CBSE Paper (All India – 2022) Term II Class 12: Mathematics Solution – ICSE / ISC / CBSE Mathematics Portal for K12 Students

MATHEMATICS

$\displaystyle \text{Series ABCD4/3} \hspace{1.0cm} | \hspace{1.0cm} \text{Q. P. Code 65/4/1 - 310A} \hspace{1.0cm} | \hspace{1.0cm} \text{Set 1 }$

$\displaystyle \text{Time Allowed : 2 hours} \hspace{5.0cm} \text{Maximum Marks : 40 }$

General Instructions :

This question paper contains three Sections – Section A, B and C.
Each Section is compulsory.
Section-A has 6 short answer type-I questions of 2 marks each.
Section-B has 4 short answer type-II questions of 3 marks each.
Section-C has 4 long answer type questions of 4 marks each.
There is an internal choice in some questions.
Q. 14 is a case study based problem with 2 sub-parts of 2 marks each.

SECTION A

Question Nos. 1 to 6 carry 2 marks each.

$\displaystyle \text{Question 1: } \text{Find: } \int \frac{dx}{x^2-6x+13}$

Answer:

$\displaystyle \int \frac{dx}{x^2-6x+13}$

$\displaystyle = \int \frac{dx}{x^2 - 6x + 9 - 9 + 13}$

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

$\displaystyle \text{Let } t = x- 3$

$\displaystyle \frac{dt}{dx} = 1 \Rightarrow dx = dt$

$\displaystyle = \int \frac{dt}{t^2 - 2^2}$

$\displaystyle = \frac{1}{4} \log \Bigg| \frac{t-2}{t+2} \Bigg| + c$

$\displaystyle = \frac{1}{4} \log \Bigg| \frac{x-3-2}{x-3+2} \Bigg| + c$

$\displaystyle = \frac{1}{4} \log \Bigg| \frac{x-5}{x-1} \Bigg| + c$

$\\$

$\displaystyle \text{Question 2: } \text{Find the general solution of the differential equation: } e^{dy/dx} = x^2$

Answer:

$\displaystyle e^{dy/dx} = x^2$

$\displaystyle \frac{dy}{dx} = \log x^2$

$\displaystyle \int dy = 2 \int \log x \ dx$

$\displaystyle y = 2 \int 1. \log x \ dx$

$\displaystyle = 2 \Big[ \log x \int 1. dx - \int \Big( \frac{d \log x}{dx} \int 1. dx \Big) dx \Big]$

$\displaystyle = 2 \Big[ x \log x - \int \frac{1}{x} . x \ dx \Big]$

$\displaystyle = 2 \Big[ x \log x - \int 1 . dx \Big]$

$\displaystyle = 2 [ x \log x - x ] + c$

$\displaystyle = 2 x [\log x - 1 ] + c$

$\\$

$\displaystyle \text{Question 3: } \text{Write the projection of the vector } ( \overrightarrow{b} + \overrightarrow{c} ) \text{ on the vector } \overrightarrow{a}, \text{ where } \overrightarrow{a} = 2 \hat{i}-2 \hat{j} + \hat{k}, \overrightarrow{b} = \hat{i}+2 \hat{j} -2 \hat{k} \text{ and } \overrightarrow{c} = 2 \hat{i}- \hat{j} + 4\hat{k}$

Answer:

$\displaystyle \overrightarrow{b} + \overrightarrow{c} = (\hat{i}+2 \hat{j} -2 \hat{k}) + (2 \hat{i}- \hat{j} + 4\hat{k} ) = 3 \hat{i}+ \hat{j} + 2\hat{k}$

$\displaystyle \therefore \text{Projection of } (\overrightarrow{b} + \overrightarrow{c}) \text{ on } \overrightarrow{a}$

$\displaystyle = \frac{(\overrightarrow{b} + \overrightarrow{c}). \overrightarrow{a}}{|\overrightarrow{a}|}$

$\displaystyle = \frac{(3 \hat{i}+ \hat{j} + 2\hat{k}).(2\hat{i}-2 \hat{j} + \hat{k})}{\sqrt{2^2+(-2)^2+ 1^2}}$

