# Oswal 36 Sample Question Papers ISC Class 12 Maths Solutions

## Section-A

(i) (b) (9, 6) ∈ R

Explanation :

Given, R = {(x, y) : x = y + 3, y > 5}

For y = 6, x = 6 + 3 = 9

(9, 6) ∈ R.

(ii) (c) 8

Explanation :

Here, n = n(A) = 3, m = n(B) = 2

No. of functions = mn = 23 = 8.

(iii) (a) 1/2

Explanation :

$$\text{We have, sin}\bigg[\frac{\pi}{2}-\text{sin}^{-1}\bigg(\frac{-\sqrt{3}}{2}\bigg)\bigg]\\\qquad\text{Let} \space\space \text{sin}^{-1}\bigg(\frac{-\sqrt{3}}{2}\bigg)=x\\\qquad⇒ \text{sin x} =\frac{-\sqrt{3}}{2}\\\qquad=\text{− sin }\frac{\pi}{3}=\text{sin}\bigg(\frac{\pi}{3}\bigg)\\\qquad\text{x}=-\frac{\pi}{3}\\\qquad⇒ \text{sin}^{-1}\bigg(\frac{-\sqrt{3}}{2}\bigg)=-\frac{\pi}{3}\\\qquad\therefore \text{sin}\bigg[\frac{\pi}{2}-\bigg(-\frac{\pi}{3}\bigg)\bigg]=\text{sin}\bigg(\frac{\pi}{2}+\frac{\pi}{3}\bigg)=\text{cos} \frac{\pi}{2}=\frac{1}{2}$$

(iv) (c) Determinant is a number associated to a square matrix

$$\text{(v) (b)} \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}$$

Explanation :

$$\text{Given,}\space \text{A}= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\\\therefore\space\space\space\space\text{A}^2= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

(vi) (b) 3

Explanation :

$$\text{Let}\space\space\space \text{A}= \begin{bmatrix} 1 & 2 \\ k & 6 \end{bmatrix}\\\qquad\therefore\space\space\space\space\text{A}^{-1}\text{does not exist if |A|}=0\\\qquad\therefore\space\space\space \begin{vmatrix} 1 & 2 \\ k & 6 \end{vmatrix}=0\\\qquad\text{⇒ 6 – 2k = 0}\\\qquad⇒ 2k = 6\\\qquad⇒ k = 3.$$

(vii) (b) 5/2

Explanation :

$$\text{Given curve}\space\space\space \text{x}^2+3y+y^2=5\\\qquad⇒\space\space\space\text{2x+3}\frac{dy}{dx}+2y\frac{dy}{dx}=0\\\qquad\text{⇒ (3 + 2y)}\frac{dy}{dx}=-2x\\\qquad\qquad\frac{dy}{dx}=\frac{-2x}{3+2y}\\\qquad\therefore\space \text{Slope of the normal at (1, 1)} = −\frac{1}{\frac{dy}{dx}}\\\qquad\qquad =-\frac{1}{\frac{-2x}{3+2y}}=\frac{3+2y}{2x}=\frac{3 +2 ×1}{2×1}=\frac{5}{2}.$$

(viii) (c) x + y = 0

Explanation :

$$\text{Given curve}\space\space\space \text{y = sin x}\\\qquad\therefore\frac{dy}{dx}=\text{cos x}\\\qquad⇒ \frac{dy}{dx}\bigg|_{(0,0)}=1\\\text{⇒ Slope of normal is – 1.}\\\qquad\therefore \text{Equation is,} \space y-0=-1(x-0).\\\qquad⇒\space x + y = 0.$$

(ix) (d) ab = 1

Explanation :

$$\text{For y = ae}^{\normalsize– x},\\\qquad\frac{dy}{dx}=-ae^{-x}\\\qquad\text{and for the curve y = be}^x \space\space\frac{dy}{dx}=be^x\\\qquad\text{If curves are orthogonal then}\text{– ae}^{\normalsize– x} × be^x = – 1\\\qquad ⇒ ab = 1.$$

(x) (d) 7/8

Explanation :

$$\text{We know,} \text{P(B|A)}=\frac{P (A \cap B)}{P(A)}\\=\frac{7/10}{4/5}=\frac{7}{8}.$$

• (xi) Consider,

$$\text{L.H.S. = sin}^{-1}\bigg(\frac{\sqrt{3}}{2}\bigg)+2\text{tan}^{-1}\bigg(\frac{1}{\sqrt{3}}\bigg)\\\qquad\frac{\pi}{3}+2×\frac{\pi}{6}=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2\pi}{3}\\\qquad=\text{R.H.S} \space\space\space\space\text{Hence Proved.}$$

$$\text{(xii) Given | A |} = 4\\\qquad\text{| – 2A | = (– 2)3.| A |}\\\qquad\text{= – 8 × 4 = – 32. \space\space \space Ans.}$$

$$\text{(xiii) We have,} \bigg(\frac{dy}{dx}\bigg)^5+\text{3xy}\bigg(\frac{d^3y}{dx^3}\bigg)+y^2\bigg(\frac{d^2y}{dx^2}\bigg)^3=0\\\qquad \therefore \text{Order = 3, degree = 2}\\\qquad \therefore\text{Sum}=3+2=5.\space\space\space\space\text{Ans}\\\text{(xiv) Here,}\space P(A)=\frac{4}{5},P(B)=\frac{1}{3}\\P(A^{‘})=\frac{1}{5},P(B^{‘})=\frac{2}{3}\\\therefore\text{The required probability} = P(A^{‘}).P(B^{‘}) =\frac{1}{5}×\frac{2}{3}=\frac{2}{15}\text{Ans.}\\\text{(xv) Multiples of 5 are 5, 10, 15 and 20}\\\text{Multiples of 7 are 7, 14}\\\text{Total favourable events = 4 + 2 = 6}\\\text{Total number of possible outcomes = 20}\\\therefore \text{Probability that the ball drawn is marked with a number multiple of 5 or 7}\\=\frac{6}{20}=\frac{3}{10}.\\$$

• (i) Here, R = {(a, b) : b = a + 1}
• R = {(a, a + 1) : a, a + 1 ∈ (1, 2, 3, 4, 5, 6)}
• ⇒ R = {(1,2,) (2,3), (3,4), (4,5), (5,6)}
• (a) R is not reflexive as (a, a) ∉ R
• (b) R is not symmetric as (1, 2) ∈ R but (2, 1) ∉ R
• (c) R is not transitive as (1, 2) ∈ R, (2, 3) ∈ R but (1, 3) ∉ R

OR

• (ii) Given, f(x) = x2 + 4
• Let f(x1) = f(x2)
• ⇒ x21 + 4 = x22 + 4
• ⇒ x1 = x2
• Thus, f(x) is one-one.
• Since, x2 + 4 is a real number. Thus, for every y in the co-domain of f, there exists a number x in R+ such that f(x) = y = x2 + 4
• Thus, we can say that f(x) is onto.
• Now, f(x) is one-one and onto. Hence, f(x) is invertible.
• Let f(x) = y ⇒ x2 + 4 = y
• ⇒ x2 = y − 4
• i.e. x = √(y − 4)
• Also, x = f −1(y)
• f –1(y) = √(y − 4) . Hence Proved.

