\displaystyle \text{For }\cos^{-1}(3x)\text{ to be defined,}
\displaystyle -1\leq3x\leq1.
\displaystyle \therefore -\frac13\leq x\leq\frac13.
\displaystyle \text{Also, }0\leq\cos^{-1}(3x)\leq\pi.
\displaystyle \therefore \cos^{-1}(3x)+\frac{\pi}{4}>0\text{ throughout the above interval.}
\displaystyle \therefore [a,b]=\left[-\frac13,\frac13\right].
\displaystyle \therefore a+b=-\frac13+\frac13=0.
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{Each triangular face of the pyramid of Peace of Kazakhstan is made up of }25
\displaystyle \text{smaller triangles as shown in figure below:} \displaystyle \text{Using the above information and concept of determinant answer the following question.}
\displaystyle \text{If }(1,2)\text{ and }(3,6)\text{ is the co-ordinates of the two vertices of one of the smaller triangles}
\displaystyle \text{and its area is }5\text{ square cm, then find the equation of line on which the third vertex of the}
\displaystyle \text{triangle lie.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the third vertex be }(x,y).
\displaystyle \text{Area of the triangle}=\frac12\left|\begin{vmatrix}1&2&1\\3&6&1\\x&y&1\end{vmatrix}\right|.
\displaystyle \therefore \frac12\left|1(6-y)-2(3-x)+(3y-6x)\right|=5.
\displaystyle \frac12|2y-4x|=5.
\displaystyle |y-2x|=5.
\displaystyle \therefore y-2x=5\quad\text{or}\quad y-2x=-5.
\displaystyle \therefore 2x-y+5=0\quad\text{or}\quad 2x-y-5=0.
\displaystyle \therefore \text{the third vertex lies on either of these two parallel lines.}
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{The figure given below shows the graph of a function }y=f(x).
\displaystyle \text{What is the derivative} \ \text{of the function?} \displaystyle \text{Answer:}
\displaystyle \text{The graph consists of two straight-line segments meeting at }(-3,0).
\displaystyle \text{For }x<-3,\text{ the slope of the graph is }-1.
\displaystyle \therefore f'(x)=-1,\qquad x<-3.
\displaystyle \text{For }x>-3,\text{ the line passes through }(-3,0)\text{ and }(0,3).
\displaystyle \therefore \text{Slope}=\frac{3-0}{0-(-3)}=1.
\displaystyle \therefore f'(x)=1,\qquad x>-3.
\displaystyle \text{At }x=-3,\text{ the left-hand derivative is }-1\text{ and the right-hand derivative is }1.
\displaystyle \therefore f'(x)=\begin{cases}-1,&x<-3\\1,&x>-3\end{cases},\quad f'(-3)\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{Find the angle between the normal to the planes:}
\displaystyle \overrightarrow{r}\cdot(\widehat{i}-\widehat{j}+\widehat{k})=3\text{ and }\overrightarrow{r}\cdot(3\widehat{i}+2\widehat{j}-\widehat{k})+5=0.
\displaystyle \text{Answer:} \displaystyle \text{The normal vectors to the two planes are}
\displaystyle \overrightarrow{n_1}=\widehat{i}-\widehat{j}+\widehat{k}\quad\text{and}\quad\overrightarrow{n_2}=3\widehat{i}+2\widehat{j}-\widehat{k}.
\displaystyle \overrightarrow{n_1}\cdot\overrightarrow{n_2}=(1)(3)+(-1)(2)+(1)(-1).
\displaystyle =3-2-1=0.
\displaystyle \therefore \overrightarrow{n_1}\cdot\overrightarrow{n_2}=|\overrightarrow{n_1}||\overrightarrow{n_2}|\cos\theta=0.
\displaystyle \therefore \cos\theta=0.
\displaystyle \therefore \theta=90^\circ.
\displaystyle \therefore \text{the angle between the normals to the two planes is }90^\circ.
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{The Average Revenue (AR) and Marginal Revenue (MR) of a company}
\displaystyle \text{on selling }x \ \text{units (on }x\text{-axis) is shown in the following graph:} \displaystyle \text{At how many units, should the manufacturer limit production so that each additional unit}
\displaystyle \text{DOES NOT bring losses?}
\displaystyle \text{Answer:}
\displaystyle \text{Marginal Revenue represents the additional revenue earned by selling one more unit.}
\displaystyle \text{For an additional unit not to bring a loss, }MR\geq0.
\displaystyle \text{From the graph, }MR=0\text{ at }x=30.
\displaystyle \text{For }x>30,\ MR<0.
\displaystyle \therefore \text{the manufacturer should limit production to }30\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 51: }f(x)=x^2+2x-3\text{ for }x\geq-1.
\displaystyle \text{Given that the minimum value of }x^2+2x-3\text{ occurs when }x=-1,\text{ explain why }
\displaystyle f(x)\text{ has an} \ \text{inverse.}
\displaystyle \text{Answer:}
\displaystyle f(x)=x^2+2x-3=(x+1)^2-4.
\displaystyle \text{The minimum value of }f(x)\text{ occurs at }x=-1.
\displaystyle f(-1)=-4.
\displaystyle \text{For }x\geq-1,\ f(x)\text{ is strictly increasing.}
\displaystyle \therefore f(x)\text{ is one-one on }[-1,\infty).
\displaystyle \text{Also, the range of }f\text{ on }[-1,\infty)\text{ is }[-4,\infty).
\displaystyle \therefore f:[-1,\infty)\to[-4,\infty)\text{ is one-one and onto.}
\displaystyle \therefore f\text{ is invertible.}
\displaystyle \text{To find the inverse, let }y=(x+1)^2-4.
\displaystyle y+4=(x+1)^2.
\displaystyle \text{Since }x\geq-1,\ x+1\geq0.
\displaystyle \therefore x+1=\sqrt{y+4}.
\displaystyle \therefore x=-1+\sqrt{y+4}.
\displaystyle \therefore f^{-1}(x)=-1+\sqrt{x+4},\qquad x\geq-4.
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{Explain why }x+4y=12,\ x,y\in N\text{ is NOT a symmetric relation.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }xRy\text{ if }x+4y=12.
\displaystyle \text{Take }x=4\text{ and }y=2.
\displaystyle 4+4(2)=12.
\displaystyle \therefore 4R2.
\displaystyle \text{Now, }2+4(4)=18\ne12.
\displaystyle \therefore 2\mathrel{\not R}4.
\displaystyle \therefore xRy\text{ does not necessarily imply }yRx.
\displaystyle \therefore \text{the relation is not symmetric.}
\displaystyle \\