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

$\\$

$\displaystyle \text{Question 4: } \text{If the distance of the point (1, 1, 1,) from the plane } x - y + z + \lambda = 0 \text{ is } \frac{5}{\sqrt{3}}, \text{ find the value(s) of } \lambda .$

Answer:

$\displaystyle \text{Given: point = (1, 1, 1 ), equation of plane } = x - y + z + \lambda = 0 \text{ and }$

$\displaystyle \text{Distance } = \frac{5}{\sqrt{3}}$

We have,

$\displaystyle D = \frac{|ax_1+by_1 + cz_1+d|}{\sqrt{a^2 + b^2 + c^2}}$

$\displaystyle \frac{5}{\sqrt{3}} = \frac{|1 \times 1 + (-1) \times 1 + 1 \times 1 + \lambda |}{\sqrt{1^2 + (-1)^2 +1^2}}$

$\displaystyle \frac{5}{\sqrt{3}} = \frac{|1-1+1 + \lambda |}{\sqrt{3}}$

Therefore,

$\displaystyle 5= 1 + \lambda \Rightarrow \lambda = 4$

$\displaystyle \text{or } -5 = 1 + \lambda \Rightarrow \lambda = - 6$

$\\$

$\displaystyle \text{Question 5: Two cards are drawn successively with replacement from a well shuffled } \\ \text{ pack of 52 cards. Find the probability distribution of the number of spade cards.}$

Answer:

$\displaystyle \text{Number of drawn cards } n = 2$

$\displaystyle \text{Let } x: \text{ Number of spade card } = 0, 1, 2$

$\displaystyle \text{Probability of a spade card } = \frac{13}{52} = \frac{1}{4}$

$\displaystyle \text{Probability of failure } = 1 - \frac{1}{4} = \frac{3}{4}$

$\displaystyle \text{Since, } x \text{ follows the binomial distribution, } \Big( 2, \frac{1}{4} \Big)$

Therefore binomial distribution.

$\displaystyle \begin{array}{ |c|c|c|c|c|} \hline x & 0 \hspace{1.0cm} & 1 \hspace{1.0cm} & 2 \hspace{1.0cm} & \text{Total} \\ \hline P(x) & \frac{9}{16} & \frac{6}{16} & \frac{1}{16} & 1 \\ \hline \end{array}$

$\\$

$\displaystyle \text{Question 6: } \text{A pair of dice is thrown and the sum of the numbers appearing on the } \\ \text{ dice is observed to be 7. Find the probability that the number 5 has appeared on at } \\ \text{ least one die. }$

$\displaystyle \text{The probability that A hits the target is } \frac{1}{3} \text{ and the probability that B hits it, is } \\ \frac{2}{5}. \text{ If both try to hit the target independently, find the probability that the target is hit. }$

Answer:

$\displaystyle \text{Event A(sum is 7) } = \{ (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3) \}$

$\displaystyle n(A) = 6$

$\displaystyle \text{Event B (5 has appeared on at least one die) } \\ = \{ (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5) \}$

$\displaystyle A \cap B = \{(2, 5), ( 5, 2) \}$

$\displaystyle \text{Therefore, required probability } = P(A/B) = \frac{P(A\cap B)}{P(A)} = \frac{2}{6} = \frac{1}{3}$

There are three ways that a target can be hit.

$\displaystyle \text{A and B does not hit it } = \frac{1}{3} \times \frac{3}{5} = \frac{3}{15}$

$\displaystyle \text{A does not hit and B hits it } = \frac{2}{3} \times \frac{2}{5} = \frac{2}{15}$

$\displaystyle \text{A and B both hit it } = \frac{1}{3} \times \frac{2}{5} = \frac{4}{15}$

$\displaystyle \text{Therefore the probability of hitting the target is } = \frac{3}{15}+\frac{2}{15}+\frac{2}{15} = \frac{9}{15} = \frac{3}{5}$

SECTION B

Question Nos. 7 to 10 carry 3 marks each.

