ISC Paper (2022) Class 12: Mathematics Solution – Semester 2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

MATHEMATICS

(Maximum Marks: 40)

(Time Allowed: One and a half hours)

(Candidates are allowed additional 10 minutes for only reading the paper.

They must NOT start writing during this time)

The Question Paper consists of three sections A, B and C.

Candidates are required to attempt all questions from Section A and all question EITHER from Section B OR Section C

All working, including rough work, should be done on the same sheet as, and adjacent to, the rest of the answer.

The intended marks for questions or parts of questions are given in brackets [ ].

Mathematical tables and graphs papers are provided.

SECTION – A [32 Marks]

Question 1: Choose the correct option to answer the following questions: [5]

$\displaystyle \text{(i) } \int \frac{\sin 2x}{\cos x} \ dx \text{ is equal to:}$

$\displaystyle (a) \ -2 \cos x + c\hspace{1.0cm} (b) \ 2 \cos x + c \hspace{1.0cm}(c) \ \frac{-\cos x}{2} + c \hspace{1.0cm}(d) \ \frac{\cos x}{2} + c$

Answer:

$\fbox{ a }$

$\displaystyle I = \int \frac{\sin 2x}{\cos x} \ dx$

$\displaystyle = \int \frac{ 2\sin x \cos x}{\cos x} \ dx$

$\displaystyle = \int \sin x \ dx$

$\displaystyle = 2 (- \cos x ) + c$

$\displaystyle = - 2 \cos x + c$

$\\$

$\displaystyle \text{(ii) If A and B are two events such that } P(A) = \frac{4}{5} \text{ and } P(B/A) = \frac{7}{8}. \\ \text{ The } P(A \cap B) \text{ is equal to: }$

$\displaystyle (a) \ \frac{7}{40}\hspace{1.0cm} (b) \ \frac{21}{40} \hspace{1.0cm}(c) \ \frac{32}{35} \hspace{1.0cm}(d) \ \frac{7}{10}$

Answer:

$\fbox{ d }$

$\displaystyle \text{Given } P(A) = \frac{4}{5} \text{ and } P(B/A) = \frac{7}{8}$

$\displaystyle \because P(B/A) = \frac{P(B \cap A)}{P(A)} = \frac{7}{8} \times \frac{4}{5} = \frac{7}{10}$

$\displaystyle \text{ Therefore, } P(A \cap B) = \frac{7}{10}$

$\\$

$\displaystyle \text{(iii) } \int e^{\sin x} \cos x \ dx \text{ is equal to: }$

$\displaystyle (a) \ e^{\cos x} + c\hspace{1.0cm} (b) \ e^{\sin x} + c \hspace{1.0cm}(c) \ \frac{\sin^2 x}{2} + c \hspace{1.0cm}(d) \ e^{\sin^ 2 x} + c$

Answer:

$\fbox{ b }$

$\displaystyle \text{Let, } I = \int e^{\sin x} \cos x \ dx$

$\displaystyle \text{Let, } \sin x = t$

$\displaystyle \Rightarrow \hspace{1.0cm} \cos x \ dx = dt$

$\displaystyle \therefore I = \int e^t \ dt = e^t + c = e^{\sin x} + c$

$\\$

$\displaystyle \text{(iv) The order and degree of the differential equation } \\ \\ \frac{d^3y}{dx^3} + \frac{d^2y}{dx^2}+ \Big( \frac{dy}{dx} \Big)^2 = 3 \text{ is: }$

$\displaystyle (a) \ \text{ order 3 and degree 1 } \hspace{1.0cm} (b) \ \text{ order 1 and degree 3 } \hspace{1.0cm} \\ \\ (c) \ \text{ order 2 and degree 1 } \hspace{1.0cm}(d) \ \text{ order 2 and degree 2 }$

Answer:

$\fbox{ a }$

$\displaystyle \text{Order is 3 because highest order derivative is } \frac{d^3y}{dx^3}.$

$\displaystyle \text{Degree is 1 because power of the highest order derivative is 1. }$

$\\$

(v) A bag contains 9 red, 7 white and 4 black balls. If two balls are drawn at random without replacement, the probability that both the balls are red will be:

