ISC Paper (2023) Class 12: Mathematics Solution

MATHEMATICS

(Maximum Marks: 100)

(Time Allowed: Three Hours)

(Candidates are allowed additional 15 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

Section A: Internal choice has been provided in three questions of four marks each and two questions of six marks each.

Section B: Internal choice has been provided in two question of four marks each.

Section C: Internal choice has been provided in two question of four marks each.

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 [65 Marks]

Question 1: [15]

In subparts (i) to (x), choose the correct option and in subparts (xi) to (xv), answer the questions as instructed.

$\displaystyle \text{(i) A relation R on } \{1, 2, 3\} \text{ is given by } R = \{(1, 1), (2, 2), (1, 2), (3, 3), (2, 3)\}. \\ \text{ Then the relation R is: }$

$\displaystyle \text{(a) } \ \text{Reflexive } \hspace{0.5cm} (b) \ \text{Symmetric } \hspace{0.5cm}(c) \ \text{Transitive } \hspace{0.5cm}(d) \ \text{Symmetric and Transitive }$

Answer:

$\fbox{ a }$

$\displaystyle \text{Given set is } A = \{1, 2, 3\} \text{ and given relation } R = \{(1, 1), (2, 2), (1, 2), (3, 3), (2, 3)\}.$

$\displaystyle R \text{is reflexive as } (1, 1), (2, 2), (3, 3) \in R$

$\displaystyle R \text{is is not transitive as } (1, 2), (2, 3)\in R \Rightarrow (1, 3) \notin R$

$\displaystyle R \text{ is is not symmetric as } (1, 2) \in R \text{ but } (2, 1) \notin R$

$\\$

$\displaystyle \text{(ii) If } A \text{ is a square matrix of order 3, then } |2A| \text{ is equal to: }$

$\displaystyle (a) \ 2|A| \hspace{1.0cm} (b) \ 4|A| \hspace{1.0cm}(c) \ 8|A| \hspace{1.0cm}(d) \ 6|A|$

Answer:

$\fbox{ c }$

$\displaystyle \text{Given A is a square matrix of order 3, Then }$

$\displaystyle |2A| = 2^3 |A| = 8 |A|$

$\\$

$\displaystyle \text{(iii) If the following function is continuous at } x = 2 \text{ then the value of } k \text{ will be: }$

$\displaystyle f(x) = \Bigg \{ \begin{matrix} 2x+1, & \text{if } x < 2 \\ k, & \text{if } x = 2 \\ 3x-1, & \text{if } x > 2 \end{matrix}$

$\displaystyle (a) \ 2 \hspace{1.0cm} (b) \ 3 \hspace{1.0cm}(c) \ 5 \hspace{1.0cm}(d) \ -1$

Answer:

$\fbox{ c }$

$\displaystyle \text{As } f(x) \text{ is continuous of } x = 2$

$\displaystyle \lim \limits_{x \to 2^+} f(x) = \lim \limits_{x \to 2^+} (3x-1) = 3 \times 2 -1 = 5$

$\displaystyle \lim \limits_{x \to 2^-} f(x) = \lim \limits_{x \to 2^-} (2x+1) = 2 \times 2 +1 = 5$

$\displaystyle \Rightarrow f(2^-) = f(2^+) = k [ \text{ Since } f \text{ is continuous at } x = 2]$

$\displaystyle \Rightarrow k = 5$

$\\$

(iv) An edge of a variable cube is increasing at the rate of 10 cm/sec. How fast will the volume of the cube increase if the edge is 5 cm long?

$\displaystyle (a) \ 75\ cm^3 /sec \hspace{1.0cm} (b) \ 750 \ cm^3 /sec \hspace{1.0cm}(c) \ 7500 \ cm^3 /sec \hspace{1.0cm}(d) \ 1250\ cm^3 /sec$

Answer:

$\fbox{ b }$

$\displaystyle \text{Let the edge of the cube be } x \text{ cm}$

$\displaystyle \text{Then, } \frac{dx}{dt} = 10 \text{ cm/s } \ldots \ldots (i)$

$\displaystyle \text{Volume of cube, } V = x^3 \text{ cm }^3$

$\displaystyle \therefore \frac{dV}{dt} = 3x^2 \Big( \frac{dx}{dt} \Big) \text{ [On differentiating w.r.t. } t ]$

$\displaystyle \frac{dV}{dt} \Bigg|_{x= \text{ 5 cm }} = 3(5)^2 \times 10 = 70 \frac{\text{ cm}^3}{\text{s}} \text{ [From equation (i) } \frac{dx}{dt} = 10 \text{ cm ] }$

$\\$

$\displaystyle \text{(v) Let } f(x) = x^3 \text{ be a function with domain } \{0, 1, ,2 , 3 \}. \text{Then the domain of is: }$

$\displaystyle (a) \ \{3, 2, 1, 0 \} \hspace{1.0cm} (b) \ \{ 0, -1, -2, -3 \} \hspace{1.0cm}(c) \ \{ 0, 1, 8,27 \} \hspace{1.0cm}(d) \ \{ 0, -1, -8, -27 \}$

Answer:

$\fbox{ c }$

$\displaystyle \text{Given, } f(x) = x^3$

$\displaystyle \text{Domain of } f(x) = \{0, 1, 2 , 3 \}$

$\displaystyle \text{Therefore Range of } f(x) = \{ 0^3, 1^3, 2^3, 3^3 \} = \{0, 1, 8, 27 \}$

$\displaystyle f \text{ can be written as, } \ \ \ f = \{ (0, 0), (1, 1), (2, 2), (3, 27) \}$

$\displaystyle \text{Now, } f^{-1} = \{ (0, 0), (1, 1), (8, 2), (27, 3) \}$

$\displaystyle \text{Thus, Domain of } f^{-1} = \{ 0, 1, 8, 27 \}$

$\\$

$\displaystyle \text{(vi) For the curve } y^2 = 2x^3 - 7, \text{ the slope of the normal at } (2, 3) \text{ is }:$

$\displaystyle (a) \ 4 \hspace{1.0cm} (b) \ \frac{1}{4} \hspace{1.0cm}(c) \ -4 \hspace{1.0cm}(d) \ \frac{-1}{4}$

Answer:

$\fbox{ d }$

$\displaystyle \text{Given curve is } y^2 = 2x^3 - 7$

$\displaystyle \text{On differentiating w.r.t } x \text{ we get }$

$\displaystyle 2y \frac{dy}{dx} = 6x^2$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{3x^2}{y}$

$\displaystyle \text{Slope of tangent at (2, 3) }= \frac{dy}{dx} \Bigg|_{at \ (2, 3) }= \frac{3(2^2)}{3} = 4$

$\displaystyle \text{Hence slope of normal at (2, 3) is } = \frac{-1}{\text{Slope of tangent at (2, 3)}} = \frac{-1}{4}$

$\\$

$\displaystyle \text{(vii) Evaluate: } \int \limits_{}^{} \frac{x}{x^2+1} dx$

$\displaystyle (a) \ 2 \log (x^2+1)+c \hspace{1.0cm} (b) \ \frac{1}{2} \log (x^2+1)+c \hspace{1.0cm}\\ (c) \ e^{x^2+1}+c \hspace{2.4cm}(d) \ \log x + \frac{x^2}{2}+c$

Answer:

$\fbox{ b }$

$\displaystyle \int \limits_{}^{} \frac{x}{x^2+1} dx = \frac{1}{2} \int \limits_{}^{} \frac{2x}{x^2+1} dx = \frac{1}{2} \log (x^2+1) + C$

$\\$

$\displaystyle \text{(viii) The derivative of } \log x \text{ with respect to } \frac{1}{x} \text{ is: }$

$\displaystyle (a) \ \frac{1}{x} \hspace{1.0cm} (b) \ \frac{-1}{x^3} \hspace{1.0cm}(c) \ \frac{-1}{x} \hspace{1.0cm}(d) \ -x$

Answer:

$\fbox{ d }$

$\displaystyle \text{Let } u = \log x \text{ and } v = \frac{1}{x}$

$\displaystyle \therefore \frac{du}{dx}= \frac{1}{x}$

$\displaystyle \text{and } \frac{dv}{dx} = - \frac{1}{x^2}$

$\displaystyle \text{Thus, } \frac{du}{dv} =\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{\frac{1}{x}}{- \frac{1}{x^2}} = -x$

$\\$

$\displaystyle \text{(ix) The interval in which the function } f(x) = 5 + 36x - 3x^2 \text{ increases will be: }$

$\displaystyle (a) \ (-\infty, 6) \hspace{1.0cm} (b) \ (6, \infty) \hspace{1.0cm}(c) \ (-6, \infty) \hspace{1.0cm}(d) \ (0, -6)$

Answer:

