\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs) - 1 Mark Each}

\displaystyle \textbf{Question 1: }\text{If }h(x)=4^x\text{ and }h^{-1}(x)=2,\text{ then the value of }x\text{ is:}
\displaystyle \text{(a) }-4\qquad\text{(b) }4\qquad\text{(c) }-16\qquad\text{(d) }16
\displaystyle \text{Answer:}
\displaystyle \text{Given, }h(x)=4^x.
\displaystyle \therefore h^{-1}(x)=\log_4 x.
\displaystyle h^{-1}(x)=2
\displaystyle \therefore \log_4 x=2
\displaystyle x=4^2=16.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }a+\frac{\pi}{2}<2\tan^{-1}x+3\cot^{-1}x<b,\text{ then }a\text{ and }b
\displaystyle \text{are respectively:}
\displaystyle \text{(a) }\frac{\pi}{2}\text{ and }2\pi\qquad\text{(b) }\frac{\pi}{2}\text{ and }-\frac{\pi}{2}
\displaystyle \text{(c) }0\text{ and }\pi\qquad\text{(d) }0\text{ and }2\pi
\displaystyle \text{Answer:}
\displaystyle \text{We know that }\cot^{-1}x=\frac{\pi}{2}-\tan^{-1}x.
\displaystyle \therefore 2\tan^{-1}x+3\cot^{-1}x
\displaystyle =2\tan^{-1}x+3\left(\frac{\pi}{2}-\tan^{-1}x\right)
\displaystyle =\frac{3\pi}{2}-\tan^{-1}x.
\displaystyle -\frac{\pi}{2}<\tan^{-1}x<\frac{\pi}{2}.
\displaystyle \therefore \pi<\frac{3\pi}{2}-\tan^{-1}x<2\pi.
\displaystyle \therefore \pi<2\tan^{-1}x+3\cot^{-1}x<2\pi.
\displaystyle \text{Comparing with }a+\frac{\pi}{2}<2\tan^{-1}x+3\cot^{-1}x<b,
\displaystyle a+\frac{\pi}{2}=\pi\quad\text{and}\quad b=2\pi.
\displaystyle \therefore a=\frac{\pi}{2},\qquad b=2\pi.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Which one of the following is true?}
\displaystyle \text{(a) }\sin(\cos^{-1}x)=\cos(\sin^{-1}x)
\displaystyle \text{(b) }\sec(\tan^{-1}x)=\tan(\sec^{-1}x)
\displaystyle \text{(c) }\cos(\tan^{-1}x)=\tan(\cos^{-1}x)
\displaystyle \text{(d) }\tan(\sin^{-1}x)=\sin(\tan^{-1}x)
\displaystyle \text{Answer:}
\displaystyle \sin(\cos^{-1}x)=\sqrt{1-x^2},\qquad -1\leq x\leq1.
\displaystyle \cos(\sin^{-1}x)=\sqrt{1-x^2},\qquad -1\leq x\leq1.
\displaystyle \therefore \sin(\cos^{-1}x)=\cos(\sin^{-1}x).
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If a matrix }A=[a_{ij}]_{2\times2},\text{ where }a_{ij}=\begin{cases}1,&i\ne j\\0,&i=j\end{cases},
\displaystyle \text{then }A^{-1}\text{ is:}
\displaystyle \text{(a) }I\qquad\text{(b) }A\qquad\text{(c) }-A\qquad\text{(d) }-I
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}0&1\\1&0\end{bmatrix}.
\displaystyle A^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0\\0&1\end{bmatrix}=I.
\displaystyle \therefore A\cdot A=I.
\displaystyle \therefore A^{-1}=A.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If the value of a }3^{\text{rd}}\text{ order determinant is }5,\text{ then the value of}
\displaystyle \text{the determinant formed by replacing its elements by their co-factors is:}
\displaystyle \text{(a) }5\qquad\text{(b) }\frac{1}{5}\qquad\text{(c) }125\qquad\text{(d) }25
\displaystyle \text{Answer:}
\displaystyle \text{Let }A\text{ be the matrix corresponding to the given determinant.}
\displaystyle |A|=5.