$$\text{To prove f is invertible we have to prove that f is one-one and onto.}\\\qquad\text{For one-one}\\\qquad\text{Let} x_1, x_2 \epsilon R_+ , then\\\qquad f (x_1) = f (x_2)\\\qquad\Rarr 5x_1^2 + 6x_1 − 9 = 5x_2^2 + 6x_2 − 9\\\qquad\Rarr 5(x_1^2 − x_2^2) + 6(x_1 − x_2) = 0\\\qquad\Rarr(x_1 − x_2)(5x_1 + 5x_2 + 6) = 0\\\qquad\Rarr x_1 − x_2 = 0\space as\space 5x_1 + 5x_2 + 6 \neq 0\\\qquad\Rarr x_1 = x_2\\\qquad\text{i.e., f is one-one function.}\\\qquad\text{For onto}\\\qquad\text{Let}\space f (x) = y\\\qquad\because y = 5x^2 + 6x − 9\\\qquad\therefore 5x^2 + 6x − (9 + y) = 0\\\qquad x=\frac{-6\pm\sqrt{36+4×5(9+y)}}{10}\\\qquad=\frac{-6\pm\sqrt{216+20y}}{10}\\\qquad\qquad=\frac{\pm\sqrt{54+5y-3}}{5}\\\qquad x=\frac{\sqrt{54+5y}-3}{5}\space\space(\because x \epsilon R_+)$$

$$\text{Clearly }\forall y\space\epsilon\space [-9,\infty],\text{the value of x} \epsilon \text{R}_+\\\Longrightarrow\text{f is onto function}\\\text{\text Hence f is one-one onto function}\\\Longrightarrow\text{f is invertiable function with }\\\text{f}^{-1}(y)=\frac{\sqrt{54+5y-3}}{5}\\\space\space\space\space \text{Hence Proved}$$

$$\text{We know that}\\\qquad\text{sin}^{\normalsize–1}(\text{sin x}) = x\\\qquad\text{So}, \text{sin}^{-1}\bigg[\text{sin}\bigg(\frac{-17\pi}{8}\bigg)\bigg]=\text{sin}^{-1}\bigg[-\text{sin}\frac{17\pi}{8}\bigg]\\\qquad=\text{sin}^{-1}\bigg[-\text{sin}\bigg(2\pi+\frac{\pi}{8}\bigg)\bigg]\\\qquad=\text{sin}^{-1}\bigg(-\text{sin}\frac{\pi}{8}\bigg)\\\qquad=\text{sin}^{-1}\bigg[\text{sin}\bigg(-\frac{\pi}{8}\bigg)\bigg]\\\qquad-\frac{\pi}{8}\space\space\space\space\space\space\space\space\text{Ans.}$$

$$\text{(i) Given,}\space\space\space y=\frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\\\qquad=\frac{e^{4x+1}}{e^{4x-1}}\\\qquad \text{On differentiating w.r.t. x, we get}\\\qquad \space \frac{dy}{dx}=\frac{4e^{4x}(e^{4x}-1)-4e^{4x}(e^{4x+1})}{(e^{4x-1})^2}$$

OR

$$\text{(ii) Given,} \\\qquad\text{x = a sin}^3 t\\\qquad\therefore\frac{dx}{dt}=\text{3a sin}^2 t.\text{cos}\space t\\\qquad\text{Also, y = a cos}^3 t\\\qquad\therefore\frac{dx}{dt}=\text{3a cos}^2 t (– \text{sin}\space t)\\\qquad\therefore\space\space\frac{dy}{dx}=\frac{-3a\space\text{cos}^2 t.\text{sin\space t}}{3a \space\text{sin}^2t.\text{cos}\space t} \bigg[\because \frac{dy}{dx}=\frac{dy/dt}{dx/dt}\bigg]\\\qquad\Rarr\frac{dy}{dx}=-\text{cot} \space t.$$

• Given, y = 2 cos (log x) + 3 sin (log x)
• On differentiating both sides w.r.t. x, we get

$$\frac{dy}{dx}=-2\text\space{\text{sin}}(\text{log} \space x).\frac{1}{x}+\text{3 cos (log x)}.\frac{1}{x}\\\qquad\Rarr x\frac{dy}{dx}=-2 \text{sin}(\text{log} x) + 3 cos (\text{log} x)\\\qquad\text{Again differentiating both sides w.r.t. x, we get}\\\qquad x\frac{d^2y}{dx^2}+\frac{dy}{dx}=-2\text{cos}(\text{log} \space x).\frac{1}{x}-3 \space\text{sin}(\text{log} \space x).\frac{1}{x}\\\qquad \Rarr x^2\frac{d^2y}{dx^2}+x.\frac{dy}{dx}=-(2\space \text{cos}\space(\text{log} x)+ 3 \space\text{sin} \space (\text{log} x))\\\qquad \Rarr x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}=-y\\\Rarr x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=0$$

$$\text{Given function is}\\\text{f(x)}=\frac{x^4}{4}-x^3-5x^2+24x+12\\\text{f}'(x)=\frac{4x^3}{4}-3x^2- 10x+ 24\\\text{For critical points, put f}^ ′(x) = 0\\\therefore\space\space x^3 – 3x^2 – 10x + 24 = 0\\(x-2)(x^2-x-12)=0\\(x – 2) (x – 4) (x + 3) = 0\\\Rarr x = 2, 4, – 3\\\text{Therefore, we have the intervals (– ∞, – 3), (– 3, 2), (2, 4) and (4, ∞).}$$

$$\text{Since f’(x) > 0 in (– 3, 2) ∪ (4, ∞).}\\\therefore\space \text{f(x) is increasing in interval (– 3, 2) ∪ (4, ∞).}\\\text{And f}^′(x) < 0 \space \text{in} (– ∞, – 3) ∪ (2, 4)\\\therefore\space \text{f(x) is decreasing in (– ∞, – 3) ∪ (2, 4).}$$

$$\text{Let S be the sample space, then S} = {HH, HT, T_1, T_2, T_3, T_4, T_5, T_6}\\\qquad\text{P(HH)}=\frac{1}{2}×\frac{1}{2}=\frac{1}{4}\\\qquad\text{P(HT)\space =}\space\frac{1}{2}×\frac{1}{2}=\frac{1}{4}\\\qquad\text{Probability of each of the elementary event} \space T_1, T_2, T_3, T_4, T_5 \space\text{and}\space T_6 \space is \space \frac{1}{2}×\frac{1}{6}=\frac{1}{12}\\\qquad\text{Let A be the event of die showing a number greater than 4 and B be the event that there is at least one tail.}\\\qquad\text{A}={\text{T}_5, \text{T}_6}\\P(B)=\frac{1}{4}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{4}\\\qquad\text{P(A ∩ B)=}\frac{1}{12}+\frac{1}{12}=\frac{1}{6}\\\qquad\text{P(A|B)}=\frac{P(A\cap B)}{P(B)}=\frac{1/6}{3/4}\\\qquad \frac{4}{6×3}=\frac{2}{9}\\\qquad \therefore \space\space\space \text{Required probability =}\frac{2}{9}.\space\space\text{Ans.}$$

$$\text{(i)}\space \text{Let}\space I=\frac{1}{2}\int( 2 \text{cos 2x cos 4x})\space \text{cos 6x dx}\\\qquad I =\frac{1}{2}(\text{cos}6x + \text{cos} 2x)\text{cos6x dx}\\\qquad[\because 2 \text{cos A cos B} = \text{cos} (A + B) + \text{cos} (A – B)]\\=\frac{1}{2}\int\text{cos 6x cos 6x dx}+\frac{1}{2}\int\text{cos 2x cos6x dx}\\\qquad=\frac{1}{4}\int\text{2 cos 6x cos 6x dx}+\frac{1}{4}\int\text{2 cos 2x cos6x dx}\\\qquad=\frac{1}{4}\int(\text{cos} 12x + \text{cos} 0)dx+\frac{1}{4}\int\text{cos8x cos 4x dx}\\\qquad=\frac{1}{4}\bigg[\int(cos 12x+1+cos8x+cos 4x)dx\bigg]\\\qquad\frac{1}{4}\bigg[\frac{\text{sin} 12x}{12}+\frac{\text{sin}8x}{8}+\frac{\text{sin}4x}{4 }+x\bigg]+\text{C}$$