$\fbox{ a }$

$\displaystyle \text{Given, } f(x) = 5 + 36x - 3x^3$

$\displaystyle \therefore f'(x) = 36 - 6x$

$\displaystyle \text{For increasing function, } 36 - 6x > 0$

$\displaystyle \Rightarrow 6-x > 0$

$\displaystyle \Rightarrow 6 > x$

$\displaystyle \text{Hence, } x \in (-\infty, 6)$

$\\$

$\displaystyle \text{(x) Evaluate: } \int \limits_{-1}^{1} x^{17} \cos^4 x \ dx$

$\displaystyle (a) \ \infty \hspace{1.0cm} (b) \ 1 \hspace{1.0cm} (c) -1 \hspace{1.0cm}(d) \ 0$

Answer:

$\fbox{ d }$

$\displaystyle \text{Let } I = \int \limits_{-1}^{1} x^{17} \cos^4 x \ dx$

$\displaystyle \text{Here, } f(x) = x^{17} \cos^{4} x$

$\displaystyle f(-x) = - x^{17} \cos^4 = -f(x)$

$\displaystyle \text{Thus, } f(x) \text{ is an odd function. }$

$\displaystyle \text{We know that, }$

$\displaystyle \int \limits_{-a}^{a} f(x) dx = 0$

$\displaystyle \therefore I = \int \limits_{-1}^{1} x^{17} \cos^4 x \ dx = 0$

$\\$

$\displaystyle \text{(xi) Solve the differential equation: } \frac{dy}{dx} =\mathrm{cosec}y$

Answer:

Given the differential equation is

$\displaystyle \frac{dy}{dx} = \mathrm{cosec} y$

$\displaystyle \Rightarrow \frac{dy}{\mathrm{cosec} y} = dx$

$\displaystyle \Rightarrow \sin y \ dy = dx$

$\displaystyle \Rightarrow \int \sin y \ dy = \int dx$

$\displaystyle \Rightarrow -\cos y = x + c$

$\displaystyle \Rightarrow x + \cos y + c = 0$

$\\$

$\displaystyle \text{(xii) For what value of } k \text{ the matrix} \begin{bmatrix} 0 & k \\ -6 & 0 \end{bmatrix} \text{ is a skew symmetric matrix?}$

Answer:

$\displaystyle \text{Let } A = \begin{bmatrix} 0 & k \\ -6 & 0 \end{bmatrix}$

$\displaystyle \text{Given, } A \text{ is skew symmetric i.e. } A' = - A$

$\displaystyle \text{Now } A' = \begin{bmatrix} 0 & -6 \\ k & 0 \end{bmatrix}$

$\displaystyle \text{Thus, } \begin{bmatrix} 0 & -6 \\ k & 0 \end{bmatrix} = -\begin{bmatrix} 0 & k \\ -6 & 0 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 0 & -6 \\ k & 0 \end{bmatrix} = \begin{bmatrix} 0 & -k \\ 6 & 0 \end{bmatrix}$

$\displaystyle \text{On comparing corresponding elements, we get } k = 6$

$\\$

$\displaystyle \text{(xiii) Evaluate: } \int \limits_{0}^{1} |2x+1| dx$

Answer:

$\displaystyle \int \limits_{0}^{1} |2x+1| dx = \int \limits_{0}^{1} (2x+1) dx = 2 \int \limits_{0}^{1} x \ dx + \int \limits_{0}^{1} dx = 2 \Bigg[ \frac{x^2}{2} \Bigg]^1_0 + \Bigg[ x \Bigg]^1_0 = 1+1+2$

$\\$

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

Answer:

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

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

$\displaystyle = \int \limits_{}^{} \frac{1}{(1 - \cos x)} dx$

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

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

$\displaystyle = \frac{1}{2} \int \limits_{}^{} \mathrm{cosec}^2 \frac{x}{2} dx$

$\displaystyle = \frac{1}{2} \Bigg( - 2 \cot \frac{x}{2} \Bigg) + c$

$\displaystyle = - \cot \frac{x}{2} + c$

$\\$

(xv) A bag contains 19 tickets, numbered from 1 to 19. Two tickets are drawn randomly in succession with replacement. Find the probability that both tickets drawn are even numbers.

Answer:

$\displaystyle \text{We have total number of tickets in bag from } 1 \text{ to } 19 = 1, 2, 3, \ldots, 19$

$\displaystyle n(S) = 19$

$\displaystyle \text{Total even number of tickets in the bag is } 2, 4, 8, \ldots, 18$

$\displaystyle n(E) = 9$

$\displaystyle \text{Probability (both tickets drawn have even numbers) } = \frac{9}{19}\times \frac{9}{19} = \Bigg(\frac{9}{19} \Bigg)^2$

$\\$

Question 2:

$\displaystyle \text{(i) If } f(x) = [4 - (x-7)^3]^{\frac{1}{5}} \text{ is a real invertible function, then find } f^{-1}(x)$

$\displaystyle \text{(ii) Let } A = R - \{2\} \text{ and } B = R - \{1\}. \text{ If } f: A\rightarrow B \text{ is a function defined by } \\ f(x) = \frac{x-1}{x-2} \text{ then show that f is one-one and an onto function. }$ [2]

Answer:

$\displaystyle \text{(i) If } f(x) \text{ is invertible then it is one-one and onto. }$

$\displaystyle \text{We have, } f(x) = [4 - (x-7)^3]^{\frac{1}{5}}$

One-One

$\displaystyle f(x_1) = f(x_2)$

$\displaystyle \Rightarrow [4 - (x_1-7)^3]^{\frac{1}{5}} = [4 - (x_2-7)^3]^{\frac{1}{5}}$

$\displaystyle \Rightarrow 4 - (x_1-7)^3 = 4 - (x_2-7)^3 \text{ [Taking 5 power both sides] }$

$\displaystyle \Rightarrow (x_1-7)^3 = (x_2-7)^3$

$\displaystyle \Rightarrow (x_1-7)= (x_2-7) \text{ [Taking cube root both sides] }$

$\displaystyle \Rightarrow x_1 = x_2$

$\displaystyle \Rightarrow f \text{ is one-one }$

Onto

$\displaystyle \text{Let } y = f(x) = [4 - (x-7)^3]^{\frac{1}{5}}$

$\displaystyle \Rightarrow y^5 = 4 - (x-7)^3$

$\displaystyle \Rightarrow (x-7)^3 = 4-y^5$

$\displaystyle \Rightarrow (x-7)= (4 - y^5)^{\frac{1}{3}}$

$\displaystyle \Rightarrow x = (4 - y^5)^{\frac{1}{3}}+7$

$\displaystyle \text{Thus, } \forall \ y \in R, \exists x = \sqrt[3]{4-y^5} + 7 \in R$

$\displaystyle \Rightarrow f \text{ is onto }$

$\displaystyle \text{Hence, } f^{-1}(y) = x = \sqrt[3]{4-y^5} + 7$

$\displaystyle \text{We get } f^{-1}(x) \text{ if we replace } y \text{ with } x \text{ in the above equation }$

$\displaystyle f^{-1}(x) = \sqrt[3]{4-x^5} + 7$

$\\$

$\displaystyle \text{(ii) Given } A = R - \{2 \}, B = R - \{ 1 \}$

$\displaystyle \text{and } f: A \rightarrow B \text{ is defined as } f(x) = \frac{x-1}{x-2}$

One-One

$\displaystyle \text{Let } x_1, x_2 \in A \text{ such that } f(x_1) = f(x_2)$

$\displaystyle \frac{x_1-1}{x_1-2}= \frac{x_2-1}{x_2-2}$

$\displaystyle \Rightarrow (x_1-1)(x_2-2)= (x_1-2)(x_2-1)$

$\displaystyle \Rightarrow x_1x_2 - 2 x_1 - x_2 + 2 = x_1x_2 - 2 x_2 - x_1 + 2$

$\displaystyle \Rightarrow -2x_1-x_2 = -2x_2 -x_1$

$\displaystyle \Rightarrow -2x_1+x_1=-2x_2 +x_2$

$\displaystyle \Rightarrow -x_1 = -x_2$

$\displaystyle \Rightarrow x_1 = x_2$

$\displaystyle \text{Therefore } f \text{ is one-one. }$

Onto

$\displaystyle \text{Let } y \in B = R- \{1\}, \text{ then } y \neq 1$

$\displaystyle \text{The function } f \text{ is onto if there exists } x \in A \text{ such that } f(x) = y.$

$\displaystyle \text{Now, } f(x) = y$

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

$\displaystyle \Rightarrow x-1 = y ( x-2)$

$\displaystyle \Rightarrow x-1 = xy - 2 y$

$\displaystyle \Rightarrow x(1-y)=1-2y$

$\displaystyle \Rightarrow x - \frac{1-2y}{1-y} \in A (y \neq 1)$

$\displaystyle \text{Thus, for any } y \in B, \exists \ x = \frac{1-2y}{1-y} \in A \text{ such that }$

$\displaystyle f(\frac{1-2y}{1-y}) = \frac{\frac{1-2y}{1-y}-1}{\frac{1-2y}{1-y}-2}$

$\displaystyle = \frac{-y}{-1} = y$

$\displaystyle \text{Therefore } f \text{ is onto. }$

$\\$

Question 3: Evaluate the following determinant without expanding. [2]