\displaystyle \text{Let }C\text{ be the matrix formed by replacing each element of }A\text{ by its co-factor.}
\displaystyle C^T=\text{adj }A.
\displaystyle \therefore |C|=|C^T|=|\text{adj }A|.
\displaystyle \text{For a }3\times3\text{ matrix, }|\text{adj }A|=|A|^{3-1}=|A|^2.
\displaystyle \therefore |C|=5^2=25.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }A=\begin{bmatrix}0&5&-y\\-5&0&x\\y&-x&0\end{bmatrix},\text{ then the value of}
\displaystyle A^{-1}\cdot(\text{adj }A)A\text{ is:}
\displaystyle \text{(a) }A^2\qquad\text{(b) }I\qquad\text{(c) }0\qquad\text{(d) }A
\displaystyle \text{Answer:}
\displaystyle \text{The given matrix }A\text{ is a skew-symmetric matrix of odd order.}
\displaystyle \therefore |A|=0.
\displaystyle \text{We know that }A(\text{adj }A)=|A|I.
\displaystyle \therefore A(\text{adj }A)=0.
\displaystyle A^{-1}\cdot(\text{adj }A)A
\displaystyle =A^{-1}\cdot0=0.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }D=\begin{vmatrix}p&p&p\\p&p+x&p\\p&p&p+y\end{vmatrix}\text{ for }p\ne0,\ x\ne0,\ y\ne0,
\displaystyle \text{then }D\text{ is divisible by:}
\displaystyle \text{(a) only }p\qquad\text{(b) }p\text{ and }x\text{ but not }y
\displaystyle \text{(c) }p\text{ and }y\text{ but not }x\qquad\text{(d) }p,\ x\text{ and }y
\displaystyle \text{Answer:}
\displaystyle D=\begin{vmatrix}p&p&p\\p&p+x&p\\p&p&p+y\end{vmatrix}.
\displaystyle \text{Applying }R_2\to R_2-R_1\text{ and }R_3\to R_3-R_1,
\displaystyle D=\begin{vmatrix}p&p&p\\0&x&0\\0&0&y\end{vmatrix}.
\displaystyle \therefore D=pxy.
\displaystyle \therefore D\text{ is divisible by }p,\ x\text{ and }y.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }\text{adj}(A)=\begin{bmatrix}2&3&5\\x&5&1\\3&3&4\end{bmatrix}\text{ and }|A|=4,\text{ then the value of }x\text{ is:}
\displaystyle \text{(a) }16\qquad\text{(b) }12\qquad\text{(c) }32\qquad\text{(d) }10
\displaystyle \text{Answer:}
\displaystyle \text{For a }3\times3\text{ matrix, }|\text{adj}(A)|=|A|^{3-1}=|A|^2.
\displaystyle \therefore |\text{adj}(A)|=4^2=16.
\displaystyle \begin{vmatrix}2&3&5\\x&5&1\\3&3&4\end{vmatrix}=16.
\displaystyle 2(20-3)-3(4x-3)+5(3x-15)=16.
\displaystyle 34-12x+9+15x-75=16.
\displaystyle 3x-32=16.
\displaystyle 3x=48.
\displaystyle x=16.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Which of the following could be a sketch of the function }y=\frac{d}{dx}(x\log x)?

\displaystyle \text{(a) Graph (a)\qquad (b) Graph (b)\qquad (c) Graph (c)\qquad (d) Graph (d)}
\displaystyle \text{Answer:}
\displaystyle y=\frac{d}{dx}(x\log x).
\displaystyle \text{Using the product rule,}
\displaystyle y=x\left(\frac{1}{x}\right)+\log x=1+\log x.
\displaystyle \text{The domain of }y\text{ is }x>0.
\displaystyle \frac{dy}{dx}=\frac{1}{x}>0\text{ for }x>0.
\displaystyle \therefore y\text{ is strictly increasing for }x>0.
\displaystyle \text{For the }x\text{-intercept, }1+\log x=0.
\displaystyle \log x=-1.