$\displaystyle (a) \ \frac{11}{95}\hspace{1.0cm} (b) \ \frac{18}{95} \hspace{1.0cm}(c) \ \frac{18}{85} \hspace{1.0cm}(d) \ \frac{18}{23}$

Answer:

$\fbox{ b }$

$\displaystyle \text{Number of red balls } = 9$

$\displaystyle \text{Number of white balls } = 7$

$\displaystyle \text{Number of black balls } = 4$

$\displaystyle \text{Total number of balls } = 9 + 7 + 4 = 20$

$\displaystyle \text{Therefore required probability } = \frac{9}{20} \times \frac{8}{19} = \frac{18}{95}$

$\\$

$\displaystyle \text{(vi) } \int a^{3x+2} \ dx \text{ is equal to: }$

$\displaystyle (a) \ \Big( \frac{a^{3x}}{3 \log_e a} \Big)+c \hspace{1.0cm} (b) \ a^2 x + \Big( \frac{a^{3x}}{3 \log_e a} \Big)+c \hspace{1.0cm}(c) \ a^2\Big( \frac{a^{3x}}{3 \log_e a} \Big)+c \hspace{1.0cm}(d) \ a^2\Big( \frac{a^{3x}}{ \log_e a} \Big)+c$

Answer:

$\fbox{ c }$

$\displaystyle \text{Let } I = \int a^{3x+2} \ dx$

$\displaystyle \text{Let } 3x + 2 = t$

$\displaystyle \Rightarrow 3 \ dx = dt$

$\displaystyle \therefore I = \int a^t \frac{dt}{3}$

$\displaystyle =\frac{1}{3} \int a^t dt$

$\displaystyle = \frac{1}{3} \frac{a^t}{\log_e a}+c$

$\displaystyle = \frac{1}{3} \frac{a^{3x+2}}{\log_e a}+c$

$\displaystyle = a^2\Big( \frac{a^{3x}}{3 \log_e a} \Big)+c$

$\\$

Question 2: [2]

$\displaystyle \text{(a) Evaluate: } \int \frac{1}{\sin^2 x \cos^2 x} \ dx$

$\displaystyle \text{(b) Evaluate: } \int \Big( \sqrt{x}+\frac{1}{\sqrt{x}}\Big)^2 \ dx$

Answer:

(a)

$\displaystyle \text{Let } I = \int \frac{1}{\sin^2 x \cos^2 x} \ dx$

$\displaystyle = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} \ dx$

$\displaystyle = \int \frac{\sin^2 x }{\sin^2 x \cos^2 x} \ dx + \int \frac{\cos^2 x}{\sin^2 x \cos^2 x} \ dx$

$\displaystyle = \int \frac{1}{\cos^2 x} \ dx + \int \frac{1}{\sin^2 x } \ dx$

$\displaystyle = \int \sec^2 x \ dx + \int \mathrm{cosec}^2 x \ dx$

$\displaystyle = \tan x - \cot x + c$

(b)

$\displaystyle \text{Let } I = \int \Big( \sqrt{x}+\frac{1}{\sqrt{x}}\Big)^2 \ dx$

$\displaystyle = \int \Bigg\{ (\sqrt{x})^2 + \Big( \frac{1}{\sqrt{x}} \Big)^2+ 2 \sqrt{x} . \frac{1}{\sqrt{x}} \Bigg\} \ dx$

$\displaystyle = \int \Big( x + \frac{1}{x} + 2 \Big) \ dx$

$\displaystyle = \int x \ dx + \int \frac{1}{x} \ dx + \int + 2 \ dx$

$\displaystyle = \frac{x^2}{2} + \log x + 2 x + c$

$\\$

Question 3: [2]

$\displaystyle \text{(a) Solve: } \frac{dy}{dx} = \sin x - x$

$\displaystyle \text{(b) Solve: } \frac{dy}{dx} + 2x = e^{ex}$

Answer:

$\displaystyle \text{(a) Given, } \frac{dy}{dx} = \sin x - x$

$\displaystyle \Rightarrow \hspace{1.0cm} \ dy = (\sin x - x ) \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} \ dy = \sin x \ dx - x \ dx$