\displaystyle \textbf{Question 53: }\text{If }\cos^{-1}\frac{x}{a}-\cos^{-1}\frac{y}{b}=\frac{\pi}{3}\text{ and} \sin^{-1}\frac{x}{a}+\sin^{-1}\frac{y}{b}=\frac{2\pi}{3},
\displaystyle \text{ then find the value of}  \  4\frac{x^2}{a^2}+\frac{y^2}{b^2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\cos^{-1}\frac{x}{a}\text{ and }B=\cos^{-1}\frac{y}{b}.
\displaystyle \therefore A-B=\frac{\pi}{3}.
\displaystyle \text{Using }\sin^{-1}t+\cos^{-1}t=\frac{\pi}{2},
\displaystyle \sin^{-1}\frac{x}{a}=\frac{\pi}{2}-A\text{ and }\sin^{-1}\frac{y}{b}=\frac{\pi}{2}-B.
\displaystyle \therefore \left(\frac{\pi}{2}-A\right)+\left(\frac{\pi}{2}-B\right)=\frac{2\pi}{3}.
\displaystyle \pi-(A+B)=\frac{2\pi}{3}.
\displaystyle \therefore A+B=\frac{\pi}{3}.
\displaystyle \text{Solving }A-B=\frac{\pi}{3}\text{ and }A+B=\frac{\pi}{3},
\displaystyle A=\frac{\pi}{3},\qquad B=0.
\displaystyle \therefore \frac{x}{a}=\cos\frac{\pi}{3}=\frac12,\qquad \frac{y}{b}=\cos0=1.
\displaystyle \therefore 4\frac{x^2}{a^2}+\frac{y^2}{b^2}=4\left(\frac12\right)^2+(1)^2.
\displaystyle =1+1=2.
\displaystyle \therefore \text{the required value is }2.
\displaystyle \\