\displaystyle \therefore x=e^{-1}=\frac{1}{e}.
\displaystyle \text{Also, as }x\to0^+,\ y\to-\infty.
\displaystyle \therefore \text{the graph crosses the }x\text{-axis at }x=\frac{1}{e}\text{ and increases thereafter.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }\int\frac{2-x}{(x-1)^2}e^x\,dx=e^xf(x)+c,\text{ then }f(x)\text{ will be:}
\displaystyle \text{(a) }\frac{1}{x-1}\qquad\text{(b) }\frac{1}{2-x}\qquad\text{(c) }\frac{1}{3-x}\qquad\text{(d) }\frac{1}{1-x}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\int\frac{2-x}{(x-1)^2}e^x\,dx=e^xf(x)+c.
\displaystyle \text{Differentiating both sides with respect to }x,
\displaystyle \frac{2-x}{(x-1)^2}e^x=e^x\{f(x)+f'(x)\}.
\displaystyle \therefore f(x)+f'(x)=\frac{2-x}{(x-1)^2}.
\displaystyle \text{Taking }f(x)=\frac{1}{1-x},
\displaystyle f'(x)=\frac{1}{(1-x)^2}.
\displaystyle f(x)+f'(x)=\frac{1}{1-x}+\frac{1}{(1-x)^2}
\displaystyle =\frac{(1-x)+1}{(1-x)^2}=\frac{2-x}{(x-1)^2}.
\displaystyle \therefore f(x)=\frac{1}{1-x}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{An integrating factor of the differential equation}
\displaystyle x\frac{dy}{dx}+yP=x\cdot e^x\cdot x^{-\frac{1}{2}\log x},\text{ where }P\text{ is a function of }x,\ x>0,
\displaystyle \text{is }(\sqrt e)^{(\log x)^2},\text{ then }P\text{ is:}
\displaystyle \text{(a) }(\log x)^2\qquad\text{(b) }\frac{1}{\log x}\qquad\text{(c) }\log x\qquad\text{(d) }e^{\log x}
\displaystyle \text{Answer:}
\displaystyle \text{Dividing the differential equation by }x,
\displaystyle \frac{dy}{dx}+\frac{P}{x}y=e^x x^{-\frac{1}{2}\log x}.
\displaystyle \text{For }\frac{dy}{dx}+Qy=R,\text{ the integrating factor is }e^{\int Q\,dx}.
\displaystyle \therefore \text{I.F.}=e^{\int\frac{P}{x}\,dx}.
\displaystyle \text{Given, }\text{I.F.}=(\sqrt e)^{(\log x)^2}
\displaystyle =e^{\frac{1}{2}(\log x)^2}.
\displaystyle \therefore \int\frac{P}{x}\,dx=\frac{1}{2}(\log x)^2.
\displaystyle \text{Differentiating both sides with respect to }x,
\displaystyle \frac{P}{x}=\frac{\log x}{x}.
\displaystyle \therefore P=\log x.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Identify the correct answer of }\int_0^\pi\sin^{2024}x\cos^{2023}x\,dx:
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }2\qquad\text{(d) }3
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^\pi\sin^{2024}x\cos^{2023}x\,dx.
\displaystyle \text{Using }\int_0^\pi f(x)\,dx=\int_0^\pi f(\pi-x)\,dx,
\displaystyle I=\int_0^\pi\sin^{2024}(\pi-x)\cos^{2023}(\pi-x)\,dx.
\displaystyle \text{Since }\sin(\pi-x)=\sin x\text{ and }\cos(\pi-x)=-\cos x,
\displaystyle I=\int_0^\pi\sin^{2024}x(-\cos x)^{2023}\,dx.
\displaystyle I=-\int_0^\pi\sin^{2024}x\cos^{2023}x\,dx=-I.
\displaystyle \therefore 2I=0.