On integrating both side, we get

$\displaystyle \int \ dy = \int \sin x \ dx - \int x$

$\displaystyle \Rightarrow \hspace{1.0cm} y = - \cos x - \frac{x^2}{2} + c$

$\displaystyle \text{(b) Given, } \frac{dy}{dx} + 2x = e^{3x}$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{dy}{dx} = e^{3x}- 2x$

$\displaystyle \Rightarrow \hspace{1.0cm} dy = ( e^{3x}- 2x) dx$

$\displaystyle \Rightarrow \hspace{1.0cm} dy = e^{3x} dx - 2x \ dx$

On integrating both side, we get

$\displaystyle \int dy = \int e^{3x} \ dx - 2 \int x \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} y = \frac{e^{3x}}{3} - 2 \frac{x^2}{2}+ c$

$\displaystyle \Rightarrow \hspace{1.0cm} y = \frac{1}{3} e^{3x} - x^2 + c$

$\\$

$\displaystyle \text{Question 4: Evaluate } \int \limits_{1}^{4} |x-2| \ dx \hspace{6.0cm} [2]$

Answer:

$\displaystyle \text{Let } I = \int \limits_{1}^{4} |x-2| \ dx$

$\displaystyle = \int \limits_{1}^{2} |x-2| \ dx+ \int \limits_{2}^{4} |x-2| \ dx$

$\displaystyle = - \int \limits_{1}^{2} (x-2) \ dx+ \int \limits_{2}^{4} (x-2) \ dx$

$\displaystyle = - \Bigg[ \frac{x^2}{2} -2x \Bigg]_{1}^{2}+ \Bigg[ \frac{x^2}{2} -2x \Bigg]_{2}^{4}$

$\displaystyle = - \Bigg[ \Big( \frac{4}{2} - 4 \Big) - \Big(\frac{1}{2} - 2 \Big) \Bigg] + \Bigg[ \Big( \frac{16}{2} - 8 \Big) - \Big( \frac{4}{2} - 4 \Big) \Bigg]$

$\displaystyle = - \Bigg[ \Big( -2+\frac{3}{2} \Big) \Bigg] + [0+2]$

$\displaystyle = - \Big( - \frac{1}{2} \Big) + 2 = \frac{1}{2}+ 2 = \frac{5}{2}$

$\\$

$\displaystyle \text{Question 5: Two horses are considered for race. The probability of selection of first horse is } \\ \\ \frac{1}{5} \text{ and that of second }\frac{2}{3}. \text{ Find the probability that: }$

(i) both will be selected

(ii) only one of them will be selected

(iii) none of them is selected

(iv) at least one of them will be selected

Answer:

$\displaystyle \text{Let Event selection of first horse be } A.$

$\displaystyle \text{Let Event selection of second horse be } B.$

$\displaystyle \text{Given, } P(A) = \frac{1}{5} and P(B) = \frac{2}{3}$

(i) Probability (both will be selected)

$\displaystyle = P(A).P(B) = \frac{1}{5} \times \frac{2}{3} = \frac{2}{15}$

(ii) Probability (only one of them will be selected)

$\displaystyle = P(A).P(\overline{B}) + P(\overline{A}) .P(B)$

$\displaystyle = \frac{1}{5} \times \Big( 1 - \frac{2}{3} \Big) + \Big( 1 - \frac{1}{5}) \times \frac{2}{3}$

$\displaystyle = \frac{1}{5} \times \frac{1}{3} + \frac{4}{5}+\frac{2}{3}$

$\displaystyle = \frac{1}{15}+\frac{8}{15}$

$\displaystyle = \frac{9}{15}$

$\displaystyle =\frac{3}{5}$

(iii) Probability (none of them is selected )

$\displaystyle = P(\overline{A}). P(\overline{B}) = \Big( 1 - \frac{1}{5} \Big) \Big( 1 - \frac{2}{3} \Big) = \frac{4}{5} \times \frac{1}{3} = \frac{4}{15}$

(iv) Probability (at least one of them will be selected – = 1 – Probability (none of them is selected )

$\displaystyle = 1 - \frac{4}{15} = \frac{11}{15}$

$\\$

Question 6: [4]

$\displaystyle \text{(a) Evaluate: } \int \frac{dx}{ x \Big[ (\log x)^2 - 6 \log x + 5 ]}$