$\displaystyle \text{Question 7: } \text{Evaluate} \int \limits_{0}^{2\pi} \frac{dx}{1+e^{\sin x} }$

Answer:

$\displaystyle \text{Let } I = \int \limits_{0}^{2\pi} \frac{1}{1+e^{\sin x} } dx$

$\displaystyle = \int \limits_{0}^{\pi} \Bigg( \frac{1}{1+e^{\sin x} } + \frac{1}{1+e^{\sin (2\pi -x)} } \Bigg) dx$

$\displaystyle = \int \limits_{0}^{\pi} \Bigg( \frac{1}{1+e^{\sin x} } + \frac{1}{1+e^{-\sin x} } \Bigg) dx$

$\displaystyle = \int \limits_{0}^{\pi} \Bigg( \frac{1}{1+e^{\sin x} } + \frac{1}{1+ \frac{1}{e^{\sin x}} } \Bigg) dx$

$\displaystyle = \int \limits_{0}^{\pi} \Bigg( \frac{1}{1+e^{\sin x} } + \frac{e^{\sin x}}{1+ e^{\sin x}} \Bigg) dx$

$\displaystyle = \int \limits_{0}^{\pi} \Bigg( \frac{1+e^{\sin x}}{1+e^{\sin x} } \Bigg) dx$

$\displaystyle I = \int \limits_{0}^{\pi} 1 \ dx = \pi - 0 = \pi$

$\\$

$\displaystyle \text{Question 8: Find the particular solution of the differential equation } \\ x \frac{dy}{dx} - y = x^2 . e^x, \text{given } y(1) = 0.$

$\displaystyle \text{Find the particular solution of the differential equation } \\ x \frac{dy}{dx} = y ( \log y - \log x + 1).$

Answer:

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

$\displaystyle \Rightarrow \frac{dy}{dx} - \frac{1}{x} y = x . e^x$

We have

$\displaystyle \frac{dy}{dx} + Py = Q$

$\displaystyle \text{Here } P = -\frac{1}{x} \text{ and } Q = x. e^x$

$\displaystyle I.F. = e^{ \int \frac{1}{x} dx } = e^{- \log x} = e^{ \log x^{-1}} = \frac{1}{x} \ \ \ \Big[ \because e^{\log x} = x \Big]$

$\displaystyle \text{Now }$

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

$\displaystyle \Rightarrow y . \frac{1}{x} = \int \Big( x . e^x . \frac{1}{x} \Big) dx + c$

$\displaystyle \Rightarrow y . \frac{1}{x} = \int e^x dx + c$

$\displaystyle \Rightarrow y. \frac{1}{x} = e^x + c \text{ or } y = e^x x + c.x$

$\displaystyle \text{Substituting } x = 1 \text{ and } y = 0$

$\displaystyle 0 = e + c \text{ or } c = - e$

$\displaystyle \text{Therefore particular solution, }$

$\displaystyle y = e^x x - ex \text{ or } x ( e^x - e)$

$\displaystyle x \frac{dy}{dx} = y ( \log y - \log x + 1 )$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{y}{x} (\log y - \log x + 1)$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{y}{x} \Bigg( \log \frac{y}{x} + 1 \Bigg) \ldots \ldots (i)$

$\displaystyle \text{Put } y = vx \Rightarrow v = \frac{y}{x}$

$\displaystyle \text{Differentiating both sides w.r.t } x$

$\displaystyle \frac{dy}{dx} = v + x \frac{dv}{dx} \ldots \ldots (ii)$

$\displaystyle \text{From (i) and (ii) we get }$

$\displaystyle v + x \frac{dv}{dx} = v ( \log v + 1 )$

$\displaystyle \Rightarrow v + x \frac{dv}{dx} = v \log v + v$

$\displaystyle \Rightarrow x \frac{dv}{dx} = v \log v$

$\displaystyle \Rightarrow \frac{dv}{v \log v} = \frac{dx}{x}$

$\displaystyle \text{Applying integration on both sides }$

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

$\displaystyle \log ( \log v) = \log x + \log c$

$\displaystyle \Rightarrow \log ( \log v) = \log x . c$

$\displaystyle \Rightarrow \log v =xc$

$\displaystyle \Rightarrow \log \Big( \frac{y}{x} \Big) = xc \ \ \ \ \Big[ \because v = \frac{y}{x} \Big]$

$\\$

$\displaystyle \text{Question 9: The two adjacent sides of a parallelogram are represented by vectors} \\ 2 \hat{i } -4\hat{ j }+ 5 \hat{k } \text{ and } \hat{i } - 2 \hat{j } - 3 \hat{k } . \text{ Find the unit vector parallel to one of its diagonals. Also,} \\ \text{find the area of the parallelogram. }$