$\displaystyle \begin{vmatrix} 5 & 5 & 5 \\ a & b & c \\ b+c & c+a & a+b \\ \end{vmatrix}$

Answer:

$\displaystyle \begin{vmatrix} 5 & 5 & 5 \\ a & b & c \\ b+c & c+a & a+b \\ \end{vmatrix}$

$\displaystyle \text{Taking 5 common from } R_1$

$\displaystyle = 5 \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ b+c & c+a & a+b \\ \end{vmatrix}$

$\displaystyle R_3 \rightarrow R_3 + R_2$

$\displaystyle \text{Taking } a+b+c \text{ common from } R_3$

$\displaystyle = 5 (a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ 1 & 1 & 1 \\ \end{vmatrix}$

$\displaystyle =5 (a+b+c) \times 0$

$\displaystyle =0$

$\\$

Question 4:

$\displaystyle \text{The probability of the event A occurring is } \frac{1}{3} \text{ and of the event } B \text{ occurring is } \\ \frac{1}{2}. \text{ If } A \text{ and } B \text{ are independent events, then find the probability of neither } \\ A \text{ nor } B \text{ occurring. }$ [2]

Answer:

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

$\displaystyle \text{Here, } P(A \cap B) = P(A).P(B) \hspace{1.0cm} \text{ [Since events are independent] }$

$\displaystyle = \frac{1}{3} . \frac{1}{2}$

$\displaystyle = \frac{1}{6}$

$\displaystyle \text{ Thus P(neither A or B occurring) = 1 - P(both A and B occurring) } \\ { \hspace{6.5cm}= 1 - \frac{1}{6} = \frac{1}{6} }$

$\\$

Question 5: Solve for $x:$

$\displaystyle 5\tan^{-1} x+ 3 \cot^{-1} x = 2\pi$

Answer:

$\displaystyle \text{Given, }5\tan^{-1} x+ 3 \cot^{-1} x = 2\pi$

$\displaystyle \text{We know that, } \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$

$\displaystyle \Rightarrow \cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x$

$\displaystyle \therefore 5\tan^{-1} x+ 3 \Big( \frac{\pi}{2} - \tan^{-1} x \Big) = 2\pi$

$\displaystyle \Rightarrow 5\tan^{-1} x+\frac{3\pi}{2} - 3 \tan^{-1} x = 2\pi$

$\displaystyle \Rightarrow 2\tan^{-1} x = \frac{\pi}{2}$

$\displaystyle \Rightarrow \tan^{-1} x = \frac{\pi}{4}$

$\displaystyle \Rightarrow x= \tan \frac{\pi}{4}$

$\displaystyle \Rightarrow x = 1$

$\\$

Question 6: [2]

$\displaystyle \text{(i) Evaluate } \int \cos^{-1} (\sin x) dx$

$\displaystyle \text{(ii) If } \int x^5 \cos (x^6) dx = k \sin (x^6) + C, \text{ find the value of } k.$

Answer:

$\displaystyle \text{(i) } I = \int \cos^{-1} (\sin x) dx$

$\displaystyle \text{Let } \cos^{-1} (\sin x) = \theta$

$\displaystyle \Rightarrow \sin x = \cos \theta$

$\displaystyle \Rightarrow \sin x = \sin \Big( \frac{\pi}{2} - \theta \Big)$

$\displaystyle \Rightarrow x =\frac{\pi}{2} - \theta$

$\displaystyle \Rightarrow \theta = \frac{\pi}{2} - x$

$\displaystyle \therefore I = \int \theta dx = \int \Big( \frac{\pi}{2} - x \Big)dx = \frac{\pi}{2} x - \frac{x^2}{2}+ C$

$\displaystyle \text{(ii) Let } I = \int x^5 \cos (x^6) dx$

$\displaystyle \text{Put } x^6 = t$

$\displaystyle \Rightarrow 6x^5 dx = dt$

$\displaystyle \Rightarrow x^5 dx = \frac{dt}{6}$

$\displaystyle \therefore I = \frac{1}{6} \int \cos t dt = \frac{1}{6} \sin t+ C = \frac{1}{6} \sin x^6+C$

According to question,

$\displaystyle \int x^5 \cos (x^6) dx = k \sin x^6+ C$

$\displaystyle \therefore \frac{1}{6} \sin x^6+C = k \sin x^6+ C$

$\displaystyle \text{On comparing we get } k = \frac{1}{6}$

$\\$

Question 7: [4]

$\displaystyle \text{If } \tan^{-1} \Big( \frac{x-1}{x+1} \Big) + \tan^{-1} \Big( \frac{2x-1}{2x+1} \Big) = \tan^{-1} \Big( \frac{23}{36} \Big) \text{ then prove that } \\ 24x^2-23x-12=0$

Answer:

$\displaystyle \tan^{-1} \Big( \frac{x-1}{x+1} \Big) + \tan^{-1} \Big( \frac{2x-1}{2x+1} \Big) = \tan^{-1} \Big( \frac{23}{36} \Big)$

$\displaystyle [\because \tan^{-1} A+ \tan^{-1} B = \tan^{-1} \Big( \frac{A+B}{1-AB} \Big) ]$

$\displaystyle \Rightarrow \tan^{-1} \Bigg( \frac{\frac{x-1}{x+1} + \frac{2x-1}{2x+1}}{1 - \frac{x-1}{x+1} \times \frac{2x-1}{2x+1}} \Bigg) = \tan^{-1} \Big( \frac{23}{36} \Big)$

$\displaystyle \Rightarrow \tan^{-1} \Bigg( \frac{(x-1)(2x+1)+(2x-1)(x+1)}{(x+1)(2x+1)-(x-1)(2x-1)} \Bigg) = \tan^{-1} \Big( \frac{23}{36} \Big)$

$\displaystyle \Rightarrow \tan^{-1} \Bigg( \frac{(2x^2-x-1)(2x^2+x-1)}{(2x^2+3x+1)-(2x^2-3x+1)} \Bigg) = \tan^{-1} \Big( \frac{23}{36} \Big)$

$\displaystyle \Rightarrow \tan^{-1} \Bigg( \frac{4x^2-2}{6x} \Bigg) = \tan^{-1} \Big( \frac{23}{36} \Big)$

$\displaystyle \Rightarrow \frac{4x^2-2}{6x} = \frac{23}{36}$

$\displaystyle \Rightarrow \frac{2x^2-1}{3x} = \frac{23}{36}$

$\displaystyle \Rightarrow 36(2x^2-1) = 23(3x)$

$\displaystyle \Rightarrow 12(2x^2-1) = 23(x)$

$\displaystyle \Rightarrow 24x^2-23x-12=0$

Hence proved.

$\\$

Question 8: [4]

$\displaystyle \text{If } y = e^{2x} \cos bx, \text{ then prove that } \frac{d^2y}{dx^2} - 2a\frac{dy}{dx}+(a^2+b^2)y = 0$

Answer:

$\displaystyle Given, y = e^{2x} \cos bx$

$\displaystyle \frac{dy}{dx} = ae^{ax} \cos bx- be^{ax} \sin bx$

$\displaystyle \therefore \frac{dy}{dx} = ay - be^{ax} \sin bx$

$\displaystyle \frac{d^2y}{dx^2}= a \frac{dy}{dx} - b(ae^{ax} \sin bx+ be^{ax} \cos bx)$

$\displaystyle \frac{d^2y}{dx^2}= a \frac{dy}{dx} -a \Big( ay - \frac{dy}{dx} \Big) - b^2y$

$\displaystyle \frac{d^2y}{dx^2}= a \frac{dy}{dx} - a^2y + a \frac{dy}{dx}- b^2 y$

$\displaystyle \frac{d^2y}{dx^2} - 2a\frac{dy}{dx}+(a^2+b^2)y = 0$

$\\$

Question 9: [4]

(i) In a company, 15% of the employees are graduates and 85% of the employees are non-graduates. As per the annual report of the company, 80% of the graduate employees and 10% of the non-graduate employees are in the Administrative positions. Find the probability that an employee selected at random from those working in administrative positions will be a graduate.

$\displaystyle \text{(ii) A Problem in Mathematics is given to the three students A, B and C. } \\ \text{ Their chances of solving the problem are } \frac{1}{3}, \frac{1}{4} \text{ and } \ \frac{1}{4} \text{ respectively. } \\ \text{ Find the probability that }$

(a) exactly two students will solve the problem.

(b) at least two of them will solve the problem.