$\displaystyle \text{(b) Evaluate: } \int x \tan^{-1} x \ dx$

Answer:

$\displaystyle \text{(a) Let, } I = \int \frac{dx}{ x \Big[ (\log x)^2 - 6 \log x + 5 ]}$

$\displaystyle \text{Let } \log x = t \Rightarrow \hspace{1.0cm} \frac{1}{x} \ dx = dt$

$\displaystyle \therefore I = \int \frac{dt}{t^2 - 6t + 5}$

$\displaystyle = \int \frac{dt}{(t-5)(t-1)}$

$\displaystyle \text{Let, } \frac{1}{(t-5)(t-1)} = \frac{A}{t-5} + \frac{B}{t-1}$

$\displaystyle \Rightarrow \hspace{1.0cm} 1 = A ( t-1) + B(t-5)$

$\displaystyle \Rightarrow \hspace{1.0cm} 1 = ( A + B) t + ( -A - 5B)$

On comparing coefficients on both sides, we get

$\displaystyle A + B = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} A = - B \ldots \ldots (i)$

$\displaystyle \text{and } -A -5B = 1 \ldots \ldots (ii)$

From equation (i) and (ii) we get $

$\displaystyle -(-B) - 5B = 1$

$\displaystyle-4B = - 1 \Rightarrow \hspace{1.0cm} B = - \frac{1}{4} \text{ and } A = \frac{1}{4}$

$\displaystyle\therefore \frac{1}{(t-5)(t-1)} = \frac{4}{t-5} + \frac{4}{t-1} = \frac{1}{4} \Big[ \frac{1}{t-5}-\frac{1}{t-1} \Big]$

$\displaystyle \text{Thus } I = \frac{1}{4} \int \Big[ \frac{1}{t-5}-\frac{1}{t-1} \Big] \ dt$

$\displaystyle= \frac{1}{4} \Big[ \int \frac{1}{t-5} \ dt- \int \frac{1}{t-1} \ dt \Big]$

$\displaystyle= \frac{1}{4} [ \log (t-5) - \log(t-1) ] + c$

$\displaystyle= \frac{1}{4} \log \Big( \frac{t-5}{t-1} \Big)$

$\displaystyle= \frac{1}{4} \log \Big( \frac{\log x-5}{\log x-1} \Big)$

(b)

$\displaystyle \text{Let } I = \int x \tan^{-1} x \ dx$

$\displaystyle = \tan^{-1} x \int x dx - \int \Big\{ \frac{d}{dx} \Big( \tan^{-1} x \Big) \int x dx \Big\} dx$

$\displaystyle = \tan^{-1} x \Big( \frac{x^2}{2} \Big) - \int \Big( \frac{1}{1+x^2} \Big) \Big( \frac{x^2}{2} \Big) dx$

$\displaystyle = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \Big( \frac{x^2+1}{1+x^2} - \frac{1}{1+x^2} \Big) dx$

$\displaystyle = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \Big( 1 - \frac{1}{1+x^2} \Big) dx$

$\displaystyle = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \Big( x - \tan^{-1} x \Big) + c$

$\displaystyle = \frac{x^2}{2} \tan^{-1} x - \frac{x}{2} + \frac{1}{2} \tan^{-1} x + c$

$\\$

Question 7: [6]

The insurance company insured 1000 scooter drivers, 2000 car drivers and 4000 truck drivers. The probability of accidents by scooter, car and truck drivers are 0.02, 0.04 and 0.03 respectively. If one of the insured persons meets with an accident, find the probability that he is a truck driver.

Answer:

$\displaystyle \text{Event Scooter driver meets with an accident: } S$

$\displaystyle \text{Event Car driver meets with an accident: } C$

$\displaystyle \text{Event Truck driver meets with an accident: } T$

$\displaystyle \text{Driver is insured: } A$

By Bayes Theorem

Required Probability :

$\displaystyle P(T/A) = \frac{P(T).P(A/T)}{P(T).P(A/T)+P(S).P(A/S)+P(C).P(A/C)} \ldots \ldots (i)$