$\displaystyle \text{If } \overrightarrow{a} = 2 \hat{i } + 2\hat{ j } + 3 \hat{k }, \overrightarrow{b} = - \hat{i } + 2 \hat{j }+ \hat{k } \text{ and } c = 3 \hat{i } + \hat{j } \text{ are such that the vector } \\ ( \overrightarrow{a} + \lambda \overrightarrow{b} ) \text{ is perpendicular to vector } \overrightarrow{c} , \text{ then find the value of } \lambda.$

Answer:

Given, adjacent sides

$\displaystyle \overrightarrow{a} = 2 \hat{i } -4\hat{ j }+ 5 \hat{k }$

$\displaystyle \overrightarrow{b} = \hat{i } - 2 \hat{j } - 3 \hat{k }$

$\displaystyle \text{Let one of the diagonal is } \overrightarrow{p}$

$\displaystyle \text{So, } \overrightarrow{p} = \overrightarrow{a} + \overrightarrow{b}$

$\displaystyle = (2 \hat{i } -4\hat{ j }+ 5 \hat{k }) + (\hat{i } - 2 \hat{j } - 3 \hat{k })$

$\displaystyle = 3\hat{i } - 6 \hat{j } +2 \hat{k }$

$\displaystyle \text{Therefore. Unit vector parallel to one of its diagonal, } p$

$\displaystyle \hat{p} = \frac{\overrightarrow{p}}{|p|}$

$\displaystyle = \frac{3\hat{i } - 6 \hat{j } +2 \hat{k } }{\sqrt{ 9 + 36 + 4 }} = \frac{3}{7} \hat{i}- \frac{6}{7} \hat{j} + \frac{2}{7} \hat{k}$

$\displaystyle \text{Now area of parallelogram } = | \overrightarrow{a} \times \overrightarrow{b} |$

$\displaystyle \therefore\overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix}$

$\displaystyle = \hat{i}( 12+10) - \hat{j} (-6-5) + \hat{k} (-4+4)$

$\displaystyle = 22\hat{i}+ 11\hat{j} + 0\hat{k}$

$\displaystyle \therefore | \overrightarrow{a} \times \overrightarrow{b} | = \sqrt{22^2+11^2+0^2} = \sqrt{605} \text{ sq. units }$

Given that

$\displaystyle (\overrightarrow{a} + \lambda \overrightarrow{b} ) \perp \overrightarrow{c}$

$\displaystyle \Rightarrow (\overrightarrow{a} + \lambda \overrightarrow{b} ) . \overrightarrow{c} = 0$

$\displaystyle \overrightarrow{a} . \overrightarrow{c} + \lambda \overrightarrow{b} . \overrightarrow{c} = 0$

$\displaystyle \lambda = -\frac{\overrightarrow{a} . \overrightarrow{c}}{\overrightarrow{b} . \overrightarrow{c}}$

$\displaystyle = - \frac{(2\hat{i}+ 2\hat{j}+ 3\hat{k}).(3\hat{i}+ \hat{j})}{(-\hat{i}+ 2\hat{j}+ \hat{k}).(3\hat{i}+ \hat{j})} = \frac{-8}{-1} = 8$

$\displaystyle \lambda = 8$

$\\$

$\displaystyle \text{Question 10: Show that the lines:} \\ \\ \frac{1-x}{2} = \frac{y-3}{4} = \frac{z}{-1} \text{ and } \frac{x-4}{3} = \frac{2y-2}{-4} = z-1 \text{ are coplanar. }$

Answer:

$\displaystyle \frac{1-x}{2} = \frac{y-3}{4} = \frac{z}{-1} \ \ \ \Rightarrow \frac{x-1}{-2} = \frac{y-3}{4} = \frac{z-0}{-1}$

$\displaystyle \frac{x-4}{3} = \frac{2y-2}{-4} = z-1 \ \ \ \Rightarrow \frac{x-4}{3} = \frac{y-1}{-2} =\frac{z-1}{1}$