OR

$$\text{(ii)}\space\space\space\int\frac{\text{cos}2x-\text{cos}2\alpha}{\text{cos}x-\text{cos}\alpha}dx=\int\frac{2\text{cos}^2x-1-2\text{cos}^2\alpha+1}{\text{cos}\space x\text{cos}\space\alpha}dx\space\space\space[\because \text{cos}2\theta =2 \text{cos}^2\theta-1]\\\qquad=\int\frac{2 \text{cos}^2x-2\text{cos}^2\alpha}{\text{cos}\space x-\text{cos}\space \alpha}dx\\\qquad=\int\frac{2(\text{cos}x- \text{cos}\space\alpha)(\text{cos x + cos}\space\alpha)}{(\text{cos} x-\text{cos}\space \alpha)}dx\\\qquad=2\int(\text{cos}\space x+\text{cos}\alpha)dx\\\qquad =2\bigg[\int\text{cos x dx}+\int\text{cos}\alpha dx \bigg]\\\qquad\text{= 2 sin x + 2x cos} \alpha + \text{C} \space\space\space\space\text{Ans.}$$

$$\text{(i) Given differential equation is}\\\qquad\frac{dy}{dx}=1 + x^2 + y^2 + x^2y^2\\= (1 + x^2) + y^2 (1 + x^2)\\\Rarr\frac{dy}{dx}=(1+x^2)(1+y^2)\\\qquad\Rarr\frac{dy}{1+y^2}=(1+x^2)dx\\\qquad\text{On integrating both sides, we have}\\\qquad\int\frac{dy}{1+y^2}=\int(1+x^2)dx\\\qquad\text{tan}^{-1}y=x+\frac{x^3}{3}+\text{C}\space\space\space…\text{(i)}\\\text{put y = 1 and x = 0 in equation (i),}\\\qquad\text{tan}^{-1}1=0+0+\text{c}\\\qquad\text{C}=\frac{\pi}{4}\\\text{Equation (i) becomes:}\\\text{tan}^{-1}y=x+\frac{x^3}{3}+\frac{\pi}{4}\\\qquad\Rarr y= \text{tan}\bigg(x+\frac{x^3}{3}+\frac{\pi}{4}\bigg)\\\qquad\text{is the required particular solution of given equation.}\\\qquad\text{OR}\\\qquad\text{The given differential equation is}\\\qquad(1-y^2)(1+log\space x)dx+2xy dy = 0\\\qquad\frac{(1+log x)}{x}dx=\frac{-2y}{(1-y^2)}dy\\\qquad\text{On integrating both sides, we have}\\\qquad\int\frac{1+\text{log}x}{x}dx=\int\frac{2y}{(1-y)^2}dy\\\qquad\text{In first integral,}\\\qquad\text{put}\space\space\space\space\text{1 + log x = t}\\\qquad\Rarr\frac{1}{x}dx=dt\\\qquad\text{Also in second integral,}\\\qquad\text{put}\space\space\space\text{1 – y}^2 = u\\\qquad\Rarr\space\space\space\space\text{– 2y dy = du}\\\qquad\therefore\int t.dt=\int\frac{1}{u}du\\\qquad\Rarr\frac{t^2}{2}-\text{log}|u|=\text{C}\\\qquad\text{or}\space \frac{1}{2}(1+\text{log} x)^2-\text{log}|1-y^2|=\text{C}\text{It is given that y = 0 when x = 1}\\\qquad\text{So},\space\space\space\frac{1}{2}(1+\text{log} 1)^2-\text{log}|1-0^2|=\text{C}\\\qquad\text{C}=\frac{1}{2}\\\qquad\therefore\space\space\space\frac{(1+\text{log}x)^2}{2}-\text{log}|1-0^2|=\text{C}\\\Rarr\text{C}=\frac{1}{2}\\\qquad\therefore\frac{(1+\text{log}\space x)^2}{2}-\text{log}|1-y^2|=\frac{1}{2}\\\qquad\text{or}\space\space\space\space(1+\text{log}x)^2-2\space \text{log}|1-y^2|=1\\\qquad\text{It is the required particular solution.}$$

$$\text{The given system of equations can be written in matrix form as :}\\\qquad\space \begin{bmatrix} 2 & -3 & 5 \\3 & 2 & -4\\1 & 1 & -2 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix}=\begin{bmatrix} 11 \\-5 \\-3 \end{bmatrix}\\\qquad\Rarr \text{AX}=\text{B}\space\space\space\text{…(i)}\\\qquad\text{where}\space \text{A}=\begin{bmatrix} 2 & -3 & 5 \\3 & 2 & -4\\1 & 1 & -2 \end{bmatrix},\text{X}=\begin{bmatrix} x \\y \\z \end{bmatrix},\text{B}=\begin{bmatrix} 11 \\-5 \\-3 \end{bmatrix}\\\qquad\text{Here}, | A| = 2(– 4 + 4) + 3(– 6 + 4) + 5(3 – 2) = – 1 \neq 0\\\qquad\therefore\text{A}^{-1}\text{exists}\\\qquad$$

$$\begin{bmatrix*}[r]\begin{vmatrix*}[r]2 & -4 \\1 & -2\end{vmatrix*} & -\begin{vmatrix*}[r]-3 & 5 \\1 & -2\end{vmatrix*} &\begin{vmatrix*}[r]-3 & 5 \\2 & -4\end{vmatrix*} \\\cr-\begin{vmatrix*}[r]3 & -4 \\1 & -2\end{vmatrix*} & \begin{vmatrix*}[r]2 & 5 \\1 & -2\end{vmatrix*} & -\begin{vmatrix*}[r]2 & 5 \\3 & 4\end{vmatrix*}\\\cr\begin{vmatrix*}[r]3 & 2 \\1 & 1\end{vmatrix*} & -\begin{vmatrix*}[r]2 & -3 \\1 & 1\end{vmatrix*} & \begin{vmatrix*}[r]2 & -3 \\3 & 2\end{vmatrix*}\end{bmatrix*}$$

$$=\begin{bmatrix*}[r]-\begin{vmatrix*}[r]0 & -1 & 2 \\2 & -9 & 23\\1 & -5 & 13\end{vmatrix*} \\\end{bmatrix*}$$

$$\therefore \space\space\text{A}^{-1}=\frac{\text{adj (A)}}{|\text{A}|}=\frac{1}{-1}\begin{bmatrix*}[r]0 & -1 & 2 \\2 & -9 & 23\\1 & -5 & 13\\\end{bmatrix*}=\begin{bmatrix*}[r]0 & -1 & 2 \\2 & -9 & 23\\1 & -5 & 13\end{bmatrix*}\\\qquad\therefore\space\space\text{From (i),}\space\space \text{X}=\text{A}^{-1}\text{B}\\\qquad\Rarr\begin{bmatrix*}[r]x \\y \\z \\\end{bmatrix*}=\begin{bmatrix*}[r]0 &1 &-2 \\-2 &9 &-23\\-1 &5 &-13 \\\end{bmatrix*}\begin{bmatrix*}[r]11 \\-5 \\-3 \\\end{bmatrix*}\\\therefore\space\space\space\space x = 1, y = 2, z = 3.\space\space\space\space\text{Ans}.$$

$$\text{(i) Let} e^{x^{3}}= t\\\qquad\therefore 3x^2 e^{x^{3}} dx= dt\\\qquad\therefore\int x^2(e^{x^3})\text{cos}(2e^{x^3})dx=\int\frac{1}{3}\text{cos 2t dt}\\\qquad=\frac{1}{3}.\frac{1}{2}\text{sin\space 2t+C}\\\qquad=\frac{1}{6}\text{sin}(2e^{x^3})+\text{C}\space\space\text{Ans.}$$