Answer:

$\displaystyle \text{(i) Probability of graduate employees } = P(E_1) = \frac{15}{100}$

$\displaystyle \text{Probability of non-graduate employees } = P(E_2) = \frac{85}{100}$

$\displaystyle \text{Probability of graduate employees in administrative position } \\ { \hspace{6.0cm} = P(\frac{A}{E_1}) = \frac{80}{100} }$

$\displaystyle \text{Probability of non-graduate employees in administrative position } \\ { \hspace{6.0cm}= P(\frac{A}{E_2}) = \frac{10}{100}}$

By Bayes’s theorem,

$\displaystyle P \Bigg( \frac{E_1}{A} \Bigg) = \frac{P(E_1) . P\Big(\frac{A}{E_1}\Big)}{P(E_1) . P\Big(\frac{A}{E_1}\Big) + P(E_2) . P\Big(\frac{A}{E_2}\Big)}$

$\displaystyle = \frac{\frac{80}{100} \times \frac{15}{100}}{\frac{80}{100} \times \frac{15}{100}+\frac{85}{100} \times \frac{10}{100}}$

$\displaystyle = \frac{1200}{1200+850}$

$\displaystyle = \frac{1200}{2050}$

$\displaystyle = 0.59$

(ii)

(a) Probabiity that exactly two students will solve the problem

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

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

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

$\displaystyle = \frac{3+2+1}{2 \times 3 \times 4} = \frac{6}{24} = \frac{1}{4}$

(b)

Probability that at least two of them will solve the problem

= Probability that exactly two students will solve the problem + Probability that all solve the problem

$\displaystyle = \frac{1}{4} + P(A \cap B \cap C)$

$\displaystyle = \frac{1}{4} + \frac{1}{2} \times \frac{1}{3} \times \frac{1}{4} = \frac{1}{4}+ \frac{1}{24} = \frac{7}{24}$

$\\$

Question 10:

(i) Solve the differential equation:

$\displaystyle (1+y^2)\ dx = (\tan^{-1} y -x) \ dy$

(ii) Solve the differential equation:

$\displaystyle (x^2-y^2)\ dx+2xy \ dy = 0$

Answer:

(i) We have

$\displaystyle (1+y^2)\ dx = (\tan^{-1} y -x) \ dy$

$\displaystyle \Rightarrow \frac{dx}{dy} = \frac{\tan^{-1} y}{1+y^2} - \frac{x}{1+y^2}$

$\displaystyle \Rightarrow \frac{dx}{dy} + \frac{x}{1+y^2}= \frac{\tan^{-1} y}{1+y^2}$

$\displaystyle \text{Which is of the form } \frac{dx}{dy} + Px = Q$

$\displaystyle \therefore I.F. = e^{\int \frac{1}{1+y^2} \ dy } = e^{\tan^{-1} y}$

$\displaystyle I.F. = e^{\int \frac{1}{1+y^2} \ dx } = e^{\tan^{-1} y}$

$\displaystyle I.F. \times x = \int I.F. \times Q \ dy$

$\displaystyle \Rightarrow e^{\tan^{-1} y} . x =\int e^{\tan^{-1} y} . \frac{\tan^{-1} y}{1+y^2} \ dy$

$\displaystyle \Rightarrow e^{\tan^{-1} y} . x = \int e^t . t \ dt$

$\displaystyle \Rightarrow e^{\tan^{-1} y} . x = t.e^t - \int 1. e^t . \ dt = t.e^t - e^t + c = e^t(t-1) + c$

$\displaystyle \Rightarrow e^{\tan^{-1} y} . x = e^{\tan^{-1} y} (\tan^{-1} y - 1) + c$

(ii)

$\displaystyle \text{Given, }(x^2-y^2)\ dx+2xy \ dy = 0$

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

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

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

$\displaystyle \Rightarrow v + \frac{x \ dv}{dx} = \frac{v^2 x^2-x^2}{2xvx}$

$\displaystyle \Rightarrow v + \frac{x \ dv}{dx} = \frac{x^2(v^2-1)}{2x^2v}$

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

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

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

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

By Superable Method on integrating both sides

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

$\displaystyle \Rightarrow \log (1+v^2) = - \log x + \log c$

$\displaystyle \Rightarrow \log (1+v^2) = \log \frac{c}{x}$

$\displaystyle \Rightarrow (1+v^2) = \frac{c}{x}$

$\displaystyle \Rightarrow x(1+v^2)=c$

$\displaystyle \Rightarrow \frac{x(x^2+y^2)}{x^2} = c$

$\displaystyle \Rightarrow x^2+y^2=cx$

$\\$

Question 11: Use matrix method to solve the following system of equations.[6]

$\displaystyle \frac{2}{x}+\frac{3}{y}+\frac{10}{z} = 4$

$\displaystyle \frac{4}{x}-\frac{6}{y}+\frac{5}{z} = 1$

$\displaystyle \frac{6}{x}+\frac{9}{y}-\frac{20}{z} = 2$

Answer:

Given the system of equations:

$\displaystyle \frac{2}{x}+\frac{3}{y}+\frac{10}{z} = 4$

$\displaystyle \frac{4}{x}-\frac{6}{y}+\frac{5}{z} = 1$

$\displaystyle \frac{6}{x}+\frac{9}{y}-\frac{20}{z} = 2$

$\displaystyle \text{Let } \frac{1}{x} = u , \frac{1}{y} = v , \frac{1}{z} = w$

The system of equations become

$\displaystyle 2u + 3 v+10w = 4$

$\displaystyle 4u - 6v + 5w = 1$

$\displaystyle 6u + 9v - 20 w = 2$

$\displaystyle \text{Writing the equation as } AX = B$

$\displaystyle \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix} \begin{bmatrix} u \\v \\ w \end{bmatrix} = \begin{bmatrix} 4 \\ 1\\2 \end{bmatrix}$

Hence,

$\displaystyle A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix} \hspace{1.0cm} X = \begin{bmatrix} u \\v \\ w \end{bmatrix} \hspace{1.0cm} B = \begin{bmatrix} 4 \\ 1\\2 \end{bmatrix}$

$\displaystyle |A| = 2 \begin{vmatrix} -6 & 5 \\ 9 & -20 \end{vmatrix} - 3 \begin{vmatrix} 4 & 5 \\ 6 & -20 \end{vmatrix} + 10 \begin{vmatrix} 4 & -6 \\ 6 & 9 \end{vmatrix}$

$\displaystyle = 2(120-45)-3(-80-30)+10(36+36)$

$\displaystyle = 2 (75)-3(-110)+10(72)$

$\displaystyle = 150+330+720$

$\displaystyle = 1200$

$\displaystyle \therefore |A| \neq 0$

So the system of equation is consistent and has a unique solution.

$\displaystyle \text{Now, } A^{-1} = \frac{1}{|A|} adj(A)$

$\displaystyle \text{adj}(A) = \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix} ' = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix} = \begin{bmatrix} 75 & 150 & 75 \\ 110 & -110 & 30 \\ 72 & 0 & -24 \end{bmatrix}$

$\displaystyle \text{Now, } A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -110 & 30 \\ 72 & 0 & -24 \end{bmatrix}$

$\displaystyle \text{Thus, } X = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -110 & 30 \\ 72 & 0 & -24 \end{bmatrix} \begin{bmatrix} 4 \\ 1\\2 \end{bmatrix}$

$\displaystyle \begin{bmatrix} u \\v \\ w \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -110 & 30 \\ 72 & 0 & -24 \end{bmatrix} \begin{bmatrix} 4 \\ 1\\2 \end{bmatrix}$

$\displaystyle \begin{bmatrix} u \\v \\ w \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 300+150+150 \\ 440-100+60 \\ 288+0-48 \end{bmatrix}$

$\displaystyle \begin{bmatrix} u \\v \\ w \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix}$

$\displaystyle \begin{bmatrix} u \\v \\ w \end{bmatrix} = \begin{bmatrix} 1/2 \\ 1/3 \\ 1/5 \end{bmatrix}$

$\displaystyle \text{Hence } u = \frac{1}{2} , \ v = \frac{1}{3} , \ w = \frac{1}{5}$

$\displaystyle \text{Hence } x=2, y=3 , z=5$

$\\$

Question 12: [6]

(i) Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface is $\displaystyle \cot^{-1} \sqrt{2}$

(ii) A running track of 440 m is to be laid out enclosing a football field. The football field is in the shape of a rectangle with a semi-circle at each end. If the area of the rectangular portion is to be maximum, then find the length of its sides. Also calculate the area of the football field.