\displaystyle \textbf{Question 54: }\text{The matrix }A=\begin{bmatrix}a&6&b\\2c&x+1&8\\-1&-8&y-3\end{bmatrix}\text{ is a skew symmetric}
\displaystyle \text{matrix, then find the value of }ab+bc-xy.
\displaystyle \text{Answer:}
\displaystyle \text{For a skew symmetric matrix, }A^T=-A.
\displaystyle \text{Hence, all diagonal elements are zero.}
\displaystyle a=0,\qquad x+1=0,\qquad y-3=0.
\displaystyle \therefore a=0,\qquad x=-1,\qquad y=3.
\displaystyle \text{Also, }2c=-6.
\displaystyle \therefore c=-3.
\displaystyle \text{And }-1=-b.
\displaystyle \therefore b=1.
\displaystyle \therefore ab+bc-xy=(0)(1)+(1)(-3)-(-1)(3).
\displaystyle =-3+3=0.
\displaystyle \therefore \text{the required value is }0.
\displaystyle \\

\displaystyle \textbf{Question 55: }\text{The displacement, }x\text{ mt of a particle from a fixed point at time }
\displaystyle t\text{ seconds is given by} \ x=6\cos\left(3t+\frac{\pi}{3}\right).
\displaystyle \text{Find the acceleration of the particle, when }t=\frac{2\pi}{3}.
\displaystyle \text{Answer:}
\displaystyle x=6\cos\left(3t+\frac{\pi}{3}\right).
\displaystyle \text{Velocity, }v=\frac{dx}{dt}.
\displaystyle v=-18\sin\left(3t+\frac{\pi}{3}\right).
\displaystyle \text{Acceleration, }a=\frac{dv}{dt}.
\displaystyle a=-54\cos\left(3t+\frac{\pi}{3}\right).
\displaystyle \text{At }t=\frac{2\pi}{3},
\displaystyle a=-54\cos\left(3\cdot\frac{2\pi}{3}+\frac{\pi}{3}\right).
\displaystyle =-54\cos\left(\frac{7\pi}{3}\right).
\displaystyle =-54\left(\frac12\right)=-27\text{ m/s}^2.
\displaystyle \therefore \text{the acceleration of the particle is }-27\text{ m/s}^2.
\displaystyle \\

\displaystyle \textbf{Question 56: }\text{Evaluate: }\int_{-\pi/2}^{\pi/2}\sin|x|\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=\sin|x|.
\displaystyle f(-x)=\sin|-x|=\sin|x|=f(x).
\displaystyle \therefore f(x)\text{ is an even function.}
\displaystyle \therefore \int_{-\pi/2}^{\pi/2}\sin|x|\,dx=2\int_0^{\pi/2}\sin x\,dx.
\displaystyle =2[-\cos x]_0^{\pi/2}.
\displaystyle =2[-\cos\frac{\pi}{2}+\cos0].
\displaystyle =2(0+1)=2.
\displaystyle \therefore \text{the value of the integral is }2.
\displaystyle \\

\displaystyle \textbf{Question 57: }\text{Rahul, a class XII student, has a probability of getting a grade A in the}
\displaystyle \text{examination of three subjects namely Mathematics, Physics and Chemistry are }0.2,\ 
\displaystyle 0.3\text{ and }0.5 \ \text{respectively. Find the probability that he gets a grade A in none of the subjects.}
\displaystyle \text{Answer:}
\displaystyle P(\text{not getting A in Mathematics})=1-0.2=0.8.
\displaystyle P(\text{not getting A in Physics})=1-0.3=0.7.
\displaystyle P(\text{not getting A in Chemistry})=1-0.5=0.5.
\displaystyle \text{Assuming the three events are independent,}
\displaystyle P(\text{getting A in none})=0.8\times0.7\times0.5.
\displaystyle =0.28.
\displaystyle \therefore \text{the required probability is }0.28.
\displaystyle \\

\displaystyle \textbf{Question 58: }\text{If }A\text{ and }B\text{ are two events such that }P(\bar A)=0.3,\ \ 
\displaystyle P(B)=0.4, \ P(A\cap\bar B)=0.5,\text{ then} \ \text{find the value of }P\left(B/(A\cup\bar B)\right).
\displaystyle \text{Answer:}
\displaystyle P(\bar A)=0.3.
\displaystyle \therefore P(A)=1-0.3=0.7.
\displaystyle P(A)=P(A\cap B)+P(A\cap\bar B).
\displaystyle \therefore P(A\cap B)=0.7-0.5=0.2.
\displaystyle P(B)=P(A\cap B)+P(\bar A\cap B).
\displaystyle \therefore P(\bar A\cap B)=0.4-0.2=0.2.
\displaystyle P(A\cup\bar B)=1-P(\bar A\cap B)=1-0.2=0.8.
\displaystyle P\left(B/(A\cup\bar B)\right)=\frac{P\{B\cap(A\cup\bar B)\}}{P(A\cup\bar B)}.
\displaystyle B\cap(A\cup\bar B)=(A\cap B)\cup(B\cap\bar B)=A\cap B.
\displaystyle \therefore P\left(B/(A\cup\bar B)\right)=\frac{0.2}{0.8}=\frac14.
\displaystyle \therefore \text{the required probability is }\frac14.
\displaystyle \\