\displaystyle \therefore I=0.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Three fair dice are thrown. What is the probability of getting a total of }15\text{ given}
\displaystyle \text{that they exhibit three different numbers that are in arithmetic progression (A.P.)?}
\displaystyle \text{(a) }\frac14\qquad\text{(b) }\frac16\qquad\text{(c) }\frac18\qquad\text{(d) }\frac1{12}
\displaystyle \text{Answer:}
\displaystyle \text{The possible sets of three distinct numbers in A.P. are}
\displaystyle (1,2,3),(2,3,4),(3,4,5),(4,5,6),(1,3,5),(2,4,6).
\displaystyle \text{Each set can occur in }3!=6\text{ different orders.}
\displaystyle \therefore \text{Number of outcomes with three different numbers in A.P.}=6\times6=36.
\displaystyle \text{For a total of }15,\text{ the only possible A.P. set is }(4,5,6).
\displaystyle \text{The numbers }4,5,6\text{ can occur in }3!=6\text{ different orders.}
\displaystyle \therefore P(\text{total }15\mid\text{three different numbers in A.P.})=\frac6{36}=\frac16.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Rohit and Vishal, two below-average students in a class, are attempting a}
\displaystyle \text{Mathematics problem during revision classes. Their respective probabilities of solving the}
\displaystyle \text{sum correctly are }\frac16\text{ and }\frac18\text{ respectively. Their previous experience shows that while}
\displaystyle \text{solving the same question, the probability of a common mistake is }\frac1{10}.\text{ What is the probability}
\displaystyle \text{that they obtain the same answer?}
\displaystyle \text{(a) }\frac34\qquad\text{(b) }\frac7{48}\qquad\text{(c) }\frac{11}{96}\qquad\text{(d) }\frac9{96}
\displaystyle \text{Answer:}
\displaystyle P(\text{Rohit correct})=\frac16,\qquad P(\text{Vishal correct})=\frac18.
\displaystyle \therefore P(\text{both correct})=\frac16\times\frac18=\frac1{48}.
\displaystyle P(\text{Rohit wrong})=1-\frac16=\frac56.
\displaystyle P(\text{Vishal wrong})=1-\frac18=\frac78.
\displaystyle \therefore P(\text{both wrong})=\frac56\times\frac78=\frac{35}{48}.
\displaystyle \text{Given that the probability of making the same mistake is }\frac1{10},
\displaystyle P(\text{both wrong and make the same mistake})=\frac{35}{48}\times\frac1{10}=\frac7{96}.
\displaystyle \therefore P(\text{same answer})=P(\text{both correct})+P(\text{same mistake}).
\displaystyle =\frac1{48}+\frac7{96}=\frac2{96}+\frac7{96}=\frac9{96}=\frac3{32}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The value of }\widehat{i}\cdot(\widehat{k}\times\widehat{j})+\widehat{j}\cdot(\widehat{i}\times\widehat{k})+\widehat{k}\cdot(\widehat{i}\times\widehat{j})\text{ is:}
\displaystyle \text{(a) }-3\qquad\text{(b) }-2\qquad\text{(c) }-1\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \widehat{k}\times\widehat{j}=-\widehat{i},\qquad\widehat{i}\times\widehat{k}=-\widehat{j},\qquad\widehat{i}\times\widehat{j}=\widehat{k}.
\displaystyle \therefore \widehat{i}\cdot(\widehat{k}\times\widehat{j})+\widehat{j}\cdot(\widehat{i}\times\widehat{k})+\widehat{k}\cdot(\widehat{i}\times\widehat{j})
\displaystyle =\widehat{i}\cdot(-\widehat{i})+\widehat{j}\cdot(-\widehat{j})+\widehat{k}\cdot\widehat{k}.
\displaystyle =-1-1+1=-1.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Four students are playing a game. In a box, there are four strips of paper with}
\displaystyle \text{four expressions written on each one. The student who picks up the meaningless expression}
\displaystyle \text{will be out of the game.}
\displaystyle \bullet\ \text{Swati picks up the expression }\overrightarrow{a}\cdot(\overrightarrow{b}\times\overrightarrow{c}).
\displaystyle \bullet\ \text{Imran picks up the expression }\overrightarrow{a}\times(\overrightarrow{b}\times\overrightarrow{c}).