Answer:

$\displaystyle \text{(i) Let } h, r \text{ and } \alpha \text{ be the height, radius and semi vertical angle of the right-angled triangle. }$

$\displaystyle \text{We know, volume of the cone is given by } V = \frac{1}{3} \pi r^2 h \Rightarrow h = \frac{3V}{\pi r^2}$

$\displaystyle \text{Also, slant height, } l = \sqrt{h^2+r^2}$

Curved surface area is given by:

$\displaystyle A = \pi r l = \pi r \sqrt{r^2+h^2}$

$\displaystyle A = \pi r \sqrt{r^2+ \frac{9V^2}{\pi^2 r^4}}$

$\displaystyle A = \sqrt{\pi^2r^4+ \frac{9V^2}{r^2}}$

Differentiating above w.r.t $r$ , we get

$\displaystyle \frac{dA}{dr} = \frac{1}{2 \sqrt{\pi^2r^4+ \frac{9V^2}{r^2}} } \Bigg( 4\pi^2r^3- \frac{18V^2}{r^3} \Bigg)$

$\displaystyle A \text{ is maximum or minimum, when } \frac{dA}{dr} = 0$

$\displaystyle \therefore \frac{1}{2 \sqrt{\pi^2r^4+ \frac{9V^2}{r^2}} } \Bigg( 4\pi^2r^3- \frac{18V^2}{r^3} \Bigg) = 0$

$\displaystyle \Rightarrow 4\pi^2r^3- 18V^2 r^{-3} = 0$

$\displaystyle \Rightarrow 4\pi^2r^3 = 18V^2 r^{-3}$

$\displaystyle \Rightarrow 4\pi^2r^3 = 18 \Bigg( \frac{1}{3} \pi r^2 h \Bigg)^2 r^{-3}$

$\displaystyle \Rightarrow \frac{h}{r} = \sqrt{2}$

$\displaystyle \text{Thus, } \cot \theta = \sqrt{2}$

$\displaystyle \text{Hence, semi-vertical angle } \theta = \cot^{-1} \sqrt{2}$

$\displaystyle \text{Also, for } r < \Bigg( \frac{3V}{\pi \sqrt{2}} \Bigg)^{\frac{1}{3}}, \ \frac{dA}{dr} < 0$

$\displaystyle \text{and for } r > \Bigg( \frac{3V}{\pi \sqrt{2}} \Bigg)^{\frac{1}{3}}, \ \frac{dA}{dr} > 0$

$\displaystyle \text{So, curved surface area for } r^3 = \frac{3V}{\pi \sqrt{2}} \ \text{ or } \ V = \frac{\pi r^3 \sqrt{2}}{3} \text{ is the least. }$

$\displaystyle \text{Here, area of rectangular portion } = 2 r x$

$\displaystyle \text{and perimeter of track } = 2x + 2\pi r$

$\displaystyle \therefore 2x+ 2 \pi r = 440 \ldots \ldots (i)$

$\displaystyle \text{So, area of rectangular portion, } A = r(440-2\pi r)$

$\displaystyle A = 440 r - 2 \pi r^2$

$\displaystyle \text{Differentiating } A \text{ w.r.t. } r, \text{ we get }$

$\displaystyle \frac{dA}{dr} = 440 - 4 \pi r$

Area to be maximum,

$\displaystyle \frac{dA}{dr} = 0$

$\displaystyle \therefore 440 - 4 \pi r = 0$

$\displaystyle \Rightarrow r = \frac{110}{\pi}$

$\displaystyle \text{ Now, } \frac{d^2A}{dr^2} = -4\pi < 0$

From (i), we get

$\displaystyle 2x = 440 - 2 \pi r = 440 - 2\pi \Big( \frac{110}{\pi} \Big) = 440 - 220 = 220$

$\displaystyle \Rightarrow x = 110$

$\displaystyle \text{ Hence, sides of rectangle are } 2r = \frac{220}{\pi} \text{ m and } x = 110 \text{ m }$

$\displaystyle \text{ Area of football field = area of triangle } + 2 \text{ ( area of semicircle) }$

$\displaystyle = 2rx + 2 \Big( \frac{\pi r^2}{2} \Big)$

$\displaystyle = 110 \times \frac{220}{\pi} + \pi \Big( \frac{110}{\pi} \Big)^2$

$\displaystyle = \frac{24200}{\pi}+ \frac{12100}{\pi}$

$\displaystyle = \frac{36300}{\pi}$

$\displaystyle = \frac{36300}{\frac{22}{7}}$

$\displaystyle = 1650 \times 7$

$\displaystyle = 11550 \text{ m}^2$

$\\$

Question 13:

$\displaystyle \text{(i) Evaluate: } \int \frac{3e^{2x}-2e^x}{e^{2x}+2e^x-8} dx$

$\displaystyle \text{(ii) Evaluate: } \int \frac{2}{(1-x)(1+x^2)} dx$

Answer:

$\displaystyle \text{(i) Let } I = \int \frac{3e^{2x}-2e^x}{e^{2x}+2e^x-8} dx$

$\displaystyle \text{Let } e^x = t$

$\displaystyle \Rightarrow e^x dx = dt$

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

$\displaystyle \therefore I = \int \frac{3t^2-2t}{t^2+2t-8} \frac{dt}{t} = \int \frac{3t-2}{t^2+2t-8} dt$

$\displaystyle \text{Now, } \frac{3t-2}{t^2+2t-8} = \frac{3t-2}{(t+4)(t-2)}$

$\displaystyle \text{So, }\frac{3t-2}{t^2+2t-8} = \frac{A}{t+4}+\frac{B}{t-2}$

$\displaystyle 3t-2 = A (t-2) + B ( t+4)$

$\displaystyle 3t-2= (A+B)t + ( 4B -2A)$

On comparing

$\displaystyle A+B = 3$

and $\displaystyle 4B - 2A = - 2$

On solving the above two equations we get

$\displaystyle A = \frac{7}{3} \text{ and } B = \frac{2}{3}$

$\displaystyle \therefore \frac{3t-2}{t^2+2t-8} = \frac{7}{3}\frac{1}{(t+4)}+ \frac{2}{3}\frac{1}{(t-2)}$

$\displaystyle \therefore I = \frac{7}{3} \int \frac{1}{t+4} dt + \frac{2}{3} \int \frac{1}{t-2} dt$

$\displaystyle = \frac{7}{3} \log |t+4| + \frac{2}{3} \log |t-2|+C$

$\displaystyle = \frac{7}{3} \log |e^x+4| + \frac{2}{3} \log |e^x-2|+C$

(ii)

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

We can write integrand as

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

Applying partial fraction,

$\displaystyle \frac{-2}{(x-1)(1+x^2)} = \frac{A}{(x-1)} + \frac{Bx+C}{(1+x^2)}$

$\displaystyle \frac{-2}{(x-1)(1+x^2)} = \frac{A(1+x^2)+(Bx+C)(x-1)}{(x-1)(1+x^2)}$

$\displaystyle - 2 = A(1+x^2)+(Bx+C)(x-1) \ldots \ldots (i)$

$\displaystyle \text{Putting } x = 1, \text{ we get }$

$\displaystyle -2 = 2A + 0 \Rightarrow A=-1$

$\displaystyle \text{Putting } x = 0, \text{ we get }$

$\displaystyle -2 = A+(-C)$

$\displaystyle \Rightarrow -2 = -1 - C$

$\displaystyle \Rightarrow C = 1$

$\displaystyle \text{Putting } A = - 1 , C = 1 \text{ in equation (i) , we get }$

$\displaystyle B = 1$

$\displaystyle \text{So, } \frac{-2}{(x-1)(1+x^2)} = \frac{-1}{(x-1)} + \frac{x+1}{(1+x^2)}$

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

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

$\displaystyle = -\int \frac{1}{(x-1)} dx+ I_1 + \int \frac{1}{(1+x^2)} dx$

$\displaystyle \text{Where } I_1 = \int \frac{x}{(1+x^2)} dx$

$\displaystyle \text{Let } x^2 + 1 = t$

$\displaystyle \Rightarrow 2x \ dx = dt$

$\displaystyle \Rightarrow x \ dx = \frac{dt}{2}$

$\displaystyle \therefore I_1 = \frac{1}{2} \int \frac{dt}{2}$

$\displaystyle = \frac{1}{2} \log |t| + C$

$\displaystyle = \frac{1}{2} \log |x^2 +1| + C_1 \ldots \ldots (ii)$

$\displaystyle \text{Thus, } I = -\int \frac{1}{(x-1)} dx+ \int \frac{x}{(1+x^2)} dx + \int \frac{1}{(1+x^2)} dx$

$\displaystyle = -\log |x-1| + \frac{1}{2} \log |x^2+1|+ \tan^{-1} x +C_1+C_2 \text{ [from equation (ii) ] }$

$\displaystyle = -\log |x-1| + \frac{1}{2} \log |x^2+1|+ \tan^{-1} x + C \hspace{1.0cm} [C= C_1+C_2]$

$\\$

Question 14: A box contains 30 fruits, out of which 10 are rotten. Two fruits are selected at random one by one without replacement from the box. Find the probability distribution of the number of unspoiled fruits. Also find the mean of the probability distribution. [6]

Answer:

$\displaystyle \text{Total number of fruits }= 30$

$\displaystyle \text{Number of spoiled fruits }= 10$

$\displaystyle \text{Number of unspoiled fruits }= 20$

$\displaystyle \text{Probability of spoiled fruits }= \frac{10}{30}$

$\displaystyle \text{Probability of unspoiled fruits }= \frac{20}{30}$

$\displaystyle \text{Let } X \text{ be the random variable of number of unspoiled fruit. }$

$\displaystyle \text{So, }X = 0, 1, 2$

$\displaystyle \text{Two fruits can be drawn from 30 fruits }$

$\displaystyle P(X=0) = \text{Two fruits are spoiled fruits }$

$\displaystyle = \frac{^{10}C_2}{^{30}C_2} = \frac{10 \times 9 \times 2}{2 \times 30 \times 29} = \frac{9}{87}$