OR

$$\text{(ii)}\int{x(\text{tan}^{-1}x)^2}dx\\\qquad\text{Let tan}^{-1}x=t\Rarr \text{x\space=\space tan\space t}\\\qquad\text{dx= sec}^2 t \space dt\\\qquad\int x(\text{tan}^{-1}x)^{2}dx=\int\text{t}^{2}.\text{tan}\space t \text{sec}^2t \space dt\text\space {[\text{Integrating\space by parts]}}\\\qquad=\text{t}^2.\frac{\text{tan}^2 t}{2}-\int\frac{2t\space \text{tan}^2t}{2}dt\\\qquad=\frac{1}{2}t^2\space \text{tan}^2t-\int\text{t}(\text{sec}^2 t-1)dt\\\qquad\space=\frac{1}{2}\space\text{t}^2\space\text{tan}^2t-\int\text{t sec}^2 t \space dt+\int t\space dt\\\qquad=\frac{1}{2}\text{t}^2\text{tan}^2t+\frac{t^2}{2}-\int\text{t sec}^2 t dt\\\qquad=\frac{1}{2}\text{t}^2(\text{tan}^2t+1)- \lbrace\text{t tant -}\int{\text{tan t dt}}\rbrace\\\qquad=\frac{1}{2}t^2(\text{tan}^2 t +1)-\text{t tan t + log|\text{sec t}|+\text{C}}\\\qquad=\frac{1}{2}(1+x^2)(\text{tan}^{-1} x)^2-\text{x tan}^{-1}x+\text{log}(\sqrt{1+x^2})+\text{C}\\\qquad=\frac{1}{2}(1+x^2)(\text{tan}^{-1}x)^2-x\space \text{tan}^{-1} x+\frac{1}{2}\text{log}(\sqrt{1+x^2})+\text{C}.\space \text{Ans}$$

$$\text{(i) Given,} \space (1+x^2)\frac{dy}{dx}=(e^{m\space\text{tan}^{-1}x}-y)\\\qquad\frac{dy}{dx}=\frac{e^{mtan^{-1}x}}{1+x^2}-\frac{y}{1+x^2}\\\qquad\Rarr\frac{dy}{dx}+\frac{1}{1+x^2}y=\frac{e^{m tan^{-1}}x}{1+x^2}\\\qquad\text{This equation is of the form}\\\qquad\frac{dy}{dx}+Py=Q(x)\\\qquad\text{Where \space P\space=}\space\frac{1}{1+x^2}\text{and }\space \text{Q(x)}=\frac{e^{mtan^{-1}x}}{1+x^2}\\\qquad\text{I.F.}=_e\int Pdx=\int e^{\bigg(\frac{1}{1+x^2}\bigg)dx}=\text{e}^{\text{tan}^{-1}x}\\\qquad\text{Hence, solution of linear differential equation is given by}\\\qquad \text{y × I.F.=}\int \text{I.F×Q(x)dx}\\\qquad\Rarr\text{y × e}^{\text{tan}^{-1}}x.\frac{e^{mtan^{-1}x}}{1+x^2}dx\\\qquad\Rarr\text{y× e}^{\text{tan}^{-1}x}=\int\frac{e^{(1+m)\text{tan}^{-1}x}}{1+x^2}dx\\\qquad\text{Let}\space\space e^{(1+m)}\text{tan}^{-1}x=t\\\qquad\Rarr(1+m)\frac{e^{(1+m)tan^{-1}x}}{1+x^2}dx=dt\\\qquad\therefore\space \text{y× e}^{\text{tan}^{-1}x}=\int\frac{1}{1+m}dt\\\qquad\Rarr\text{y × e}^{\text{tan}^{-1}x}=\frac{1}{(1+m)}t+\text{C}\\\qquad=\frac{1}{1+m}e^{(1+m)\text{tan}^{-1}x}+\text{C}\\\qquad\Rarr\text{y× e}^{tan^{-1}0}=\int\frac{e^0}{1+x^2}dx=\int\frac{dx}{1+x^2}\\\qquad\Rarr y× e^{tan^{-1}}x=\text{tan}^{-1}x+\text{C}\\\qquad\text{When x = 0, then y = 1,}\\\qquad\therefore\space \text{1×e}^{\text{tan}^{-1}0}=\text{tan}^{-1}0+\text{C}\\\qquad\Rarr 1 × e^{0} = 0 + \text{C}\\\qquad\Rarr \text{C}=1\\\text{Hence, particular solution of differential equation is}\\\qquad\text{y × e}^{\text{tan}^{-1}x}=\tan^{-1}x + 1. \space \text{Ans}.$$

$$\Rarr\text{y× e}^{tan^{-1}0}=\int\frac{e^0}{1+x^2}dx=\int\frac{dx}{1+x^2}\\\qquad\Rarr y× e^{tan^{-1}}x=\text{tan}^{-1}x+\text{C}\\\qquad\text{When x = 0, then y = 1,}\\\qquad\therefore\space \text{1×e}^{\text{tan}^{-1}0}=\text{tan}^{-1}0+\text{C}\\\qquad\Rarr 1 × e^{0} = 0 + \text{C}\\\qquad\Rarr \text{C}=1\\\text{Hence, particular solution of differential equation is}\\\qquad\text{y × e}^{\text{tan}^{-1}x}=\tan^{-1}x + 1. \space \text{Ans}.$$