Comparing given equations with standard form, we get

$\displaystyle x_1 = 1, y_1 = 3, z_1 = 0; \ \ \ x_2 = 4, y_2= 1 , z_2 = 1$

$\displaystyle a_1 = -2, b_1 = 4, c_1 = -1 ; \ \ \ a_2 = 3, b_2 = -2, c_2 = 1$

Lines are co-planer if

$\displaystyle \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\a_1 & b_1 & c_ 1 \\ a_2 & c_2 & c_2 \end{vmatrix} = 0$

$\displaystyle LHS = \begin{vmatrix} 4-1 & 1-3 & 1-0 \\-2 & 4 & -1 \\ 3 & -2 & 1 \end{vmatrix} = \begin{vmatrix} 3 & -2 & 1 \\ -2 & 4 & -1 \\ 3 & -2 & 1 \end{vmatrix}$

$\displaystyle \text{Expanding along } R_1$

$\displaystyle = 3 (4-2) + 2 ( -2+3) + 1 ( 4-12)$

$\displaystyle = 3 \times 2 + 2 \times 1 + 1 (-8)$

$\displaystyle = 6 + 2 - 8 = 0$

Therefore both lines are coplanar.

SECTION C

Question Nos. 11 to 14 carry 4 marks each.

$\displaystyle \text{Question 11: Find the area of the region bounded by the curve } 4x^2 = y \text{ and the line } \\ y = 8x + 12, \text{ using integration.}$

Answer:

Given equations

$\displaystyle y = 4x^2 \ldots \ldots (i)$

$\displaystyle y = 8x+12 \ldots \ldots (ii)$

Eliminating y from equation (ii)

$\displaystyle 4x^2 = 8x+12$

$\displaystyle \Rightarrow 4x^2 - 8x - 12 = 0$

$\displaystyle \Rightarrow 4x^2 - 12x + 4x - 12 = 0$

$\displaystyle \Rightarrow 4x(x-3) + 4(x-3) = 0$

$\displaystyle \Rightarrow (x-3)(4x+4) = 0$

$\displaystyle \text{Therefore } x = 3 \text{ and } x = -1$

$\displaystyle \text{The Area bounded by the curve and (3, 36) and (-1, 4) }$

$\displaystyle = \int \limits_{-1}^{3} (y_1-y_2) dx$

$\displaystyle = \int \limits_{-1}^{3} \Big\{ (8x + 12)- 4x^2 \Big\} dx$

$\displaystyle = \Bigg[ \frac{8x^2}{2} + 12 x - 4 . \frac{x^3}{3} \Bigg]_{-1}^{3}$

$\displaystyle = \Bigg[ 4x^2 + 12 x - 4 . \frac{x^3}{3} \Bigg]_{-1}^{3}$

$\displaystyle = \Bigg[ 4(3)^2 + 12 (3) - 4 . \frac{3^3}{3} \Bigg] - \Bigg[ 4(-1)^2 + 12 (-1) - 4 . \frac{(-1)^3}{3} \Bigg]$

$\displaystyle = [36+ 36- 36] - [ 4 - 12 + \frac{4}{3} ]$

$\displaystyle = 36 - \Big( - \frac{20}{3} \Big)$

$\displaystyle = 36 + \frac{20}{3}$

$\displaystyle = \frac{128}{3} \text{ sq. units }$

$\\$

$\displaystyle \text{Question 12: Find:} \int \frac{x^2}{(x^2+1)(3x^2+4)} dx$

$\displaystyle \text{Evaluate: } \int \limits_{-2}^{1} \sqrt{5-4x-x^2} \ dx$

Answer:

$\displaystyle Given \int \frac{x^2}{(x^2+1)(3x^2+4)} dx$

$\displaystyle Put x^2 = y$

$\displaystyle Let \frac{y}{(y+1)(3y+4)} = \frac{A}{y+1} + \frac{B}{3y+4}$

$\displaystyle y = A(3y+4) + B ( y+1)$

$\displaystyle \text{Comparing coefficient of y \& constant term from both side }$

$\displaystyle 1 = 3A + B \ldots \ldots (i)$

$\displaystyle and 0 = 4A + B \ldots \ldots (ii)$

$\displaystyle \text{From equation (i) and equation (ii) }$

$\displaystyle A = - 1 and B = 4$

$\displaystyle \int \frac{y}{(y+1)(3y+4)} = -\int \frac{1}{y+1} + \int \frac{4}{3y+4}$