$\displaystyle P(X=1) = \text{One fruit is unspoiled and One fruits is spoiled fruits }$

$\displaystyle = \frac{^{20}C_1 . ^{10}C_1 }{^{30}C_2} = \frac{20 \times 10 \times 2}{30 \times 29} = \frac{40}{87}$

$\displaystyle P(X=2) = \text{Two fruits are unspoiled fruits }$

$\displaystyle = \frac{^{20}C_2}{^{30}C_2} = \frac{20 \times 19 \times 2}{ 30 \times 29 \times 2} = \frac{38}{87}$

$\displaystyle \text{Mean of probability distribution }= \Sigma X_i P(X_i)$

$\displaystyle = X_0 P(X_0) + X_1P(X_1)+ X_2 P(X_2)$

$\displaystyle = 0 \times \frac{9}{87} + 1 \times \frac{40}{87} + 2 \times \frac{38}{87}$

$\displaystyle = \frac{116}{87} = 1.33$

$\\$

SECTION – B [15 Marks]

Question 15: In subparts (i) and (ii), choose the correct options, and in subparts (iii) to (v), answer the questions as instructed. [5]

$\displaystyle \text{(i) If } |\overrightarrow{a}| = 3, |\overrightarrow{b}| = \frac{\sqrt{2}}{3} \text{ and } \overrightarrow{a}\times \overrightarrow{b} \text{ is a unit vector then the angle between } \\ \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ will be: }$

$\displaystyle (a) \ \frac{\pi}{6} \hspace{1.0cm} (b) \ \frac{\pi}{4} \hspace{1.0cm}(c) \ \frac{\pi}{3} \hspace{1.0cm}(d) \ \frac{\pi}{2}$

$\displaystyle \text{(ii) The distance of the point } 2 \hat{i} + \hat{j} - \hat{k} \text{ from the plane } \overrightarrow{r}.(\hat{i} -2 \hat{j} +4 \hat{k})= 9 \text{ will be }$

$\displaystyle (a) \ 13 \hspace{1.0cm} (b) \ \frac{13}{\sqrt{21}} \hspace{1.0cm}(c) \ 21 \hspace{1.0cm}(d) \ \frac{21}{\sqrt{13}}$

$\displaystyle \text{(iii) Find the area of the parallelogram whose diagonals are } \hat{i} -3\hat{j} + \hat{k} \text{ and } \hat{i} + \hat{j} + \hat{k}$

(iv) Find the equation of the plane passing through the point (2, 4, 6) and making equal intercepts on the coordinate axes

$\displaystyle \text{(v) If the two vectors } 3\hat{i} +\alpha\hat{j} + \hat{k} \text{ and } 2\hat{i} -\hat{j} + 8\hat{k} \\ \text{ are perpendicular to each other, then find the value of } \alpha$

Answer:

(i)

$\fbox{ b }$

$\displaystyle \text{Given } |\overrightarrow{a}| = 3, |\overrightarrow{b}| = \frac{\sqrt{2}}{3}$

$\displaystyle \text{and } |\overrightarrow{a} \times \overrightarrow{b} | = 1$

$\displaystyle \text{We know that,} |\overrightarrow{a} \times \overrightarrow{b} | = \Bigg| |\overrightarrow{a}| |\overrightarrow{b}| \sin \theta \hat{n} \Bigg|$

$\displaystyle \text{where } \theta \text{ is the angle between } \overrightarrow{a} \text{ and } \overrightarrow{b}$

$\displaystyle |\overrightarrow{a} \times \overrightarrow{b} | = |\overrightarrow{a}||\overrightarrow{b}| \sin \theta . 1$

$\displaystyle \Rightarrow 1 = 3 \times \frac{\sqrt{2}}{3} \sin \theta$

$\displaystyle \Rightarrow \sin \theta = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4}$

$\displaystyle \Rightarrow \theta = \frac{\pi}{4}$

(ii)

$\fbox{ b }$

$\displaystyle \text{We know that distance } = \frac{|\overrightarrow{a}.\overrightarrow{n}-d|}{|\overrightarrow{n}|}$

Given

$\displaystyle \overrightarrow{a} = 2 \hat{i} + \hat{j} - \hat{k} \hspace{1.0cm} \overrightarrow{n} = \hat{i} -2 \hat{j} +4 \hat{k} \hspace{1.0cm} d=9$

$\displaystyle \therefore \text{Distance } = \frac{|(2 \hat{i} + \hat{j} - \hat{k}).(\hat{i} -2 \hat{j} +4 \hat{k})-9|}{|\hat{i} -2 \hat{j} +4 \hat{k}|} = \frac{|2-2-4-9|}{\sqrt{1+4+16}} = \frac{13}{\sqrt{21}}$

$\displaystyle \text{(iii) Let } d_1 = \hat{i} -3\hat{j} + \hat{k} \text{ and } d_2 = \hat{i} + \hat{j} + \hat{k}$

$\displaystyle \text{Area of parallelogram } = \frac{1}{2} | \overrightarrow{d_1} \times \overrightarrow{d_2} |$

$\displaystyle \overrightarrow{d_1} \times \overrightarrow{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 1 & 1 & 1 \end{vmatrix}$

$\displaystyle = \hat{i} ( - 3 01) - \hat{j} (1-1) + \hat{k} (1+3)$

$\displaystyle = -4 \hat{i} + 4 \hat{k}$

$\displaystyle \therefore |\overrightarrow{d_1} \times \overrightarrow{d_2}| = \sqrt{(-4)^2+(4)^2} = \sqrt{16+16} =\sqrt{32}$

$\displaystyle \text{Thus, area of parallelogram }= \frac{1}{2} \sqrt{32} {\text{ unit}}^2 = \sqrt{8} {\text{ unit}}^2$

$\displaystyle \text{(iv) The equation of the plane intercepts on the coordinate axes are } a, b, \text{ and } c \text{ is }$

$\displaystyle \frac{x}{a} + \frac{y}{b} +\frac{z}{c} = 1$

$\displaystyle \text{Given, } a = b = c$

$\displaystyle \therefore \frac{x}{a} + \frac{y}{a} +\frac{z}{a} = 1$

$\displaystyle \Rightarrow x + y + z = a$

$\displaystyle \text{This plane passes through } (2, 4, 6, ).$

$\displaystyle \therefore 2 + 4 + 6 = a$

$\displaystyle \Rightarrow a = 12$

$\displaystyle \text{Thus the required equation of the plane is } x + y + z = 12.$

(v)

$\displaystyle \text{For perpendicular vectors } \overrightarrow{a} . \overrightarrow{b} = 0$

$\displaystyle \therefore (3\hat{i} +\alpha\hat{j} + \hat{k}).(2\hat{i} -\hat{j} + 8\hat{k})=0$

$\displaystyle \Rightarrow 6 - \alpha + 8 = 0$

$\displaystyle \Rightarrow \alpha = 14$

$\\$

Question 16: [2]

$\displaystyle \text{(i) If } A(1, 2, - 3) \text{ and } B(- 1, - 2, 1) \text{ are the end points of a vector } \overrightarrow{AB} \\ \text{ then find the unit vector in the direction of } \overrightarrow{AB}.$

$\displaystyle \text{(ii) If } \hat{a} \text{ is a unit vector and } (2 \overrightarrow{x} - 3 \hat{a}).(2\overrightarrow{x}+ 3 \hat{a}) = 91, \text{ find the value of } |\overrightarrow{x}|.$

Answer:

$\displaystyle \text{(i) We have } A(1, 2, - 3) \text{ and } B(- 1, - 2, 1)$

$\displaystyle \overrightarrow{AB} = (-1-1)\hat{i}+(-2-2)\hat{j}+(1+3) \hat{k}$

$\displaystyle \therefore \overrightarrow{AB} = -2\hat{i}-4\hat{j}+4\hat{k}$

$\displaystyle \text{Unit vector in the direction of } \overrightarrow{AB}$

$\displaystyle = \frac{-2\hat{i}-4\hat{j}+4\hat{k}}{\sqrt{(-2)^2+(-4)^2+(4)^2}}$

$\displaystyle = \frac{-2\hat{i}-4\hat{j}+4\hat{k}}{\sqrt{4+16+16}}$

$\displaystyle = \frac{-2\hat{i}-4\hat{j}+4\hat{k}}{6}$

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

$\displaystyle \text{(ii) Given, } (2 \overrightarrow{x} - 3 \hat{a} ).(2 \overrightarrow{x}+3 \hat{a}) = 91$

$\displaystyle \Rightarrow 4 |\overrightarrow{x}|^2 + 6 \overrightarrow{x} . \hat{a} - 6 \hat{a} . \overrightarrow{x} - 9 |\hat{a}|^2 = 91$

$\displaystyle \because \hat{a} = 1$

$\displaystyle \Rightarrow 4 |\overrightarrow{x}|^2 + 6 \overrightarrow{x} . 1 - 6 .1 . \overrightarrow{x} - 9. 1 = 91$

$\displaystyle \Rightarrow 4 |\overrightarrow{x}|^2 = 100$

$\displaystyle \Rightarrow |\overrightarrow{x}|^2 = 25$

$\displaystyle \Rightarrow |\overrightarrow{x}| = 5$

$\\$

Question 17:

$\displaystyle \text{(i) Find the equation of the plane passing through the point } (1, 1, - 1) \\ \\ \text{ and perpendicular to the planes } x + 2y + 3z = 7 \text{ and } 2x - 3y + 4z = 0.$

$\displaystyle \text{(ii) A line passes through the point } (2, - 1, 3) \text{ and is perpendicular to the line }$

$\displaystyle \overrightarrow{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda (2\hat{i} -2 \hat{j} + \hat{k}) \text{ and }$

$\displaystyle \overrightarrow{r}= (2\hat{i} - \hat{j} - 3\hat{k}) + \mu (\hat{i} +2 \hat{j} + 2\hat{k}) .$

Obtain its equation.