\displaystyle \bullet\ \text{Aryan picks up the expression }(\overrightarrow{a}\cdot\overrightarrow{b})\times(\overrightarrow{c}\cdot\overrightarrow{d}).
\displaystyle \bullet\ \text{Maria picks up the expression }(\overrightarrow{a}\times\overrightarrow{b})\cdot(\overrightarrow{c}\times\overrightarrow{d}).
\displaystyle \text{Who is out of the game?}
\displaystyle \text{(a) Swati}\qquad\text{(b) Imran}\qquad\text{(c) Aryan}\qquad\text{(d) Maria}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{a}\cdot(\overrightarrow{b}\times\overrightarrow{c})\text{ is a scalar triple product and is meaningful.}
\displaystyle \overrightarrow{a}\times(\overrightarrow{b}\times\overrightarrow{c})\text{ is a vector triple product and is meaningful.}
\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}\text{ and }\overrightarrow{c}\cdot\overrightarrow{d}\text{ are both scalars.}
\displaystyle \text{The cross product of two scalars is not defined.}
\displaystyle \therefore (\overrightarrow{a}\cdot\overrightarrow{b})\times(\overrightarrow{c}\cdot\overrightarrow{d})\text{ is meaningless.}
\displaystyle (\overrightarrow{a}\times\overrightarrow{b})\cdot(\overrightarrow{c}\times\overrightarrow{d})\text{ is a scalar product and is meaningful.}
\displaystyle \therefore \text{Aryan is out of the game.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }L\text{ is the foot of the perpendicular drawn from a point }P(a,b,c)\text{ on the}
\displaystyle XY\text{ plane, what is the co-ordinate of }L?
\displaystyle \text{(a) }(a,b,0)\qquad\text{(b) }(0,0,c)\qquad\text{(c) }(0,b,0)\qquad\text{(d) }(a,0,0)
\displaystyle \text{Answer:}
\displaystyle \text{The }XY\text{ plane is given by }z=0.
\displaystyle \text{The perpendicular from }P(a,b,c)\text{ to the }XY\text{ plane is parallel to the }z\text{-axis.}
\displaystyle \text{Hence, the }x\text{- and }y\text{-coordinates remain unchanged and the }z\text{-coordinate becomes }0.
\displaystyle \therefore L=(a,b,0).
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Find the equation of the plane passing through the point }(2,-1,3)\text{ and}
\displaystyle \text{containing the line }\frac{x-1}{3}=\frac{2-y}{4}=\frac{z+1}{2}.
\displaystyle \text{(a) }2x-2y+z=9\qquad\text{(b) }2x+2y+z=5
\displaystyle \text{(c) }11x+2y+z+21=0\qquad\text{(d) }11x-2y+z+17=0
\displaystyle \text{Answer:}
\displaystyle \text{The given line passes through }A(1,2,-1).
\displaystyle \text{Its direction ratios are }3,-4,2.
\displaystyle \therefore \overrightarrow{d}=3\widehat{i}-4\widehat{j}+2\widehat{k}.
\displaystyle \text{Let }P(2,-1,3)\text{ be the given point.}
\displaystyle \overrightarrow{AP}=(2-1)\widehat{i}+(-1-2)\widehat{j}+(3+1)\widehat{k}.
\displaystyle \therefore \overrightarrow{AP}=\widehat{i}-3\widehat{j}+4\widehat{k}.
\displaystyle \text{A normal vector to the plane is }\overrightarrow{d}\times\overrightarrow{AP}.
\displaystyle \overrightarrow{d}\times\overrightarrow{AP}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&-4&2\\1&-3&4\end{vmatrix}.
\displaystyle =-10\widehat{i}-10\widehat{j}-5\widehat{k}.
\displaystyle \therefore \text{the normal vector is proportional to }2\widehat{i}+2\widehat{j}+\widehat{k}.
\displaystyle \text{Hence, the equation of the plane through }(2,-1,3)\text{ is}
\displaystyle 2(x-2)+2(y+1)+(z-3)=0.