OR

$$\text{(ii) Let ABC be an isosceles triangle with AB = AC and a circle of radius r unit with centre I is inscribed in} \space \Delta \text{ABC}.\\\qquad\text{Now,}\space \text{AD}\perp \text{BC}\\\qquad\therefore \text{BD\space=\space DC}\\\qquad\text{Let AE = AF = x units, [Tangents drawn from an external point]}\\\qquad\text{CE = CD = BD = y units}\\\qquad\therefore\text{BD = BF = y units}\\\qquad\text{Perimeter of triangle = AB + BC + AC}\\\qquad\text{= x + y + 2y + x + y}\\\qquad\text{P = 2x + 4y …(i)}\\\qquad\text{In} \space\Delta\text{AIE},\space \angle\text{AEI}=90\degree\\\qquad\therefore\frac{\text{AE}}{\text{IE}}=\text{cot}\space\theta\\\qquad\text{AE = r cot θ = x …(ii)}\\\qquad\text{In} \Delta \text{ADC}, ∠\text{ADC} = 90°\\\qquad\therefore\frac{\text{DC}}{\text{AC}}=\text{sin}\space\theta\\\qquad\qquad\frac{y}{x+y}=\text{sin}\space\theta\\\qquad\Rarr \text{x sin} \theta + \text{y sin} \theta =\text{y}\\\qquad\Rarr\text{r cot} \theta \text{sin}\space \theta = y(1 – sin \theta) \space\space\space[\text{Using (ii)}]\\\qquad\Rarr\frac{\text{rcos}\space\theta}{1-\text{sin}\space\theta}=y\space\space\space\space\space\text{…(iii)}\\\qquad\text{From equations (i), (ii) and (iii),}\\\qquad\text{P=2x+4y}\\\qquad\text{P=2r\text{cot}}\space \theta+\frac{4r\text{cos}\space\theta}{1-\text{sin}\space\theta}\\\qquad\therefore\frac{dP}{d\theta}=\frac{d}{d\theta}\text{2r cot}\space \theta+\frac{d}{d\theta}\frac{4r\text{cos}\space\theta}{1-\text{sin}\space \theta}\\\qquad=-\text{2r cosec}^2\theta+4r\bigg[\frac{(1-\text{sin}\space\theta)(-\text{sin}\space\theta)-\text{cos}\space \theta(-\text{cos}\theta)}{(1-\text{sin}\theta)^2}\bigg]\\\qquad\text{=\space -2r\space\text{cosec}}^2\theta+4r\bigg[\frac{-\text{sin}\space\theta+\text{sin}^2\theta+\text{cos}^2\theta}{(1-\text{sin}\space\theta)^2}\bigg]\\\qquad=-\text{2r cosec}^2\theta+4r\bigg[\frac{1-\text{sin}\theta}{(1-\text{sin}\theta)^2}\bigg]\\\qquad=-\text{2r cosec}^2\theta+4r\bigg[\frac{1-\text{sin}\space\theta}{(1-\text{sin\space}\theta)^2}\bigg]\\\qquad=\text{-2r\space\text{cosec}}^2\theta+\frac{4r}{1-\text{sin}\space\theta}\\\qquad=2r\bigg[\frac{-1}{\text{sin}^2\theta}+\frac{2}{1-\text{sin}\space\theta}\bigg]\\\qquad=2r\bigg[\frac{-1+\text{sin} \space\theta+2\text{sin}^{2}\space\theta}{\text{sin}^{2}\theta(1-\text{sin}\space\theta)}\bigg]\\\qquad\therefore\frac{dP}{d\theta}=\frac{2r(2 \text{sin}\theta-1)(\text{sin}\theta+1)}{\text{sin}^2\theta(1-\text{sin}\theta)}\\\qquad\text{For maxima and minima,}\\\qquad\text{Put},\frac{dP}{d\theta}=0\\\qquad\therefore \frac{2r(2 \text{sin}\space\theta-1)(\text{sin}\space\theta+1)}{\text{sin}^2\theta(1-\text{sin}\space\theta)}=0\\\qquad\text{2r}(2 \text{sin}\space\theta-1)(\text{sin}\space\theta+1)=0\\\qquad\because r\neq0 \\\qquad\therefore\space\space \text{2 sin}\space\theta-1=0\\\qquad\Rarr\text{sin}\space\theta=\frac{1}{2}\\\qquad\\\qquad \theta=30\degree \text{or}\frac{\pi}{6},\\\qquad \Biggm\vert\text{or sin}\space \theta + 1 = 0\\\qquad\Rarr\text{sin}\space \theta = – 1\\\qquad\Rarr\theta=-\frac{\pi}{2}\\\qquad\qquad\because\theta=-\frac{\pi}{2}\text{=is not possible}\\\qquad\therefore\theta=30\degree\text{or}\space\frac{\pi}{6}\\\qquad\text{Now},\space\space\space\frac{d^2P}{d\theta^{2}}=\frac{d}{d\theta}\bigg[-2r\space\text{cosec}^2\theta+\frac{4r}{1-\text{sin}\space\theta}\bigg]\\\qquad=-\text{2r}(2\text{cosec}\space \theta)(-\text{cosec}\theta\space \text{cot}\theta)-4r.\frac{1}{(1-\text{sin}\theta)^2}(-\text{cos}\theta)\\\qquad=4r\bigg[\text{cosec}^2\theta\text{cot}\theta+\frac{\text{cos}\space\theta}{(1-\text{sin}\theta)^2}\bigg]\\\qquad =\text{4r}\bigg[\text{cosec}^2\frac{\pi}{6}\text{cot}\frac{\pi}{6}+\frac{\text{cos}\frac{\pi}{6}}{\bigg(1-\text{sin}\frac{\pi}{6}\bigg)^2}\bigg]\\\qquad=\text{4r}(4. \sqrt{3} + 2 \sqrt{3})\\\qquad\therefore\bigg(\frac{d^2P}{d\theta^2}\bigg)_{\theta=\pi/6}\text{\textgreater}\text\space0\\\qquad\text{Hence, perimeter is minimum at π/6.}\\\qquad\text{Now, Perimeter of} \Delta ABC = 2x + 4y\\\qquad= 2r\text{cot}\theta+\frac{4r\text{cos}\space\theta}{1-\text{sin}\space\theta}\\\qquad=\text{2r\text{cot}}\frac{\pi}{6}+\frac{4r\text{cos}\frac{\pi}{6}}{1-\text{sin}\frac{\pi}{6}}\\\qquad=2r.\sqrt{3}+\frac{4r.\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}\\\qquad =r(2\sqrt{3}+4\sqrt{3})\\\qquad\therefore\text{Least perimeter of} \Delta\text{ABC is}\space 6\sqrt{3r}\space \text{units}.\space\space\space\text{Hence Proved.}$$

$$\text{According to question,}\\\qquad\text{P(H) = 3P(T)}\\\qquad\text{But}\space\space\space\text{P(H) + P(T) = 1}\\\qquad\Rarr\text{3P(T) + P(T) = 1}\\\qquad\Rarr\text{4P(T) = 1}\\\qquad\Rarr P(T)=\frac{1}{4}\\\qquad\text{So},\space\space\space \text{P(H)}=\frac{3}{4}\\\qquad\qquad\text{Let X denotes number of tails obtained when the coin is tossed twice. Then, X can be 0, 1or 2.}\\\qquad\text{P(X = 0) = P(HH)}\\\qquad=\bigg(\frac{3}{4}\bigg)^2=\frac{9}{16}\\\qquad\text{P(X = 1) = P(HT or TH)}\\\qquad=\frac{3}{4}×\frac{1}{4}+\frac{1}{4}×\frac{3}{4}\\\qquad =\frac{6}{16}\\\qquad \text{P(X=2)}=\text{P(TT)}\\\qquad=\frac{1}{4}×\frac{1}{4}=\frac{1}{16}\\\qquad\therefore\text{Probability distribution table is}\\\qquad \text{Now, Mean}\space\space\space\space=\text{E(X)}=\displaystyle\sum_{i=1}^n\space X_iP(X_i)\\\qquad=0×\frac{9}{16}+1×\frac{6}{16}+2×\frac{1}{16}\\\qquad\qquad=\frac{8}{16}=\frac{1}{2}\\\qquad\therefore\text{Mean}=\frac{1}{2}$$

 Xi 0 1 2 P(Xi) $$\frac{9}{16}$$ $$\frac{6}{16}$$ $$\frac{1}{16}$$

$$\text{Now, Mean}\space\space\space\space=\text{E(X)}=\displaystyle\sum_{i=1}^n\space X_iP(X_i)\\\qquad=0×\frac{9}{16}+1×\frac{6}{16}+2×\frac{1}{16}\\\qquad\qquad\qquad\qquad=\frac{8}{16}=\frac{1}{2}\\\qquad\therefore\text{Mean}=\frac{1}{2}$$

## Section-B

$$\text{(i)}\space [a-b,b-c,c-a]=[(a-b)×(b-c)].(c-a)\\\qquad=[a×b-a×c-b×b+b×c].(c-a)\\\qquad=[a×b-a×c+b×c].(c-a)\\\qquad=(a×b).c-(a×c).c+(b×c).c-(a×b).a-(a×c).a-(b×c).a\\\qquad\text{= (a × b).c – (b × c)a}\\\qquad\text{= (a × b).c – (a × b).c}\text{= 0.} \space\space \text{Ans. (a)}$$

$$\text{(ii)}\space \text{Angle between two planes}\space a_1 x + b_1 y + c_1 = 0 \space\text{and}\space a_2x + b_2y + c_2 = 0 \text{is given by},\\\qquad\text{cos}\space \theta=\bigg|\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\bigg|\\\qquad\text{Here}\space a_1 = 2, b_1 = – 1, c_1 = 1\\\qquad\text{and}\space\space\space a_2 = 1, b_2 = 1, c_2 = 2\\\qquad\therefore\space\space\text{cos}\space \theta=\bigg|\frac{2(1)+(-1)(1)+(1)(2)}{\sqrt{2^2+1^2+1^2}\sqrt{1^2+1^2+2^2}}\bigg|\\\qquad \text{cos}\space \theta=\bigg|\frac{2-1+2}{\sqrt{6}\sqrt{6}}\bigg|=\frac{1}{2}\\\qquad\qquad\therefore\space \theta=60\degree=\frac{\pi}{3}\space\space\text{Ans. (b)}$$