Answer:

$\displaystyle \text{(i) Let the equation of the plane passing through point } (1, 1, -1) \text{ be }$

$\displaystyle a(x-1) + b(y-1) + c(z+1) = 0 \ldots \ldots (i)$

$\displaystyle \text{Equation (i) is perpendicular to the plane } x +2y + 3z-7=0$

$\displaystyle \therefore 1.a+2.b+3.c = 0$

$\displaystyle \Rightarrow a + 2b + 3c = 0 \ldots \ldots (ii)$

$\displaystyle \text{Again equation (i) is perpendicular to plane } 2x - 3y + 4 z = 0$

$\displaystyle \therefore 2.a-3.b+4.c = 0$

$\displaystyle \Rightarrow 2a - 3b + 4c = 0 \ldots \ldots (iii)$

Solving equation (ii) and (iii), we get

$\displaystyle \frac{a}{8+9}= \frac{b}{6-4}=\frac{c}{-3-4}$

$\displaystyle \Rightarrow \frac{a}{17} = \frac{b}{2}= \frac{c}{-7} = k$

$\displaystyle \Rightarrow a = 17k, b = 2k \text{ and } c = -7k$

$\displaystyle \text{Putting the values of } a, b, \text{ and } c \text{ in equation (i), we get }$

$\displaystyle 17k(x-1)+2k(y-1)-7k(z+1)=0$

$\displaystyle \Rightarrow 17(x-1)+2(y-1)-7(z+1)=0$

$\displaystyle \Rightarrow 17x+2y-7z-17-2-7=0$

$\displaystyle \Rightarrow 17x+2y-7z-26=0$

$\displaystyle \text{(ii) Let equation of required line passing through } (2, -1, 3) \text{ is }$

$\displaystyle \overrightarrow{r}= (2\hat{i} - \hat{j} - 3\hat{k}) + \mu (\hat{i} +2 \hat{j} + 2\hat{k})$

Equation (i) is perpendicular to

$\displaystyle \overrightarrow{r} = (\hat{i} - \hat{j} - \hat{k}) + \lambda (2\hat{i} + \hat{j} + \hat{k})$

$\displaystyle \therefore 2x-2y+2 = 0$

Again equation (i) is perpendicular to

$\displaystyle \overrightarrow{r}= (2\hat{i} - \hat{j} - 3\hat{k}) + \pi (\hat{i} +2 \hat{j} + 2\hat{k})$

$\displaystyle \therefore k + 2y + 22 = 0$

On solving equation (ii) and (iii)

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

$\displaystyle \Rightarrow \frac{x}{-6} = \frac{y}{-3} = \frac{z}{2} =k$

$\displaystyle \Rightarrow x = -6k, y = 3k \text{ and } z = 2k$

$\displaystyle \text{Putting the values of } x, y, \text{ and } z \text{ in equation we get}$

$\displaystyle \overrightarrow{r}= (2\hat{i} - 3\hat{j} + 3\hat{k}) + \lambda_1 (-6\hat{i} -3 \hat{j} + 3\hat{k})$

$\\$

$\displaystyle \text{Question 18: Find the area of the region bound by the curve }x^2 = 4y \\ \text{ and the line } x = 4y-2.$ [4]

Answer:

$\displaystyle \text{Given curve } x^2 = 4y \ldots \ldots (i) \text{ represents an upward parabola with vertex } (0, 0) \\ \text{ and axis along y-axis. }$

$\displaystyle \text{Given equation of the line is } x = 4y-2 \ldots \ldots (ii)$

$\displaystyle \text{On solving equation (i) and (ii), we get }$

$\displaystyle x^2 = x + 2$

$\displaystyle \Rightarrow x^2 - x - 2 = 0$

$\displaystyle \Rightarrow (x-2)(x+1) = 0$

$\displaystyle \Rightarrow x = 2, -1$

$\displaystyle \text{When } x = 2, y = 1$

$\displaystyle \text{and when }x = -1, y = \frac{1}{4}$

$\displaystyle \text{Thus the line meets the parabola at the points } A(2, 1) \text{ and } B \Big(-1, \frac{1}{4} \Big)$

$\displaystyle \text{Required area = ( Area under line } x = 4y - 2) - ( \text{Area under the parabola } \\ x^2 = 4y)$

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

$\displaystyle \text{[From equation (ii) } y = \frac{x+2}{4} \text{ and from equation (i) } y = \frac{x^2}{4}$

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

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

$\displaystyle = \frac{1}{4} \Big( 6 +\frac{3}{2} \Big) - \frac{1}{12} (9)$

$\displaystyle = \frac{15}{8}- \frac{9}{12}$

$\displaystyle = \frac{9}{8} \text{ sq. units}$

$\\$

Question 19: | In subparts (i) and (ii) choose the correct options and in subparts (iii) to (v), answer the questions as instructed. [5]

$\displaystyle \text{(i) If the demand function is given by } p = 1500 - 2x - x^2 \text{ then find the } \\ \text{ marginal revenue when } x = 10$

$\displaystyle (a) \ 1160 \hspace{1.0cm} (b) \ 1600 \hspace{1.0cm}(c) \ 1100 \hspace{1.0cm}(d) \ 1200$

$\displaystyle \text{(ii) If the two regression coefficients are } 0.8 \text{ and } 0.2, \text{ then the value of coefficient} \\ \text{ of correlation } r \text{ will be: }$

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

$\displaystyle \text{(iii) Out of the two regression lines } x + 2y -5 = 0 \text{ and } 2x + 3y = 8, \\ \text{find the line of regression of } y \text{ on } x.$

$\displaystyle \text{(iv) The cost function } C(x) = 3x^2- 6x + 5. \text{ Find the average cost when } \\ x = 2.$

$\displaystyle \text{(v) The fixed cost of a product is Rs. } 30,000 \text{ and its variable cost per unit is Rs. } 800. \\ \text{ If the demand function is } p(x) = 4500 - 100x. \text{ Find the break-even values. }$

Answer:

(i)

$\fbox{ a }$

$\displaystyle \text{Given, } p = 1500 - 2x - x^2$

$\displaystyle \text{Revenue Function, } R = px$

$\displaystyle \Rightarrow R = 1500x - 2x^2 - x^3$

$\displaystyle \text{Marginal Revenue } = \frac{dR}{dx}$

$\displaystyle = \frac{d}{dx} (1500x - 2x^2 - x^3)$

$\displaystyle = 1500 - 4x - 3x^2$

$\displaystyle \text{Marginal revenue at } x = 10 \text{ is }$

$\displaystyle = 1500 - 4(10) - 3(10)^2$

$\displaystyle = 1500 - 40 - 300$

$\displaystyle = 1160$

(ii)

$\fbox{ c }$

$\displaystyle r = \sqrt{0.8 \times 0.2} = \sqrt{0.16} = \sqrt{(0.4)^2} = 0.4$

Here, correlation coefficient will be positive because both the coefficients are positive.

(iii)

$\displaystyle x+2y-5 = 0 \ldots \ldots (i)$

$\displaystyle 2x+3y= 8 \ldots \ldots (ii)$

$\displaystyle \text{From equation (i), be regression of } y \text{ on } x \text{ and equation (ii) be regression of } x \text{ on }y$

$\displaystyle \text{Slope of equation (i) }= -\frac{1}{2}$

$\displaystyle \text{Slope of equation (ii) }= -\frac{2}{3}$

$\displaystyle \Rightarrow b_{yx} = \frac{-1}{2}$

$\displaystyle \Rightarrow b_{xy} = \frac{-3}{2}$

$\displaystyle \text{Since both }b_{yx} \text{ and }b_{xy} \text{ are of same sign and }$

$\displaystyle b_{yx} \times b_{xy} = \frac{-1}{2} \times \frac{-3}{2} = \frac{3}{4} < 1$

Therefore, our assumption is true.