$\displaystyle \text{Given } P(S) = 0.2, P(C)=0.05 \text{ and } P(T) = 0.03$

$\displaystyle P(A/S) = \text{ Probability that a scooter driver is insured } \\ = \frac{1000} {1000+2000+4000} = \frac{1000}{7000} = \frac{1}{7}$

$\displaystyle P(A/C) = \text{ Probability that a car driver is insured } \\ = \frac{2000}{1000+2000+4000} = \frac{2000}{7000} = \frac{2}{7}$

$\displaystyle P(A/T) = \text{ Probability that a truck driver is insured } \\ = \frac{4000}{1000+2000+4000} = \frac{4000}{7000} = \frac{4}{7}$

Now, from equation (1)

$\displaystyle P(T/A) = \frac{0.03 \times \frac{4}{7}}{0.03 \times \frac{4}{7} +0.02 \times \frac{1}{7} + 0.05 \times \frac{2}{7}} = \frac{0.12}{0.12+0.02+0.10} = \frac{0.12}{0.24} = \frac{1}{2}$

$\\$

Question 8:

$\displaystyle \text{(a) Write a particular solution of the differential equation: } \\ \\ \frac{dy}{dx} = \frac{y^2}{xy-x^2}, \text{ when } x = 1 \text{ and } y = 1$

$\displaystyle \text{(b) Write a particular solution of the differential equation:} \\ \\ (1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}, \text{ when } y = 0 \text{ and } x = 0$

Answer:

(a) Given differential equation is

$\displaystyle \frac{dy}{dx} = \frac{y^2}{xy-x^2}$

The given differential equation is homogeneous in degree 2

$\displaystyle \text{Put, } y = vx \Rightarrow \hspace{1.0cm} \frac{dy}{dx} = v + x \frac{dv}{dx}$

$\displaystyle \therefore v + x \frac{dv}{dx} = \frac{(vx)^2}{x(vx)-x^2}$

$\displaystyle \Rightarrow \hspace{1.0cm} v + x \frac{dv}{dx} = \frac{v^2x^2}{vx^2-x^2}$

$\displaystyle \Rightarrow \hspace{1.0cm} v + x \frac{dv}{dx} = \frac{v^2}{v-1}$

$\displaystyle \Rightarrow \hspace{1.0cm} x \frac{dv}{dx} =\frac{v^2}{v-1} - v$

$\displaystyle \Rightarrow \hspace{1.0cm} x \frac{dv}{dx} = \frac{v^2-v^2+v}{v-1}$

$\displaystyle \Rightarrow \hspace{1.0cm} x \frac{dv}{dx} =\frac{v}{v-1}$

$\displaystyle \Rightarrow \hspace{1.0cm} \Big( \frac{v-1}{v} \Big) dv = \frac{1}{x} dx$

On integrating both sides we get

$\displaystyle \int dv - \int \frac{1}{v} dv = \int \frac{1}{x} dx$

$\displaystyle \Rightarrow \hspace{1.0cm} v-\log v = \log x + \log c$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{y}{x} = \log \Big( \frac{y}{x} \Big) + \log x + \log c$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{y}{x} = \log (cy)$

$\displaystyle \text{When } x = 1, y = 1$

$\displaystyle \therefore \frac{1}{1} = \log c$

$\displaystyle \text{or } \log c = 1$

Therefore particular solution of given differential equation is

$\displaystyle \frac{y}{x} = \log y + \log c$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{y}{x} = \log y + 1$

$\displaystyle \Rightarrow \hspace{1.0cm} \log y = \frac{y}{x} -1$

$\displaystyle \Rightarrow \hspace{1.0cm} y = e^{\Big( \frac{y}{x} - 1 \Big)}$

(b) Given differential equation is

$\displaystyle (1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{dy}{dx} + \Big( \frac{2x}{1+x^2} \Big) y = \frac{1}{(1+x^2)^2}$

$\displaystyle \text{ [On dividing both sides by } (1+x^2)]$

$\displaystyle \text{Which is a linear differential equation on the form } \frac{dy}{dx} + Py = Q$

$\displaystyle \text{Here, } P = \frac{2x}{1+x^2} \text{ and } Q = \frac{1}{(1+x^2)^2}$

$\displaystyle \therefore I.F. = e^{\int P dx} = e^{\int \frac{2x}{1+x^2} dx} = e^{\log (1+x^2)}= (1+x^2)$