$$\text{(iii) For}\space \bigg(\frac{1}{\sqrt{2},}\frac{1}{2},k\bigg)\text{to represent direction cosines, we should have}\\\qquad\bigg(\frac{1}{\sqrt{2}}\bigg)^2+\bigg(\frac{1}{2}\bigg)^2+k^2=1\\\qquad\qquad\text{or}\space\frac{1}{2}+\frac{1}{4}+k^2=1\\\qquad\Rarr\space k=\pm\frac{1}{2}.$$

$$\text{(iv)}\space\space\vec{a}=2\hat{i}-\hat{j}+\hat{k},\space\vec{b}=\hat{i}+2\hat{j}-3\hat{k}\space \text{and}\space\vec{c}=3\hat{i}+x\hat{j}+5\hat{k}\\\qquad\text{If}\space\space\vec{a},\space\vec{b}\space\text{and}\space\vec{c}\space\text{are coplanar, then}\\\qquad\begin{vmatrix}2 & -1 & 1 \\1 & 2 &-3\\ 3 & x & 5\end{vmatrix}=0\\\qquad\Rarr\space\space2 \begin{vmatrix} 2 & -3 \\ x & 5 \end{vmatrix}+1\begin{vmatrix} 1 & -3 \\ 3 & 5 \end{vmatrix}+\begin{vmatrix} 1 & 2 \\ 3 & x \end{vmatrix}=0\\\qquad\Rarr2(10+3x)+(5+9)+(x-6)=0\\\qquad\Rarr\space 20 + 6x + 14 + x – 6 = 0\\\qquad\space\Rarr\text{7x=28}\\\qquad\therefore\text{x = 4.}\space\space\space\text{Ans.}$$

$$\text{(v) Given : Distance of the plane from origin is 5 units and normal vector is}\space 2\hat{i}-3\hat{j}+6\hat{k}.\\\qquad\text{Equation of the plane is}\space \vec{r}.\hat{n}=d\\\qquad\text{where},\vec{n}=2\hat{i}-3\hat{j}+6\hat{k}\\\qquad|\vec{n}|=|\sqrt{2^2+(-3)^2+6^2}|=|\sqrt{4+9+36}|=7\\\qquad\therefore\space\space\space{\hat n}=\frac{\vec{n}}{|\vec{n}|}=\frac{2\hat{i}-3\hat{j}+6\hat{k}}{7}\\\qquad\therefore\space\space\space\vec{r}.\hat{n}=d\\\qquad\Rarr\vec{r}.\bigg(\frac{2\hat{i}-3\hat{j}+6\hat{k}}{7}\bigg)=5\\\qquad\therefore\space\space\space\vec{r}.(2\hat{i}-3\hat{j}+6\hat{k})=35\\\qquad\text{Hence, required equation of the plane}\space\vec{r}.(2\hat{i}-3\hat{j}+6\hat{k})=35.\space\space\text{Ans}$$

$$\text{(i) The given lines are}\space\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\space\text{and}\space\frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}\\\qquad\text{Direction ratios of the given lines are (– 3, 2, 1) and (1, – 3, 2) and these are not proportional.}\\\qquad\therefore\text{Given lines are not parallel.}\\\qquad\text{Hence, these lines are intersecting.}\space\space\textbf{Hence Proved.}\\\qquad\text{If lines are coplanar then}\\\qquad\begin{vmatrix}x_2-x_1 &y_2-y_1 &z_2-z_1\\a_1 & b_1 &c_1\\a_2 & b_2 &c_2\end{vmatrix}=0\\\qquad\qquad\begin{vmatrix}0-(-1) &7-3 &-7-(-2)\\-3 & 2 &1\\1 & -3&2\end{vmatrix}=0\\\qquad \begin{vmatrix}1 &4 &-5\\-3 & 2 &1\\1 & -3 &2\end{vmatrix}=0\\\qquad \Rarr1 \begin{vmatrix}2 & 1 \\-3 & 2\end{vmatrix}-4\begin{vmatrix}2 & 1 \\-3 & 2\end{vmatrix}+(-5)\begin{vmatrix}-3 & 2 \\1 & -3\end{vmatrix}=0\\\qquad\Rarr\space (4+3)-4(-6-1)-5(9-2)=0\\\qquad\Rarr 7+28-35=0\\\qquad\text{Hence, the given lines are coplanar.}\space \textbf{Hence Proved.}\\\text{Equation of the plane containing these lines is}\\ \begin{vmatrix}x-x_1 & y-y_1 & z-z_1 \\x_2-x_1 & y_2-y_1 & z_2-z_1\\a_1&b_1 &c_1 \end{vmatrix}=0\\\qquad \begin{vmatrix}x-(-1) & y-3 & z-(-2) \\0-(-1) & 7-3 & -7-(-2)\\-3 &2 &1 \end{vmatrix}=0\\\qquad\qquad \begin{vmatrix}x+1 & y-3 & z+2 \\1 & 4 & -5\\-3 &2 &1 \end{vmatrix}=0\\\qquad \Rarr\space(x+1)\begin{vmatrix}4 & -5\\2 & 1\end{vmatrix}-(y-3)\begin{vmatrix}1 & -5\\-3 & 2\end{vmatrix}+(z+2)\begin{vmatrix}1 & 4\\-3 & 2\end{vmatrix}=0\\\Rarr(x+1)(4+10)-(y-3)(1-15)+(z+2)(2+12)=0\\\qquad\Rarr 14(x+1)-(-14)(y-3)+14(z+2)=0\\\Rarr14x + 14 + 14y – 42 + 14z + 28 = 0\\\Rarr 14x + 14y + 14z = 0 \\\Rarr x + y + z = 0\\\text{Hence, equation of the plane is x + y + z = 0.} \space \textbf{Ans.}$$

OR

$$\text{(ii) Given line is}\\\qquad 5x – 25 = 14 – 7y = 35z\\\Rarr 5(x – 5) = – 7(y – 2) = 35z\\\Rarr\space \frac{x-5}{1/5}=\frac{y-2}{-1/7}=\frac{z-0}{1/35}\\\Rarr\space \frac{x-5}{7}=\frac{y-2}{-5}=\frac{z-0}{1}\\\text{Direction ratios of this line are 7, – 5, 1.}\\\therefore\text{Vector equation of the line which passes through the point A(1, 2, – 1) and whose direction ratios are proportional to 7, – 5, 1 is}\\\qquad\vec{r}=\hat{i}+2\hat{j}-\hat{k}+\lambda(7\hat{i}-5\hat{j}+\hat{k}).$$

$$\text{(i)\space Given : Coordinates of vertices of DABC are A(1, 2, 3), B (2, – 1, 4) and C(4, 5, – 1)}\\\qquad\therefore\space\space\overrightarrow{\text{AB}}=\hat{i}-3\hat{j}+\hat{k}\\\qquad\text{and}\space\overrightarrow{\text{AC}}=3\hat{i}+3\hat{j}+4\hat{k}\\\qquad\text{We know that,}\\\qquad\text{ar}(\Delta\text{ABC})=\frac{1}{2}\bigg|\overrightarrow {\text{AB}}×\overrightarrow {\text{AC}}\bigg|\\\qquad\text{Now},\overrightarrow{\text{AB}}×\overrightarrow{\text{AC}}=\begin{vmatrix}\vec{i} & \vec{j} & \vec{k} \\1 & -3 & 1\\3 & 3 & -4\end{vmatrix}\\\qquad=\vec{i}(12-3)-\vec{j}(-4-3)+\vec{k}(3+9)\\\qquad=9\vec{i}+7\vec{j}+12\vec{k}\\\qquad\therefore\space\space|\overrightarrow{\text{AB}}×\overrightarrow{\text{AC}}|=\sqrt{9^2+7^2+(12)^2}=\sqrt{274}\\\qquad\text{So},\space\space\text{ar}(\Delta\text{ABC})=\frac{1}{2}\sqrt{274}\space\text{sq}.\text{units}$$