\displaystyle \textbf{Question 59: }\text{There are }10\text{ cookies in a box. Six have chocolate centres and}
\displaystyle \text{four have jam-filled centres. Shweta randomly chooses a cookie from the box and eats it. }
\displaystyle \text{Then, Ali randomly chooses and eats one of the remaining cookies.}
\displaystyle \text{What is the probability that Shweta and Ali choose cookies with different centres?}
\displaystyle \text{Answer:}
\displaystyle P(\text{Shweta chooses chocolate and Ali chooses jam})=\frac6{10}\times\frac4{9}.
\displaystyle =\frac{24}{90}=\frac4{15}.
\displaystyle P(\text{Shweta chooses jam and Ali chooses chocolate})=\frac4{10}\times\frac6{9}.
\displaystyle =\frac{24}{90}=\frac4{15}.
\displaystyle \therefore P(\text{different centres})=\frac4{15}+\frac4{15}=\frac8{15}.
\displaystyle \therefore \text{the required probability is }\frac8{15}.
\displaystyle \\

\displaystyle \textbf{Question 60: }\text{A biased dice is tossed and the respective probabilities for various faces to} \\ \text{turn up are:}
\displaystyle \begin{array}{c|cccccc}\text{Face}&1&2&3&4&5&6\\\hline\text{Probability}&0.10&0.24&0.19&0.18&0.15&0.14\end{array}
\displaystyle \text{If an odd face has turned up, then what is the probability that it is face }1\text{ or face }3?
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\{\text{face }1\text{ or face }3\}\text{ and }B=\{\text{odd face}\}.
\displaystyle P(A)=P(1)+P(3)=0.10+0.19=0.29.
\displaystyle P(B)=P(1)+P(3)+P(5).
\displaystyle =0.10+0.19+0.15=0.44.
\displaystyle \text{Since }A\subset B,\quad A\cap B=A.
\displaystyle P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{0.29}{0.44}.
\displaystyle =\frac{29}{44}.
\displaystyle \therefore \text{the required probability is }\frac{29}{44}.
\displaystyle \\

\displaystyle \textbf{Question 61: }\text{If }|\overrightarrow{a}|=1,\ |\overrightarrow{b}|=2,\ |\overrightarrow{c}|=3\text{ and }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0},\text{ then verify}
\displaystyle \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are NOT mutually perpendicular.}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}.
\displaystyle \therefore \overrightarrow{c}=-(\overrightarrow{a}+\overrightarrow{b}).
\displaystyle |\overrightarrow{c}|^2=|\overrightarrow{a}+\overrightarrow{b}|^2.
\displaystyle |\overrightarrow{c}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2\overrightarrow{a}\cdot\overrightarrow{b}.
\displaystyle 3^2=1^2+2^2+2\overrightarrow{a}\cdot\overrightarrow{b}.
\displaystyle 9=1+4+2\overrightarrow{a}\cdot\overrightarrow{b}.
\displaystyle \therefore \overrightarrow{a}\cdot\overrightarrow{b}=2\ne0.
\displaystyle \text{Hence, }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are not perpendicular.}
\displaystyle \therefore \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are not mutually perpendicular.}
\displaystyle \\

\displaystyle \textbf{Question 62: }\text{If }\overrightarrow{p}\times\overrightarrow{q}=\overrightarrow{0}\text{ and }\overrightarrow{p}\cdot\overrightarrow{q}=0,\text{ then what conclusion can we draw?}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{p}\times\overrightarrow{q}=\overrightarrow{0}.
\displaystyle \text{Therefore, }\overrightarrow{p}\text{ and }\overrightarrow{q}\text{ are parallel, provided both are non-zero vectors.}
\displaystyle \text{Also, }\overrightarrow{p}\cdot\overrightarrow{q}=0.
\displaystyle \text{If both vectors are non-zero and parallel, then}
\displaystyle \overrightarrow{p}\cdot\overrightarrow{q}=|\overrightarrow{p}||\overrightarrow{q}|\cos0^\circ\ne0
\displaystyle \text{or }|\overrightarrow{p}||\overrightarrow{q}|\cos180^\circ\ne0.
\displaystyle \text{This contradicts }\overrightarrow{p}\cdot\overrightarrow{q}=0.
\displaystyle \therefore \text{at least one of }\overrightarrow{p}\text{ and }\overrightarrow{q}\text{ must be the zero vector.}
\displaystyle \\