Now, solution is

$\displaystyle y (I.F.) = \int Q. I.F. \ dx + c$

$\displaystyle \Rightarrow \hspace{1.0cm} y(1+x^2) = \int \frac{1}{(1+x^2)^2} (1+x^2) \ dx + c$

$\displaystyle \Rightarrow \hspace{1.0cm} y(1+x^2) = \int \frac{1}{(1+x^2)} \ dx + c$

$\displaystyle \Rightarrow \hspace{1.0cm} y(1+x^2) = \tan^{-1} x + c$

$\displaystyle \text{When } x=0, y=0$

$\displaystyle \therefore 0 = \tan^{-1} 0 + c$

$\displaystyle \Rightarrow \hspace{1.0cm} c = 0$

Particular solution is:

$\displaystyle y(1+x^2) = \tan^{-1} x \text{ or } y = \frac{\tan^{-1} x}{(1+x^2)}$

$\\$

SECTION – B [8 Marks]

Question 9: Choose the correct option to answer the following questions: [2]

$\displaystyle \text{(i) If the intercept form of the equation of the plane } 2x-3y+4z=12 \\ \\ \text{ is } \frac{x}{a} +\frac{y}{b} + \frac{z}{c} =1, \text{ then the values of } a, b, c \text{ are respectively: }$

$\displaystyle (a) \ a=6, b=-4, c=3 \hspace{1.0cm} (b) \ a=-6, b=-4, c=3 \hspace{1.0cm} \\ (c) \ a=6, b=4, c=3 \hspace{1.0cm}(d) \ a=6, b=4, c=-3$

Answer:

(i)

$\fbox{ a }$

$\displaystyle \text{Given equation of plane } 2x-3y+4z=12$

Rewriting the equation of plane as:

$\displaystyle \frac{2x}{12} - \frac{3y}{12} + \frac{4z}{12} = \frac{12}{12}$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{x}{6} - \frac{y}{4}+ \frac{z}{3} = 1$

$\displaystyle \text{Comparing the above equation with } \frac{x}{a} +\frac{y}{b}+ \frac{z}{c} = 1 \\ \text{ we get } a = 6, b = -4, c = 3$

$\displaystyle \text{(ii) The distance of the plane, whose equation is given by } 3x-4y+12z = 3, \text{ from the origin will be }$

$\displaystyle (a) \ \frac{3}{13} \text{ units } \hspace{1.0cm} (b) \ \frac{-2}{13} \text{ units } \hspace{1.0cm} (c) \ -3 \text{ units } \hspace{1.0cm}(d) \ \frac{13}{19} \text{ units }$

Answer:

(ii)

$\fbox{ a }$

$\displaystyle \text{Distance from origin } (0,0,0) \text{ to the plane } 3x-4y+12z-3 = 0$

$\displaystyle d = \frac{|3(0)-4(0) +12(0) - 3|}{\sqrt{(3)^2+(-4)^2+(12)^2}}$

$\displaystyle \Rightarrow \hspace{1.0cm} = \frac{3}{\sqrt{9+16+144}} = \frac{3}{\sqrt{169}} = \frac{3}{13} \text{ units }$

$\\$

$\displaystyle \text{ Question 10: Find the equation of the plane passing through (-2, 6, 6) , (1, -1, 0)} \\ \\ \text{ and ( 1, 2, -1). } \hspace{3.0cm} [2]$

Answer:

$\displaystyle \text{ The equation of the plane passing through } (-2, 6, 6) , (1, -1, 0) and ( 1, 2, -1) \text{ is given by }$

$\displaystyle \begin{vmatrix} x-(-2) & y-6 & z-6 \\ 1-(-2) & -1-6 & 0-6 \\ 1-(-2) & 2-6 & -1-6 \end{vmatrix} = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} \begin{vmatrix} x+2 & y-6 & z-6 \\ 3 & -7 & -6 \\ 3 & -4 & 7 \end{vmatrix} = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} (x+2)(49-24)-(y-6)(-21+18)+(z-6)(-12+21) = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} 25(x+2)+3(y-6)+9(z-6) = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} 25x + 3y + 9z-22=0$