OR

$$\text{(ii) Here,}\space\vec{a}=\hat{i}+\hat{j}+\hat{k},\hat{n}\space \text{is unit vector}\\\qquad\vec{b}=2\hat{i}+4\hat{j}-5\hat{k}\\\qquad\vec{c}=\lambda\hat{i}+2\hat{j}+3\hat{k}\\\qquad\vec{b}+\vec{c}=(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}\\\qquad\text{Then},\hat{n}=\frac{(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(2+\lambda)^2+36+4}}\\\qquad\text{Given,}\space\space\space\vec{a}.\hat{n}=1\\\qquad\bigg(\hat{i}+\hat{j}+\hat{K}\bigg).\bigg(\frac{(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(2+\lambda)^2+40}}\bigg)=1\\\qquad\Rarr(2+\lambda)+6-2=\sqrt{(2+\lambda)^2+40}\space\space\bigg[\because\space\hat{i}.\hat{i}=1,\hat{j}.\hat{j}=1,\hat{k}.\hat{k}=1\bigg]\\\qquad\Rarr(2+\lambda)+4=\sqrt{(2+\lambda)^2+40}\\\qquad\Rarr\lambda+6=\sqrt{(2+\lambda)^2+40}$$

$$\text{On squaring both sides, we get}\\\qquad(6+\lambda)^2=(2+\lambda)^2+40\\\qquad\Rarr 36 + \lambda^2 + 12\lambda = 4 + \lambda^2 + 4\lambda + 40\\\qquad\Rarr 36 + 12\lambda − 4 − 4\lambda − 40 = 0\\\qquad\Rarr8\lambda-8=0\\\qquad\Rarr\lambda=1\\\qquad\text{Then}, \vec{b}+\vec{c}=(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}\\\qquad=3\hat{i}+6\hat{j}-2\hat{k}\\\qquad\text{unit vector along}\bigg(\vec{b}+\vec{c}\bigg)=\frac{3\hat{i}+6\hat{j}+2\hat{k}}{\sqrt{9+36+4}}\\\qquad=\frac{3\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{49}}\\\qquad=\frac{3\hat{i}+6\hat{j}-2\hat{k}}{7}$$

$$\text{(i) Given,}\\\qquad \lbrace(x,y):0\le y \le x^2,0 \le y \le x,0\le x \le 2\rbrace\\\qquad\Rarr y\le x^2…(i)\\\qquad y\le x\space…(ii)\\\qquad x \le 2 \space …(iii)\\\qquad x\ge 0\space…(iv)\\\qquad y\ge 0\space…(v)\\\qquad\text{Considering inequalities as equation :}\\\qquad\text{y} = x^2, y = x, x = 2, x = 0, y = 0 \\\qquad\text{Solving\space y=x}^2\text{and}\space y=x\\\qquad\Rarr x^2=x\\\qquad\Rarr x(x-1)=0 \\\qquad\Rarr x=0, x=1 \\\qquad\therefore y=0,y=1\\\qquad\therefore\text{Points of intersection of curve (i) and the (ii) are (0, 0) and (1, 1)}\\\qquad\text{Required area}=\int^1_0\text{(y of the parabola) dx}+\int^2_1\text{(y of the line)dx}\\\qquad\text{Required area}=\int^1_0 x^2dx + \int^2_1 x dx\\\qquad=\bigg[\frac{x^3}{3}\bigg]^1_0+\bigg[\frac{x^2}{2}\bigg]^2_1\\\qquad =\bigg(\frac{1}{3}-0\bigg)+\bigg(\frac{4}{2}-\frac{1}{2}\bigg)\\\qquad=\frac{1}{3}+\bigg(2-\frac{1}{2}\bigg)\\\qquad=\frac{1}{3}+\frac{3}{2}=\frac{2+9}{6}\\\qquad\qquad=\frac{11}{6}\text{square units.}$$

## Section-B

$$\text{(i) C(x)} = 0·005x^3 – 0·02x^2 + 30x + 5000\\\qquad\text{M.C.}=\frac{d}{dx} \lbrace\text{C}(x)\rbrace=\frac{d}{dx}(0·005x^3 − 0.02x^2 + 30x + 5000)\\\qquad = 0·015x^2 – 0·04x + 30 + 0 \\\qquad\text{M.C.} =0·015x^2 – 0·04x + 30\\\qquad\text{(M.C.)}_{x=3}=0·015(3)^2 – 0·04(3) + 30\\\qquad = 0·135 – 0·120 + 30 \\\qquad = 30·015 \\\qquad\therefore\text{(M.C.)}_{x=3}= ₹30·015. \space\space\textbf{Ans. (b)}$$

$$\text{(ii) R(x)} = 3x^2 + 36x + 5\\\qquad \text{M.R.}=\frac{d}{dx}(\text{M.R.})=\frac{d}{dx}(3x^2 + 36x + 5)\\\qquad\text{M.R.} = 6x + 36 \\\qquad \text{MR at}\space x=5,\space\space\space\space\text{M.R.} = 6(5) + 36 = ₹ 66. \space\space\textbf{Ans. (d)}$$

$$\text{(iii) C(x)} = 0·007x^3 – 0·003×2 + 15x + 4000\\\qquad\text{M.C.}=\frac{d}{dx}\lbrace \text{C(x)}\rbrace=\frac{d}{dx}(0·007x^3 − 0·003x^2 + 15x + 4000)\\\qquad\text{M.C.} = 0·021x^2 – 0·006x + 15 + 0 \\\text{MC at x = 17}\space\space\space\space\text{M.C.}=0·021(17)^2 – 0·006(17) + 15\\\qquad\text{M.C.}=0·021(289) – 0·006(17) + 15\\\qquad\text{M.C.}=6·069 – 0·102 + 15\\\qquad\text{M.C.}=₹20·967.$$

$$\text{(iv) (c) Regresion}\\\qquad\text{(v) Given, X = 0.85Y and Y = 0.89X}\\\qquad\therefore\space b_{xy}=0.85 \space\text{and} \space Y=0.89X\\\qquad\therefore\space b_{xy}=0.85\space\text{and}\space b_{yx}=0.89\\\qquad\text{Coefficient of correlation is given as,}\\\qquad r=\pm\sqrt{b_{xy}×b_{yx}}\\\qquad=\pm\sqrt{0.85×0.89}\\\qquad=\pm\sqrt{0 7565}\\\qquad=\pm0.87\\\qquad \because\space b_{xy},b_{yx} \text{\textgreater}\space 0\\\qquad\therefore r=0.87\space \textbf{Ans.}$$