$\displaystyle \text{Hence the equation (i) } x+ 2y -5 = 0 \text{ is a line of regression of } y \text{ on } x.$

$\displaystyle \text{(iv) Given, } C(x) = 3x^2- 6x + 5$

$\displaystyle \text{Average cost } = \frac{C(x)}{x} = 3x-6+\frac{5}{x}$

$\displaystyle \text{(Average cost)}_{\text{ at } x= 2} = 3(2) - 6 + \frac{5}{2} = \frac{5}{2} = 2.5$

(v)

$\displaystyle \text{Total cost = fixed cost + variable cost }$

$\displaystyle C(x) = 30000+ 800x \text{ where } x = \text{ total units }$

$\displaystyle \text{Also, revenue function, } R(x) = p.x = ( 4500 - 100x) x = 4500x - 100x^2$

$\displaystyle \text{Profit function } P(x) = R(x) - C(x) = 4500x - 100x^2 - 30000 - 800x$

$\displaystyle = -100x^2+3700x - 30000$

$\displaystyle \text{At break even pont } P(x) = 0$

$\displaystyle \therefore -100x^2 + 3700x - 30000 = 0$

$\displaystyle \Rightarrow x^2 - 37x + 300 = 0$

$\displaystyle x = \frac{37 \pm \sqrt{(-37)^2-4 (1)(300)}}{2 \times 1}$

$\displaystyle = \frac{37 \pm \sqrt{1369-1200}}{2}$

$\displaystyle = \frac{37 \pm 13}{2}$

$\displaystyle = \frac{37 + 13}{2} \text{ and } \frac{37 - 13}{2}$

$\displaystyle = \frac{50}{2} \text{ and } \frac{24}{2}$

$\displaystyle = 25 \text{ and } 12$

$\displaystyle \text{(So, breakeven values are 25 and 12 }$

$\\$

Question 20:

$\displaystyle \text{(i) The total cost function for } x \text{ units is given by } C(x) = \sqrt{6x+5}+2500. \\ \text{ Show that the marginal cost decreases as the output } x \text{ increases. }$

$\displaystyle \text{(ii) The average revenue function is given by } AR = 25-\frac{x}{4}. \\ \text{ Find total revenue function and marginal revenue function. }$

Answer:

$\displaystyle \text{ (i) Given, } C(x) = \sqrt{6x+5}+2500$

$\displaystyle MC = \frac{dC}{dx}$

$\displaystyle = \frac{d}{dx} [ \sqrt{6x+5}+2500]$

$\displaystyle \Rightarrow MC = \frac{1}{2} (6x+5)^{\frac{-1}{2}} (6)$

$\displaystyle \Rightarrow MC = \frac{3}{\sqrt{6x+5}}$

$\displaystyle \text{ Now, put } x = 2,$

$\displaystyle MC = \frac{3}{\sqrt{6(2)+5}} = \frac{3}{\sqrt{17}} = \frac{3}{4.12} = 0.72$

$\displaystyle Put x= 3,$

$\displaystyle MC = \frac{3}{\sqrt{6(3)+5}} = \frac{3}{\sqrt{23}} = \frac{3}{4.79} = 0.62$

$\displaystyle \text{ Therefore, it is clear, as we increase output } x, MC\text{ decreases. }$

(ii)

$\displaystyle \text{ Given, average revenue } = AR = 25 - \frac{x}{4}$

$\displaystyle \text{ Since, } AR = \frac{R}{x} = \frac{px}{x} = p$

$\displaystyle \therefore P = 25 - \frac{x}{4}$

$\displaystyle \text{ Total revenue } R(x) = p.x = 25x - \frac{x^2}{4}$

$\displaystyle \text{ Marginal revenue, } MR = \frac{d}{dx} R(x) = \frac{d}{dx} \Big( 25x - \frac{x^2}{4} \Big) = 25 - \frac{x}{2}$

$\\$

Question 21:

Solve the following Linear Programming Problem graphically.

$\displaystyle \text{ Maximize } Z = 5x + 2y \text{ subject to: }$

$\displaystyle x - 2y \leq 2,$

$\displaystyle 3x + 2y \leq 12,$

$\displaystyle - 3x + 2y \leq 3,$

$\displaystyle x \geq 0, y \geq 0$

Answer:

Given LPP is:

$\displaystyle \text{ Maximize } Z = 5x + 2y \text{ subject to: }$

$\displaystyle x - 2y \leq 2,$

$\displaystyle 3x + 2y \leq 12,$

$\displaystyle - 3x + 2y \leq 3,$

$\displaystyle x \geq 0, y \geq 0$

Converting the inequations into equation, we get:

$\displaystyle x - 2y = 2 \ldots \ldots (i)$

$\displaystyle 3x + 2y =12 \ldots \ldots (ii)$

$\displaystyle - 3x + 2y = 3 \ldots \ldots (iii)$

$\displaystyle x = 0, y = 0 \ldots \ldots (iv)$

$\displaystyle \text{ On plotting the above equations, we get the cornet points as } \\ A(0, 1.5), B(3.5, 0.75), C(2, 0), D(1.5, 3.75), O(0.0).$

The value of the objective functions are:

$\begin{array}{ |l|l|} \hline \text{Point } (x, y) & z=5x+2y \\ \hline \text{A(0, 1.5) } & 5\times 0 + 2 \times 1.5 = 3 \\ \hline B(3.5, 0.75) & 5\times 3.5 + 2 \times 0.75 = 19 \text{ Max } \\ \hline C(2, 0) & 5\times 2 + 2 \times 0 = 10 \\ \hline D(1.5, 3.75) & 5\times 1.5 + 2 \times 3.75 = 15 \\ \hline O(0, 0) & 5\times 0 + 2 \times 0 = 0 \\ \hline \end{array}$

$\displaystyle \text{ So, the maximum value of } z \text{ is } 19.$

$\\$

Question 22:

(i) The following table shows the Mean, the Standard Deviation and the coefficient of correlation of two variables x and y. [4]

$\begin{array}{ |l|l|l|} \hline \text{Series } & x \hspace{1.0cm} & y \hspace{1.0cm} \\ \hline \text{Mean } & 8 & 6 \\ \hline \text{Standard Deviation } & 12 & 4 \\ \hline \text{Coefficient of correlation } & 0.6 & \\ \hline \end{array}$

Calculate:
$\displaystyle \text{ a) the regression coefficient } b_{xy} \text{ and } b_{yx}$
$\displaystyle \text{ b) the probable value of } y \text{ when } x = 20$

$\displaystyle \text{ (ii) An analyst analyzed 102 trips of a travel company. He studied the relation } \\ \text{ between travel expenses } (y) \text{ and the duration } (x) \text{ of these trips. He found } \\ \text{ that the relation between } x \text{ and } y \text{ was linear. Given the following data,} \\ \text{ find the regression equation of } y \text{ on } x.$

$\displaystyle \Sigma x = 510, \hspace{0.5cm} \Sigma y = 7140, \hspace{0.5cm} \Sigma x^2 = 4150, \hspace{0.5cm} \Sigma y^2 = 740200, \hspace{0.5cm} \Sigma xy = 54900$

Answer:

$\displaystyle \text{ Given, } r = 0.6$

$\displaystyle \text{ Mean of } x = \overline{x} = 8$

$\displaystyle \text{ Mean of } y = \overline{y} = 6$

$\displaystyle \text{ S.D. of } x = \sigma_x = 12$

$\displaystyle \text{ S.D. of } x = \sigma_y = 4$

(i)

$\displaystyle b_{xy}= \frac{r \sigma_x }{\sigma_y} = \frac{0.6 \times 12}{4} = 1.8$

$\displaystyle b_{yx}= \frac{r \sigma_y }{\sigma_x} = \frac{0.6 \times 4}{12} = 0.2$

(ii)

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

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

$\displaystyle y-6 = 0.2(x-8)$

$\displaystyle \Rightarrow y = 0.2x-1.6 + 6$

$\displaystyle \Rightarrow y = 0.2x+4.4$

$\displaystyle \text{ at } x = 20$

$\displaystyle y = 0.2 \times 20 + 4.4 = 4+4.4= 8.4$

$\displaystyle \text{ Given } n = 102, \hspace{0.2cm} \Sigma x = 510, \hspace{0.2cm} \Sigma y = 7140, \hspace{0.2cm} \Sigma x^2 = 4150, \hspace{0.2cm} \Sigma y^2 = 740200, \hspace{0.2cm} \\ \Sigma xy = 54900$

$\displaystyle \text{ We know, Regression equation of } y \text{ on } x \text{ is given by }$

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

$\displaystyle \text{ So, } \overline{x} = \frac{\Sigma x}{n} = \frac{510}{102} = 5$

$\displaystyle \overline{y} = \frac{\Sigma y}{n} = \frac{7140}{102} = 70$

$\displaystyle b_{yx} = \frac{\Sigma xy - n \overline{x} . \overline{y} }{\Sigma x^2 - n(\overline{x} )^2} = \frac{54900-(102)(5)(70)}{4150-102(5^2)} = \frac{54900-35700}{4150-2550} = \frac{19200}{1600} = 12$

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

$\displaystyle y-70 = 12(x-5)$

$\displaystyle \Rightarrow y = 12 x - 60+70$

$\displaystyle \Rightarrow y = 12 x + 10$

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