$\displaystyle \Rightarrow \hspace{1.0cm} 25x + 3y + 9z = 22$

$\displaystyle \text{ Thus the required equation of the plane is } 25x + 3y + 9z = 22$

$\\$

$\displaystyle \text{ Question 11: Find the area of the region bounded by the curves } \\ \\ y = x^2 + 2, y = x, x=0 \text{ and } x = 3. [4]$

Answer:

$\displaystyle \text{ Let }y = x^2+2 = f(x) \text{ and } y = g(x) = x$

$\displaystyle \text{ Here, coordinates of point are }O(0,0), A(0, 2), B (3, 3) \text{ and } C(3, 11).$

$\displaystyle \text{ Therefore, Required area }= \int \limits_{0}^{3} [f(x) - g(x) ] dx$

$\displaystyle = \int \limits_{0}^{3} (x^2+2-x) dx$

$\displaystyle = \int \limits_{0}^{3} x^2 dx + 2 \int \limits_{0}^{3} dx - \int \limits_{0}^{3} x \ dx$

$\displaystyle = \Big[ \frac{x^3} {3} \Big]_{0}^{3} + 2 \Big[ x \Big]_{0}^{3} - \Big[ \frac{x^2}{2} \Big]_{0}^{3}$

$\displaystyle = \frac{27}{3} + 6 - \frac{9}{2}$

$\displaystyle = 9+6-\frac{9}{2}$

$\displaystyle = 15 - \frac{9}{2}$

$\displaystyle = \frac{21}{2} \text{ square units }$

$\\$

SECTION – C [18 Marks]

Question 12: Choose the correct option to answer the following questions: [2]

$\displaystyle \text{ (i) If two regression coefficients } b_{xy} \text{ and } b_{yx} \text{ are } -0.8 \text{ and } -0.2 \\ \\ \text{ respectively, then the value of coefficient of correlation } (r) \text{ will be, }$

$\displaystyle (a) \ 0.16 \hspace{1.0cm} (b) \ -0.16 \hspace{1.0cm}(c) \ 0.4 \hspace{1.0cm}(d) \ -0.4$

$\displaystyle \text{ (ii) The line of regression of } y \text{ on } x is 4x-5y+33=0 \text{ and the line }\\ \\ \text{ of regression of } x \text{ on } y \text{ is } 20x-9y-107=0, \text{ then the value of } x \text{ when } \\ \\ y = 7 \text{ is, }$

$\displaystyle (a) \ 8.5 \hspace{1.0cm} (b) \ -8.5 \hspace{1.0cm}(c) \ 0.5 \hspace{1.0cm}(d) \ -0.5$

Answer:

(i)

$\fbox{ c }$

$\displaystyle \text{ Given } b_{xy} = -0.8 \text{ and } b_{yx} = -0.2$

$\displaystyle r = \sqrt{b_{xy} . b_{yx} } = \sqrt{(-0.8)(-0.2) } = \sqrt{0.16} = 0.4$

$\displaystyle [b_{xy} < 0, b_{yx} < 0 , then r > 0]$

(ii)

$\fbox{ a }$

$\displaystyle \text{ Given, regression line of } x \text{ on } y \text{ is }$

$\displaystyle 20x-9y-107 = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} 20 x = 9y + 107$

$\displaystyle \Rightarrow \hspace{1.0cm} x = \frac{9}{20} y + \frac{107}{20}$

$\displaystyle \text{ when } y = 7$

$\displaystyle x = \frac{9}{20} (7) + \frac{107}{20} = \frac{63}{20} + \frac{107}{20} = \frac{170}{20} = 8.5$

$\\$

$\displaystyle \text{ Question 13: The mean and standard deviation of the two variables } x \text{ and } \\ y \text{ are given as } \overline{x} = 6 \text{ and } \overline{y}= 8, \sigma_x=4, \sigma_y = 12. \text{ The correlation coefficient } \\ \text{ is given as } r= \frac{2}{3}. \text{Find regression line of } x \text{ on } y. [ 2]$

Answer:

$\begin{array}{ |l|l|l|} \hline \text{Variable } & x \hspace{1.0cm} & y \hspace{1.0cm} \\ \hline \text{Mean } & \overline{x} = 6 & \overline{y} = 8 \\ \hline \text{Standard Deviation } (\sigma ) & \sigma_x= 4 &\sigma_y= 12 \\ \hline \end{array}$