$\displaystyle = - \int \frac{1}{x^2+1} dx + \int \frac{4}{3x^2 + 4} dx$

$\displaystyle = - [\tan ^{-1} x ] + 4 \int \frac{1}{3x^2+4} dx - [\tan^{-1} x ] + 4 \int \frac{1}{3(x^2 + \frac{4}{3} ) } dx$

$\displaystyle = - \tan^{-1} x + \frac{4}{3} \int \frac{1}{x^2 + \Big( \frac{2}{\sqrt{3}} \Big)^2} dx$

$\displaystyle = - \tan^{-1} x + \frac{4}{3} \Bigg[ \frac{1}{\frac{2}{\sqrt{3}}} \tan^{-1} \frac{x}{\frac{2}{\sqrt{3}}} \Bigg] + c$

$\displaystyle = - \tan^{-1} x + \frac{4}{3} \times \frac{\sqrt{3}}{2} \tan^{-1} \frac{\sqrt{3} x}{2} + c$

$\displaystyle = - \tan^{-1} x + \frac{2}{\sqrt{3}} \tan^{-1} \frac{\sqrt{3} x}{2} + c$

$\displaystyle = \frac{2}{\sqrt{3}} \tan^{-1} \frac{\sqrt{3}x}{2} - \tan^{-1} x + c$

$\displaystyle \int \limits_{-2}^{1} \sqrt{5-4x-x^2} \ dx$

$\displaystyle = \int \limits_{-2}^{1} \sqrt{-(x^2+ 4x-5)} \ dx$

$\displaystyle = \int \limits_{-2}^{1} \sqrt{-(x^2+ 4x + 4 - 4 -5)} \ dx$

$\displaystyle = \int \limits_{-2}^{1} \sqrt{- \Big((x+2)^2 - 9 \Big)} \ dx$

$\displaystyle = \int \limits_{-2}^{1} \sqrt{(9-(x+2)^2 )} \ dx$

$\displaystyle \text{Let } t = x+2 \Rightarrow \frac{dt}{dx} = 1 \Rightarrow dt = dx$

$\displaystyle \text{Limit when } x = -2 \text{ is } t=0 \text{ and when } x = 1 \text{ is } t= 3$

$\displaystyle = \int \limits_{0}^{3} \sqrt{9 - t^2} dt$

$\displaystyle = \int \limits_{0}^{3} \sqrt{3^2 - t^2} dt$

$\displaystyle = \Big[ \frac{t}{2} \sqrt{3^2 - t^2 } + \frac{9}{2} \sin^{-1} \frac{t}{3} \Big]_{0}^{3}$

$\displaystyle = \Big[ 0 + \frac{9}{2} \sin^{-1} \Big( \frac{3}{3} \Big) \Big] - [0+0]$

$\displaystyle = \frac{9}{2} \sin^{-1} (1)$

$\displaystyle = \frac{9}{2} \sin^{-1} \Big( \sin \frac{\pi}{2} \Big) = \frac{9\pi}{4}$

$\\$

$\displaystyle \text{Question 13: Find the distance of the point (1, -2, 9) from the point of intersection of } \\ \\ \text{the line } \overrightarrow{r }= 4 \hat{i } + 2 \hat{j } + 7 \hat{k } + \lambda (3 \hat{i }+ 4 \hat{j } + 2 \hat{k } ) \text{ and the plane } \overrightarrow{r} . (\hat{i }-\hat{j }+\hat{k } )=10.$

Answer:

$\displaystyle \text{Given line } \overrightarrow{r }= 4 \hat{i } + 2 \hat{j } + 7 \hat{k } + \lambda (3 \hat{i }+ 4 \hat{j } + 2 \hat{k } ) \ldots \ldots (i)$

$\displaystyle \text{Given plane } \overrightarrow{r} . (\hat{i }-\hat{j }+\hat{k } )=10 \ldots \ldots (ii)$

From equation (i) and equation (ii)

$\displaystyle (4 \hat{i } + 2 \hat{j } + 7 \hat{k } + \lambda (3 \hat{i }+ 4 \hat{j } + 2 \hat{k } ). (\hat{i }-\hat{j }+\hat{k } )=10$