\displaystyle 2x+2y+z=5.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Rohit joins a career counselling institute as a counsellor. The manager says,}
\displaystyle \text{``Within a year I want a breakeven point. For that, I will give you Rs. }24,000\text{ fixed salary}
\displaystyle \text{per month and the variable salary will be }25\%\text{ of the revenue recovered on hiring students}
\displaystyle \text{at the rate of Rs. }800\text{ charged from every student.'' Find how many students should be admitted}
\displaystyle \text{by Rohit in a year in the institute to fulfil his manager's condition.}
\displaystyle \text{(a) }30\qquad\text{(b) }40\qquad\text{(c) }80\qquad\text{(d) }100
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of students admitted per month be }x.
\displaystyle \text{Monthly revenue}=800x.
\displaystyle \text{Variable salary}=25\%\text{ of }800x=200x.
\displaystyle \text{Total monthly salary}=24000+200x.
\displaystyle \text{At the breakeven point, revenue}=\text{total salary}.
\displaystyle 800x=24000+200x.
\displaystyle 600x=24000.
\displaystyle x=40.
\displaystyle \therefore \text{The monthly breakeven point is }40\text{ students.}
\displaystyle \text{Hence, the yearly breakeven number}=40\times12=480\text{ students.}
\displaystyle \therefore \text{As printed, none of the given options is correct.}
\displaystyle \text{Option (b) is correct only if the intended question asks for students per month.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The fixed cost of a new product is Rs. }200\text{ and the variable cost per unit is Rs. }x.
\displaystyle \text{If the demand function }p(x)=30,\text{ then what are the value(s) of }x\text{ that result in a loss}
\displaystyle \text{given that }x\text{ units of the product are sold?}
\displaystyle \text{(a) }10\leq x\leq20\qquad\text{(b) }x>20
\displaystyle \text{(c) }x<10\qquad\text{(d) }x<10\text{ or }x>20
\displaystyle \text{Answer:}
\displaystyle \text{Fixed cost}=200.
\displaystyle \text{Variable cost per unit}=x.
\displaystyle \therefore \text{Variable cost for }x\text{ units}=x^2.
\displaystyle \therefore \text{Total cost}=200+x^2.
\displaystyle \text{Since }p(x)=30,\text{ selling price per unit}=30.
\displaystyle \therefore \text{Revenue}=30x.
\displaystyle \text{For a loss, total cost}>\text{revenue}.
\displaystyle 200+x^2>30x.
\displaystyle x^2-30x+200>0.
\displaystyle (x-10)(x-20)>0.
\displaystyle \therefore x<10\text{ or }x>20.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{In statistical modelling, regression analysis is a set of statistical processes for}
\displaystyle \text{estimating the relationships between a dependent variable and one or more independent}
\displaystyle \text{variables. The most common form of regression analysis is linear regression, in which one}
\displaystyle \text{finds the line that most closely fits the data according to a specific mathematical criterion.}
\displaystyle \text{If plotted on a graph, the independent variable is represented along }\underline{\hspace{2cm}}.
\displaystyle \text{(a) Depends on the dataset}\qquad\text{(b) Y axis}
\displaystyle \text{(c) X axis}\qquad\text{(d) None of the above}
\displaystyle \text{Answer:}
\displaystyle \text{In a regression graph, the independent variable is generally plotted along the }X\text{ axis.}
\displaystyle \text{The dependent variable is generally plotted along the }Y\text{ axis.}
\displaystyle \therefore \text{The independent variable is represented along the }X\text{ axis.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{The function represented by the given graph is not differentiable at which of}
\displaystyle \text{the following points:} \displaystyle \text{(a) }-1,0,1\qquad\text{(b) }-1,1\qquad\text{(c) }0\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \text{From the graph, there are sharp corners at }x=-1,\ 0\text{ and }1.