$\displaystyle \text{ Correlation Coefficient, } r = \frac{2}{3}$

$\displaystyle \text{ Now, } b_{xy} = r \frac{\sigma_x}{\sigma_y} = \frac{2}{3} . \frac{4}{12} = \frac{2}{9}$

$\displaystyle \text{ Regression line } x \text{ on } y \text{ is given by }$

$\displaystyle (x - \overline{x}) = b_{xy}(y - \overline{y})$

$\displaystyle \Rightarrow \hspace{1.0cm} (x-6 )=\frac{2}{9} (y-8)$

$\displaystyle \Rightarrow \hspace{1.0cm} 9x - 54 = 2y - 16$

$\displaystyle \Rightarrow \hspace{1.0cm} x = \frac{2y}{9}+\frac{38}{9}$

$\\$

Question 14: A manufacturer has two machines X and Y that may run at most 360 minutes in a day to produce two types of toys A and B.

To produce each Toy A, machine X and Y need to run at most 12 minutes and 6 minutes respectively.

To produce each Toy B, machine X and Y need to run at most 6 minutes and 9 minutes respectively.

By selling the toys A and B, the manufacturer makes a profit of Rs. 30 and Rs. 20 respectively.

Formulate a linear programing problem and find the number of toys A and B that should be manufactured in a day to get maximum profit. [4]

Answer:

$\displaystyle \text{ Let } x \text{ and } y \text{ toys of type } A \text{ and type } B, \text{ respectively be manufactured in a day. }$

According to the question, we construct the following table:

$\begin{array}{ |c|c|c|} \hline \text{Type of Toys } & \text{Machine X} \hspace{1.0cm} & \text{Machine Y} \hspace{1.0cm} \\ \hline A & 12 & 6 \\ \hline B & 6 &9 \\ \hline \end{array}$

The given problem can be formulated as follows:

$\displaystyle Max Z = 30 x + 20 y$

Subject to constraints:

$\displaystyle 12x + 6y \leq 360$

$\displaystyle 6x+9y \leq 360$

$\displaystyle x, y \geq 0$

$\displaystyle \text{ Now, } 12x+6y = 360 \Rightarrow 2x+y = 60$

$\begin{array}{ |c|c|c|} \hline x \hspace{1.0cm} & 0 \hspace{1.0cm} & 30 \hspace{1.0cm} \\ \hline y \hspace{1.0cm} & 60 \hspace{1.0cm} &0 \hspace{1.0cm} \\ \hline \end{array}$

$\displaystyle \text{ and } 6x+9y = 360 \Rightarrow 2x+3y = 120$

$\begin{array}{ |c|c|c|} \hline x \hspace{1.0cm} & 0 \hspace{1.0cm} & 60 \hspace{1.0cm} \\ \hline y \hspace{1.0cm} & 40 \hspace{1.0cm} &0 \hspace{1.0cm} \\ \hline \end{array}$

$\displaystyle \text{ Also, interaction point of } 2x+y = 60 \text{ and } 2x+3y = 120 \text{ is } (15, 30)$

The feasible region determined by the constraints is shown in the above graph.

$\displaystyle \text{ The corner points of the feasible region are } O(0,0), A(30, 0), B(15, 30) \text{ and } C (0, 40)$

$\displaystyle \text{ The value of } Z \text{ at the corner points is as follows: }$

$\begin{array}{ |c|l|} \hline \text{Corner Points} & Z=30x+20y \\ \hline O(0,0) & Z=30(0)+20(0) = 0 \\ \hline A(30, 0) & Z=30(30)+20(0) = 900 \\ \hline B(15, 30) & Z=30(15)+20(30) = 1050 \\ \hline C (0, 40) & Z=30(0)+20(40) = 800 \\ \hline \end{array}$

$\displaystyle \text{ The maximum value of } Z \text{ is } 1050 \text{ at } (15, 30)$

Thus, the manufacturer should manufacture 15 toys of type A and 30 toys of type B to maximize the profit.

$\\$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

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ISC Paper (2022) Class 12: Mathematics Solution – Semester 2

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