$$\text{(i) Given, Cost price of x items =}\space₹\bigg(\frac{x}{5}+500\bigg)\\\qquad\text{Selling price of x items =}\space ₹\bigg(5-\frac{x}{100}\bigg)x\\\qquad=₹\bigg(5x-\frac{x^2}{100}\bigg)\\\qquad\therefore \text{Profit = S.P. – C.P.}\\\qquad=₹\bigg(5x-\frac{x^2}{100}-\frac{x}{5}-500\bigg)\\\qquad \text{P(x)}=\frac{24x}{5}-\frac{x}{50}-500\\\text{Differentiating w.r.t. x, we get}\\\qquad\frac{dP}{dx}=\frac{d}{dx}\bigg(\frac{24x}{5}\bigg)-\frac{d}{dx}\bigg(\frac{x^2}{100}\bigg)-\frac{d}{dx}(500)\\\qquad\frac{dP}{dx}=\frac{24}{5}-\frac{x}{50}-0\\\qquad\text{For P to be maximum,}\\\qquad\frac{dP}{dx}=0\\\qquad\qquad\qquad\qquad\therefore\frac{24}{5}-\frac{x}{50}=0\\\qquad\Rarr\frac{x}{50}=\frac{24}{5}\\\qquad\Rarr x=\frac{24×50}{5}=240\\\qquad\Rarr\frac{d^2P}{dx^2}=\frac{d}{dx}\bigg(\frac{24}{5}\bigg)-\frac{d}{dx}\bigg(\frac{x}{50}\bigg)\\\qquad\Rarr\frac{d^2P}{dx^2}\text{\textless}\space0\\\therefore\text{Profit is maximum when 240 items are sold.}\\\qquad\text{Maximum profit at x = 240,} \textbf{Ans.}\\\qquad\text{P(240) =}\frac{24}{5}×240 – \frac{240×240}{100}-500\\\qquad\qquad= 24 × 48 – 576 – 500\\\qquad = 1152 – 1076\\\qquad=₹76\\\qquad\text{Hence, maximum profit is ₹ 76.}\space \textbf{Ans}.$$

OR

$$\text{(ii)}\space\text{Let the annual subscription be increaesed by ₹ x.}\\\qquad\therefore\text{Charges per subscriber\space= ₹(300+x)}\\\qquad\text{and}\space\text{Number of subscribers\space = (500-x)}\\\qquad\text{Annual income (I)}= ₹(500-x)(300-x)\\\qquad=₹(150000+200x-x^2)\\\qquad \text{I}=150000+200x-x^2\\\qquad\text{Differentiating\space w.r.t.x, \text{we get}}\\\qquad\frac{dl}{dx}=\frac{d}{dx}(150000)+\frac{d}{dx}(200x)-\frac{d}{dx}(x^2)\\\qquad\qquad\Rarr\frac{dl}{dx}=\text{0+200-2x}\\\qquad\text{For income to be maximum,}\space \frac{dl}{dx}=0\\\qquad\Rarr\text{200 – 2x=0}\\\qquad\Rarr x=100\\\qquad \Rarr\space 200-2x=0\\\qquad \Rarr x=100\\\qquad\Rarr\frac{d^2t}{dx^2}=\frac{d}{dx}(200-2x)\\\qquad\qquad\Rarr\frac{d^2t}{dx^2}=0-2=-2\\\qquad\Rarr\frac{d^2t}{dx^2}=\text{\textless}\space 0\\\qquad\text{Hence, annual income is maximum at increment of ₹ 100.}\\\qquad\text{New annual income = }₹ 400 × 400 = ₹160000\\\qquad\text{Old annual income = ₹500 × 300 = ₹150000}\\\qquad\text{Increase in annual income = ₹10000.}\space\space\textbf{Ans.}\\\qquad$$

 Marks in Physics x Marks in Chemistry y dx = x – 40 dx dy = y – 36 dy (dx)2 (dy)2 dxdy 46 40 6 4 36 16 24 42 38 2 2 4 4 4 44 36 4 0 16 0 0 40 35 0 -1 0 1 0 43 39 3 3 9 9 9 41 37 1 1 1 1 1 45 41 5 5 25 25 25 Σdx = 21 Σdy = 14 Σ(dx)2 = 91 Σ(dy)2 = 56 Σdxdy = 63

$$\text{Here}, n = 7 b_{xy} =\frac{n\sum dx.dy-\sum dx \sum dy}{n\sum (dy)^2-(\sum dy)^2}\\\qquad\qquad\qquad b_{xy}=\frac{7×63-21×14}{7×56-(14)^2}\\\qquad b_{xy}=\frac{441-294}{392-196}\\\qquad b_{xy}=\frac{147}{196}=0.75\\\qquad b_{yx}=\frac{n\sum dx.dy-\sum dx\sum dy}{n\sum(dx)^2-(\sum dx)^2}\\\qquad b_{yx}=\frac{7×63-21×4}{7×91-(21)^2}\\\qquad=\frac{441-294}{637-441}=\frac{147}{196}=0.75\\\qquad\bar{x}=A+\frac{\sum dx}{n},\space\bar{y}=A+\frac{\sum dy}{n}\\\qquad \bar x=40+\frac{21}{7}=43,\bar y=36+\frac{14}{7}=38\\\qquad\text{Equation of X on Y}\\\qquad x − \bar x = b_{xy} (y − \bar y)\\\qquad x-43=0.75(y-38)\\\qquad x = 0·75y – 28·5 + 43\\\qquad x = 0·75y + 14·5\\\qquad\text{Equation of Y on X}\\\qquad y-\bar{y}=b_{yx}(y-\bar{y})\\\qquad x – 43 = 0·75(y – 38)\\\qquad x = 0·75y – 28·5 + 43\\\qquad x = 0·75y + 14·5\\\qquad\text{Equation of Y on X}\\\qquad y − \bar y = b_{yx} (x − \bar x)\\\qquad y – 38 = 0·75(x – 43)\\\qquad y = 0·75x – 32·25 + 38\\\qquad y = 0·75x + 5·75 \\\qquad \text{Hence}, \text{the regression coefficients are} 0·75 \space\text{and}\space 0·75 \text{and the regression equations are x} = 0·75y + 14·5 and y = 0·75x + 5·75.\space \textbf{Ans}.$$

$$\text{(i) Let the number of type A cake made be x and the number of type B cake made be y.}\\\qquad\text{To maximise the number of cakes.}\\\qquad\because\space Z = x + y \\\qquad\text{Subject to constraint :}\\\qquad\text{200x + 100y} \le 5000\\\qquad\Rarr 2x+y\le50\space\space\space…(i)\\\qquad\text{Also}\space 25x + 50y \le 1000\\\qquad\Rarr x + 2y \le 40 …(ii)\\\qquad \text{and} x \ge 0, y \ge 0\\\qquad\text{Consider,} 2x + y = 50 x + 2y = 40$$

 x 0 25 10 y 50 0 30
 x 0 40 10 y 20 0 15

The feasible region is the shaded region.

 Corner Points Z = x + y A (25, 0), Z = 25 + 0 = 25 B (20, 10), Z = 20 + 10 = 30 C (0, 20), Z = 0 + 20 = 20

Hence, maximum number of cakes = 20 + 10 = 30.

OR

$$\text{(ii) We have,}\\\qquad \text{Maximise Z = 34x + 45y}\\\qquad \text{Subject to the constraints:}\\\qquad x + y \le 300\\\qquad 2x + 3y \le 70\\\qquad x \ge 0, y \ge 0 \\\qquad\text{We have,\space Maximise}Z = 34x + 45y\\\qquad\text{Subject to the constraints :}\\\qquad x + y \le 300 \\\qquad 2x + 3y \le 70 \\\qquad x \ge 0, y \ge 0\\\qquad\text{Converting the given inequalities into equations, we obtain the following equations:}\\\qquad x + y = 300\\\qquad 2x + 3y = 70\\\qquad\text{Then,} x + y = 300 \space\text{and}\space 2x + 3y = 70$$

 x 0 300 y 300 0
 x 0 35 y 70/3 0

Plotting these points on the graph, we get the shaded feasible region i.e., OCDO.

 Corner point Value of Z = 34x + 45y O (0, 0) 34(0) + 45(0) = 0 C (35, 0) 34(35) + 45(0) = 1190 D (0, 70/3) 34(0) + 45 (70/3) = 1050

Clearly, the maximum value of Z is 1190 at (35, 0).

#### ISC 36 Sample Question Papers

All Subjects Combined for Class 12 Exam 2023

#### ISC 36 Sample Question Papers

All Subjects Combined for Class 12 Exam 2023

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