\displaystyle \textbf{Question 63: }\text{A line makes angles }\alpha,\ \beta\text{ and }\gamma\text{ with the }x,\ y\text{ and }z
\displaystyle \text{ axes respectively. Given that} \ \alpha+\beta=\frac{\pi}{2},\text{ then what is the value of }\gamma?
\displaystyle \text{Answer:}
\displaystyle \text{For the direction angles of a line,}
\displaystyle \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1.
\displaystyle \text{Given, }\alpha+\beta=\frac{\pi}{2}.
\displaystyle \therefore \beta=\frac{\pi}{2}-\alpha.
\displaystyle \therefore \cos\beta=\sin\alpha.
\displaystyle \cos^2\alpha+\sin^2\alpha+\cos^2\gamma=1.
\displaystyle 1+\cos^2\gamma=1.
\displaystyle \therefore \cos^2\gamma=0.
\displaystyle \therefore \cos\gamma=0.
\displaystyle \therefore \gamma=\frac{\pi}{2}.
\displaystyle \\

\displaystyle \textbf{Question 64: }\text{A variable plane at unit distance from the origin cuts the co-ordinate}
\displaystyle \text{axes at } A,\ B\text{ and }C. \ \text{The centroid of the triangle formed by joining the points }
\displaystyle A,\ B\text{ and } C\text{ satisfies the relation}  \ \frac1{x^2}+\frac1{y^2}+\frac1{z^2}=k,\text{ then what is the value of }k?
\displaystyle \text{Answer:}
\displaystyle \text{Let the plane cut the co-ordinate axes at }A(a,0,0),\ B(0,b,0)\text{ and }C(0,0,c).
\displaystyle \text{Its equation in intercept form is}
\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.
\displaystyle \text{Distance of this plane from the origin is}
\displaystyle \frac{1}{\sqrt{\frac1{a^2}+\frac1{b^2}+\frac1{c^2}}}=1.
\displaystyle \therefore \frac1{a^2}+\frac1{b^2}+\frac1{c^2}=1.
\displaystyle \text{The centroid of triangle }ABC\text{ is}
\displaystyle \left(\frac a3,\frac b3,\frac c3\right).
\displaystyle \therefore x=\frac a3,\qquad y=\frac b3,\qquad z=\frac c3.
\displaystyle \frac1{x^2}+\frac1{y^2}+\frac1{z^2}=9\left(\frac1{a^2}+\frac1{b^2}+\frac1{c^2}\right).
\displaystyle =9(1)=9.
\displaystyle \therefore k=9.
\displaystyle \\

\displaystyle \textbf{Question 65: }\text{The angle between the line }\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}
\displaystyle \text{ and the plane} \ 2x-y+\sqrt{k}z+4=0\text{ is }\alpha.
\displaystyle \text{Given }\sin\alpha=\frac13,\text{ what will be the value of }k?
\displaystyle \text{Answer:}
\displaystyle \text{The direction ratios of the line are }1,\ 2,\ 2.
\displaystyle \text{The direction ratios of the normal to the plane are }2,\ -1,\ \sqrt{k}.
\displaystyle \text{If }\alpha\text{ is the angle between the line and the plane, then}
\displaystyle \sin\alpha=\frac{|1(2)+2(-1)+2\sqrt{k}|}{\sqrt{1^2+2^2+2^2}\sqrt{2^2+(-1)^2+k}}.
\displaystyle =\frac{2\sqrt{k}}{3\sqrt{k+5}}.
\displaystyle \text{Given, }\sin\alpha=\frac13.
\displaystyle \therefore \frac{2\sqrt{k}}{3\sqrt{k+5}}=\frac13.
\displaystyle \frac{2\sqrt{k}}{\sqrt{k+5}}=1.
\displaystyle 4k=k+5.
\displaystyle 3k=5.
\displaystyle \therefore k=\frac53.
\displaystyle \\