\displaystyle \text{At a sharp corner, the left-hand derivative and right-hand derivative are unequal.}
\displaystyle \therefore \text{the function is not differentiable at }x=-1,\ 0\text{ and }1.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{In an examination, a candidate takes three tests namely }\alpha,\beta,\gamma\text{ in succession}
\displaystyle \text{and the probability of failing the first test }\alpha\text{ is }\frac12.\text{ The probability of passing each}
\displaystyle \text{succeeding test is }\frac12\text{ or }\frac14\text{ according to whether he passes or fails in the preceding one.}
\displaystyle \text{The candidate is selected if he passes at least two tests. What is the probability that the}
\displaystyle \text{candidate is selected?}
\displaystyle \text{(a) }\frac38\qquad\text{(b) }\frac18\qquad\text{(c) }\frac58\qquad\text{(d) }\frac34
\displaystyle \text{Answer:}
\displaystyle P(P_\alpha)=P(F_\alpha)=\frac12.
\displaystyle P(P_{\text{next}}\mid P_{\text{previous}})=\frac12,\qquad P(F_{\text{next}}\mid P_{\text{previous}})=\frac12.
\displaystyle P(P_{\text{next}}\mid F_{\text{previous}})=\frac14,\qquad P(F_{\text{next}}\mid F_{\text{previous}})=\frac34.
\displaystyle \text{For selection, the candidate must pass at least two of the three tests.}
\displaystyle \text{The favourable sequences are }PPP,\ PPF,\ PFP\text{ and }FPP.
\displaystyle P(PPP)=\frac12\times\frac12\times\frac12=\frac18.
\displaystyle P(PPF)=\frac12\times\frac12\times\frac12=\frac18.
\displaystyle P(PFP)=\frac12\times\frac12\times\frac14=\frac1{16}.
\displaystyle P(FPP)=\frac12\times\frac14\times\frac12=\frac1{16}.
\displaystyle \therefore P(\text{selected})=\frac18+\frac18+\frac1{16}+\frac1{16}.
\displaystyle =\frac6{16}=\frac38.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{In the picture given above, take a look at the double-arrowed lines drawn}
\displaystyle \text{on the overpass. This is an example of skew lines in the real world. Based on this, which}
\displaystyle \text{of these statements is INCORRECT?} \displaystyle \text{(a) These lines are not parallel.}\qquad\text{(b) These lines are intersecting.}
\displaystyle \text{(c) These lines are not coplanar.}
\displaystyle \text{(d) These lines can only exist in }3\text{ or higher dimensional space.}
\displaystyle \text{Answer:}
\displaystyle \text{Skew lines are lines which are neither parallel nor intersecting.}
\displaystyle \text{They are non-coplanar and hence cannot lie in the same plane.}
\displaystyle \text{Therefore, skew lines can exist only in three or higher dimensional space.}
\displaystyle \text{Hence, the statement ``These lines are intersecting'' is incorrect.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{The graph below shows the two lines of regressions for a certain set of }x
\displaystyle \text{and }y\text{ values. What is the mean of }x\text{ and mean of }y? \displaystyle \text{(a) }5\text{ and }8\qquad\text{(b) }8\text{ and }5\qquad\text{(c) }(1,0)
\displaystyle \text{(d) The mean of }x\text{ and mean of }y\text{ cannot be read from the graph.}
\displaystyle \text{Answer:}
\displaystyle \text{The two lines of regression intersect at the point }(\bar{x},\bar{y}).
\displaystyle \text{From the graph, the two regression lines intersect at }(5,8).
\displaystyle \therefore \bar{x}=5\quad\text{and}\quad\bar{y}=8.
\displaystyle \therefore \text{The mean of }x\text{ is }5\text{ and the mean of }y\text{ is }8.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{A point }x=c\text{ is called the critical point of a function if:}
\displaystyle \text{(a) }f'(c)=0
\displaystyle \text{(b) }f\text{ is not differentiable at }x=c.
\displaystyle \text{(c) both (a) and (b).}\qquad\text{(d) none of the above.}
\displaystyle \text{Answer:}
\displaystyle \text{A point }x=c\text{ in the domain of }f\text{ is a critical point if }f'(c)=0
\displaystyle \text{or if }f'(c)\text{ does not exist.}
\displaystyle \therefore \text{both conditions (a) and (b) can define a critical point.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

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