$\displaystyle \Rightarrow ( 4 -2 + 7) + \lambda ( 3 - 4 + 2) = 10$

$\displaystyle 9 + \lambda = 10$

$\displaystyle \Rightarrow \lambda = 1$

Therefore the point of intersection of the line and the plane is

$\displaystyle 4 \hat{i } + 2 \hat{j } + 7 \hat{k } + 1 (3 \hat{i }+ 4 \hat{j } + 2 \hat{k } ) = 7 \hat{i } + 6 \hat{j } + 9 \hat{k } \Rightarrow (7, 6, 9)$

$\displaystyle \text{Other given point is } ( 1, -2, 9)$

Therefore the distance between the two points

$\displaystyle = \sqrt{ (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2 }$

$\displaystyle = \sqrt{ (7-1)^2 + ( 6 - ( -2))^2 + ( 9-9)^2 }$

$\displaystyle = \sqrt{36 + 64 + 0 } = \sqrt{100} = 10 \text { units }$

$\\$

$\displaystyle \text{Question 14: A shopkeeper sells three types of flower seeds A1, A2, A3. They are } \\ \text{sold in the form of a mixture, where the proportions of these seeds are 4 : 4 : 2, respectively. } \\ \text{The germination rates of the three types of seeds are 45\%, 60\% and 35\% respectively.}$

$\displaystyle \text{Based on the above information: }$

$\displaystyle \text{(a) Calculate the probability that a randomly chosen seed will germinate; }$
$\displaystyle \text{(b) Calculate the probability that the seed is of type A2, given that a randomly } \\ \text{chosen seed germinates. }$

Answer:

$\displaystyle \text{We have } A_1: A_2: A_3: = 4 : 4: 2$

$\displaystyle \therefore P(A_1) = \frac{4}{10} , \ \ \ P(A_2) = \frac{4}{10} \ \ \text{ and } P(A_3) = \frac{2}{10}$

$\displaystyle \text{Let } E \text{ be the event that a seed germinates and let } \overline{E} \\ \text{ be the event that the seed does not germinate. }$

Then

$\displaystyle P\Big( \frac{E}{A_1} \Big) = \frac{45}{100} \text{ and } P\Big( \frac{\overline{E}}{A_1} \Big) = \frac{55}{100}$

$\displaystyle P\Big( \frac{E}{A_2} \Big) = \frac{60}{100} \text{ and } P\Big( \frac{\overline{E}}{A_2} \Big) = \frac{40}{100}$

$\displaystyle P\Big( \frac{E}{A_3} \Big) = \frac{35}{100} \text{ and } P\Big( \frac{\overline{E}}{A_3} \Big) = \frac{65}{100}$

(a) Probability that a randomly chosen seed germinates

$\displaystyle P(E) = P(A_1) . P\Big( \frac{\overline{E}}{A_1} \Big) + P(A_2) . P\Big( \frac{\overline{E}}{A_2} \Big) + P(A_3) . P\Big( \frac{\overline{E}}{A_3} \Big)$

$\displaystyle = \frac{4}{10} \times \frac{45}{100} + \frac{4}{10} \times \frac{60}{100} + \frac{2}{10} \times \frac{35}{100}$

$\displaystyle = \frac{180}{1000} + \frac{240}{1000} + \frac{70}{1000}$

$\displaystyle = \frac{490}{1000} = 0.49$

(b)

$\displaystyle P \Big( \frac{A_2}{E} \Big) = \frac{ P(A_2) P\Big( \frac{E}{A_2} \Big) }{ P(A_1) P\Big( \frac{E}{A_1} \Big) +P(A_2) P\Big( \frac{E}{A_2} \Big) +P(A_3) P\Big( \frac{E}{A_3} \Big) }$

$\displaystyle = \frac{ \frac{4}{10} \times \frac{60}{100} }{\frac{4}{10} \times \frac{45}{100} + \frac{4}{10} \times \frac{60}{100} + \frac{2}{10} \times \frac{35}{100}}$

$\displaystyle = \frac{\frac{240}{1000}}{\frac{180}{1000}+ \frac{240}{1000} + \frac{70}{1000} } = \frac{\frac{240}{1000}}{\frac{490}{1000}} = \frac{240}{490} = \frac{24}{49} = 0.48$

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CBSE Paper (All India – 2022) Term II Class 12: Mathematics Solution

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