\displaystyle \textbf{Question 66: }\text{The cost function of a product is given by } \\ C(x)=\frac{x^3}{3}-45x^2-900x
\displaystyle \text{ where }x\text{ is the} \ \text{number of units produced. What will be the slope of marginal cost?}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }C(x)=\frac{x^3}{3}-45x^2-900x.
\displaystyle \text{Marginal cost }=C'(x).
\displaystyle C'(x)=x^2-90x-900.
\displaystyle \text{The slope of marginal cost }=\frac{d}{dx}\{C'(x)\}=C''(x).
\displaystyle C''(x)=2x-90.
\displaystyle \therefore \text{the slope of marginal cost is }2x-90.
\displaystyle \\

\displaystyle \textbf{Question 67: }\text{A company paid Rs. }8475\text{ towards the rent of the building and Rs. }7625
\displaystyle \text{ interest on the loan.} \ \text{The cost of producing one unit of a product is Rs. }20.
\displaystyle \text{ If each unit is sold for Rs. }27,\text{ find the}  \ \text{break-even point.}
\displaystyle \text{Answer:}
\displaystyle \text{Fixed cost}=8475+7625=\text{Rs. }16100.
\displaystyle \text{Let }x\text{ units be produced and sold.}
\displaystyle \text{Total cost}=16100+20x.
\displaystyle \text{Total revenue}=27x.
\displaystyle \text{At the break-even point, Total Revenue}=\text{Total Cost}.
\displaystyle 27x=16100+20x.
\displaystyle 7x=16100.
\displaystyle \therefore x=2300.
\displaystyle \text{Break-even sales}=27\times2300=\text{Rs. }62100.
\displaystyle \therefore \text{the break-even point is }2300\text{ units, corresponding to sales of Rs. }62100.
\displaystyle \\

\displaystyle \textbf{Question 68: }\text{The total cost of producing and marketing }x\text{ units of bulbs by a whole}
\displaystyle \text{seller is given by} \ C(x)=\frac{1}{3}x^2+e^{2x}+3e.
\displaystyle \text{What will be the average cost of producing }3\text{ units of bulbs?}
\displaystyle \text{Answer:}
\displaystyle \text{Average cost}=\frac{C(x)}{x}.
\displaystyle C(3)=\frac{1}{3}(3)^2+e^{2(3)}+3e.
\displaystyle =3+e^6+3e.
\displaystyle \therefore \text{Average cost of producing }3\text{ units}=\frac{C(3)}{3}.
\displaystyle =\frac{3+e^6+3e}{3}.
\displaystyle =1+\frac{e^6}{3}+e.
\displaystyle \therefore \text{the average cost of producing }3\text{ units is }1+\frac{e^6}{3}+e.
\displaystyle \\

\displaystyle \textbf{Question 69: }\text{Nikhil is investigating the growth patterns of a certain species of slug and}
\displaystyle \text{measures their thickness, }t,\text{ and length, }l.\text{ The results are shown below:}
\displaystyle \begin{array}{c|c}\text{Thickness }t\text{ (mm)}&\text{Length }l\text{ (cm)}\\\hline9&5.9\\8&4.1\\3&2.1\\6&1.8\\4&3.0\\10&7.8\end{array}
\displaystyle \text{Using scatter diagram, explain why a line of best fit should not be used for this data?}
\displaystyle \text{Answer:}
\displaystyle \text{Plot the points }(9,5.9),(8,4.1),(3,2.1),(6,1.8),(4,3.0)\text{ and }(10,7.8).
\displaystyle \text{The scatter diagram shows that the points do not lie sufficiently close to a straight line.}\displaystyle \text{In particular, the point }(6,1.8)\text{ lies well away from the general increasing trend.}
\displaystyle \text{Hence, the data does not show a sufficiently strong linear relationship.}
\displaystyle \therefore \text{a line of best fit should not be used for this data.}
\displaystyle \\

\displaystyle \textbf{Question 70: }\text{If the covariance of }x\text{ and }y\text{ is }58.08,\text{ the variance of }
\displaystyle x\text{ is }121\text{ and the standard deviation} \ \text{of }y\text{ is }8\text{ then, find the regression coefficient} \\ \text{of }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\text{Cov}(x,y)=58.08,\qquad \sigma_x^2=121,\qquad \sigma_y=8.
\displaystyle \therefore \sigma_y^2=8^2=64.
\displaystyle \text{The regression coefficient of }x\text{ on }y\text{ is}
\displaystyle b_{xy}=\frac{\text{Cov}(x,y)}{\sigma_y^2}.
\displaystyle =\frac{58.08}{64}=0.9075.
\displaystyle \therefore \text{the regression coefficient of }x\text{ on }y\text{ is }0.9075.
\displaystyle \\

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